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R be defined by by yandy e LP; furthermore, j is continuous. Statements (i) and (ii) are obvious; (iii) and (iv) require Theorem (C.2.14). Exercise 1 If, in addition, for some integer m > 1, l i m ^ ^ sm+1h(s) = c, a constant. State additional conclusions that you can draw. [Hint: Consider h,h, . . . , A(m).] oo; if w G L2, then w * w e L2 n L00 is uniformly continuous and goes to zero as t-> oo (observe that w * ue L1). Exercise 6 Verify in detail that Theorem (14) applies when u(i) e Un and W(i) e UnXn. Be sure to justify every step by invoking properties of norms, induced norms, etc.
Show that if// e [1, oo) and if/?' #/?, then/ p <£Z/, b u t / , eL p . 11
Remark In later applications, we shall consider functions from U+ into IR". In this case, all above definitions hold, except that \f(t)\ is interpreted as the chosen norm of the vector f(t) e Un; one norm in Un is chosen once and for all for the whole development. Since all norms in Un are equivalent, norms in Rn are selected for numerical convenience and for obtaining best bounds.
12
Notation Sometimes we consider functions on IR (or IR+) to IR (or Un). To keep track of these cases we use the symbol
and a similar definition for L/(U). In case n = 1, we drop the subscript n. Once it is understood that we consider IR (or IR+) exclusively, we then use the abbreviation Lnp.
4
Geometric interpretation of norms
In order to obtain a more intuitive interpretation of norms, we show below that to every norm is associated a certain type of convex set. Conversely, to any such convex set is associated a norm. We start by defining some concepts. 1 Let £ be a linear space, over IR or C. A set K in E is said to be convex iff x,y e K^> Xx + (1 - X)y e K, MXe [0, 1]. A set K in E is said to be balanced (equilibre) iff x e K=> ocx e K, V | a | < 1. A set K in E is said to be absorbing iff x e E =>3A(x) > 0 such that xeX(x)K. The set {y\y = Xx, XeU+) is called the ray Ox. 2
Theorem Let E be a linear space. (A) If N: E-> U+ is a norm on E, then the unit ball associated with N( •), namely, the set B = {xeE\N(x)<
1}
is convex, balanced, absorbing and B intersects every ray Ox on a finite interval (i.e., \fx e E, 3XN < oo such that Ox n B = {y | y = Xx, 0 < X < XN}).
4
GEOMETRIC
INTERPRETATION
OF
NORMS
19
(B) If the set K c E is convex, balanced, absorbing and if K intersects every ray Ox on a finite interval, then the function pK: E-+ U+ defined by 3
PK(X)
4 inf{A | A > 0 and x e XK}
is a norm on E. 4
Remark The ray condition is indispensable; to wit, let K c U2 and K = {(x,y)\xe R, \y\ < 1}. Clearly K is convex, balanced, and absorbing, but pK is not a norm [pK(x0) = 0 for x 0 = (1, 0)]. Proof (A) By assumption N is a norm. (i) B is convex because x, y e B imply JV(X) < 1, N(y) < 1. Now VAe[0, 1]. iV[Ax + (1 - % ] <
JV(AJC)
+ JV[(1 - X)y] = XN(x) + (1 - A)iV(>0 < 1
Hence Ax + (1 — A)y is in B, for all 2 e [0, 1]. (ii) B is balanced because x e B=>N(x) < 1. Hence V | a | < 1, N(ocx) = |a|iV(jc) < |a| < 1; i.e., axeB for all |a| < 1. (hi) B is absorbing because for any xe E, N(x) < oo and x/N(x) eB; i.e., xe N(x)B. (iv) OxnB = {Xx\X > 0 and N(Xx) < 1}. Now N(Xx) < 1 <=> \X\ < l/N(x); hence Ax e Ox n £ iff 0 < A < l/7V(x). (B) By assumption K has the four properties; we have to show that the function pK satisfies the axioms of a norm. (i) Obviously by (3), pK(0) = 0. Next we show that x ^ 0 =>pK(x) > 0 by contradiction. Suppose x0 ^ 6 and pK(x0) = 0 = inf{A | A > 0, x0 e AK}. Hence VA > 0, x 0 e AX; i.e., (1/A)x0 e K. Hence the whole ray Ox0 is contained in K. (ii) Since K is absorbing, pK(x) < oo for any XE E. (iii) pK(ocx) = \oc\pK(x) because K is balanced. Indeed, axeXK for some A implies | a | x e XK or x e (A/1 a | )X. Hence (iv) ^ ( x + y) < pK(x) + follows from the convexity of K and the definition of pK. By definition of pK, for any e > 0, —— and —— eK pK(x) + e pK(y) + 8 Since K is convex (we drop the subscript K for simplicity), I"
Pix) + 8
1
x
+
f
p(y) + e
So, Ve > 0, x + y c [/?(x) + /?O0 + 2e]K; hence /?(x + .y) < i?(x) + Xy).
II
20
5
NORMS
Induced norms of linear maps
5.1
The space of linear maps
Let E be a linear space over the field IK, which is U or C. Let &(E, E) be the class of all linear maps from E into E. 3{E, E) is a linear space if the addition of A and B is defined by 1
{A + B)x = Ax + Bx,
VJC
e E,
VA, B e S(E, E)
and the scalar product of a and A is defined by 2
(<xA)(x) = a(Ax),
VaeK,
Vxe£,
VAe&(E,E)
In addition, the product AB can be defined as the composition of the maps A and B: 3
(AB)(x) = A(Bx),
Vx e E,
VA, B e ${E, E)
Note that the product is not commutative. It is easy to check that (i) (ii) (iii) (iv)
multiplication defined by (3) is associative; addition and multiplication are distributive; multiplication and scalar multiplication commute; /, the identity map from E into E, is the unit of the multiplication.
Thus with the three operations defined by (1), (2), and (3), &(E, E) is a (noncommutative) algebra with a unit. (Refer to Appendix D for a discussion of algebras.) 4
Remark With obvious but simple modifications, the considerations of this chapter apply to linear maps from a linear space E to a linear space F. In Section 5.2, where we introduce induced norms, it becomes then crucial to distinguish between the norms on the domain and the norms on the range. To avoid cluttering the notation, we restrict ourselves to J^(£, E), i.e., to the case where the domain and the range are the same spaces.
5.2
Induced norms
Let | • | be a norm on E and A e S(E, E). Define the function || • || from a subset of ££{E, E) into U+ by 5 6 7
Fact Definition (5) is equivalent to \\A\\ = sup \Az\ |z| = l
5
8
INDUCED
NORMS
OF LINEAR
MAPS
21
\\A\\ is called the induced norm of the linear map A or the operator norm induced by the vector norm | • |. To show that (5) and (7) are equivalent, let z = x/\x\, then !!
To interpret (7) geometrically, let K be the convex, balanced, absorbing set corresponding to the vector norm | • |; i.e., the unit ball K±{xeE\\x\
< 1}
Let AK denote the image of K under the map A. Then (7) is equivalent to 9 Roughly speaking, \\A\\ is the smallest magnification factor X for which XK includes AK. 10
Theorem Let | • | be any norm on C". Let A be a nonsingular linear map from C" to Cn. Let ||-|| denote the induced norm on n x n matrices with elements in C. U.t.c. (a) There is a constant P which depends only on n and | • | such that
11 (b) If Euclidean norms are used, then ft < 1. Proof (a) From the equivalence of norms and (2.3), there are constants az and ocu such that
Then (11) follows from Cramer's formula. (b) By polar decomposition A = UH where U is unitary (i.e., UU* = I) and H is Hermitian. Note that |det >41 = |det U\ \det H\ = | d e t / / | Since A is nonsingular and H2 = A*A, H is positive definite. Call hu h2, . . . , hn its eigenvalues and order them so that 0 < ht < h2 < • • • < hn. By the properties of the (operator) norm induced by the Euclidean norm [see (1.18)],
12
22
II
NORMS
So
Note that the inequality (13) holds with equal sign if h2 = hn. In fact, the right-hand side of (13) is equal to the left-hand side times Y\k=2 (hjhk).
5.3
Continuous linear maps
Let (E, |-|) be a normed space over the field IK, and let ||-|| be the induced norm on some subspace of &{E, E). We define 14
&{E, E)^{Ae
&(E, E)\\\A\\ < oo}
Theorem U.t.c. VA, B e JS?(£, E), Va e IK, Vx e £
Proof
All these inequalities follow directly from Definition (7). For example, M2?|| = sup \ABz\ < \\A\\ sup \Bz\ = \\A\\ \\B\\ \z\ = l
ss
\z\ = l
The inequalities above show that S£{E, E) is a normed algebra. It can be shown that if (E, | • |) is complete (i.e., if E is a Banach space), then J£(E, E) is also a Banach algebra with the norm (7). It is, in fact, a noncommutative Banach algebra with a unit. Exercise 1 Let A e 3{E, E). Show that the following three statements are equivalent: (i) The linear function A is continuous at 6 e E. (ii) The linear function A is continuous on E. (iii) The induced norm of A, \\A\\, is finite. This exercise justifies our saying that ££{E, E) is the class of all continuous linear maps from the normed space E into E. Some authors call the elements of J£(E, E) " bounded " linear maps. This terminology conflicts with the general definition of a bounded map from a normed space into another normed space. What they mean is that A e J£(E, E), iff A e 2(E, E), and the restriction of A to the unit ball is a bounded map.
6
TWO EXAMPLES
23
Remark Suppose that on 3?(E, E) we define a norm N(-) which, in addition to the axioms of the norm, satisfies 19
N(AB) < N(A)N(B) and suppose that we have a vector norm | • | such that
20
\Ax\
VxeE,
VAe£(E,E)
Then we say that N(-) is an (operator) norm on S{E, E) compatible with the (vector) norm | • | on E. Exercise 2 (i) If ||-|| is the norm induced by | • | and if N(-) is any norm compatible with | • |, show that 21 \\A\\
6
Two examples
The two examples that follow are designed to illustrate the fact that the induced norm depends on the vector norm. The results of these examples will be useful later. Example 1 Let (E, \'\j = L°°([R+) = {/: U+ -> Rlll/IL < oo}. Let H be a linear map defined on E in terms of an integrable function h: U -> U; 1
H:ut-+Hu^h*u,
Vw e L00
i.e., 2
(Hu)(t) = f hit- T)W(T) dT,
Vr e U+
We assume that \\hWi = Jo \h(t)\ dt < oo. 3
Theorem (a) (b)
U.t.c.
H: L00 -> L00 \\H\\, the induced norm of the linear map / / , is given by
II
24
NORMS
and ||A*w||oo can be made arbitrarily close to ||/?Hi||w||oo by appropriate choice of u. Proof We start by calculating the induced norm of H. We drop the subscript oo throughout; e.g., \\u\\ denotes the L00 norm of u: U+ -* M. We have three norms in this proof: the absolute value of real numbers, e.g., |w(r)|; the norm on L°°, e.g., ||w||; the induced norm on ££(E, E), namely, \\H\\. 4
\\H\\ = sup \\h * u\\ = sup sup | (h * u)(i)\ ||u|| = i
||«|| = i r > 0
Since \\u\\ = 1,
Hence
This inequality shows that H is a continuous linear map from L°° into L00. Inequality (5) shows also that ||>^||± is an upper bound on the induced norm of the map H: L00 -* L00. Let us show that \\hWi is the induced norm. Consider a sequence of inputs ul9 u2, . . . , with \\ut\\ = 1, where for t = 1, 2, 3, . . . , we define ut: x\->ut(i) = sgn[h(t — T)],
T e U+ ,
t el
+
and we take /*(f) = 0 for t < 0. Consider now the value at time t of the output due to ut(')
where
denotes the norm on £. Hence for t = 1, 2, 3, . . . , we have by (5)
Letting £ -> oo, these inequalities imply that \H\ = p l ^ .
"
Example 2 Let E = L2(U) = {/: R -• R | ||/|| 2 < oo}. Let # be a linear map defined on L2 by H: u\-+Hu, where 6
6
7
TWO
EXAMPLES
25
Theorem If the linear map His defined by (6), where he I}, then (a) (b)
H:L2-+L2, || # | | 2 , the induced norm of the linear map H e J£?(L2, L2), is given by
8
|| H || 2 =max\h(jco)\ coeM
Proof Since he I}, its Fourier transform J^(/z) = h(jco) is uniformly continuous on U and -»0 as \co\ -• oo. (See Appendix B.1.1.) Now, with all integrals below being over U, we have \\Hu\\22 = |A * t/||22 = J (A * u)(t)(h * w)(0 A = — \{h * u)(jco)(h * u)(j(o)* Jco
(Parseval)
271 J
= —
I n(jco) | 2 1fl(jcu)| 2
(convolution theorem)
By Parseval,
Hence, Vw such that ||w||2 = 1,
Thus, the induced norm satisfies 9
\\H\\2 < max | h(joj) \ aeU
Note that since coh-» \h(jco)\ is continuous on R and ->0 as |co| -> oo, the maximum exists. We are going to show that \\H\\2 is actually equal to the right-hand side of (9). Observe that for X > 0, &(e-**2) = (n/X)l/2 e x p [ - co2/4A] and [exp(-Ar 2 ) cos a>0t] = K^M) 1/2 {exp[-(c» - co0)2/4A] + exp - [(co + co0)2/U]} As A->0, this expression tends to n[S(co — a>0) + 8(co + co0)], where <$(•) denotes the Dirac delta (generalized) function. Pick co0 to be the abscissa of the maximum of coi-> | ft(jco)\; for each X pick a normalization «(A) such that ux(i) = n(X)[exp( — Xt2)] cos co01
26
II
NORMS
has unit norm. Since | h(jco) | is continuous, we see that \\h * uAi ->max|/J(jco)|
as
2->0
CO
Consequently, we have shown that ||#|| 2 =max|fi(/c>)|
ss
CO
10
Remark These results can be generalized to the case where u and h * u are vector valued; i.e., map M+ into Un. We state the result as an exercise. Exercise 1 Let u: U+ ->Un and let u be locally integrable. Let H be the matrix impulse response; so H: U+ -• UnXn. Assume throughout that the elements of H, namely, the h^-ys, are in L1, for i,j=l,2,...,n. Denote by H the linear operator defined by H: M H H u = H * u
where (H * u)(t) = J
o
H(t - T)W(T) dx
Establish the following induced norms: (a) w eL„°°(IR+): \\u\\^ = max,- sup,> 0 |u t (i)|; show that the induced norm of H is
(b) weL n 2 ((R + ): ||w||22 = Jg5 £j»=1 l M *(0| 2 <&; show that the induced norm of H is ||H|| 2 = (i m a x ) 1 / 2
12 where 13
Amax = max max ^t[li(jco)*ti(j(o)] where ^(Af) denotes the (necessarily real) iih eigenvalue of the Hermitian matrix M. (c) ueLn\U+): IMU = Jo E"=i lw*(0l
14
(column sum)
7
7
NORMS
AND
SPECTRAL
RADIUS
27
Norms and spectral radius Letv4eC n X M and
where r(A) is called the spectral radius of A. 2
Exercise 1 Show that for any induced norm, r(A) < \\A\\. For the given matrix A, we may try to find a norm on Cn that gives the "minimum" induced norm for A. In fact, we shall prove the following theorem.
3
Theorem Let Jf denote the set of all norms on C"; then for any A e CnXn, inf [ s u p ( M * | / | x | ) l = r(A) J
UeC"
\-\eJT
By the definition of the infimum, (4) is equivalent to inequality (2) and the statement that for any e > 0 and any A eCnXn, there is a (vector) norm on C" such that the corresponding induced norm satisfies \\A\\
Let J be the Jordan form of A; then there is a nonsingular matrix P J =PAP~X
=A+U
where A is diagonal and all the elements of U are either 0 or 1 with all nonzero elements located on the diagonal above the main diagonal. For some small 3 > 0, define the nonsingular matrix D by
Then DJD'1
=A + 3U
Define a norm | • | on C" by xi-> \DPx\2 where | • 12 is the Euclidean norm on Cn. The corresponding induced norm is ||^41| =max|^4x| =
max
|x| = l
|DPx|2=l
\DPAx\2
II
28
With z 4 DPx, we have
Z>P^(JC
NORMS
= DPAP~xD~xz
= (A + 8U)z; consequently,
2
M|| = max <(A + SU)z\(A + SU)z} < max{|Az| 2 2 + 231
+ <S2|l/z|22}
< r(^) 2 + 2(5r(^) + (52 or \\A\\
::
The trouble with this result is that the vector norm must be specially tailored for the matrix A. It is, however, useful in applications where the problem involves only one matrix; for in that case the norm can be tailored to that matrix. Exercise 2 Let N be a norm on C". Show that xt-+N(Px) is a norm on Cn if and only if P is a nonsingular matrix eCnXn. Show that if P is singular, then p: x\-^N(Px) satisfies axioms (ii) and (iii) of the norms [see (II. 1.1)], but (i) is replaced by (i7) p ( * ) > 0 ,
VieC".
(In such a case, p is called a seminorm.) Exercise 3 (contraction mapping theorem) The following is a statement of the contraction mapping theorem in a form particularly useful for applications. The proof is essentially the same as the usual textbook case. Let (P, d) be a metric space and (^,|| • ||) be a Banach space. Let p0eP and XQ e J1. Consider the closed balls : BP =
{peP\d(p-p0)
B® = {xe &\\\x-x0\\
If (i) f: BP x B<%-> & and/is continuous in BP x BM; (ii) there is some k < 1 such that \\f(P,x)-f(p,x')\\ (iii)
\\f(p, x0) - x01| < (1 - k)rm,
VpeBP, ty?
Vx, x' e Bm
e BP
then (a) Vp G BP, the iteration scheme *n+i(p) =fU>> xn(p)]9
with x0(p) = x0
8
THE M E A S U R E
OF A
MATRIX
29
converges to a unique continuous function x: BP -> Bm such that
x(p)=flP,x(p)l
^peBP
(b) the convergence is uniform inp over BP; i.e., for the given k and the given BP, \\x(p) ~ xa(p)\\ < knrm,
VpeBP,
V/i e Z +
(Hint: Be sure to verify that xn{p) e i?^ for all n > 1, and a l l p e BP.) In general, any function such as /(/?, •)> which maps a set i?^ into a Banach space ^ and which satisfies an inequality like (ii) above, is called a contraction over Bm. Exercise 4 Let A be a continuous linear map from a Banach space E into itself. We say that X e C is a spectral value of ^ iff ^4 — XI does not have a continuous inverse on E. Call the spectrum of A, Sp(^4), the set of all spectral values of A, and the spectral radius of A, the positive number *(A) ^
sup \X\ XeSp(A)
(a)
Show that
(A-Xiy1^
00
-^X^k+l)Ak
k= 0
and that the series converges absolutely [in ££{E\ E)] for |A| > ||^4||. (b) Show that (A - Xiy1 e &(E; E)for \X\> Sp(A). Exercise 5 Let A, B e £P(E\ E), E a Banach space. Show that it is not generally true that *(AB) < \\A\\*(B)9
*(AB) < *(A)*(B)
(Hint: Consider 2 x 2 matrices, one of which is nil potent.)
8
The measure of a matrix
Let | • | be some norm on Cn. Let || • || denote the corresponding induced norm on the n x n matrices in CnX\ The function ||-||: CnXn -> U+ is a convex function; therefore, at every point I e C " x " , ||-|| has a one-sided directional derivative in any direction AeCnXn; i.e., the limit Km(\\X+OA\\e\o exists for all X and all A.
\\X\\)/6
II
30
NORMS
The one-sided directional derivative of ||-|| at / e C nX " in the direction A is called the measure of the matrix A and is denoted by fi(A). Thus, 1
ii(A) ^ l i m ( | | / + 6U|| - l ) / 0 We shall prove the existence of this limit in Lemma (4) below. The concept of the measure of a matrix appears naturally in the study of differential equations. Consider x(t) = A(i)x(i), where A(-) is continuous. Let us bound the one-sided derivative in the positive direction of t H» |(0I> where 0(-) is a solution of the differential equation: D+\$(t)\
2
Mim[|0(* + 0)| - |0(O|]/0
= iim[|M) +0.4OMOI
-\Ht)\]/e
e\o
< lim[||/ + 6A(t)\\ \
/)+|
3
Inequality (3) is sharper than the standard one D+\$(t)\<\\A(t)\\
|0(O|
Indeed, fi(A(t)) may be negative, whereas \\A(t)\\ is always nonnegative. 4
Lemma For any A eCnXn, n{A) is well defined. Proof Call/(0) the ratio in (1); we show that f(6) decreases as 6 \ 0 and that it is bounded below. Hence as 6 decreases to O,/(0) tends to a well-defined limit. Let k e (0, 1). k0f(k6) =\\I + k6A\\ - 1 = |Jfc(J + 6A) + (1 - k)I\\ - 1 < A: ||7 + 0A\\ + l - k - l = k(\\I + 0A\\ - 1)
= W(0) So for ke(0, 1), f(kQ)
::
Properties of n For convenience, we list a number of properties of \i in the following theorem.
8
5
Theorem
6
(a)
THE MEASURE
OF A M A T R I X
31
Let A, B e C"x", and \i be given by (1). Then KI)=h
K-I)=~h
M0) = 0
(Note: fi(A) = 0 does not imply that A — 0.)
7
(b)
8
(c)
fi(cA) = cfi(A)9
9
(d)
/i(4 + d) = fi(A) + c,
10
(e) max[/jL(A) - fi(-B\ (f)
11
-fi(-A)
nXn
fi:C
-^U
+
JI(B)]
Vc e R < fi(A + B) < p{A) + fi(B)
is convex on CnXn
/*[A4 + (1 - A)5] < Xii(A) + (1 - A)/x(£), (g)
12
| ii{A) - KB) \fi(A)-KB)\
13
Mc > 0
(h)
14
\<\fi(A-B)\<\\A-B\\ < \KB-A)\
<\\A-B\\
-fi(-A)
(i) —/i( —v4)|x| < \Ax\
VA G [0, 1]
and
V/e{l, 2, . . . , n) - ^ ) | x | < |^x|, V ^ e C
(Afate: | Ax | w not bounded above by ju(^4) \x\.) (j) Let | • | be a norm on Cn and P e CnXn be nonsingular. Call fiP the measure defined in terms of the induced norm corresponding to the vector norm |-| P defined by x\-> \x\P = \Px\. Then fiP(A) = ix(PAP-1)
15 (k)
If A is nonsingular,
16 Proof
(a) Immediate from Definition (1). (b) The triangle inequality and 9 > 0 give
and
Finally, the relation between fi(A) and — fi{ — A) follows from 0 = ||/ + 6(A - A)\\ - 1 < (||/ + 28A\\ - 1 + ||/ - 26A\\ - l)/2 (c) (8) is true for c = 0 in view of (6). For c> 0, observe that (||/ + 6cA\\ - l)/0 = c(\\I + cOA\\ - l)/cd
32
II
NORMS
Since c > 0, as 0 \ O , so does cO. (d) Consider 11/ + 6(A + cl)\\ - 1 = (1 + eg) 11/ + [9/(1 + c6)]A\\ - 1
Finally, Vc e R, as 0 \ O , 0/(1 + c 0 ) \ O . (e) For the second inequality ||/ + 0(A + B)\\ - 1 = i(l|/ + 20A+I
+ 26B\\ - 2)
<>m + 20A\\ - I) + i(\\I + 20B\\ - I) With 0 \ O , we conclude \i(A + B) < fi(A) + ii(B). The other inequality in (10) follows from the one just proved: 18 19
/JL(A)
= fi(A+B-B)<
fx(-B) + \i(A + B)
fi(B) = ii(A+B-A)< fi(-A) + \i(A + B) (f) Convexity is immediate from (8) and the second inequality (10). (g) The second inequality (12) follows from (7). To obtain the first, replace B by —B in (18) and obtain fi(A) - n(B) < ii{A - B) Replacing — A by A in (19), we obtain li(B) - ii(A) < ti(B - A) and (12) follows. (h) Let e eCn be a normalized eigenvector of A associated with the eigenvalue Xt; so Ae = Xte and ||/ + 0( — A)\\ > \e — 0Xte\. Therefore,
As 0 \ O , the right-hand side tends to Re X{ and the left to —n( — A). To obtain the second inequality, consider
Again as 0 \ O , we obtain the second inequality. (i) For 0 > 0 |,4x| = \x-(x-6Ax)\/6
= \x-(I-6A)x\/9<(\x\
- ||/-0^||
|JC|)/0
8
THE M E A S U R E
OF A
33
MATRIX
where in the last step we noted that the left-hand side is independent of 6. The second inequality (i) follows from the derivation above by changing A into — A and noting that \Ax\ = \ —Ax\. (j)
| | / + 0.4||p ^ sup \x + 6Ax\P
#
|x|P=l
= sup \Px + 0PAP~1Px\ ^ \\I+ ePAP~1\\ \Px\ = l
(k) By (14) -p(-A)\x\
< \Ax\
and
inf \Ax\ =
fl^-1!!)-1
\x\ = l
the first inequality follows. For the second, take the norm of AA~ * = J. •• ••
24
Theorem Let x = (xt, x2,...,
xn) e Cn and A = {atj) e C n X " ; then
Proof The calculation of pt(A) (i = 1, 2, oo) is a simple exercise using the definition. (A* denotes, as usual, the complex conjugate of A.) s: Exercise 1 Show that if A ^ 0 and if A is skew Hermitian, or skew symmetric, then fi2(A) = 0. Exercise 2 Let AeUnXn be a singular, positive semidefinite, symmetric matrix. Show that fi2( — A) = 0. 26
Comments The formulas of Theorem (24) show that p(A) is easy to calculate forp = 1, oo or to estimate for/? = 2. Also fi(A) may actually be smaller than the corresponding ||^||. Also, fi(A) may be negative. As applications of the notion of measure we give two facts. The first one gives upper and lower bounds on solutions of linear differential equations, and the second gives a sufficient condition for the existence and uniqueness of operating points in circuit theory.
34
27
II
NORMS
Theorem Let t n-> A(t) be a regulated function from R+ to CMXn. Then the solution of
28 satisfies the inequalities 29
|x0 0 )|exp - f / i [ - y 4 ( 0 ] * ' l < 1401 ^ l*('o)l exp f /x|>4(0] dt' Proof Let x(0 denote any nonzero solution of (28), let n(i) = \x(t)\ and D+n(t) denote the right-hand derivative of /?(•) at t; then by definition, for all t$ D, where D is the at most countable set of points at which A(-) is discontinuous, D+n(t) = lim[n(t + 0) - n(t)]/0 = lim[ | x(t) + 9A(t)x(t) \ - | x(t) \ ]/0
30 But
Inserting this inequality in (30) and using the definition of /x, we obtain D+n(t) < fi[A(t)]n(t)
31
Since n(f) > 0 for all t, (31) becomes D+n(t)/n(t) < fi[A(t)] and the second inequality (29) follows by integration. The first inequality is proved in a similar manner. :: In many applications one has to find the operating point (or equilibrium point) for a system described by 32
x =f(x) + u n
where w, xeU a n d / : U" -> R . The problem is to find an x e Un such that f(x) = u. It is important to know whether this equation has solutions for any ueUn and whether such solutions are unique. The theorem below gives sufficient conditions for this to be the case. Df(x) denotes the derivative off at x (equivalently, the Jacobian matrix off at x). 33 34
n
Theorem Let/: Un -» Un be continuously differentiable. U.t.c, if there exists a function rn: U+ -> U+ such that li[Df{x)}< - m ( | x | ) < 0 , where m(a) > 0 for all a > 0 and -00
35
m(oc) dx = oo
VxeK"
NOTES A N D
REFERENCES
35
then X K / ( X ) is a C 1 diffeomorphism of Un onto itself [equivalently, / is a continuously differentiate bijection of Un onto Un with a continuously differentiable inverse]; thus, for any u e Un, (32) has a solution x which depends in a C 1 fashion on x. Proof By Palais' theorem our claim will be established if we show that dct[Df(x)] j - 0, V x e r and that | x | ->oo implies that | / ( * ) | -» oo. The first condition follows from the observation that Vy e W with y #= 0, where we used (14) and assumption (34). The second requirement holds true because by using Taylor's theorem about 0, the origin of C", we obtain
f(x)=f(0)+ \f Df(*x)dUx So
\f(x)\ >\^D/ax)d^x\
-|/(0)|
> -n^Df(Xx)
d^\x\
> - fn[Df(Xx)]
dX\x\-
> JCm(X\x\)dX\x\ o
- |/(0)|
by (14)
|/(0) |
by (11)
-|/(0)|
by (34)
= f'lm(a)da- |/(0)| and the second requirement holds as a consequence of (35). Remark The theorem still holds if (34) is replaced by fi[-Df(x)]<
-m(\x\)<0,
Notes and references The discussion of norms and equivalent norms is standard [Yos.2, Die.l, Edw.l, Hou.l]. The examples may be illuminating. The calculations of induced norms were introduced in the engineering literature by Sandberg, in his many papers starting in 1963, and by Zames at about the same time. (See reference list.) The notion of measure of a matrix is due to Dahlquist [Dah.l]. Inequality (8.29) can be found in Coppel's book [Cop.l]. For Theorem (8.32) see the paper by Desoer and Haneda [Des.ll].
III
GENERAL THEOREMS
This chapter is abstract because abstraction saves time. Indeed, suppose that we solve a dozen specific problems and that we detect a common pattern in their solution. It would obviously be useful to abstract from these solutions the essence of those characteristics that lead to the solution; then when encountering a new problem, we would know what to look for. This is, of course, very beneficial, but what is the cost ? The cost is that this process of abstraction requires care. Indeed, in a specific concrete problem, certain features of the problem are well known and well understood; so one takes advantage of them without comment. (In other words, familiarity makes you believe that it is simpler than it actually is.) In an abstract setup, all required properties must be precisely stated, and every step of the reasoning must be carefully scrutinized. On the other hand, it often happens that some extremely simple abstract reasonings lead to very general and useful theorems. A number of such theorems are collected in this chapter. In the first section, we set up the general framework which will underlie our future work. Next we state, prove, and discuss the small gain theorem in its general and in its incremental form. In Section 4, we prove another 36
1
SETTING
OF THE
PROBLEM
37
boundedness result, which has an interesting converse. In the next section, we prove a very general existence and uniqueness result with which we can prove that most of the feedback systems that we shall encounter are determinate in the sense that to a specific input they have a unique response. Section 6 is devoted to the loop transformation theorem, which can be thought of as a " stability equivalence" result. The chapter ends by the general definition of if stability and a general feedback property.
1 Setting of the problem A large number of feedback systems can be put in the form shown in Fig. III. 1. The symbols wl9 u2 , denote the inputs, yt, y2 the outputs, and e1,e2 the errors. The w/s, j/ f 's, and et9s (i = 1, 2) are functions of time; usually they
Figure III.l
are defined for t > 0 or for t e Z+; usually they take values in R, Rn (sometimes, C"), or in a normed space or a Hilbert space. The general problem under investigation is: Given some assumptions on Hl9 H2, show that if ul9 u2 belong to some class, then eu e2 and yt, y2 also belong to the same class. The equations of the feedback system are 1 2
ut = eY + H2 e2 u2=e2H1e1
H1 (H2) is an operator which acts on its input ex (e2, resp.) to produce the output yt (y2, resp.). Equations (1) and (2) have an obvious control interpretation shown in Fig. III.l. It is worthwhile to point out that they have also an w-port interpretation (Fig. III.2 shows the case of two-ports). Let the ut take values in R2 with components (ui)1, (ut)2. The operator H1 is an "impedance" type operator; it operates on currents and gives out voltages.
Figure III.2
III
38
GENERAL
THEOREMS
The operator H2 is an " admittance "-type operator; it operates on voltages and gives out currents. Equation (1) expresses the Kirchhoff current law, and Eq. (2) expresses the Kirchhoff voltage law.
General framework Since we wish to cover continuous-time and discrete-time systems and also systems in which the w(.'s, et9s, and yt's are either real valued or take values in Un or even in normed spaces (for distributed systems), we set up the following general framework. Let ZT\ subset of R+ (typically,
where Ov- is the zero vector in ir. The linearity of PT is obvious from its definition. PT is a projection on 3F since PT2 = PT. We say t h a t / r is obtained by truncating / at T. 5 Introduce a norm || • || on $F\ this defines a normed linear subspace ££ of the linear space ^ :
(Typically, ||/|| = Jg> ||/(0II A or ||/|| = £ | ||/(*)||.) Associated with the normed space «£? is the extended space ££e defined by 6 We shall impose throughout the following requirements on the norm || • || that is used to define if and if e [see (5) and (6) above]: 7 8
(i) (ii)
V/e S£e, the map T\-+ \\fT\\ is monotonically increasing; V/e if,
1
SETTING
OF THE P R O B L E M
39
The following notation will turn out to be useful:
With these two assumptions, we may reinterpret Definitions (5) and (6) of if and S£e as follows:/belongs to if iff the real-valued function T H - ||/ r || is bounded on «^~, a n d / e <£e iff T\-> \\fT\\ maps ^T into U (i.e., there are no growth constraints!). If i^ is either U or Un, and if || • || is an If norm, we denote if by If or L„p; similarly, S£e is denoted by L / or L%e. For the discrete case, we write P, / / , / / , and l*e. Exercise 1 (a) Take if to be L00. L e t / : f M>exp(f2). D o e s / e L 0 0 ? Does (b)
Give examples of sequences in IJ- but not in Z1, in /e°° but not in /°°.
Exercise 2 Consider the restriction of PT to S£ and call it still PT. Show that in terms of induced norms, ||P r || < 1. Thus, PT is a continuous linear operator on ^£, and its operator norm is at most 1. Most models used in system theory are nonanticipative; so we define nonanticipative maps. Let H: if\ -> if' e ; H is said to be causal {nonanticipative) iff 9
PTHPT = PTH,
VJe^
Exercise 3 Let Hl9 H2: i f \ - » if'e and be nonanticipative. Show that HlH2 is also nonanticipative. {Hint: Use the fact that the composition of two functions is associative.) Exercise 4 Let h e l}(U) and u e L2{U). Show that the map H: L2{U) -» L2{U) defined by i.e., (#K)(0 =
h{t-
T)W(T)
rfr,
fe U
is nonanticipative if and only if h{i) = 0 , almost everywhere on (—oo, 0). Exercise 5 Assumption (7) on the norms ||-|| in 3F is not necessarily fulfilled by any norm. Construct an example. [Hint: Consider the space of measures JJ, on M+ ; define the norm as in Appendix (C.3.3); show that for s o m e / / , \\fiT\\ > \\fi\\.]
Exercise 6 Within the general framework above, assumption (7) need not always hold. Let ^£ be the class of functions having well-defined Fourier
40
Ml
GENERAL
THEOREMS
transforms [by Appendix B, J ^ Z D L 1 , if D L 2 , and l e t / r = PTfe S£\ define the norm of/ r by
JS?9<5(0, . . . ] . Let fe
&e
where / r is the Fourier transform of fT. (a) Show that f o r / e L2(IR), the relation above defines a norm. (b) Let/(f) = e?l(-t)-0.25(t - 10~3), where 1(0 denotes the unit step. Show that ||P 0 /ll=ll/o II > 11/11 • Exercise 7 Some interconnections of nonanticipative subsystems are not nonanticipative. Give an example. [Hint: Consider a discrete system with unity feedback; uu eu yt: Z+ -> U, u2 = 0, H2 = identity map, (H^Jin) = — ex{n) + et(n — 1), for all neZ+ .] Discuss what happens if Hx is perturbed a little so that (H^Jin) = - ( 1 + &)e1(n) + ex(« - 1), with £ <^ 1. Exercise 8 Let HUH2: &e-* &e. If Hx is a linear map and if (/ + H1H2Yl and (I + H2H1)~l are well-defined maps from J2?e into J^ e , show that HX(I + # 2 HJ-1
= (J +
H.H.y'H,
Give a system theoretic interpretation in terms of block diagrams. Give an example to show that this relation is false if Hx is not a linear map. Exercise 9 In the general framework above, suppose that u2 is identically zero. Suppose that (J -f H2H1)~1 is a well-defined map from S£e into <£e, then show that, for the feedback system shown in Fig. III.l and described by (1) and (2), we have
Exercise 10 Consider an alternate definition of nonanticipative map. Again let H: !*£'e-> ££'e; then H is said to be nonanticipative iff MT e $~ and Vx, y e <£e. 10
PTx=PTy=>PTHx=PTHy Show that definitions (9) and (10) are equivalent.
2
Small gain theorem
The small gain theorem is a very general theorem, which gives sufficient conditions under which a "bounded input" produces a "bounded output." Its formulation is chosen so that the question of boundedness is completely disconnected from the questions of existence, uniqueness, etc.
2
SMALL GAIN
THEOREM
41
1
Theorem Consider the system shown in Fig. III.l. Let Hl9 H2 : J^ e -» S£e. Let ex, e2 e ^ e a n dflfe/z^eWi and i/2 by
2
ut = et + H2 e2
3
u2= e2 — H1e1 Suppose that there are constants
U.t.c, if y^2 < 1, then
(ii) if, in addition II " i l l , II " 2 II < ° ° then et, e2, yu y2 have finite norms, and the norms of the errors, namely, ||et\\ and ||e 2 ||, are bounded by the right-hand sides of (6) and (7), provided all subscripts T are dropped. 8 Given an operator Hi: S£e -* S£\, suppose that there are real numbers px and ft such that 8a Clearly ft is not uniquely defined by the above inequality. Intuitively, it is clear that we are interested in the smallest ft that "works." More precisely, we call gain of Hl the number y{H^) defined by 8b
yiHJ = inf {ft e U+ | 3j5x s.t. inequality (8a) holds} This infimum will often be denoted by y(Hx) or yv With this terminology, Theorem (1) can be interpreted to state that if y(Hx)y the gain of Hl9 andy(// 2 ), the gain of H2, have a product smaller than 1, then, provided a solution exists, any bounded input pair (wl9 u2) produces a bounded output pair, and the map (wl5 u2) -> (yl9 y2) has also finite gain. Exercise
Show that if H± is causal, then condition (8a) can be replaced by
8c Note that, in (8c), since x ranges only on ££, T drops put from the picture. 9
Comments (a) Generality Theorem (1) applies to continuous-time or discrete-time systems. The system may be single-input-single-output (i.e.,
42
III
GENERAL
THEOREMS
if = U) or multiple-input multiple-output (i.e., if = Un) or any normed space (distributed systems). (b) Statement The theorem assumes the bare minimum; it assumes ei, e2e <£\ and defines ux and u2 accordingly. In practice, ul9 u2 are given and et, e2 are calculated. Thus, the formulation above avoids the question of existence of solutions of (2) and (3). In practice, the constants pt and yt are obtained by bounding
For this purpose, inequality (II.8.29) and the Bellman-Gronwall lemma Appendix (E.l) are very useful. (c) The form of the assumptions are improved versions of results of Sandberg [San. 6, 7, 8, 9, 10, 11], Zames [Zam. 3], Lee-Desoer [Lee. 1]. The usual formulation sets /?x = 0 and therefore does not apply to the memoryless nonlinearity (f): U -> R, where
Nor does it apply to some cases of relays with hysteresis, and to saturating nonlinearities, whereas the present one does. (d) The results extend to the case where H1 and H2 are relations (i.e., "multivalued functions"). In that case, (4) and (5) must hold for all the "images" Hte t of each et e <£e (i = 1, 2). This fact is very useful in the case of feedback systems with hysteresis. (e) Theorems (II.6.3), (II.6.8), and Exercise 1 of Section II.6 give very useful evaluations of gains for convolution maps. Exercise 1 With Theorem (1) in mind, construct an example (say, of the type studied in Chapter I) for which existence and/or uniqueness and/or continuous dependence fails even though the product of the gains 7(// 1 )y(// 2 ) < 1. Proof of Theorem (7) 10
From (2), we see that VTe &~, e1T = uiT — (H2e2)T
Since all vectors e S£e, 11 We have a similar calculation from (3) 12 Hence, using the fact that y2 > 0, 13
2
SMALL GAIN
THEOREM
43
Since yly2 < 1, 14
\\e1T\\ <(\\uiT\\ + y2||w2T|| + £ 2 + y 2 &)(l -
y1y2y1
The remainder follows immediately. Exercise 2 Consider a feedback system where w1? el9 yx, e 2 , y 2 : //"i and H2 map L°°([R+) into itself; the input u2 is identically zero. More precisely, H1e1 = g * el9 where g is a given function in L*(IR+); (H2e2)(t) = (p[e2(t), t]9 where 0 : [ R x [ R + - » I R i s continuous and belongs to the sector [0, k). Assume that for all uy eL°°([R+), there is a unique solution el9 yx e L°°([R+). Obtain conditions on the constant k and the function g(-) such that u1eLx>(R+) implies j ^ e L°°(IR+). Under the conditions found, exhibit a number m such that
Exercise 3 Repeat Exercise 2 for discrete systems, i.e., uu eu yi: Z + -» [R; ^ and // 2 map /°° into itself;... Exercise 4 Consider a feedback system where u2 = 6 and // 2 = J, the identity; thus (I + H1)el = wx. Suppose that / ^ : if -• if and that ( J + / / i ) - 1 : S£-+y?. Show that under these conditions, y[(7 + / / J - 1 ] < oo if and only if 3 constants s, and /?/ with e > 0 such that If u2 = 0, there is no longer need to separate the gain of H1 and the gain of H2; one can consider the "loop gain" H2Hl. The following corollary illustrates the idea. 15
Corollary Consider the system shown in Fig. III.I, where u2 = 0. Let Hu H2 : S£e -> if e . Let ^ e if e and define wx by
16
ux = e± + H2e2
17
Suppose that there are constants y21, y1? /? 21 , and /?1? with y21>0, s.t.
U.t.c, if "y2i < 1J then
yx> 0,
44
III
GENERAL
THEOREMS
Note that Comments (9) above apply also to the corollary. What the corollary shows is that once the "loop gain" is smaller than 1, the closedloop system has finite gain. Compare the corollary with Theorem (1.2.7). Proof
3
Exercise for the reader.
n
Small gain theorem: incremental form
We now consider a theorem that guarantees existence, uniqueness, boundedness, and continuity. 1
Theorem Consider the feedback system shown in Fig. III. 1 and described by Eqs. (1.1) and (1.2). Assume that for each T, PT^e ls a complete normed linear space (i.e., a Banach space). Let Hx, H2\ $£e-+ $£e. Suppose there are constants yx and y2 such that VTe &~ and V{, £'e S£e
U.t.c, if 7i72 < 1> then (i) Vwl5 u2 e J£\ 3 a unique solution eu e2, yu y2e <£e, which can be obtained by iteration; (ii) the map (wl5 u2)\-^(el, e2) is uniformly continuous on PTJ£ex PT£>e<md on if x J^7; (iii) If, in addition, the solution corresponding to wx = u2 = 9 is in $£\ t h e n ul9 U2G ££ => ex, e2 e «£?. 4
Comments (a) If H1 is linear, assumption (2) is very closely related to assumption (2.4). More precisely, if Ht is a linear map and if for some yt
then
(b) The incremental small gain theorem answers all four questions: existence, uniqueness, boundedness, and continuous dependence. (c) It is a simple exercise to prove that (2) and (3) imply that H1 and H2 are nonanticipative. 5 Given a nonanticipative operator H1: S£e -> $£e which satisfies (2), the inflnium of all real numbers yt which satisfy (2) is called the incremental gain of Hv It is often denoted by y(#iX °r for simplicity, yt.
3
Proof
(i)
SMALL GAIN T H E O R E M : INCREMENTAL
FORM
45
From the feedback Equations (1.1) and (1.2), we obtain e2 =u2 + i/1(w1 -
H2e2)
and by truncation e2T = u2T + [H^u, - H2e2)]T, Using nonanticipativeness of Hx and H2, e2T = u2T + {i/Jt/xr This equation is of the form e2T =f(e2T). on indeed,
(H2e2T)T]}T We claim that f is a contraction
\\ where we used successively assumptions (2) and (3). By assumption y ^ < 1, consequently / is a contraction. Therefore, V T e ^ , Vul9 u2eJ£e, the resulting e2T is uniquely defined element of PT^e. The same holds for e1T since by (1.1) and nonanticipativeness of H2, we have
(ii) For any Te^~, for any (ul9 u2), {ux\ u2')eJ£e unique solution (ex, e2), (ex\ e2) e $£e x £ge. Also
e\T = w'1T -
x «Sfc, there is a
(H2e'2T)T
Subtracting, taking norms, and using the triangle inequality, we obtain \\e1T - e'lT\\ < \\u1T - u[T\\ + y2\\e2T - e2T\\ Similarly, \\e2T ~ e2T\\ < \\u2T - u2T\\ + yilki T - e'1T\\ 6
||e 1T - e'lT\\ < (\\uiT - u'lT\\ + y 2 ||w 2r - u2T\\)(l - yiy 2 ) _1 , Hence (ii) follows. (iii) follows from (6) and the corresponding inequality for e2 when w/ = u2 — 9 is inserted. » Exercise 1 Formulate and prove a corollary to Theorem (1) which is the analogue of Corollary (2.15) to Theorem (2.1). Exercise 2 Consider a system that satisfies all the conditions of Theorem (1). Suppose that ul9 uu i/ 2 , u2, e <£\ but that ux — ii1 and u2 — u2e $£'. Show that et — e2 e <£ and
III
46
4
GENERAL
THEOREMS
A boundedness result
We may think of the small gain theorem and its incremental form as two extreme forms of boundedness results. The former assumes the least— the finiteness of the gains [see (2.4) and (2.5)]; the latter assumes the most— finite incremental gains for Hx and H2 [see (3.2) and (3.3)]. We now state and prove a theorem that uses somewhat intermediate assumptions. It emphasizes assumptions on the composite function H2H1, which, in a very rough way, can be thought of as the loop gain. 1
Theorem Consider the system shown in Fig. III.l and described by (1.1) and (1.2). Assume that H1 and H2 are nonanticipative, that the gain yiH^ < oo and the incremental gain y(H2) < oo. U.t.c. (a)
if 3 8 > 0 s.t.t
2 then (wl5 u2) e if x j£? implies that any solution (el9 e2) that is in ££\ x S£e is also in S£ x 5£ and for some k < oo 3
\\et\\
i = l,2,
(b) If the relation between (uu u2) and (ex, e2) is a nonanticipative map from if x if into <£ x <£ which has finite gain, then (3) implies (2). Proof (a) Consider fixed wl9 u2 e if and a corresponding solution (eu e2) of Eqs. (1.1) and (1.2) with eu e2 e ££e. Then e1 + H2 H1e1 =ul — [H2(H1e1 + u2) - H2 HYe^\ Upon truncation, since PT is linear PT(I + H2 HJe, =PTu1-
PT[H2(Hiei
+ u2) - H2 H&]
Using the fact that the incremental gain y(H2) < oo,
Now using the assumption (2) and letting T-> oo in the right-hand side, we obtain with the help of (1.8)
Thus, 5 t To lighten the notation, we write ||«2llr instead of ||«2rll-
5
AN
EXISTENCE A N D
UNIQUENESS
THEOREM
47
i.e., e1 e ££. Now from e2 = u2 + H1el, we obtain 6 Therefore, (3) is established, and k may be taken to be
(b)
Let u2 = 0; then to wx e if, there is a unique ex e J£ such that (I + H2H1)e1
=Ul
From (3) it follows that y[(I + H2H1)~1] < oo; therefore, VTe ^
from which (2) follows.
::
Exercise 1 Consider the following special case to show that condition (2) is closely related to the Nyquist criterion. Let uu eu yt map U+ -> U; let u2 = 0 and // 2 ^ e the identity map. Let H1el = g * eiy where ^eL 1 (IR + ) and its Laplace transform g(s) is a rational function which is analytic in Re s > 0 and which tends to zero as \s\ -^ oo. Show that if the Nyquist diagram of g [i.e., the map co)-^g(jco), co e U] encircles one or more times the critical point (—1, 0), then (2) does not hold. [Hint: Use a state representation for H1\ choose appropriate initial conditions and an appropriate exponential form for et( •).]
5
An existence and uniqueness theorem
In this section, we restrict ourselves to !T = R+ . Indeed, for the discretetime case, the problem of existence and uniqueness is completely different. Instead of solving for a function from IR+ to i^, we have to solve, at each sampling time, for one point in ir. An example of result required for the discrete-time case, where x(n) =f[x(n)] + u(n) x(n), u(n) E Un,f: Un -• Un, is given by Theorem (II.8.33). We wish to show that the existence and uniqueness of the solution (ei9 e2) of Eqs. (1.1) and (1.2) in the space S£e x <£e can be guaranteed on the basis of some quite reasonable assumptions, which are usually far less restrictive than those required for guaranteeing finite gain.
48
III
GENERAL
THEOREMS
To simplify the notation, it is convenient to think of the pairs (w1? u2) and (el9 e2) as single elements of if e x if e , denoted, respectively, by u and e. Similarly, instead of the maps Ht and H2, let us think in terms of the map defined by Then Eqs. (1.1) and (1.2) become 1
u=e + He Furthermore, no confusion will arise if we abbreviate <£e x $£e by ^£ 2. The following theorem will require a Lipschitz-type condition on H and will guarantee that for any u e $£ 2, there is one and only one solution ee<£? of(l).
2
Theorem Let i/: if e2 -» if e 2 and be nonanticipative. Let 5" = IR+ . Let P r be the projection defined on if e 2 as in (1.4). Define a new projection by
3 U.t.c, if for all compact intervals I c U+9 there are numbers y(I) < 1 and A(J) > 0 such that W e I and Ve, e' e ifc2 subject to Pte = Pte\
(where for simplicity we have suppressed the dependence of A and y on J) and if W and A, Pt A ife2 is complete, then for any u e ife2, the equation u = e + He has one and only one solution. Proof Suppose we have computed a solution up to some t > 0, let us show that the solution can be extended uniquely on all M+ . Let / be a compact interval including t and such that J n (t, oo) is nonempty. Now for the A and y corresponding to the chosen /, let 5
eA^PttAe=Pt+Ae-Pte and uA be similarly defined. Now apply Pt A to u = e -f He,
6
eA = uA-Pt>AH(Pte
+ eA)
where we used the nonanticipativeness of H. Equation (6) is of the form eA =f(eA), since uA is given and Pte is known. Thus,/: Pt A ife2 -» Pt A ifc2. We claim that / is a contraction on PtiA^e2- Indeed, for all eAt eA e Pt,A^e2^ 7
\\f(eA) -f(eA')\\
= \\Pt,A[H(Pte + eA) - H(Pte + 2A')]||
5
AN EXISTENCE A N D
UNIQUENESS
THEOREM
49
where we used successively the linearity of Pt A , (5), (6), and inequality (4). From (7), we conclude t h a t / i s a contraction on the (complete) normed space PtAS£'e2 since y(I) < 1 by assumption. Therefore, the iteration scheme eAn+1)=f(eAn)),
i = 0 , 1,2, . . .
starting from any initial guess, eA°\ will converge to the unique solution on (t, t + A]. Thus, by starting at t = 0 and by repeating this process, we can construct piece by piece the unique solution of Eq. (1) on (0, oo). :s Corollary Suppose that, in addition to the assumption of the theorem, we assume that His linear; then using the notation of (5) to (7), the condition (4) becomes: For any compact interval /, there exist y(I) < 1 and A(7) > 0 such that for all t e /, and for all eA e PtA &e2. Proof
Follows directly from (7).
::
Comments (a) The most important example of an engineering system that does not fulfill such condition is the chattering phenomenon in servos [Flu. 1 ]. The simplest example is where Ht is an integrator (i.e., Hx is defined by its transfer function \/s) and H2 is a memoryless nonlinearity with characteristic (j)(e2) = 1 for e2 > 0 and <j)(e2) = — 1 for e2 < 0. (b) In the assumptions of the theorem above, the order of the quantifiers is very important. For any chosen compact interval /, the same y(I) and A(J) must " w o r k " for all t e I. The solution is constructed over successive intervals of length A(J) > 0; so, eventually, the process moves out of the compact interval /. Then a new compact interval is chosen, say, /', for which there is a A(T) and y(If).... Clearly, the process will never stop because, by assumption, for any compact interval J, there is a A(J) > 0 which " works " for all t e L (c) If H is time invariant, successive compact intervals can be chosen to be translates of the first one, and the same A and y can be used throughout. For some time-varying problems, as the solution is continued further to the right, A(7) must be chosen smaller and smaller in order to achieve y(I) < 1 in (4). Then there is no unique A that works for all t e U+. Exercise 1 Consider the special case where (a) the feedback loop is open (i.e., H2 maps everything into the zero vector); (b) wx(r) =
III
50
GENERAL
THEOREMS
( c ) Ji = Hxex is defined in terms of a differential equation yx =f(yl9 t) where / i s continuous o n R " x H " x R + .
eu
Display assumptions o n / s u c h that the input output map ul\->y1 satisfies (4). Exercise 2 Let ul9el9 e2 map IR+ into U. Take u2 = 0. Suppose that (H^^if) = 4>(ei9 t), where <£(•, •) is continuous and the conditions for some y < oo, satisfies, Suppose that (H2e2)(t) = (g * ^)(0» where # eL e 1 ([R + ). Show that for any wx eLe°°, there is one and only one solution el and e2 eL e °°. [Hint: Consider the equation ex(t) = ux(t) — (g * e2)(t), where e 2 (0 = ^[^i(0? *] • • • Note that / : /I'—> Jo \g(r)\ dx is continuous with/(0) =0.] Exercise 3
Consider the equation e(t) = u(t) + f G(t •'o
T>(T)
rft
where w: U+ -> [Rn, ueLpne, for some/? 6 [1, oo]; also G: [R+-^[RnXn and all elements of GeLj^. Show that Eq. (10) has one and only one solution e(-)eLpnefor each ueLpne. (Hint: t\-+\G(t)\ is in Lelm9 fh->J{, | G ( T ) | rfr is continuous, monotonically increasing, and equal to zero at t = 0, . . . .) Exercise 4 Consider the equation
where w: R+ -> R", ue L™; also / : IR+ x IR" x Rn is continuous and satisfies a global Lipschitz condition, namely, there is a /c e U+ such that
Show that Eq. (11) has, for each ueL™, one and only one solution e e L™ .
6
Loop transformation theorem First, let us make the following observation.
Exercise. 1 Let H1 and K be maps from S£\ into S£e. (Note that neither Ht nor # need be linear.) U.t.c, (/ + KH^1: <£e -* seeoH1(J + KH^'1: <£e-+ <£e. [=> is immediate. For <=, consider the identity
6
LOOP T R A N S F O R M A T I O N
THEOREM
51
Exercise 2 Let E, F, and G be linear spaces. Let g: G-+ E; let fx and f2 map E into F. Show that (fi + fi)09=fi
°g+f2°g
[Hint: By definition (/x + / 2 ) 0 ) = / i W + / 2 (*)-] Show that if E = F = G, in order to have 9°(fi
+fz) = 9°fi
+ 9 °fi
g must, except for trivial cases, be a linear map from E into Zs. Consider the feedback system S shown in Fig. III. 1 and described by 1
ux = e1 + H2 e2
2
u2=e2-
H1e1
We shall derive from the given feedback system S another feedback system SK such that S is " stable " if and only if SK is " stable." It will turn out that some straightforward theorems, such as (2.1) and (3.1) above, applied to SK lead to very interesting stability conditions for S. The system SK is shown in Fig. III.3 and will be derived in the proof of the theorem in the following.
Figure III.3
Theorem Let // 1? i / 2 , K, and (I + KH^'1 linear, U.t.c,
map if e into jSfc. Let ^ be
(a) if wx, w2, el9 e 2 a r e m &\ a n d a r e solutions of S [i.e., of Eqs. (1) and (2)], then uY — Ku2, w2, >7i — (/ + KH{)eu and
Since K is linear, we obtain from (2) Ke2 = K(u2 + Hxe^) = Ku2 + KH^e^
52
III
GENERAL
THEOREMS
Using this result in (1), we obtain successively ex = ut — (H2 — K)e2 — Ku2 — KH1el and 5
> / ! - ( / + KH^
= (ut - Ku2) - (H2 - K)e2
Writing ex in terms of rjl9 6
ex =(I +
KHlYlr\l
we obtain from (2) 7
e2=u2
+ Hl(I + KH1Yir}l
Equations (5) and (7) represent the system SK shown in Fig. III.3; indeed, by inspection, ex is related to rjx by the first equality of (5). (a) follows from the derivation of (5) and (7) from (1) and (2). (b) is obtained by tracing this derivation in the reverse direction and by invoking Exercise 1. (c) is immediate. Finally, (d) follows, once it is recognized that the linearity of K was used only once, namely, in obtaining (4). 8
Comment Strictly speaking, it is not necessary to assume that (7 + KH^'1 is defined on S£e. An examination of the proof will show that we need only that the feedback system of Fig. III.3 be such that rju e2, yxe S£\ for any input pair (wl9 u2) e =£?e2, equivalently, that the equations m = (ui ~
Ku
i) ~ (H2 - * > 2 ,
e2 = u2 + yt,
yt = Hl(rjl - Ky\)
have a unique solution (rjt, e2 , yx) e <£e x $£e x if c . Exercise 3 Give a circuit theoretic interpretation of the loop transformation shown in Fig. III.3 in terms of the «-port model of Fig. III.2.
7
Stability
Throughout the remainder of the book, we shall consider systems of the form shown in Fig. III. 1 and described by (1.1) and (1.2). Let us note that the stability theorems above such as Theorem (2.1); Theorem (3.1); Exercise 1, Section 3; and Theorem (4.1) not only guarantee that whenever wx and u2 belong to =£? then eu e2, yu y2 also belong to i>?, but more than that they guarantee that there is a constant k such that
The important point is that k is independent of ut and u
7
3
<£ S T A B I L I T Y
53
Therefore, from now on we shall say that a system described by (1.1) and (1.2) is j£? stable whenever there is a constant k < oo such that (1) and (2) hold. For example, the loop transformation theorem (6.3) essentially gives conditions under which the system S is <£ stable if and only if the system SK is S£ stable; it is an <£ stability equivalence theorem.
Relation between stability and instability Provided the term "instability" is suitably narrowly defined, the observation in the following exhibits when "instability" occurs. 4
Observation Let H be nonanticipative and map J2?e2 into 5£ 2. Let (J + H)~l be nonanticipative and map ife2 into J£?e2. U.t.c. the feedback system (J + H)e = u is "stable" in the sense that it has the properties: (a) e is given in terms of u by nonanticipative map from <£2 into 5£2; (b) 3y < oo s.t. \\e\\ < y\\u\\, for all u e if2, if and only if (/ + H) has a nonanticipative inverse which maps if 2 into J£?2, and (J + H)~l has finite gain. <= Immediate from (/ 4- H)~1u — e. => the map u\-*e is, by assumption, nonanticipative and sends u e S£2 into an e e if2. ::
Proof
The relationship between stability and instability can be clarified by the following tree of dichotomies; to emphasize these dichotomies, we label the cases by sequences of zeros and ones as in Boolean algebra. 0:
(/ + Hy1
cannot be defined on all of if2.
Then the problem is not well formulated because for some u e if2, one cannot define e, i.e., the equations have no solution for some inputs e<£2. 1: (/ + H)'1 is defined on all of i^ 2 . 1.0: (/ + i/) _ 1 [if 2 ] is a strict superset of ^2. Thus, for some 1.1: 1.1.0: y[(I + H)~1] = oo. Then there is a sequence (w,) c S£2 with the corresponding et e <£2, for all /, but . 1.1.1: 7[(/ + / / ) _ 1 ] < oo. 1.1.1.0: (/ + H) ~i is noncausal. 1.1.1.1: (I + H)-1 is causal.
54
III
GENERAL
THEOREMS
Only for the case 1.1.1.1 do we have «£? stability. In Chapter V, some classes of unstable feedback systems will be demonstrated by showing that either 1.0 is the case or that 1.1.1.0 is the case.
8
General feedback formula
If H1 is linear, there are two ways of writing the relation between wx and yt; H2, however, may be nonlinear, This flexibility is very useful in many computations. 1
Theorem Consider the system shown in Fig. III. 1. Let/f1? H2 , (/ + H1H2)~1, and (J + H2 # i ) - 1 map S£\ into $£e. U.t.c, if u2 = 0 and if Ht is linear, then
2
= (I + H1H2y1H2
yt= Ht(I + H2 H^u,
u,
Proof 3
H±+ H1H2 H1=I
H.+ (H.H^H,
= (I +
H.H^H,
where we used the fact that the composition of two functions is associative and then the definition of the sum of the two functions. Furthermore, since H1 is linear, 4
H1+HiH2H1
=Hl(I + H2H1)
Combining (3) and (4) and, composing it on the left with (7 + and on the right with (/ + H2 # i ) ~ \ we obtain (2).
H^H^'1 ::
Exercise Consider a feedback system as shown in Fig. III.l when uu ex, y1: IR+ ->1RM; u2(t) =6, Vt. Suppose that Hx consists of a given "plant" P preceded by a "compensator" G. Thus H1 — PG. We assume throughout that P and G are linear. Since 5
y1=PG[I
+ H2PG]-1u1
consider a "comparison system" consisting of the same "plant" P preceded by a compensator G0 ; call y0 the output of this open-loop system: 6
y0= PG0 u If we choose
7
G0=G(I
+
H2PnGy1
Then, when P =Pn, y0 = y1 for all u. Consider now an arbitrary change in P; more precisely, let P change from Pn to P = Pn + AP. Then, by (6), Ay0 = APG0 u
NOTES AND
(a) 8
REFERENCES
55
Show that Ay, =(I + PGH2y1
Ay0
(b) Prove (8) directly, for AP small, by taking derivatives in function spaces. [Hint: Note that PG(I + H2PG)~1 = (I + PGH2)~~lPG.] 9
Comment This formula is important because it shows that if the " loop gain" PGH2 is large, then a given change AP in the plant will cause a much smaller change in the closed-loop system (5) than in the open-loop system (6). This fact is fundamental in the engineering usefulness of feedback systems.
Notes and references The general setting of the problem is due to Sandberg [San.3, 4, 5, 6, 9] and Zames [Zam.l, 2, 3]. A detailed discussion of causality can be found in Saeks [Sae.l, 2]. The small gain theorem has been used in many ways by many people with varying degrees of generality. In its two general forms, it appears in Sandberg [San.9] and Zames [Zam.3]. The existence and uniqueness result is a formalization of the conventional argument for Volterra integral equations of the second kind (see, for example, Zames [Zam.l]). The loop shifting theorem has a long history. It was known to and used by Sandberg [San.2, 9] and Zames [Zam.3]; the fact that linearity is no longer required if u2 = 0 is due to Sandberg [San. 10]. Equation (8.8) is a wellknown result of sensitivity theory [Per.l]. The boundedness result of Section 2 is due to Willems [Wil. 5].
IV
0
LINEAR SYSTEMS
Introduction
Most of the results concerning nonlinear feedback systems are based, in some manner or other, on properties of liner feedback systems. Therefore, it is important to develop the best possible results for linear feedback systems and to have them in a form that makes them readily applicable to the nonlinear case. This is the main purpose of this chapter. In the development that follows, the input «(•) and the output y(-) of the feedback systems under consideration take values in Un.lfn = l, we refer to these systems as single-loop feedback systems or as scalar systems, or, more precisely, as single-input-single-output systems. If n > 1, we refer to them as multivariate systems or as multiple-loop feedback systems, or, more precisely, as w-input-«-output systems. Wherever possible, all results are stated for multivariable systems, and results that are specifically applicable to scalar systems are stated only to the extent that the main contrasts with the multivariable case can be exhibited. 56
1
SYSTEMS WITH
RATIONAL TRANSFER
FUNCTIONS
57
Sections 1, 2, and 3 consider exclusively time-invariantlumpedsystems, i.e., systems with rational transfer functions. The thrust of Section 1 is to contrast the properties of scalar systems versus those of multivariable systems [see Theorem (1.12)]. Section 2 develops, for the same class of systems, the factorization method for obtaining necessary and sufficient conditions for the stability of multivariable systems and provides much of the insight that is later needed to study multivariable distributed systems. Section 3 considers the case where the feedback path contains dynamics and investigates the problem of " p o l e " and " z e r o " cancellation. Section 4 turns to distributed systems and develops the basic necessary and sufficient conditions for stability that are used throughout the rest of the book. Important notions such as pseudo-right-coprime factorizations are introduced and comprehensive results are obtained for the practically significant case of a multivariable system whose transfer function has a finite number of poles in the closed right-half-plane, but is otherwise stable. The graphical test for the stability of distributed systems is given in Section 5. Three cases of increasing generality are considered in succession, and the most general results are given without proof; the proof would require too long an excursion into the theory of almost periodic functions. Section 6 develops the main results concerning discrete-time systems, and the parallels as well as the contrasts with the continuous-time case are indicated. In Section 7 we present the basic results concerning linear time-varying systems, including necessary and sufficient conditions for "open-loop" stability, and a perturbational result for both continuous-time and discrete-time systems. Section 8 uses Liapunov theory to give bounds on the rate of variation of a nonlinear differential system that insure overall exponential stability when the "frozen" system is exponentially stable. This formulation has the advantage that it reduces to well-known conditions in case the system is linear. Finally, Section 9 studies the problem of linearization: Under what conditions is it true that the linearized version of a nonlinear differential system gives good results on all of R+ ? Among other things, the results obtained justify the use of small-signal equivalent circuits in analysis and design and the use of variational equations in control and optimization problems.
1
Linear feedback systems with rational transfer functions
Notation U[s] = commutative ring of polynomials in the (complex) variable s with coefficients in U. It can be shown that U[s] is a principal ideal domain. U(s) = (commutative) field of rational functions in s with coefficients in U.
58
IV
LINEAR
SYSTEMS
R[.s]mX*, (R(^)WM) = class of allmxq matrices whose elements are in U[s] (U(s), resp.). If N then N(s) is called a polynomial matrix. C denotes the extended complex plane (i.e., the complex plane including the point at infinity.) It is understood, once and for all, that the ratios of polynomials s+l s(s + 2)
and
(s + l)(s - 1) s(s + 2)(s - 1)
represent the same rational function / and that / has, in C, — 1 and oo as zeros and 0 and —2 as poles. The two expressions above represent the same rational function precisely in the same way as the expressions 0.5 and \ represent the same real number. We consider two cases: first, the single-input-single-output, linear, timeinvariant feedback system whose feedback is / and whose open-loop transfer function is g(s)eU(s); second, the n-input-n-output, linear, timeinvariant feedback systems whose feedback is the matrix FeUnXn and whose open-loop transfer function is G(s) e UnXn(s). As shown in Fig. IV. 1, u is the input, y is the output, and e is the error; the basic equations are 1
y = G*e
2
e =u — Fy
Figure IV.l
We say that G(s)eUnXn(s) [g(s)eR(s)] is proper (strictly proper) iff all elements of G(s) [g(s) itself] are bounded at oo (tend to zero at oo, resp.). 5
A rational transfer function H(s) is said to be exponentially stable iff (i) it is proper, and (ii) all its poles have negative real parts.
6
Fact If H(s) e U(s)n x n is exponentially stable and strictly proper, then the corresponding impulse response H(t) and its derivative H(t) have the following properties:
1
SYSTEMS
WITH
RATIONAL TRANSFER
FUNCTIONS
59
(a) There are constants hm > 0 and a > 0 which depend on the H considered such that (b) There are constants hmd > 0 and a > 0 such that denotes the norm of the matrix H(t) e UnXn, this norm being induced by some chosen norm in Rn.) Proof
Immediate by partial fraction expansion.
::
To summarize the input-output properties of systems with exponentially stable transfer functions, we state the following theorem. Theorem Let the closed-loop transfer function H(s) eUnX n(s) be exponentially stable and strictly proper; then (a) for any minimal representation of H(s) (i.e., any minimal state representation of the form x = Ax + Bu, y = Cx), the equilibrium point x = 0 is globally, exponentially, uniformly (in 0 stable: (b) if u e Ln1, then y = H * ue L* r^ Ln™, y eL„, y is absolutely continuous and y(t) -• 0 as t -> oo; (c) if u e L„2, then y = H * ue Ln2 n Ln°°, y e L„2, y is continuous and y(t) -> 0 as t -> oo; (d) if u e L„°°, then y = H*ue L^, y e Lrt°°, and y is uniformly continuous; (e) if u e Ln°° and if, as t -> oo, i/(0 -> w^ , a constant vector in IR", then, in addition y(t) -> Hifyu^ as f -> oo and the convergence is exponential; (f) if w e L / and 1 < p < oo, then y = H * ue Lnp and 3) e L / . Proof Follows immediately from the results of Appendix C. The asymptotic results of (e) follow from Laplace transform. :: Exercise 1 In the case of z transforms, we say that the rational transfer function H(z) is exponentially stable iff it is proper (bounded at infinity) and has all its poles with absolute value smaller than 1 (i.e., in the open unit disk). Formulate and prove the inequality that corresponds to (7) and a theorem analogous to Theorem (9). We consider now the relation between the properties of the open-loop transfer function G(s) and the poles of the closed-loop transfer function. This subject is basic to stability questions. Fact Let H(s) and G(s) e UnXn(s) and be related by (4); then / - FH(s) = [I + FG(s)]~l
60
IV
LINEAR
SYSTEMS
If, in addition, det F =£ 0, pc e C is a pole of H(s) if and only if pc is a pole of
[I + FG(s)Y\ Proof Equation (11) is immediate by calculation from (4). The conclusion follows since det F # 0. :: 12
Theorem Let g(s) e U(s) and G(s) e U(s)n x". Let C be the extended complex plane. Define h(s) and H(s) by (4). U.t.c.: (i) pc e C is a pole of h(s) if and only if pc e C is a zero of 1 + g(s). (ii) If pc e C is a zero of det[7 + FG(s)], then pc e C is a pole of /?(V). (But the converse is not true.) (iii) If pc E C is a pole of H(s), then either /?c is a zero of det[Z + FG(s)] orpc is a pole of G(s). (iv) If G(.s') is exponentially stable, then H(s) is exponentially stable if and only if det [J + FG(s)] has no zero at oo and has all its zeros with negative real parts.
13
Remarks 1. Note that we did not assume that g{s) or G(s) were proper. 2. Here we encounter for the first time the extended complex plane C. This is crucial because when H(s) has a pole at infinity (equivalently, is not proper, or is not bounded at infinity), it takes some bounded inputs into some unbounded outputs. For example, letn = l,g(s) = —s/(s + 1), then h(s) = — s. Now h takes the bounded input t H-> sin(t2) into the output t h-• 2t cos(t2) which is unbounded. Thus, (i) and (ii) show that if 1 + g(co) = 0 (or det [J + FG(oo)] = 0), then the closed-loop system is not L00 stable. 3. It is very important to keep in mind that the set of poles of H(s) (in C) may be larger than the set of zeros of det[Z + FG(s)] (in C). For example, consider F = I and J + G(s) = diag[s, l/s, (s - 1 )/(s + 1), (s + l)/s - 1)], clearly det[/ + G(s)] = 1, V^eC, and [I + G(s)]~l = diag[l/^, s, (s + 1)/ (s - 1), (s - l)/s + 1)]. Therefore, by (10), H(s) has a pole at 0, 1, - 1 , oo even though det [J + G(s)] has no zeros whatsoever.
14 Example Consider a simple example where G(s) is proper and, in fact, tends to zero as s -• oo. Let
1
SYSTEMS
WITH
RATIONAL TRANSFER
FUNCTIONS
61
Then
and det[I +
G(s)]=(s+l)/(s-l)
Hence det[/ + G(s)] has no zeros in the closed right half-plane nor at infinity; G(s) has a double pole at s = 1; det [I + G(s)] has a simple pole at .y = 1; here H(s) has a simple pole at s = 1. So the closed-loop system is unstable even though det [J + G(s)] has no zeros in the closed right half-plane and is equal to 1 at infinity. Proof of Theorem 12 (i) Note that if p0 is a pole of g(s), so that | g(s) | -* oo as s^p0, then, by (4), h(s) -> 1 as s -+p0. Therefore, the only way for the scalar rational function /z^) to have a pole at /?c e C is that 1 + g(pc) = 0. Conversely, if 1 + g(pc) = 0, the numerator of (4), namely, $(V), tends to — 1 as s->pc; hence no cancellation can occur at pc, and pc is a pole of h(s). (ii) By assumption det[7 + FG(pc)] = 0. For a proof by contradiction, using Fact (10), assume that [J + FG(s)]
15
1
is bounded in some neighborhood of pc
Then s ^ det{[/ + FGis)]-1} = {det[/ + FGis)]}'1 is bounded in a neighborhood of pc. This contradicts the assumption det[I + FG(pc)]=0. Therefore, by (11), I - FH(s) has a pole at pc e C whenever det[7 + FG(pc)] = 0. Hence H(s) also has a pole at pc. (iii) Follows immediately from (10) and Cramer's rule. (iv) => By assumption H(s) is exponentially stable, i.e., proper, and has no poles in the closed right half-plane. Hence by (iii), det[/ + FG(s)] cannot have a zero at infinity, nor in the closed right half-plane. <= Given the assumptions on G(s) and det[7 + FG(s)], (iii) implies that all the poles of H(s) are finite and have negative real parts. :: Exercise 2 Formulate the statements analogous to those of Theorem (12) for the z-transform case. Exercise 3 Give an example where F is singular and the converse of conclusion (ii) of Theorem (12) fails (e.g., F = diag(l, 0), G(s) = diag[(s + 1) _1 ,
IV
62
2
LINEAR
SYSTEMS
Necessary and sufficient conditions: factorization method
The purpose of this section is to obtain necessary and sufficient conditions for the stability of feedback systems described byf H(s) = G(s)[I + FG(s)]-1
1
where G(s) e UnXn(s) and Fe UnX\ The system is depicted in Fig. IV. 1. The single-input-single-output case is no problem since, w h e n / V 0, 1a
h(s)=g(s)/[l+fg(s)] is exponentially stable if and only if 1 +fg(s) has no zeros in Re s > 0, including the point at infinity. In view of Example (1.14), we know that the obvious extension of this result is false for the «-input-«-output case. On the other hand, if we write the rational function g(s) as n(s)/d(s), where the polynomials n(s) and d(s) are coprime, we know that h(s) [given by (la)] is exponentially stable if and only if the polynomial d(s) + fn(s) has no zeros in the closed right half-plane. We propose to derive in this section a similar criterion for the «-input-«-output case. This will require the factorization of the rational function G{s) as a product N(s)D(s)~1, where N(s) and D(s) are polynomial matrices and N(s) and D(s) are "coprime" in a suitable sense. Then H(s) will be exponentially stable if and only if the polynomial det [D(s) + FN(s)] has no zeros in the closed right half-plane.
2.1
Factorization
Let N(s) and D(s) e UnX"[s]. The polynomial matrix A is said to be a common right divisor of TV and D iff there are polynomial matrices Nl and Z)x such that 2
N = NtA,
D = D^
Note that all matrices in (2) are in [R"Xn[s]; therefore, (2) asserts that the polynomial matrices vV and A^A are equal for all s e C; ditto for D and Z^A. If P, Q, Re UnXn[s] and if P = QR, we say that P is a left-multiple of R, and R is a right divisor of P. Given two matrices TV and DeUnXn[s], we say that AelR nX,, |>] is a greatest common right divisor (abbreviated gcrd) of TV and D iff f i n this section since we deal exclusively with transfer functions, we drop the " ~ " which we use everywhere else to denote Laplace transforms. Even the best notations have to be abused!
2
NECESSARY
AND
SUFFICIENT
CONDITIONS
63
(a) A is a common right divisor of TV and D and (b) A is a left multiple of every common right divisor of TV and D. A square matrix with elements in U[s] is said to be unimodular iff its determinant is a nonzero real number. By Cramer's rule, the inverse of a unimodular matrix is a matrix in [R"x"[s], namely, a polynomial matrix. In general, the inverse of a matrix in R" *"[.?] is not in IRnXn[s], but is in (R"x%s). When the polynomial matrix A, a greatest common right divisor of the polynomial matrices TV and D, is unimodular, we say that TV and D are right coprime. Left coprime and greatest common left divisor are defined similarly.
Factorization Theorem Let H(s) be a q x m matrix with elements in U(s). Then there exists two polynomial matrices N(s) and D(s) such that H(s) = N(s)D(s)-1 such that N(s) and D(s) are right coprime [i.e., every greatest common right divisor of N(s) and D(s) is a unimodular matrix].
Remarks (a) The analogy between (4) and the scalar case is clear. In the scalar case, the polynomials n(s) and d(s) are determined modulo a common factor, which is a nonzero number. In the present case, the polynomial matrices TV and D are determined modulo a right factor, which is a unimodular polynomial matrix. (b) If TV and D are right coprime, then it does not follow that the polynomials det N(s) and det D(s) are coprime. Consider, for example, TV = diag (s-l,s2); D(s) = diagO - 2, s - 1). (c) It is clear that we could just as well have selected a representation in the opposite order: H(s)=D(sylN(s) where N(s) and D(s) are left coprime. This representation will be useful later. (d) It can be shown that if H(s) is bounded at infinity, then det D(s) = k det(sl — A), where A is the matrix of any minimal realization [A, B, C, D] of H(s), and k a nonzero constant. (e) All the operations that follow are legitimate because R is a field. No change would be required if we used C as the field of interest. This occurs, in practice, when polynomials in U[s] are factored; the factors (s — zt) belong often to C[s].
64
IV
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SYSTEMS
The analogy between the scalar case and (4) is immediate. We establish this canonical form by giving a procedure for calculating N and D. Procedure
Let H(s) be of the form
where the ntj{s) and dj(s) are polynomials and where dj(s), for/ = 1, 2, . . . , m, is the least common multiple of the denominators of the elements in the yth column of H(s). From (5)
Call these polynomial matrices N(s) and D(s), respectively. Note that det[D(V)] ^k 0. Extracting from N and D a greatest common right divisor R(s), we have N(s) = N(s)R(s)
D(s) = D(s)R(s),
V^GC
where all the matrices have elements in U[s]. Since det[D(s)] ^ 0, R(s) is necessarily nonsingular; hence the canonical form (4) follows. :: The extraction of a greatest common right divisor is obtained as the solution of this. 8
Problem Given two polynomial matrices N(s) and D(s), where N(s) eU[s]qXm D(s) e R[j] mXM , with det D(s) # 0. Find (a) two polynomial matrices Un(s) and Ul2(s) that satisfy (9) below; (b) a greatest common right divisor R(s) of N(s) and D(s); (c)
two polynomial matrices Vu(s) and V2i(s), which are right coprime
such that 9
Ux! (s)D(s) + Ut 2(s)N(s) = R(s), and
10
N(s)D(s)~l = V21(s)Vn(s)-\
except at poles
2
11
NECESSARY
AND
SUFFICIENT
CONDITIONS
65
Procedure We shall use elementary row operations on matrices whose elements belong to U[s]. There are three types of elementary row operations: (i) (ii) (iii) add the
Multiply any row by a nonzero constant. Interchange any two rows. For / ^ /', multiply rowy by an arbitrary polynomial p(s) e U[s] and result to row /.
The effect of any such elementary row operation on a matrix M(s) e ^r^-jmxm j s e q U i v a i e n t to a multiplication on the left by a matrix that is the m x m unit matrix upon which one has performed the desired elementary row operation. Note that any such elementary row matrix has a determinant that is a nonzero constant] hence it is a unimodular matrix, and its inverse is also a polynomial matrix.
Step 1 M(s) = F(s). Step 2 If all elements of the first column of M(s), except the (1, 1) element are zero, go to Step 6; otherwise, go to Step 3. Step 3 From the elements in the first column of M(s), pick one nonzero element with least degree. If necessary, interchange two rows to bring this element in the (1, 1) position. Let M = (mik) be the resulting matrix. Step 4 For i = 2, 3, . . . , j , where j is the number of rows of M(s), divide mn by mil; thus, 13
mn =qilmll+ril
(/ = 2, 3, ... J)
where d°(rn) < J°(w 11 ). For / = 2, 3, . . . , j \ subtract from the /th row, the first row of M(s) multiplied by qn. The result is a new matrix, M(s), whose first column is ( m n , r 21 , r 31 , . . . , r 7l ). Step 5 M(s) = M(s); go to Step 2. Step 6 If M(s) has only one row or one column, go to Step 9. Otherwise, go to Step 7. Step 7 Delete the first row and first column of M(s) eU[s]jXk, thus obtaining a new matrix M(s) e Step 8 M(s) = M(s); go to Step 2. Step 9 Stop.
IV
66
LINEAR
SYSTEMS
The result of this procedure is to transform F(s) into an upper triangular matrix e U[s](m+q)Xm. Equivalently, there exists a matrix U(s) e U[s](m+q)x (m+q) such that for all s e C,
14 Since U(s) eU[s](m+q)x(m+q) and is a product of elementary row matrices, det[U(s)]= constant ^ 0 and V(s) ± Uisy1 e U[s](m+q)x(m+q\ Note that since det D(s) =£ 0, det R(s) ^ 0. Multiplying (14) on the left by V(s) gives 15
D(s) = Vli(s)R(sl
N(s) = V21(s)R(sl
Furthermore, 16 and hence, by (15), 17 From (15), 7?(s) is a common right divisor of N(s) and D(s). From (16), (17), and lemma (20) in the following, R(s) is a greatest common right divisor of N(s) and D(s). From (15) 18 where Vll and V2l are right coprime as is shown in lemma (20) in the following. :: 19
Exercise 1 (a) Write down the form of the elementary row matrices for the three operations. (b) Write down the matrices corresponding to the inverses of these operations. (c) Show that the calculation of f/(^) -1 amounts to the product of known elementary row matrices.
20
Lemma Let N(s) e U[s]qXm, D(s)eU[s]mXm. Then there exists two polynomial matrices P(s) e R[s]mXq9 Q(s) eU[s]mXm such that
21
P(s)N(s) + Q(s)D(s) = Im x m,
V GC
if and only if N(s) and D(s) are right coprime. Proof => Suppose not; then there is a right common divisor which is not unimodular, say R(s): 22
N = NR,
D=DR
2
NECESSARY
AND
SUFFICIENT
CONDITIONS
67
where N, D are polynomial matrices and det R(s) is a polynomial of degree > 1 . Then (21) gives 23
[P(s)N(s) + Q(s)D(s)]R(s) = /,
V^eC
This is a contradiction. Indeed, the bracketed factor is a polynomial matrix; hence dGt[P(s)N(s) + Q(s)D(s)] is a polynomial, but det[7?(V)] is also a polynomial of degree > 1 , hence contradiction with (23). <= The procedure (9) applied to N(s) and D(s) gives a greatest right common divisor R(s) and two polynomial matrices P(s), Q(s) such that [see (16)] 24 P(s)N(s) + Q(s)D(s) = R(s) Since N and Z) are right coprime by assumption, R(s) is unimodular; hence Ris)'1 e M[s]mXm. So multiplying (24) on the left by R(s)~\ we obtain 25 which is of the form (21).
::
Exercise 2 Show that N(s) e U[s]qXm and D(s) e tR^] mXm are right coprime if and only if the matrix [%{SJ}] e M[s](q+m)Xm is of the full rank (in the commutative ring R[,$]); after viewing this matrix as an element of Q('?+m>Xm5 it must have rank /??, \/s e C.
2.2
Necessary and sufficient conditions
We consider again the #-input-/7-output feedback system with G(s) e M(s)nX" as forward transfer function and FeUnXn as feedback gain. (See Fig. IV. 1.) We shall require now that G(s) be proper, i.e., bounded at infinity. 30 31
Theorem
Let G(s) e U(s)n x n and be proper; let F e M"x n and det[/ + K7(oo)]^0
Let G{s) = N(s)D(s)~l, where N(s) and D(s) are polynomial matrices, which are right coprime. U.t.c, the closed-loop transfer function H(s) = G(s) [I + FG(s)]~l is exponentially stable if and only if the polynomial det[D(s) + FN(s)] has all its zeros with negative real parts. (In fact, we show that H(s) is proper and that pc is a pole of H(s) if and only if pc is a zero of the polynomial det[£>(.s) + FN(s)].) Exercise 3 case. Proof 32
State and prove the corresponding theorem for the z-transform
Let us note that H(s) is proper:
H(s) = G(s)[I + FGis)]'1 = G(s) Adj[/ + FG(s)]{det[I + FGis)]}'1
68
IV
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SYSTEMS
Since G(s) is proper, the two matrices in the extreme right-hand side of (32) are bounded at infinity. By (31), the last factor is also bounded at infinity; hence H(s) is proper. Consequently, H(s) can only have poles in the finite plane. The theorem will be proved if we show that 33 o
pc e C is a pole of H{s) pc e C is a zero of the polynomial det[D(s) + FN(s)]
=> Let pc e C be a pole of H(s). Using the factorization of G(s), we obtain H(s) = N(s)[D(s) + FN(s)]~l = N(s) Ad)[D(s) + FN(s)] {det [£>(.*) + FN(s)]}~1
34
Now N(s) and D(s) are polynomial matrices and F is constant matrix; hence by (34), if H(s) has a pole at />c, then pc is a zero of the polynomial s — i> det[Z)(j) + /W(j)]. <= Let /7C e C be a zero of det[D(s) + FA^^)]. We have to show that pc is a pole of i/(.s). Since N(s) and £)(.?) are right coprime, by lemma (20), there are polynomial matrices P(s), Q(s) e U[s]nXn such that 35
P(s)N(s) + Q(s)D(s) = /,
V.y G C
Adding and subtracting the polynomial matrix Q(s)FN(s), we obtain 36
[P(s) - Q(s)F]N(s) + Q(s)[D(s) + fW(j)] = /,
V^ G C
So #(5) and D(s) + FA^(^) are right coprime. Using (34) in (36), we obtain 37
{[P(s) - Q(s)F]H(s) + Q(s)}[D(s) + FN(s)] = /,
V^eC
Hence 38
det{[/>(5) - Q(s)F]H(s) + 2(^)} det[Z)(5) + FN(s)] = 1,
V^eC
By assumption, /?c e C is a zero of the second factor of (38); hence for (38) to hold at pc, the first factor must have a pole at pc. Now P(s), Q(s), and F are polynomials; hence have no poles in C. Therefore, H(s) must have a pole at pc. Thus, (33) is established, and the theorem is proved. :: 40
Remark
From (35) we can write dQt[P(s)G(s) + Q(s)] det D(s) = 1,
Hence 41
p0 G C is a pole of G^) <=> /?0 is a zero of the polynomial det[D(s)]
V^GC
2
43
Corollary
44
NECESSARY
AND
SUFFICIENT
CONDITIONS
69
Under the assumptions of Theorem (30), we have
det[/
+
FG(S)] = £ § > = det[/ + FG(oo)] ^ ~ ^ det H(s) l\T=i(s-Poi)
where {p0i}im and {pci)im are the poles of G(s) and H{s) (counting multiplicities), respectively. Proof
By (32), taking determinants
The last step follows from (33), (34), and (41). k = det[/ + FG(oo)] # 0 as a consequence of (31). Furthermore, the number of poles of G(s) and of H(s) are equal; indeed, if they were not, (45) would either go to zero at infinity [contradicting (31)] or would be unbounded at infinity [contradicting the fact that G(s) is proper]. :: 46
Remark Equation (44) shows that as a result of possible cancellations, the set of zeros of det[7 + FG(s)] may be a. proper subset of the set {pci} of closedloop poles. [See Example (1.14.)] This shows that, under the assumptions of Theorem (30), if G{s) is exponentially stable, then H{s) is exponentially stable if and only if all the zeros of det[l + FG(s)] have negative real parts.
50
Alternate Derivation of Corollary (43) Suppose that the forward transfer function G(s) has R0 = [A, B, C, D] as a minimal representation. Clearly, then
51
G(s) = C(sI -A)~1B+
D
Referrring to Fig. IV. 1, note that e is the input to G(s) and e = u — Fy. A little calculation leads to the state equations of the closed-loop system: FDylFC]x
+ B(I+ FD)~1u
52
x = [A -B(I+
53
y = [C- D(I + FDy'FClx + D(I + FD)~lu = (I + DF)-xCx + (/ + DFYlDu This is a representation of the closed-loop system. It is a minimal representation because, as can easily be checked, by a reasoning based on assuming the contrary, the output feedback cannot destroy the complete controllability and the complete observability of the system in the forward loop. Therefore, the natural frequencies of the closed-loop system are precisely
70
IV
LINEAR
SYSTEMS
the eigenvalues of A - B(I + FD)~lFC. Now using det(PQ) = det P det g, det(7 + MN) = det(7 + NM), we obtain successively: 54
55
det[.sJ -A + B(I + FD)~1FC] = det(sl - A) det[/ + (si - A)~lB(I + FD)~lFC] = det(s7 - ^) det[/ + (/ + FD)~lFC(sI - A)~lB] = det<>/ - ^) det[(/ + FD)'1] det[/ + FC(.s7 - A)~lB + FT)] Now the last two determinants are, respectively, equal to det[/ + FG(co)] and det [7 + FG(s)] as can be seen from (51). Therefore, we conclude from (55)
56
det[7 + FG(s)] = det [7 + FG(oo)]
det [si -A+ B(I + FDYlFC] det(sl - A)
Equation (44) follows from (56) because the representations R0 = [A, Z?, C, D] and that of the closed-loop system in (52) and (53) are minimal. 58 59
Remark on Inverses of Matrices in U(s)n x "
Let 77(s) be factored as
H(s) = N(s)D(s)'1 where N and D are in [R[^]"x" and are right coprime. We have
60 61
p e C is a pole of H(s)op
e C is a zero of det [D(s)]
z e C is a pole of His)'1 o z e C i s a zero of det [N(s)] Note that no assertion whatsoever is made concerning poles at infinity. Furthermore, since det D(s) and det N(s) may have common zeros, it may happen that both H(s) and H(s)~l have a pole at some s e C. For example, H{s) = diag[(* + \)l(s - 1), (s - l)/(s + 1)].
62
Remark on Zeros of the 7 + FG(s) Let the assumptions of Theorem (30) hold. The matrix 7 + FG(s), by analogy with the scalar case, will be called the return difference matrix. Its factored form is (D + FN)D~l, where the two polynomial matrices are right coprime, provided det F / 0; then we know that p is a pole of the return difference if and only if det D(p) = 0. For simplicity assume that det(7) + FN) ^ 0. Then, by analogy with the scalar case, we say that z is a zero of the return difference matrix I + FG(-) ifr det[7)(z) + FN(z)] =0. Note that, as indicated above, it may happen that some p e C is at the same time a pole and a zero of 7 + FG(). Exercise 4 Suppose that the return difference is factored in two ways 7 + FG(s) = Nr(s)Dr(sy1
=
Dtisy'N^s)
2
NECESSARY
AND
SUFFICIENT
CONDITIONS
71
where Nr, Dr (Nt, Dt, resp.) are right coprime (left coprime), polynomial matrices. Show that, for z e C, det Nt(z) = 0 if and only if det Nr{z) = 0. In the scalar case, the zeros and poles of transfer functions have wellknown dynamical characterizations. Such characterizations generalize to the matrix case. For simplicity, we consider only the square matrix case. The dynamical characterization is given by the following theorem. 63
Theorem
Let the assumptions of Theorem (30) hold, then
(i) p e C is a pole of / + FG() if and only if there is an input u to the matrix transfer function / + FG( •) of the form a
with each ua e C", for which the corresponding zero-state response of I + FG() has the property that e(t) = rept
64 where (ii)
for all
t>0
0n^reCn; if z e C is a zero of / + FG( •), then
(a) there exists a nonzero vector g e C" and a polynomial £ a masa such that for the input u(t) = l(t)eztg + X ma6\t)
65
a
the corresponding zero-state reponse of / + FG{) satisfies the relation 66
e[f;O-,0n;w(-)]=0„,
for
t>0
(b) if, in addition, z is not a pole of I + FG(-), then there exists a nonzero vector y e Cn such that 67
\l/(t) ± y'y[t\ 0 - , 0n; «(•)] = 0,
for all
t>0
where w(-) is defined in (65) with g arbitrary in C" and the w a 's are appropriate vectors in C which depend on g; (iii) if r e C is neither a zero nor a pole of / + FG(), then for all nonzero vectors k e C there is a polynomial £ a masa such that the input 68
M(0
= l ( 0 ^ + X/w/(0 a
produces a zero-state response with exponential form 69
e[t;0-99tt;u(-)]
= [I+ F6(v)]kevt*0„,
for all
r>0
::
72
IV
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SYSTEMS
Proof We shall only sketch the proof of (i) and (ii). First use Exercise 4 and write I + FG(s) = Dl(s)~lNl(s); p is a pole if and only if det Dt(p) = 0, and z is a zero if and only if det Nt(z) = 0. Since N{{-) and £>/(•) are left coprime, we have Nt(s)P(s) = I - Dt(s)Q(s), Vs e C. (i) If p is a pole, there is a vector r e C" such that Dl(p)r = 0n. Thus, DL(s)r = (s — p)k(s), where £(•) e C"[s], i.e., is a polynomial vector. Choose an input u(s) = P(s)k(s), then
= [r/(^Jp)]-G(^(*) Since the last term is a polynomial in $, (64) follows. The converse follows easily by contradiction. (ii) (a) If z is a zero, then there is a nonzero vector g such that Nt(z)g = 0n. Hence Nt(s)g/(s — z) is a polynomial vector, say, /?(s). Choose /w(s) in (65) to be -P(s)/?(s);then
Hence e(t) = Qn for all t > 0. (b) The proof follows similar lines by choosing y' = c Dt(z), where c'Nt(z) = 0n\ :: With this concept of zero and with Theorem (30) in mind, the analogy between the scalar case and the /7-input-tf-output case is complete. 70
Theorem (a) pc e C is a pole of h = g(\ + fg)~{ if and only if pc is a zero of the return difference 1 +fg, i.e., 1 +fg(pc) = 0 . (b) Suppose that det F ^ 0 , G(s)eUnXn(s) and is proper, and finally that det[/ + FG(oo)] ^ 0; then pc e C is a pole of ft = G(I + FG)~l if and only if pc is a zero of the return difference matrix / + FG. (c) If G (g, resp.) is proper, i.e., bounded at infinity, and if det[7 + FG(oo)] # 0 [1 +fg(oo) # 0], then /? (/?, resp.) is proper. Proof (a) and (c) are immediate. For variety, let us use a left-coprime decomposition of G: G = D~lN. Hence H =(/
+ < 5 F ) _ 1 G =(D
+
NF)XN
Noting that Z) + /VT and N are left coprime [indeed, NP + DQ — I if and only if N(/> - F 0 + (D + NF)£> = / ] , we conclude that /?c is a pole of // if and only if pc. is a zero of det[D + FN]. Since f is nonsingular, D + /VFand D are left coprime [indeed, NP + DQ = / if and only if (D + NF)(F~lP) +
3
LINEAR
FEEDBACK
SYSTEMS WITH
DYNAMICS
73
D(Q - F~lP) = / ] , and since I + GF = D~\B + NF), we conclude that pc is a zero of / + GF [and of / + FG = F(I + GF)F~l ], if and only if pc is zero of det[D -f NF]. Thus, /?c is a pole of H if and only if it is a zero of / -f FG.
71
Remark If F is singular, the equivalence (b) does not hold. To wit G = diag[(s — l ) " 1 , (s + l)~l], F' = diag(0, 1), H has a pole at s = 1 and 5 = —2, and J -h FG has a zero only at s = —2.
3
Linear feedback systems with dynamics in the feedback path (rational transfer functions case)
Consider a continuous-time, linear, time-invariant, fl-input-/7-output feedback system with input w, output y, and error e. Call G0 and Gf the transfer functions of the open-loop system and of the feedback system; we assume that they are rational matrix functions and are proper: 1 2
e
y = G0e;
G0eRnXn(s),
G0 proper
= u - Gf y;
Gf e Wx "(s),
Gf proper
We assume throughout that 3 As a consequence, we know that (/ + GfG0)~l is a well-defined member of Unxn(s) because det[/ + Gf(s)G0(s)] # 0. Our goal is to exhibit necessary and sufficient conditions for the closedloop transfer functions 4
He: w n e , He(s) = [I + G/^Gote)]" 1
5
//, : ut-*y, Hy(s) = (70(5)[/ + G ^ C o ^ ) ] " 1 to be exponentially stable. Exercise Given that G 0 , G r e U"x "(s) are proper (bounded at infinity), show that det[/ 4- G/(oo)G0(oo)] ^ 0 if and only if He and 7/y are proper. One approach to the problem is to pick some minimal realization for G0 and for Gf, derive from them minimal realizations of He and Hy, and require that the characteristic polynomials of the resulting A matrices, Ae and Ay, be stable. We shall use, instead, the factorization technique of Section 2.1.
74
IV
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SYSTEMS
To start with, consider how one would approach the problem in the single-input-single-output case. First choose irreducible representations of the rational transfer functions g0 and gf : 6
go — noldo,
7
gf = rif/df,
« 0 ^ o coprime polynomials n
f?df
coprime polynomials
Then 8
he = (1 + gjgoY1 = d0df/(nfn0
9
hy =g0(l + gfQoY1 = n0df/(nfn0
+ dfd0) + dfd0)
The numerator and denominator polynomials in the right-hand sides of (8) and (9) are not necessarily coprime. Indeed, for he, common factors may cancel out between d0, nf and df, n0, respectively. For hy, cancellation can occur only between df and n0. For this reason it is false to assert that the zeros of nfn0 + dfd0 are the poles of he and hy. Intuitively, it is clear that the same type of phenomenon will occur in the «-input-«-output case; however, the noncommutativity of the product of matrices requires a bit more care. Consider the /z-input-/?-output system defined by (1) and (2). We assume that (3) holds. We seek necessary and sufficient conditions under which He and Hy, defined in (4) and (5), are exponentially stable. In fact, we shall obtain more, namely, polynomials whose roots are the poles of He and Hy, respectively. Let us factor G0 and Gf. 10 11
G0=N0D0~1; Gf = DjlNf;
N0, D0eUnXtt[s];
N0, D0 right coprime
Nf, DfeUnXn[s];
Nf, Df left coprime
We, of course, have det D0 # 0, det Df £ 0. (t) 12
Q = DfDQ + NfN0 Then by (4) and (5)
13
He = D0Q-1Df9
Hy=N0Q-1Df
f In most of the analysis of Section 3, the explicit dependence on s is not indicated because it is not necessary. For example, Eq. (12) asserts the equality of two polynomial matrices, i.e., the values taken by the left- and right-hand side, are equal for all se C; Eq. (10) asserts the equality of two matrices of rational functions, i.e., the values taken by the left- and right-hand side are equal for all s e C, except at the poles. Similarly, det D0 # 0 means that the polynomial det DQ is not the zero element of the ring U[s]; equivalently, the function s >-> det[Z)0W] is not equal to zero for all s e C.
3
LINEAR
FEEDBACK
SYSTEMS
WITH
DYNAMICS
75
Again note that det Q ^ 0, because, by (3), He tends to a nonsingular constant matrix as|s| -> oo. Let Lbe a greatest common left divisor (geld) of Q and Df; then 14
Q=LQ,
Df=LDf
x
where L. Q, 5 / e[R" "[^], and Dy and Q. are fe/jf coprime. Equivalently, there are polynomial matrices P and Q s.t. 15
DfP + QQ = I Thus, 7/c can be written // e = J D 0 Q - 1 5 /
16
Let J?e be a gcrd of D0 and Q. Then 17
Q=QeRe
18
D0=D0Re where Re, Qe, D0e MnXn[s], and Qe and D 0 are r/#/?f coprime. Equivalently, there are polynomial matrices Pe, Qe s.t.
19 Substituting (17) in (15), we obtain
DfP + neReQ = i Hence Df and Qe are left coprime. Hence 20
with
He = tiQn-elBf
l&e,Bf left coprime [D0, Qe right coprime
Operating similarly on H , we obtain 21 Let Ry be gcrd of N0 and Q, then 22
=QyRy
23
N0=N0Ry where Q y , i? y , N0 e [R"x "[>?], and Q^ and N 0 are n#/?r coprime. Hence there are polynomial matrices P y and Qy such that
24 Again Df and Qy can be shown to be left coprime. Thus, we obtain 25
Hy=N0n;lDf
where
\Qy and Df are left coprime N0 and Qy are right coprime
76
26
IV
LINEAR
SYSTEMS
Theorem For the system defined by (1), (2), with assumption (3) and the notations above, we have
27
pe e C is a pole of Heope
is a zero of det Qe
28
py e C is a pole of Hyopy
is a zero of det Qy
Theorem (26) gives immediately the necessary and sufficient conditions for He and Hy to be exponentially stable; namely, det Qe and det Qy must have all their zeros in the open left-half plane. Proof Note that (3) and the assumption that G0 and Gf are proper imply that He and Hy are proper. Hence we concern ourselves only with poles in C. We prove (27) only; the proof of (28) is similar. <= By multiplying (19) on the right, we obtain 29
Pe D0 Q; lDf +QeDf=
Pe He +QeDf=
Q; xDf
By assumption det Qe(/?e) = 0. Since Qe and Df are left coprime, the righthand side of (29) has a pole at pe [by the proof of Theorem (2.30)]. On the left-hand side, Pe, Qe, and Df e IR/,X|,[.s], hence are bounded in any neighborhood of pe; hence He has a pole at pe. => By Cramer's rule in (20) 30
He = 5 0 (Adj QJD^det Qe By assumption^ is a pole of He. But 5 0 , Adj D c , and Df e tR"x"[1s'] (hence have no finite poles); hence (30) requires that det Qe(pe) = 0. ::
31
Interpretation Let p = 1, 2, or oo. If det Qe and det Qe have all their zeros in the open left half-plane, then (a) He and Hy are exponentially stable, (b) for any u e Lnp, e and y are in L / , (c) for any ue L / , and for any minimal representation of G0 and Gf, the corresponding state trajectories x0(-) and Xy(-) (starting from the zero state) are in Lnp. Exercise Prove the above statement. [Use Theorem (1.9) and note that any minimal realization is completely observable.]
35
Remark With the notations above,
36
{zeros of det[/ + GfG0]} c {poles of He)
37
{zeros of det [I + GfG0]} c {poles of Hy} u {zeros of det Ry)
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DYNAMICS
77
These inclusion relations show how dangerous it would be to try to ascertain the stability of the closed-loop transfer functions He and Hy by considering only the zeros of the rational function s\~>det[/ + Gf(s)G0(s)]. Proof (36) and (37) follow directly from (27), (20), and (4) and from (28), (25), and (5). Indeed, 38 39 41
Examples The examples in the following are purposefully simple. Their purpose is to illustrate the relations between the poles of G0, Gf, Hy, Hy and the zeros of Q and det(7 + GfG0).
42
Example 1 Neither G0 nor Gf is exponentially stable; He is unstable and Hy is exponentially stable.
78
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43
Example 2 Neither G0 nor Gf is exponentially stable; He is exponentially stable and Hy is unstable.
50
Interpretation of the Zeros of det Q(s) = 0 For the system described by (1) and (2), statements (27) and (28) assert that the stability of the transfer functions He and Hy is completely characterized by the location of the zeros of det Qe(s) = 0 and det Qy(s) = 0, resp. The transfer functions He and Hy, however, characterize only the zero-state response of the feedback system. To consider responses that do not start from the zero state, we have to adopt a specific state representation. Let (At, Bt, Ct, D() = Rt where / = 0, / , be some minimal state representations of G0 and Gf, resp. Call the corresponding states x0 and xf, resp. The vector (x0\ x/)' is a suitable state vector for the feedback system. Of course, it is not necessarily of minimal dimension. Note that det Dt(s) = kt det(sl - At), for i = 0,/. We claim that
51
det Q(s) = 0 is the characteristic equation of the feedback system described by (1) and (2).
3
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FEEDBACK
SYSTEMS
WITH
79
DYNAMICS
This means, in particular, that if, for some X9 det Q(A) = 0, then for some initial state c and for u = 0, the state trajectory is of the form ceu. To prove (51) we use the factorization (10) and (11) to rewrite the feedback equation as follows: „ (D0v=e, (w = Nfy u J y = G0e u-e = Gfy { _ ' Ar fJ \y=N0v, \DfuDfe = w where we introduced the new variables v and w. The four equations on the right are differential equations since N0, D09 Nf9 Df are polynomial matrices. Let us eliminate e, w, and y by substitution; then we obtain the following equivalent system (NfN0 + DfD0)v = Dfu9 y = N0v, e = D0v, w = Nfy =
NfN0v
So given u and the appropriate initial conditions, the solution t -> v(t)9 for t > 0, is obtained by the first equation; y, e9 and w are obtained by substitution. Since these substitutions involve only products by polynomial matrices, no new modes are introduced. Clearly then, the characteristic equation of the feedback system is det(Nf N0 + Df D0) = det Q = 0. s: 59
State-Space Representation Formulation The feedback system described by (1) and (2) consisted of subsystems prescribed by their transfer functions. This allowed us to consider only the zero-state responses of the subsystems and of the feedback configuration. In contrast, suppose now that the subsystems are prescribed by their state representations R0 and Rf9 respectively:
60
R0 = (A0,B0,C09
D0);
hence
G0(s) = C0(sl - ^ 0 ) _ 1 ^ o + A)
61
Rf=(Af9Bf,Cf9
Df);
hence
Gf(s) = Cf(sl-Af)~1Bf+
62
Df
These state representations are not assumed to be minimal. We wish to be able to study the behavior of the state trajectory of the feedback system when both the input and the initial state are arbitrarily chosen. We assume that det [J + Gy(cx))Go(oo)] = det(J + Df D0) # 0 Let x0 and xf denote the states associated with R0 and Rf9 respectively; let us choose x, defined by x = (x0\ x/)', to be the state of the representation Rc of the closed-loop system. We shall see that under assumption (62), Rc takes the standard form Rc = (A9B9C9D). Once the representation Rc is obtained, one has a complete description of the relation between input, initial state, and output. In fact, the zerostate response y of the closed-loop system is given by
64
$(s) = Hy(s)u(s) = [C(sl - AyxB
+ D]vi(s)
IV
80
LINEAR
SYSTEMS
and the state trajectory starting from x(0) at 0 is given by x(s) = (si - Aylx(0)
65
+ (si -
AylBu(s)
In order to discuss the boundedness of the state trajectory t\-+x(t), we recall the definition (1.5) of exponentially stable transfer functions and their properties [Theorem (1.9)]; thus, we see that we need to relate the closedloop eigenvalues Xcl (defined as the eigenvalues of A) to the eigenvalues of R0 and Rf (respectively defined as the eigenvalues of A0 and Af). Thus, we need to calculate A. The matrices A, B, C, D which define Rc are obtained from the equations describing the subsystems and from the equations describing their interconnections. As in (1) and (2) above, let u and y denote the input and output of the closed-loop system. Call e the error; then we have: (a)
For the subsystems x0 =A0x0 + B0e, y = C0x0 + D0e,
66
xf =Afxf + Bfy y2 = Cfxf + Dfy
(We used here the fact that the error e and the output y are, respectively, the input and output of the forward subsystem, and also that y is the input of the feedback subsystem.) (b) For the interconnection 67
e — u — y2=u—
CfXf — Dfy
We shall obtain the representation Rc by eliminating e from the first three equations of (66) and from (67). In fact, (67) and the second equation of (66) give 68
e=(I
+ DfD0)~\u-
DfC0x0-Cfxf)
Using (68) in the first three equations of (66), we obtain, after a few manipulations, RQ
A =
\A0-B0(I
L
71
C=[(I
+ DfDoy1DfC0
*f(! + A) DfV'Co
+ D0DfylC0
-(I +
-B0(I
+ DfDoy'Cf
1
Af - Bf(I + D0 Z),)"1 A A J
DoDfy'DoCf]
Equations (69)—(71) specify the representation Rc of the closed-loop system. Note the symmetry of the elements of A; by interchanging subscripts 0 and/, one obtains one row from the other. This type of symmetry is not present in
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SYSTEMS WITH
DYNAMICS
81
B, C, and D because of the lack of symmetry of the point of application of u as seen from the block diagram representation of the feedback system. To obtain the desired relation between the eigenvalues of A and those of A0 and Af is conceptually very simple. The manipulations are messy; they involve elementary row and column operations performed on blocks of si - A, the observation that D0(I + DfDo)'1 = (J + D0 Df)~1D0, and if M and TV are, resp., p x q and q x p matrices, then det(/ p + MN) = det(J, + NM) For the details the reader is referred to Hsu and Chen [Hsu.l]. The result can be expressed as follows: Let lci denote the closed-loop eigenvalues (i.e., the eigenvalues of A); let koi and lfi denote those of A0 and Af, resp. Then
or equivalently
Note that (72) is a generalization of (2.56), and (73) is a generalization of (2.44). We note again that the zeros of det[J + Gf(s)G0(s)] may be a, proper subset of {kci}, the set of closed-loop eigenvalues; indeed, cancellations may occur in the right-hand side of (73). On the other hand, (73) shows that, in the case where Re X0i < 0 and Re Xfi < 0, for all /, the kci will all have negative real parts if and only if all the zeros of det[/ + Gf(s)G0(s)] have negative real parts. Concerning state trajectories, we can state the following fact. 74
Fact If all the closed-loop eigenvalues kci have negative real parts, then for all initial states x(0) = [x0(0)', x^O)']' and for all bounded inputs, i.e., u e L,,00, the state trajectory of the closed-loop system, i.e., th^x(t), is bounded; this is also true for e(-) and y(-). Exercise 5 Derive (69), (70), and (71) in the case where D0 and Df are zero. Exercise 6 Show that if R0 and Rf are minimal, it does not follow that Rc is minimal.
82
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Exercise 7 Let Hy denote, as in (5), the closed-loop transfer junction. Show that 75
[/ - Gf(s)Hy(s)][I + Gf(s)G0(s)] = J,
(a) (b)
If det(/ + Df D0) = det(J + D0 Df) = 0
then (i) as s -> oo, Hy(s) -* oo, and (ii) the closed-loop system does not have a state representation in the standard form (A, B, C, D).
4
Convolution feedback systems
4.1
General results
In the previous sections, we considered only rational transfer functions and hence were restricted to lumped models. We consider now feedback systems made of subsystems whose input-output properties are represented by convolution operators. Thus, our present study includes distributed models. We shall make use without comment of the results of Appendixes C and D. To appreciate the generality of the convolution assumption, recall that L. Schwartz has shown that any linear time-invariant operator that satisfies some slight continuity properties has a convolutional representation, more precisely, a representation of the form u\-*T * u where T is a distribution (generalized function). If the operator is, in addition, nonanticipative, then the support of T is in U + , and conversely.
Figure IV.2
The system under study is shown in Fig. IV.2. The input w, the error e, and the output y are functions from U+ into Un. We shall also allow them to be generalized functions: ^-functions will appear frequently. The linear time-invariant subsystems G and F are characterized by their impulse responses (?(•) and F(-). The equations are y = G * e, u = e + F * G * e,
1 2
equivalently,
' y = Ge u = e + FGe
Hence
3a
H(s) = G(s)[I + F(s)G(s)Y
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83
and / - F(s)H(s) = [I + P(s)G(s)]'1
3b
over the common half-plane, where the Laplace transforms P and G are both defined. In this section we shall require for stability that the closed-loop impulse response //(•) e s/nXn(0). When this is the case, 4
\\y\\P<\\H\L'\M\P
for
l
i.e., the closed-loop system is LP stable for allp e [1, GO]. The converse is not true, because the class of impulse responses could be enlarged by considering those with singular measure [Tho. 1]. We shall abbreviate J / ( 0 ) by $0 \ \H\tf denotes the norm of H(-)e stfnXn. We take the position that the increased technical difficulties are not worth the slight extension. Throughout this section, we assume that the feedback transfer function matrix F(s) belongs to the algebra s/n x n, for reasons that are mostly technical. This assumption unfortunately rules out all "unstable" feedbacks, including the commonly found case of rate feedback. This is evidently the price we have to pay for allowing distributed, possibly unstable elements in the forward path. The first theorem in the following shows that, under very broad conditions, if the closed-loop impulse response //(•) e s$nXn{o) for some a > 0, then G(-) must satisfy certain conditions. 5
Theorem Consider a system described by (1) and (2), where the feedback transfer matrix F( •) e stfn x". Let G( •) be a matrix of distributions with support in IR + , and suppose that in some neighborhood of the origin, G contains at most impulse functions. Assume that the closed-loop impulse response H(-) is uniquely determined by the equation
6
H+H*F*G
=G,
or
H=G-H*F*G
U.t.c, if H(-) e jtfnxn(a) for some a > 0, then (i) (/(•) is Laplace transformable; for some a > 0, G(-) is analytic in Re s > a, G(-) e <s/nXn(a) and can be continued to a meromorphic function in Re s > o. (ii) G( •) is of the form 7
G{s) =
rf(s)l@(s)Y1
where
JP(S),
(iii) 8
inf Re s>
\det[I + F(s)G(s)]\ > 0
$(s) e
JnXn(a)
84
IV
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SYSTEMS
Remark 1. Note that we do not assume that F(-) is constant, or even if F(-) is constant, that it is nonsingular. This allows us to consider systems whose number of inputs need not necessarily equal the number of outputs. If F() is rectangular, it can be made square by adding rows or columns of zeros, as appropriate. 2. This theorem shows that under the mild assumptions on G(-), the solvability of the closed-loop system, and the assumption that H() is of exponential order, we can deduce a great deal about the form of (/(•)• Proof
(i)
Balancing impulses in (6) gives H0=G0
— H0 F0 G0,
or
H0(I + F0 G0) = G0
where F0, G0, and H0 denote, respectively, the strengths of the impulses at t =0 of F(-), G ( ) , anc * H(-). In order for H0 to be uniquely determined from (9), it is necessary for I + F0G0 to be nonsingular. Now define E(s) = [I + P(s)G(s)]'1 = I - F(s)H(s) Since E(s) = I — F(s)H(s) (see Exercise 2) and since F and H both belong to jtfnXn(a), clearly so does E. Thus, E(s) is bounded in Re s > cr, analytic in Re s > a, and tends to I + F0G0 as Re s-> ao. Hence sh-^l/det E(s) is meromorphic in Re s > a and is analytic and bounded away from zero in some half-plane Re s > a for a sufficiently large. Consequently, by Appendix D.3.3, Eis)'1 e JnXn(ot). But by (10), this means that IS(-) + F(-) * G() e ^nXn(oc). Thus, F * G is Laplace transformable, and FG is analytic for Re s > a. Furthermore, FG can be analytically continued to a meromorphic function in Re s > a, since det E(s) can have almost isolated zeros in Re s > a. Since G ( ) = H(s)[I + F(s)G(s)], the same conclusion holds for (/(•)• Finally,
G=HE-1eJnXn(a).
(ii) Since G( •) is Laplace transformable, we may write H = G(I + FG) ~ *. Routine matrix algebra can now be used to show that G(s)=H(s)[E(s)]-1 where E(s) is defined in (10). (iii) As we have seen, H{ •) e stfn x n{o) implies that E( •) e stfn x "(a), whence det E(-) e s$(o). Thus, there exists a finite constant k such that | det E(s) | < k whenever Re s > a. But this, in turn, implies that 1/|det E(s)\ = |det[/ + F(s)G(s)]\ > \\k which is enough to prove (8).
whenever
Re s > a s:
The factorization (7) is the first hint of a similarity between the theory for systems with rational transfer functions and the general case of convolution feedback systems.
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FEEDBACK
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85
Exercise 1 Suppose that n = 1, that g(0 = g0S(t) + gfl(0> with 0 a (') e L1, and /(s) = / 0 , a nonzero constant. Show that (8) is equivalent to the statement: s H-> 1 +f0g(s) has no zeros in Re s > a. This holds if and only if the Nyquist diagram of g [i.e., the graph of co \-+§(a + jco), COGU] does not intersect or encircle the point (—1// 0 , 0) in C. For other graphical tests, see Section 5. [The significance of this exercise is as follows. There are clearly two possible ways in which the condition (8) can be violated: (i) det[J + F(s0) G(s0)] = 0 for some s0 with Re s0 > cr, or (ii) no such s0 may exist, but there may exist a sequence {st} with Re st > a such that det[7 + Fis^GiSi)] -* 0 as / -> oo. This exercise shows that under the stated conditions, the second possibility cannot occur. Furthermore, it provides a means for verifying whether or not the first possibility occurs by studying only the behavior of the function co i—• g(a +jco).] Exercise 2 Show that the following relationships hold for the system described in Theorem (5): 13a
H =G(I + FG)'1 = ( / + GF)~ 1 G
13b
G=H(I-FHY1
={I-HFY1H
13c
(/ - FH)(I + FG)=I
13d
(/ - HF)(I + GF) = I The relation between (13a) and (13b) is made clearer by the observation that H is obtained from G by putting a feedback F around G, while G is obtained from H by putting a feedback of — F around H. Exercise 3 Show that if ft e Jn x n and F e i n X " , then so does E. Show that if in addition F ~J e Jn x \ then ft e Jn x n if and only if E e Jn x \ Knowing that in order for the closed-loop impulse response to be of exponential order, G ( ) must be of the form (7), we can now formulate a very general theorem that can, in turn, be specialized to various practically significant situations.
14
Theorem Consider a feedback system described by (1) and (2). Suppose that F( •) E stfn x", that the system has a uniquely defined closed-loop impulse response H(), that G(-) is Laplace transformable, and hence that ft(s) = G(s)[I + F(s)G(s)]~1 over the common half-plane of convergence. U.t.c, # e j t f n x n i f andonlyif (i) there exist «#(•) and Qj(')'m^nXn
15
G{S) =
such that
JV°(S)QJ-1(S)
86
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SYSTEMS
Moreover, the following non-cancellation condition is satisfied: Whenever (s^ is a sequence with Re st > 0 such that lim^^ldet @>(s?)\ = 0> w e n a v e lim i n f ^ | det[$(Si) + i ^ V P f o ) ] I > 0. [This condition is hereafter referred to in abbreviated form as follows: The ordered pair of functions [det 3, det(^ 4- FJ^)] satisfy condition (N).] (ii) inf \det[I + F(s)G(s)]\ > 0 Res>0
Proof
=> Rewrite the relation between G and H as follows G=H(I
-FHy1
Hence with Jf = H, 3 = I — FH, we have (15), Furthermore, the no-cancellation condition (N) also holds since det(# + PJf) = det J - 1,
Vs in Re s > 0
Thus, statement (i) is established. Consider now (13c) and take determinant of both sides; then det [J + FG] = l/det(7 - FH) nXn
Since I — FH e ja/ , it is bounded in Re s > 0, and so is its determinant. Hence (16) follows from (17). <= We show first of all that (16) and (15) together imply that inf \det[&(s) + P(s)rf(s)]\
>0
Re s > 0
To prove (18), suppose by way of contradiction that (18) is false, and let (st)f be a sequence with Re st > 0 such that det[S(st) + Fis^Jf (jf)] -> 0. Since det(^ + FJT) = det § • det(J + FG), we see that if det(# + PJf) -> 0, then either det §) -» 0 or det(7 + FG) -• 0 [after taking subsequences of (stf\ if necessary]. The latter possibility is ruled out by (16). As to the former possibility, suppose det ^(st) -• 0. Then since the pair (det Q), det(it + PJ*)) satisfies condition (AT), it follows that lim inf |det[^(^ t ) + F(si)^(si)]\ > 0, which is a contradiction. Thus, (18) is established. Now, by matrix algebra, we have H = G(I + FG)'1 = Jf{§ + / t f ) _ 1 . In view of (18) and Appendix D.3.3, we see that (S + PJf)'1 eJnXn, whence HeJnXn. :: Exercise 4 Prove the following alternate version of Theorem (14). Let PeJnXn. Then H = G(I + FG)'1 E JnXn if and only if there exist Jf and § in JnXn such that (i) G = Jf^\ and (ii) inf \dQt[3(s) + F(s)J*(s)]\ > 0 Res>0
4
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87
Corollary Suppose F eJnXn and that G =tyV'^-\ where Jt, 3eJnXn, and the pair [det Q), det {0} + PJF)\ satisfies condition (N). Then H = G(I + FGy1eJnXn if and only if (16) is satisfied. Theorem (14), though quite easy to state and prove, has many interesting implications. Suppose we are given F e j$n Xn and G, and we wish to determine whether H = G(I + FG)'1 e JnXn or not. How would we proceed? We would first check the condition (16). If (16) fails, then definitely H <£ j
Then
and hence (16) is satisfied. However.
and clearly H $ sinXn. Thus, in order to conclude that H e stf"Xn, in addition to (16), we use the added assumption that G is of the form (15), where the appropriate condition (N) is satisfied. This brings us to the following question: Suppose we express G in the form (15), where «#, l e i " x " , but suppose the condition (N) fails to hold. What, if anything, can we conclude? The answer is, in general, nothing—one simply has to try to find a different factorization for G. On the other hand, if JV* and §) constitute a so-called pseudo-rightcoprime factorization, then definite conclusions can be drawn. We now proceed to study this last concept in detail. Definition Two elements Jf, <3) e srfnXn are said to be pseudo-right-coprime (pre) if there exist $ , -f, # e JnXn such that (i) det # ( s ) =?* 0 for all s with Re s > 0, \s\ < GO, and
88
21
IV
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Definition Given a Laplace transform G(-), the pair (Jf, 3) is said to be a pseudo-right-coprime factorization (prcf) of G if
(i) (ii) Jf and # are pre, and (iii) whenever ( s ^ 0 0 is a sequence with Re 5*; > 0 and | ^ | -» oo, we have l i m i n f ^ J d e t ^ > ; ) | > 0. 22
Fact
Let G(s) e UnXn(s) and be proper. Then G has a prcf.
Proof If G(s) e IR"x n(s) and is proper, then there exist polynomial matrices N and D such tht iV and D are right coprime and such that ND'1 = G. Moreover, we can assume that D is column proper. Let det D(s) = I7?=iC?— /?£)m', and let <5; denote the highest power of s appearing in the /th column of D(s). Since D is column proper, we have YJ = I ^i = Z?=i m «- Let M(s) = diag[(j + l)*1, . . . , 0 + I)*5"], and define 23
/ = M "
1
,
3 = DM"1
Then <# and ^ are proper rational matrices with poles at £ = — 1, and hence Jf.Se JnXn. Moreover, G = Jf®-1. Thus, (i) of Definition (21) is satisfied. Since N and D are right coprime, there exist polynomial matrices P and Q such that 24
PN + QD= /,
for all
j
Now let a be an integer greater than or equal to the degree of any element of P and Q. Then the rational matrices 25
€ =P'(s+iy*,
-f' =
Q(s+iy*
nXn
both belong to jtf 26
. Moreover, from (24) we have
%Jf + r® = (PN + QD)M~\s
+ l)~ a = M _ 1 ( J + l ) " a = # ( * ) , say
Clearly, # ( 5 ) e i " x " . Moreover, det # ( 5 ) = (J + 1)"', where / is an integer. (In fact, / = ocn + £f = x mi.) So det #"(5) / 0 whenever Re.? > 0 and \s\ < 00. Thus, the condition (ii) of Definition (21) is satisfied. Finally, we have k
det S(s) = det D(s) • [det Mis)]-1 = ["] (s- p^
n
' Y[(s-
\ydi
So clearly det S(s) -• 1 whenever \s\ -> 00. Thus, the pair {Jf, §)) defined by (23) constitutes a prcf of G, according to Definition (21). :: Exercise 5 Show that Jf defined in (23) is proper.
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FEEDBACK
SYSTEMS
89
Exercise 6 Let (<#, $)) be a prcf of G. Show that G is a meromorphic in C + and that p, with Re p > 0 is a pole of G if and only if det Q)(p) = 0. Exercise 7 Let G possess a prcf (yT, # ) . Show that whenever {s^ is a sequence with Re ^ > 0 and lim i ^^ \st\ = GO, we have lim sup| G(s-)\ < oo. Loosely speaking, if G possesses a prcf, then G(s) must remain bounded as | s | -> oo with Re s > 0. The importance of pseudo-right-coprime factorizations is brought out in the following theorem. 27
Theorem Let FeJnXn, and let G = Jf$)~l, where (yf, ^ ) s a prcf of G. 1 U.t.c., H = G(7 + FG)~ e <*/"x n if and only if
28
(i)
inf | det[/ + F(s)G(s)] \ > 0 Res>0
and 29
(ii) det[^0) + F(S)J?(S)]
# 0 whenever
det &(s) = 0 and Re s > 0
nXn
Prflfl/ => Suppose fteJ . Then (28) follows, by Theorem (14). To prove (29), we show that if (29) fails, then H $JnXn. So let .y0 be such that det[@(s0) + F(s0).#(,s0)] = 0 and Re ^0 > 0. Since Jf and<3)are pre, there exist $ , -T, and # in JnXn such that
30 and det iP~(s) ^ 0 whenever Re s > 0. Now, starting from (30), we get
Now as iT-^o, the right side of (31) becomes unbounded, because det {0 + FJf) -• 0, while det it(s0) # 0. Thus, the left side of (31) must also become unbounded. But % — i^F and i^ are both bounded over the region Re s > 0. Hence H must become unbounded as s -> s0, so that H $ jtfnXn. Hence HeJnXn implies (29). <^= Suppose (28) and (29) hold. We must show that H e JnXn. We begin by showing that the pair (det 3, det(^ + FU^) satisfies condition (AT). Let (^j)!00 be a sequence with Re st > 0 such that det 3{sj) -• 0 as / -• 00. Since lim inf I det ^Csf) | > 0 as | st | -> 00, it follows that the sequence (si) is bounded. Therefore, it contains a convergent subsequence, which we renumber again as (si), and denote the limit of (s,) by s0. Then Re ,s0 > 0 and det 3(s0) = 0. But by (29), this implies that det[3(s0) + F(s0)J^(s0)] # 0. Thus, the pair (det S, det(^ + F # ) ) satisfies condition (JV). This fact, together with (28), implies by Theorem (14) that H eJnXn. ::
IV
90
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Exercise 8 Prove the following alternate version of the Theorem (27). Given that(^,^)isprcfofGandthatFe^nX^wehavethatH=G(/+FG)_1e^n><" if and only if inf \dtt[@(s) + P(S)JP(S)]\
>0
Re s > 0
32
Remark As mentioned after Corollary (19), the condition (28) [or (16), which is the same thing] is necessary for H to belong to s/nXn, but is not sufficient. The present theorem more clearly brings out when (28) is sufficient for stability. Let (Jf, §) be a prcf of G and suppose F e s£nXn and that (28) holds. Then H e stfnXn if and only if (29) is satisfied. This condition is to be interpreted as follows. We have det {0 + tJf) = det Q) • det(7 + FG). Now suppose det @(s) = 0 with Re s > 0. If det(J 4- FG) is finite, then det(# + i v f ) = 0, and H$jrfnXn. [This is, in fact, the case in the example following Corollary (19).] On the other hand, if det(J + FG) has a pole at s, then the product det Q) • det(7 + FG) is indeterminate, and one has to evaluate det(# + FJf) directly. This theorem shows the importance of finding a prcf of a given transfer function G(-). This problem is made simpler by the following fact, which shows that it is only necessary to find a prcf of the " unstable part" of G ( ) .
33
Fact Let (Jf,®) be a prcf of G(-), and suppose G1(-)eJnXn. (Jf + Gx3, 3) is a prcf of G + Gt.
Then
Proof By hypothesis, G = Jr@-\ so clearly G + G1 = (Jf + Gx§) • S'1. Also by hypothesis, lim inf|det S>(s)\ > 0 whenever 1^1 -» oo with Re s > 0. So all that remains to be shown is that Jf + GXQJ and Q) are pre. Since Jf and Q) are pre by hypothesis, there exist %, f^, and iff in s3nXn such that det \iP"(s)\ ^ 0 whenever Re s > 0 and \s\ < oo, and such that 34
€Jf + ^
= #
So from (34) it follows that 35
$(Jf + Gx<2)) + {T - mG^Q) = # which shows that/T + GXQ) and S> are pre. As an application of Theorem (27), we derive completely explicit, necessary, and sufficient conditions for a class of multivariate feedback systems to be stable.
36
Theorem Let G(-) be of the form nti
k
37
G(s) =
n
RtJl(s -
J
Pi)
+ Gb(s) = Gu(s) + Gb(s)
::
4
CONVOLUTION
FEEDBACK
SYSTEMS
91
where Re Pi > 0 for all i, and Gb e i " x " . Let Gu(s)=N(s)D(sy1
38
where N and D are right-coprime polynomial matrices, and further D is column proper. Suppose F e JnXn. Then ft = G(I + FG)'1 eJnXn if and only if 39
(i)
inf | det[J + F(s)G(s)] | > 0 Res>0
and 40
(ii) dQt[D(Pl)+P(Pi)N(pd+
^1)6^)0^)]^
for i = l , . . . , *
Proof Given N and D, define Jf and ^ as in (23). Then (Jf, # ) is a prcf of GM, and therefore, by Fact (33), {Jf + GhQ), Q)) is a prcf of G. Hence by Theorem (27), H e ^nXn if and only if (39) holds, and in addition 41
det[#(s) + F(S)JP(S) + F(s)G 6 (s)^(s)] 7^ 0 whenever det §(s) = 0 and Re s > 0 However, it is clear from the way that Jf and Q) are constructed that det Q)(s) = 0 with Re s > 0 if and only if s = Pi, / = 1, . . . , k. Furthermore, jf(s) = N(s)M-1(sl 9(s) = D{s)M~\s\ and det M(Pi) ^ 0 for / = 1, . . . , k. Therefore, (41) simplifies to (40). ::
42
Corollary
Suppose n = 1 (scalar case), that k
43
0(s)=t.
mi
T.rijl(s-Pi)J
+db(s)=gu(s)+gb(s)
• = i j = \
and suppose/, gb e s>2. Then h = gj(\ +fg) e s£ if and only if 44
inf \l+f(s)g(s)\
>0
Res>0
45
f(Pi)*0,
i = l,...,fc
Proof Let gu(s) = n(s)jd(s), where «(£) and d{s) are polynomials with no common zeros. Then (39) reduces to (44), while (40) becomes 46
d(Pi) +f(Pi)n(Pi)
+fiPdgb(pdd(Pd
* 0,
1 = 1, . . . , k
However, since d(Pi) = 0 and n(Pi) ^ 0, (46) reduces to (45). 48
::
Remark When G has several right half-plane poles, the right-coprime factorization (38) may be quite cumbersome. Since the condition (40) is purely local, it is possible to deal with each pole successively. Call R^s) the singular part of the Laurent expansion of G about Pi, i = 1, 2, . . . , k. Call Mt the
92
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difference between G and Rt; Mt is analytic in a small disk centered a t / v We have G(s) = [Us) + Mt(Pi)] ^lMt(s)
- Miipd]
The first bracket is a proper rational matrix, hence has a right-coprime factorization, say, A^D; -1 ; the second bracket, say $ f , is locally analytic and
6,0*,) = 0. Thus, G(s)=Ni(s)Di(sy1
+ ®i(s)
We leave it as an exercise to show that (40) can be replaced by 50
detld&d+PipdfiiiPiK^O,
for
i = l,2,...,fc
Exercise 9 With the notations and assumptions of Remark (48), show that H has a pole at pt if and only if (50) fails.
5.
Graphical test
It is well known that the Nyquist graphical test is important for three reasons: (i) It is based on directly measured experimental data, which can be obtained with great accuracy. It is not like the Routh-Hurwitz test or the Lienard-Chipart test, which are applied to data derived by interpolation from the measurements. (ii) It applies to distributed systems. (iii) In case the criterion is not satisfied, the required design modifications are readily apparent. In this section we present the graphical test for the distributed «-input-«-output case and also for the scalar case. We consider a particular case of the system of Theorem (4.36) in that the feedback transfer function is a constant matrix F. As before, fc
G(*) = I
m,-
ZRijKs-Piy
+ G.is)
The graphical test is used to check the inequality (4.39). The following equation describes our notation OO
det[/ + FG(s)] 4 1 + g{s) 4
1 + ga(s) + £ $r"' i= 0
I
+ I k=l
jUfc
E rj(s a=l
- pkf
5
GRAPHICAL
TEST
93
where ga e L1, Y,o \9i\ < oo, Re /?fe > 0 for /: = 1, 2, . . . , /, and pk, the order of the pole pk of det[7 + FG(s)], need not be equal to mk. In fact, 0 < \ik < nmk. We assume 0 = t0, tt > 0 for all / > 0. In the scalar case, as in Corollary (4.42), the expression to be tested is of the same form as (2), except for the fact that we have taken / = 1, for simplicity. Roughly speaking, the graphical test is based on the image of the jco axis under the map s h-> 1 + g(s). We must first define the phase of 1 -f g(jco) for all co EU: 4
6(jco) 4 arg[l + g(jco)] = Im{log[l + g(jco)]} There are two difficulties with this expression. First, 1 + g(jco) may be zero for some co. In that case, the test (4.39) fails; i.e., the closed-loop transfer function $ <stfnXn, and no further work is required. Second, one or more of the poles pk may be on theya>axis. In that case, we " indent" the jco axis at every pole. More precisely, let Ik denote the line segment centered on the yco-axis pole pk and of length 2p, where p > 0 and sufficiently small; for all jco-axis poles we replace these line segments Ik by half-circles of radius p, centered on the pole and lying in the right half-plane. Along the indented jco axis, the function s i—• 1 + g(s) is uniformly continuous, and the phase 9 is well defined. To every point on the indented jco axis associate its ordinate co; this defines a bijection between the real line, — oo < co < oo, and the indented jco axis. For this reason, we abuse notations and denote by 6(jco) the phase of 1 + g(s) at the point s of the indented jco axis whose ordinate is co. Thus, as co increases and goes around the polepk = jcok of order mk, we have 9[j(cok + p)] — 6 [j(cok + p)] = — mkn+ O(p); i.e., 6(-) decreases by mkn. At this point, we have O(jco) well defined as a continuous function of co except that we have to specify the branch of the logarithm we use in (4). We choose it as follows: if 0 < 11 + $(/0) | < oo,
then 6(j0) is equal to zero or n
according as 1 + g(jO) is positive or negative, respectively; if
1 + g(-) has a pole at 0,
then #0*0) is equal to zero or n
according as whether 1 + g(o) is positive or negative for all <J restricted to (0, p) where p is chosen sufficiently small. Thus, we have a well-defined continuous function co H-> 6(jco), which is the phase of 1 + g(s) along the indented jco axis. Let V0 denote the jco axis duly indented in the manner described above. The graphical test is based on the image of V0 under the map 1 4- g(-). Let Jf (for Nyquist diagram) denote this image. It will help us to consider in succession three increasingly complicated cases.
94
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Case I All the g/s in (2) are zero Then 1 + g(s) is analytic for Re s > 0, except at the poles pk, and, by the Riemann-Lebesgue lemma, it tends to 1 as | s | -* oo with Re s > 0. This limit is attained uniformly in arg s for |arg s\ < njl. Therefore, for any £ > 0, there is an r > 0 sufficiently large so that (a) all poles pk are included in B(0; r) n C+ , namely, the right half of the disk centered on the origin and of radius r\ (b) for all |a| < TT/2, 11 + g(r^)\ < 1 + e. Call c€r the right half-circle, which is on the boundary of the above halfdisk. The function 1 + g(') is meromorphic in the interior of B(0; r) n C + and uniformly continuous on its boundary, namely, V0 u <€r. Therefore, the principle of the argument applies: 1 + g{ •) has no zeros in the interior of V0 u c€r if and only if the net increase in its argument 6 as we traverse V0 u <€r in the clockwise direction is A0 = £ x fc \ik. Since by statement (b) above, 1 + g is bounded away from zero outside B(0, r), we conclude with the following theorem. 5
Theorem With the notations of (1), (2), and (3), and assuming that gt = 0 for all i > 0,
6
inf \det[I + F6(s)]\ > 0 Re s>o
if and only if 7
1 + g(jco) ^ 0,
for all
OJGU
and 8
A6 = lim [00'") " 0( -JQ)] =^^k fi->oo
= np
k
where the summation is over all poles pk that lie in the open right half-plane. The interpretation of (8) is the following: As co increases from — oo to + oo, following the required indentations at the jco-axis poles of 1 + #(•), the corresponding point traverses Jf and encircles the origin Yj^k=nP times in the counterclockwise direction. Note that np denotes the number of poles in the open right half-plane, where multiple poles are counted according to their multiplicity. Note also that (7) asserts that Jf does not go through the origin. Case II The g/$ are not required to be zero, but we assume that there is a sequence of nonnegative integers O^o00 and a positive number x > 0 such that n0 = 0 and 10
tt = ntT,
V7 e Z +
5
GRAPHICAL
TEST
95
This is an important special case because it occurs (a) in circuits which include one uniform lossless transmission line or a number of such transmission lines whose lengths are rationally related; (b) in control systems and in bioengineering models where there is only one transportation lag or a number of transportation lags whose time delays are rationally related. Condition (10) has two interpretations: (a) The tt9s are rationally related: tt = tknjnk, V7, k e Z + . ex n co 1 s (b) The function cov-^Y^Qi P(~~ J i ' ) * & periodic function of period 2%\x. Let OC
00
11 By the Riemann-Lebesgue lemma, as \s\ -> oo with Re s > 0, we have 1 2a
1 + g(s) -*?(s)
uniformly in arg s on
[ — n/2, n/2]
and 1 2b
1 + g(s) -> 1 + g0
for
s -» CXD along the real axis
Let 13
0(co) * arg/t/G)) Since cDt-*f(jco) is periodic with period
14
ITZ/T,
we have
(J)((JO + 27i/i) - <j)((o) = m2n, where m is an integer. If m =0, there is no increase in 0 over one period and coi->(/)(co) is periodic. If m ^ 0, then there is a net increase in phase of m2n per period, and we have
15 where X =mx and 4>p(') is periodic. Graphically, (12a) implies that the image of V0 under the map s\-+l + g(s) approaches, as | co \ -> oo, a closed curve J ^ , which is represented by 16
^oo^{f(jco)\0
96
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SYSTEMS
The graphical test for Case II is a special instance of Case III in the following. Case II can also be established independently by conformal mapping techniques using the transformation z = e x p ( — ST) [Wil. 5]. The test can be expressed as follows. 18
Theorem With the notations (1), (2), and (3) and assuming that the tt's are rationally related [(10)], we have
19
inf \det[I + FG(s)]\ > 0 Re s > 0
if and only if 20
l+£o
21
1 + g(jco) j= 0,
^0 for all
weR
22 where the summation is taken over all open right-half-plane poles, counting their multiplicity. Condition (22) implies that X = 0, or equivalently m = 0; graphically, it means that the closed curve Jf ^ does not encircle the origin. Furthermore, (21) and (12a) require that both Jf and Jf ^ do not go through the origin. Case III General Case: No Specializing Assumptions We shall not prove this case. We shall only describe what the test involves and in what way it differs from the preceding cases. In the present case, the function/defined, as in (11), by OO
turns out to be an almost periodic function in the right half-plane. (For definition see [Cor.l].) The limiting conditions (12a) and (12b) still hold. Since co\->f(ja)) is almost periodic, given any e > 0, there are e-translation numbers T£ which satisfy the condition 26
| f(jco) - f(j(D + JT£) | < 8,
for all
coeU
Furthermore, these translation numbers are 1(E) relatively dense on R. This means that any interval of length 1(e) contains at least one e-translation number. For these facts consult [Cor.l]. These facts have a very important consequence—consider the image under f(j') of the compact interval [0, /(e)], namely, 27
Jf{ 4 {z e C | z =f(j(o), 0 < oo < /(e)}
5
G R A P H I C A L TEST
97
Let a e Z and consider 28
Jfal ^ {z e C | z = / ( » , a/(e) < co < (a + l)/(e)} By definition of /(e), we have the important fact that for all a e Z , e/Taj is within an s neighborhood of Jfx. Consequently, the image of the jco axis under / ( • ) lies within an s neighborhood of Jf{. Using the same £ as in (26) and (28) and invoking the Riemann-Lebesgue lemma, there is an Q£ > 0 such that for all | co \ > QE, the image of V0 under 1 + g( •) lies within a 2s neighborhood of Jf t. As \co\ becomes very large, Jf enters and remains within 2e neighborhood of Jfx. It can be shown that the phase <j> of the almost periodic function f(jco) is necessarily of the form
29
(j>(co) =Xco + h(co)y
co EU
where XeU and /?(•) is almost periodic, hence bounded on U. Note that k ^ 0 if and only if, for s sufficiently small, the corresponding curve Jf x encircles the origin. To state the graphical test, it is convenient to denote by N{Jf i; 2s) the 2s neighborhood of Jf u with Jf', being the curve defined by (27). Using these facts, we can state the graphical test as follows. Theorem With the notations of (1), (2), and (3), inf
\det[I + FG(s)]\ > 0
Res>0
if and only if (0 l + ^ o ^ 0 (ii) the. position of the curves {z e C | z =f(jco), co e U} and {z e C|z = 1 + g(jco), co eU} with respect to the origin O of the complex plane is such that (a) for an s > 0, sufficiently small, the point O does not belong to N{Jfx\ 2s) and to {z e C|z = 1 + g(jco), 0 < co < Q(e)}, where Q(s) is defined above; (b) for the same s, with Te denoting a corresponding translation number, it is required that |#T,)-#0)|<jt/2
(c)
lim [6(ja>) ~
where np is the number of open right-half-plane poles counting multiplicities of §, 0 is the phase of 1 + g, and
98
32
IV
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SYSTEMS
Remarks 1. The graphical test (30) requires the knowledge of the sequences ($1)0°° a n d (^)o°° m order to plot Jf x and obtain (•). 2. Condition (b) essentially says that Jf x does not encircle the origin. Indeed, by (13), is the phase off; by (26), | f(jtE) — f(0) \ < s; by condition (a) I f(JTE) I and | /(0) | are both larger than 2s. So condition (b) says that as co increases from 0 to T £ , the point f(jco) does not encircle the origin. 3. In simple cases, having chosen an s > 0, it is easy to obtain an estimate for Q(e) and /(e). Then, only images of compact intervals need to be plotted. In complicated cases, it is not easy to give a lower bound on 1(e); there is the possibility that in (29) X is, say, positive and very small so that over [0, /(e)], the contribution of Xco is not apparent. Using (^i)0°° and (^)o°° and Diophantine analysis, we can obtain a lower bound on X if X were positive; hence this gives a lower bound on 1(e) so that the condition (b) will guarantee X=0. Exercise Use the graphical test to discuss the stability of the following system: uu eu yt: U+ -• U; u2 = 0; the forward gain is a real constant k; the feedback subsystem has e~s as transfer function; the constant k takes values in (0, 00). Check your answer by calculating the impulse response of the system.
6
Discrete-time systems Discrete-time systems arise
(a) from modeling systems that are inherently digital such as digital filters and computer-controlled systems where the inputs and outputs are periodically sampled; (b) from approximations of continuous-time models where the inputs and outputs are approximated as piecewise constant functions; for example, u(i) = ak, for kT < t < (k + 1)T, ak constant, and k = 0, 1, . . . ; (c) in computer algorithms for solving continuous-time problems. In all these cases, the inputs and outputs can be viewed as sequences defined on Z+ (the set of nonnegative integers) taking values in a Banach space E (i.e., complete normed space). Typically this Banach space will be R, C, Un, C", Unxn, etc., . . . .
6.1
Existence and uniqueness
We consider once again a system whose configuration is as shown in Fig. IV. 1. Since we are dealing now with the discrete-time case, w, e, and y represent sequences in W. We study the conditions under which there exists
6
DISCRETE-TIME
SYSTEMS
99
a unique sequence e corresponding to each sequence u. In what follows, we allow both F and G to be time-varying operators with memory, but we restrict them to be linear and nonanticipative. So, to establish our notation, we have Sn = the set of all maps x from Z + into Un
1 2
u, e, y e S",
or equivalently,
F: Sn -± Sn,
3
u, e, y: Z+ ->W, e.g., u = (u0, ul9 w2, ...)
and
(Fx)t = £ Fu
Xj
y=o
G: S" -• Sn,
4
and
(Gx), = £ G lJjci
Notice that there is no real restriction in assuming that the input, error, and output all have the same number of components; for if this were not the case, then one can always add rows or columns of zeros, as appropriate, to the matrices Ftj and Gtj. 5
Theorem There exists a unique e e S" satisfying the equation
6
e = u — FGe n
corresponding to each u e S , if and only if 7
det(I + i v m G m , w ) ^ 0 , Proof
VmeZ +
Since the system in Fig. IV. 1 is characterized by the equations
8a
e = u — Fy
8b
y = Ge
it is clear that there exists a unique pair (e, y) in Sn x Sn satisfying (8a) and (8b) corresponding to each u e Sn, if and only if (6) has a unique solution e e Sn corresponding to each u e Sn. Thus, we justify concentrating on the existence and uniqueness of solutions to (6). <= If we expand out the equations forming (6) and write them out as simultaneous equations in the infinite number of unknowns e0, el9 . . . , em, . . . , we see that the matrix of coefficients multiplying (e0, el9 ...) is lower triangular, and that the diagonal elements are J + F 0 0 G 0s0 , . . . , I + Fm>m G m m , . . . . Thus, if (7) holds, then for any u e Sn, we can find a unique e e S" satisfying (6). => By contradiction. Suppose (7) is false, and let k be the smallest integer m such that det(7 + Fm>mGm>m) = 0. Then, given any u e Sn, the
100
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quantities e0, eu . . . , ek_x can be uniquely determined. Now the equation for ek is k
9
k
YFkjGijej
e* = w * - E i=0j=0
or, collecting all unknown quantities on one side and all known quantities on the other, k-l
(I + J7k,kGk,k)ek = uk-YJ
10
i
k-l
!Lfk,iGijej-Y
i=0;=0
F
k,kGkjej
j=0
Depending on the particular sequence u under consideration, the right side of (10) may not be in the range of / + FkkGkk, in which case (10) has no solution; or it may be in the range of / + FkkGkk, in which case (10) has infinitely many solutions. But in either case, (10) does not have a unique solution, whence neither does (6). :: The importance of Theorem (5) lies in the fact that it provides a very simple, necessary, and sufficient condition for the closed-loop system characterized by (8a) and (8b) to be solvable, under relatively mild assumptions on F and G. The only crucial assumptions on F and G are that they are (i) linear, and (ii) nonanticipative. But other than that, F and G may be time invariant or time varying, and they may have memory or be memoryless. Exercise 1 Based on Theorem (5), one can readily derive a large number of criteria for various classes of discrete feedback systems to be solvable. We give one of the most important cases as an exercise. Let G: Sn -* Sn be time invariant, i.e., Gttj = ^(i — j) = &i-j, say (the causality assumption on G, of course, implies that (Si = 0 for / < 0), and let F: Sn -» S" be a memoryless time-varying gain, i.e., Fiti = # ' ; , Fitj = 0 for i>j. Then (6) has a unique solution e e S" corresponding to each u e Sn if and only if 11
det(J + &m &0) # 0,
6.2
Vw e Z +
Z-transforms
The main properties of z transforms are a direct consequence of properties of power series. Given a sequence a = (a0, au a2, ...) with the afs in a Banach space E, we define its z transform by OO
12
%{a) =a(z) = Yiakz~k fc = 0
6
DISCRETE-TIME
SYSTEMS
101
This is a power series in z _ 1 ; it has a radius of convergence ra and converges outside B(0; ra), the closed disk of center 0, and radius ra. It is a well-known result that ra =limsup||tfj 1 / f c k-+ oo
Furthermore, (a) the right-hand side of (12) converges absolutely for any \z\ > ra ; (b) for any S > 0, it converges uniformly in the complement of B(0; ra 4- (5), the open disk of center 0 and radius ra + S. (c) the right-hand side of (12) is an analytic function in the open set
\A >ra. In the following we shall use the superscript tilde(~) to denote z transforms; viz., a = &{a\ G = 3T(G). The connection between z transforms and the Laplace transform is the following: To the sequence a = (a0, al9 ...) associate the time "function" Y,o ak 5(t — k). Its Laplace transform is Xfc°=o ^ke~ks- By comparison with (12), we see that z corresponds to es. Thus, the open left half-plane, Re s < 0, corresponds to B(0; 1), the open unit disk \z\ < 1; the closed right halfplane C + , Re s > 0, corresponds to the complement of i?(0; 1), i.e., \z\ > 1. This observation is very useful in translating a proof of the continuoustime case into one for the discrete-time case. To recover the sequence from its z transform, we use a classical result on Laurent expansions as in the following theorem. Theorem Let a = (a0, al9 ...) be a sequence in a Banach space E. Let a be its z transform and ra its radius of convergence. Then, for any simple closed rectifiable curve y which lies in \z\ >ra and which encircles the origin once in the counter clockwise direction, ak = (1/2TT/) & a(z)zk~1 dz,
k = 0, 1, . . .
These results are to be found in Dieudonne [Die.l, Chap.9], Hille and Phillips, [Hil.l, Chap.5.4].
6.3
Stability of discrete-time feedback systems
In this subsection, we leave the proofs of the various theorems as exercises to the reader because they can be easily constructed from those of the corresponding theorems for the continuous-time case. It is a fact, however, that in many instances the theorems for the discrete case are more intuitively
102
IV
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clear, and less cluttered, than their continuous analogues. Where this happens, it is pointed out. We consider a discrete-time linear feedback system as shown in Fig. IV.2. The subsystems G and Fare both time invariant, and as such they are characterized by their "unit pulse responses" G and F, respectively. Specifically, we have 18a 18b
SnXn = {set of all mappings from Z + into UnXn} G, FeSnXn;
e.g., G = (G 0 , G„ . . . .), G f eK n X ",
V/eZ +
The equations characterizing the feedback system are m
19a
e=u-F*y,
or em = um - £
K-iyi
i= 0 m
19b
y=G*e,
or ym =
Y,Gm.iei i= 0
If w, e, y, G, and F are all z transformable, then (19a) and (19b) can also be rewritten as 20a
e(z) = u(z) - F(z)e(z)
20b
y(z) = G(z)e(z)
Note that if both G and F are z transformable, then the closed-loop transfer function satisfies
21a 21b
R = G(I + FGy1 (/ - FH)(I + FG) = (/ + FG)(/ - FH) = I But in any case, / / satisfies the time domain equation
22
H = G-
H*F*G
If G(z) is rational and F(z) equals a constant F, we can state a theorem analogous to Theorem (2.30) and Corollary (2.43). 23 24
Theorem Let G{z) be rational and proper; let F(z) = F be a constant and suppose det[/ + FG(oo)] = det(/ + FG0) * 0 Let G(z) = N(z)D(z)~1, where the polynomial matrices N(z) and D(z) are right coprime. U.t.c,
6
DISCRETE-TIME
SYSTEMS
103
(a) the closed-loop transfer function H(z) is exponentially stable [i.e., all of its poles are in B(0; p) for some p < 1] if and only if the polynomial det[5(z) + FN(z)] has all of its zeros in B(0; 1), the open unit disk; 25
(b)
det[/ + FG(z)] =
~ = det H(z)
~ det D{z)
where (p0i)in and (/>c;)i" are the poles of G ( ) [i.e., the zeros of det D(z)] and the poles of H() (i.e., zeros of det[5(z) + FN(z)]), respectively. In both cases, multiple poles are counted according to their multiplicity. Note that the remarks (2.46), (2.58), (2.62), and Theorem (2.70) also apply in the present case. Note also that the assumption (24) is quite reasonable since, by Theorem (5), (24) is necessary and sufficient for the closed-loop system to have a unique solution. 26
Corollary
If H( •) e U(zf
xn
and if it is exponentially stable, then
(a) for any minimal representation [of the form xk + l = Axk + Buk, yk = Cxk + Duk] of H(z), the equilibrium point x = 0 is globally, exponentially, uniformly (in k) stable; (b) for any p e [1, oo], if u e / / , then y = H * uelnp, and both uk and yk -> 0 as k -> oo whenever p < oo; (c) if u e ln™ and w* -• &/<», a constant vector in Rn, then yk -> H^u^ as k -> oo, and the convergence is exponential. :: In the case where G(z) is not necessarily rational, our results are based on a straightforward extension of Wiener's theorem (Appendix D.3.6.). Let | • | denote any suitable norm on Un and also the corresponding induced norm on UnXn. A sequence (a0, ax, ...) of vectors in Un (resp. of matrices in UnXn) is said to be in lnl (resp., l*Xn) if £ f * 0 I ai I < oo, and its z transform is said to be in l„ (resp., VnXn). The following lemma is an important tool in the subsequent work. 27
Lemma Let M = (M 0 , Mu ...) be a sequence in UnXn. Assume that Me tfx^i.e., \\M\U ^ X r = o \Mk\ < o o . U.t.c, (a) its z transform zi—>Af(z) is analytic in \z\ > 1, uniformly continuous on the unit circle e[\ 0<9<2n; \M(z)\ < HA/^ for all \z\ > 1, and M(z) -> M0 as | z | -> oo;
IV
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(b) M has an inverse in l*Xn [i.e., there is a P = (P 0 , Pu ...), with \\P\\X < oo, such that M * P = P * M = (I, 0, 0, ...)] if and only if 28a
inf |detM(z)| > 0 (c) for any u e / / [i.e., u = (u0, uu ...) with £ o \uk\p ± (\\u\\p)p < oo],
28b First, we state a general set of necessary conditions. 29
Theorem Consider the system characterized by (19a) and (19b). Suppose Fel„Xn, and that the closed-loop system has a unique solution (e, y) in Sn x Sn corresponding to each input u e Sn [or, equivalently, that H is uniquely determined by (22)]. U.t.c, if He l^Xn then
30 31 32 33
(i) G is z transformable; (ii) G is analytic outside some finite ball B(0; r), where r < oo; (iii) G is of the form G(z) = N(z)[D(z)]-\ where JV(-), D(')el^n\ (iv) all singularities of G in the region {z: 1 < \z\ < r} are isolated poles;
34
(v)
inf | det[/ + F(z)G(z)] | > 0.
::
\z\ = l
35
Remarks (i) Note that Theorem (29) is significantly more comprehensive than its continuous-time analogue, namely, Theorem (4.5), because there is no assumption in the present case corresponding to " G is a distribution of order at most zero in some neighborhood of the origin," as in Theorem (4.5). The only assumptions on G in the present case are linearity and time invariance, and nothing else. (ii) In Theorem (4.5), we conclude that all singularities of G must be in some half plane, while in the present case we conclude that all singularities of G must lie in some disk in C. As shown later, the fact that all singularities of G lie in a compact set has the effect of making the pseudoright-coprime factorization technique much more pertinent and valuable to the discrete-time case than the continuous-time case. We now state the discrete analogue of Theorem (4.14).
36 37
Theorem Consider the feedback system characterized by (19a) and (19b). Suppose G is z transformable, and that Fe ljlXn. Then H e l„Xn if and only if (i) There exist N, D in Vn x „ such that (a) det D0 ^ 0
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105
(b) det [D(z) + F(z)N(z)] # 0 whenever \z\ > 1 and det D(z) = 0, and (c) G(z) =N(z)[D(z)]-\ and 38
(ii)
inf | det[/ + F(z)G(z)] | > 0 ki>i
39
Remark It is easy to verify that (i) (a) and (i)(b) merely ensure that the pair (det D, det(D + FN)) satisfies the no-cancellation condition (N) in the region {z: \z\ > 1}. In the present case, we have exploited the fact that D(z), N(z), and F(z) all have definite limits as \z\ -> oo, namely, D0, N0, and F0, resp. Next we turn to pseudo-right-coprime factorizations and to stability results based on them.
40
Definition Two elements Jf, S) in 7,JX|I are said to be pseudo right coprime (pre) if there exist 4?, IT, # in Vnxn such that (i) det iT(z) ^ 0 whenever \z\ > 1, \z\ < oo, and
41 42
(ii)
for all | * | > 1 .
Definition Given a z transform G, the pair (^/T, ^ ) is said to be a pseudoright-coprime factorization (prcf) of G if (i) (ii) JV" and # are pre, and (iii) det 30 # 0.
43
Remark As is the case with Laplace transforms, in order for G to have a prcf, it is necessary that all poles of G outside the unit disk be contained in some compact set. But in contrast with the continuous-time case, this places no real restriction on G. because if G has poles of arbitrarily large magnitude, then we know by (29) that H^ /,JX „, and further analysis would be unnecessary.
44
Theorem Let FellnXn, and let G = J?9~\ Then H e l\ x „ if and only if
45
(i)
where (<#, §) is a prcf of G.
inf | det [J + F(z)G(z)] \ > 0
and 46a
(ii) det[5(z) + F(z)J?'(z)] # 0
whenever det D(z) = 0 and
\z\ > 1
106
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46b Theorem Let FellnXn, and let G = JfQ)~\ Then H e l„xn if and only if 47a
where (J*9 3) is a prcf of G.
inf | det[^(z) + F(z)Jf(z)\ \ > 0 Finally, we turn to the case where G(z) has a finite number of poles in the region { z e C : \z\ > 1}.
47b Theorem
Let Fe l*Xn, and suppose G{z) is of the form
48
G(z) = £ where
E * 0 -/(* - p,-y + C6(z) = SM(z) + 5*(z)
GbellXn, 5 I I (z)=iV(z)[5(z)]- 1
49
where iV and 5 are right-coprime-polynomial matrices, and further D is column proper. Then H e /,JX|I if and only if 50
(i)
inf | det[/ + ^(z)S(z)] | > 0
and (ii) d e t f e ) + F(pdN(Pi) + F(Pi)Gb(Pi)B(Pi)] ^ 0 51
Corollary
52
for
i = 1, . . . , &
Suppose n = 1, that g(z) = Y,
ZrijKz-pty'+faiz)
i=\
j=l
and suppose/, gbe ll. Then // e Z1 if and only if 53
(i)
inf | l + / ( z ) £ ( z ) | > 0
and 54
(ii)
/(/>,) # 0 ,
for
i = l , . . . , A:
We conclude with a note on the graphical test for discrete-time systems Let G(z) be of the form (48), and let Fe l*Xn. The objective is to derive a means of verifying the condition (50) by examining G(el9), 6 e [0, 2n]. For this purpose, define 55
1 + g(z) = det[J + F(z)G(z)] Note that g(z) is of the form (52).
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107
The graphical test is applicable to the image of the set D = {z = eie, 9e [0, 2n]} under the map ZI->1 + g(z). In case g(z) has a pole at some point eldo, we indent D by "removing" from it the set {z = eid, 9 = [90 — 3, 90 + 3]}, and "adding" the set {z = [1 + f](9 - 00)]ew, 9e[603, 90 + 3]}, where rj: U -» U is the "triangular pulse" function (0 n(x) = lx + 3 ((5 - x
56
if |JC| >3 if x e [ - < 5 , 0) if * e [0, (5]
Furthermore, we choose (5 > 0 sufficiently small so that inf 11 + g(z) | > 0 zeSo0
where Sdo is the set {z = re{\ 1 < r < 1 + rj(6 - 0o), 9 e [0O -3,60 + 3]} and such that Sdo contains no poles of g other than exp(7#0). Let us again refer to the simple closed curve that results from these indentations as D. Then with each 6 e [0, 2n] we can associate a unique element of D, which we denote by z(6). Now define 57
§i(0)=g[z(0)] It remains to define the argument of 1 + ^i(#). It is clear that as 6 varies from 0 to 27i, the function gt(9) is uniformly continuous in 9. Moreover, we can assume without loss of generality that 1 +#i(0) ^ 0 , because if 1 + ^i(0) = 0, the condition (50) is automatically violated, and further analysis is unnecessary. So we define
58
[l+g1(9)]
where the particular branch of the logarithm function is so chosen that 0(0) = 0 (resp., TT) if 1 + ^ ( 0 ) > 0 [resp, 1 + ^ ( 0 ) < 0]. With these definitions and conventions, we are ready to state the graphical test for discrete-time systems. 59 Theorem Under the above conditions, 60
inf 11 + g(z) \ > 0 if and only if
61
(i)
62
(ii)
l+<7o^0 inf
|1 +£ 1 (0)| > 0
06[O,2TT]
63
(iii)
(j)(2n) - 0(0)
=2n-mp
where mp is the number of poles of g(z) outside the closed unit disk.
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Exercise 2 (a) Show that (60) holds if and only if (i) 1 + g0 ^ 0, and (ii) 1 + g(z) =£ 0 whenever \z\ > 1. (b) Define £ = z _ 1 , Vze C, and define h(0 ^giC1). Show that (60) holds if and only if 1 + h(£) =£ 0 whenever | £ \ < 1. (c) Prove Theorem (59) by applying the principle of the argument to the function £ -> 1 +£(£)• Exercise 3 (a)
Let z = es. Show that (60) holds if and only if inf | 1 + g(es) \ > 0
64
Re s > 0
(b)
Let 00
§(z) = Y.9iz~l
00
so that
9(es) = X 9ie~ls
1=0
1=0
Prove Theorem (59) by applying Theorem (5.18) to the condition (64).
7
Linear time-varying systems
Linear time-varying systems have several features that distinguish them from linear time-invariant systems. To name only two: (1) A linear time-invariant system is represented by a convolution integral, while a linear time-varying system is characterized only by an integral operator. As a result, linear time-invariant systems can be conveniently studied in the Laplace domain or in the time domain, while time-varying systems can be studied only in the time domain. (2) Unlike a time-invariant system, a time-varying system can be L°° stable but not L1 stable, or vice versa. However, a linear time-invariant system is L00 stable if and only if it is L1 stable. In many of the stability studies in this section, we do not assume that the systems under study are nonanticipative, since this assumption usually does not affect the development one way or the other.
7.1
L00 Stability Consider a system whose zero-state response is given by 00
1
y(t) = I i =
«00
w,(0«(( -tt)+
— GO
where w, y: U -> U. We assume that
w(t, T)«(T) dx = {Wu)(i) ^ — CO
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109
(i) the Wj(*) and u(-, •) are all measurable [this implies, in particular, that w(t, •) is measurable for almost all t e U]; (ii) for all t e U, we have
2 -
Throughout this section || • || will denote the L00 norm, e.g., ||w||. Let ^ , the input space, be L°°(lR), and let ^ , the output space, be the set of all measurable functions from R into U. Then the conditions (2) and (3) ensure that ^ m a p s % into ^ , and clearly W is a linear map. Moreover, the domain of W is all of °ll\ given any «(•) such that ||w|| < oo, the quantity >>(0 is well defined by (1), and \y(t)\ < GO, Vf e IR. Our objective is to derive conditions under which a system whose zero-state or input-output response is characterized by (l)isL 0 0 stable. There are two possible ways of interpreting the sentence, "The system is L00 stable"; namely, (I) an input u e L00 produces an output y = Wu e L00; (II) there exists a constant c < oo such that ||^|| = || Wu\\ < c\\u\\ whenever weL 00 . The first statement simply states that a bounded input produces a bounded output. Statement (II) is more restrictive; it includes statement (I) as well as the added assertion that there is a maximum possible finite ratio between the norm of the input u and that of the output y = Wu. Now let us define the quantity 4
Coo
= sup ( £ reR
| w,(01 + f
\» = - oo
J
| w(f,
T)|
dx) /
-oo
With these definitions, we can state the following theorem. 5 6
Theorem Statements (I), (II), and (III)
c 0 0
are all equivalent. Proof We proceed in the following order: (III) (III) => (II) Given u e L00, we have
=> (II)
=> (I)
=> (III)
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But since the extreme right side of (7) is independent of t, it follows that y e L00, and that \\y\\ < c a Jw||. Thus, statement (II) holds with c = c^. (II) => (I) Obvious (I) => (III) The proof is based on the principle of uniform boundedness, which states: Let A be an arbitrary index set, and let (% , t e A) be a family of continuous linear functional mapping a Banach space X into a normed linear space Y. If for each xe X the set {rjt(x), t e A} is a bounded subset of Y, then {\\rjt||, t e A} is a bounded subset of U. To apply this principle for our present purposes, we let X = °U = L°°, Y = R, and define a family of linear functional (t]t, t e U), as follows. For each t e M, r\t is the map % -» U defined by o
-
8 o
Clearly ^r is linear, and it is continuous by virtue of the assumptions (2) and (3). Next, let us determine ||^f||. On the one hand, we have
so that 00
.00
We show now that the inequality in (9) is, in fact, an equality by showing that for any e > 0, we can find a function weL 00 such that \\u\\ = 1 and rjt(u) > bt — &, where bt is the right-hand side of (11). So given 8, pick an open interval 11 containing t — tt such that 10
f |vv(r, T)|
Let u be the element of L00 defined by
Then \\u\\ = l, and
Hence
rfT<(e/6)2~|f|
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111
Now suppose statement (I) holds; i.e., suppose that for all w e L00, the set {nt{u), t e U} is bounded. Then by the principle of uniform boundedness, the set {11*7,||, t e U} is bounded. But by (11), this means that QO
..GO
which is precisely statement (III). 13
ss
Remarks (a) Notice that W is not required to be nonanticipative. (b) If u: U-+W1 and y: U -> Km, then wt(-) and w(-, •) would all be m x n matrices, and (12) would stand as is, except that | • | would have to be interpreted as any norm on RmXn. (c) The number c^ is, in fact, the induced norm of the linear map W from L00 into L00. Exercise 1 Consider a feedback system where ul9 eu yx\ U+ -> U; Hu H2 map L00 into itself; and u2 is identically zero. Suppose that
where vv(-, •) satisfies (3) above; suppose also that
where (/>: U x U+ -• R is continuous and belongs to the sector [0, k). (a) Obtain conditions on the constants cM and k such that u1 e L00 implies yt e L°°. (b) Show that if w(-, •) satisfies (3) above and 0 is saturating [i.e., for each 7 > 0, 7 arbitrarily small, there exists a /? < oo such that 10(o-, t) | < y\<j\ + /?, Vcre R, Vr e R + ] , then ul e L00 implies j ^ e L00, regardless of the value of c^. In the case of discrete systems, we have a similar characterization. 14
Theorem Consider a system whose input u and output y map ^+ into R". Let 00
where w(k, i) e UnXn and satisfies
112
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U.t.c., the following three statements are equivalent:
(i)
.
17 (ii) 3c
7.2 L Stability Consider a system represented by r00 18
y(t)
=
w(t,
T)W(T) dxy
•'-oo
where Suppose w e L1 and that w(f, •) is bounded for all t e IR. Then y(t) is unambiguously defined by (18) for each t e U. We now derive conditions for the L1 stability of the system represented by (18). 19
Theorem Let y(-) and u(-) be related by (18). U.t.c, (I) ue L1 => y e L1, and moreover, there exists a constant c < oo such that || j || x < cllwlli if and only if
20
(II)
sup
| w(r, x) | dt = c^ < oo
Proof <= Suppose (20) holds, and let w(-) e L1. Then for y(-) defined by (18), we have
where we use Fubini's theorem to justify interchanging the order of integration. Thus (I) follows, with c = c1. => Let u(x) = d(x — X). Then the corresponding y(-) is given by y(t) = w{t, X)
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113
If (I) holds, then this y(-) must belong to L1. [Note that even though 3(T — X) does not belong to L1, it can be approximated arbitrarily closely in the weak * sense by an L1 function. ] In other words. r°°
| w(t, X) | dt < oo J -oo
Now let X vary over R, and let u{x) = d(x — X) as before. Then (I) implies that the corresponding functions w(t, X) must belong to L1 for each X e U, and moreover that the family of functions [w(t, X), XeU] must be uniformly bounded. Hence, we must have -00
sup A E R
| w(t, X)\ dX < oo ^ -CO
which is clearly equivalent to (20).
::
21
Remark Note that c{ is the induced norm of the operator W when viewed as a mapping from L1 into itself. To study If stability for p e (1, oo), we consider once again a system represented by (18), where w(t, •) and u(-) are locally integrable.
22
Theorem
Under these conditions, if
(a) for some constant c^ / . + CO
23
\w(t9 T)| dx Kc^ < oo,
VrelR
(b) for some constant cx 24
| w(t,
T)
Jr < d < oo,
VT G
R
•'-oo
Then, for any fixed p e [1, oo], uelf{— oo, oo) => .y elf( — oo, oo) In fact, 25 26
||j;|| p < c i ^ ^ t l ^ l l p ,
where
\\p + l/g = I
Remarks I. Assumption (23) is the necessary and sufficient condition for L00 stability, while (24) is the necessary and sufficient condition for L1 stability. II. Assumptions (23) and (24) can be rewritten as 3 constants c^ and cx s.t.
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For the time-invariant case, w(t, T) =f(t — T), the two conditions are equivalent and reduce t o / ( - ) e L1. III. (23) does not imply (24), or vice versa. To see this, consider w(t9T) = l(t-T)-e-x\l
+
\t\yl
Then (23) holds, but not (24). IV. The results still hold if u takes values in IR", y takes values in lRm, and w(t, T) is an m x n matrix. Then | w(t, x) | should be read as the induced norm of the matrix w(t, T). Proof In view of Theorem (19), we need only consider the case where 1
(by Holder) (by 23) dt
by (24) Hence
Exercise 3 (23).
Give an example of a function w(-, •) that satisfies (24) but not
Exercise 4 Prove the following stronger version of Theorem (22); consider a system of the form (18). U.t.c, this system is IF stable for all p e [1, oo] if and only if it is L1 stable and L°° stable.
7
27
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115
Application Consider the linear time-varying system where uu el9 e2, yl9 y2 U+-+U and
28
y1=g4:ei,
29
gestf
y2(t) = k(t)e2(t)9
k(-) continuous on R.
Assume u2 = 0 for simplicity. The equations for the systems can be written as 30 31
et(t) = ux(t) - k(t)(g * ex){t) e2(t) = (g* Ul)(t) - (g * ke2){t) Since (28) and (29) are linear, we may apply the incremental small gain theorem; then, in addition to bounds, we have existence, uniqueness, and continuous dependence. (a) L00 Stability
We have immediately «! e L00
=> et and e2 e L00
if either 33a
sup \k(t)\ f \g(x)d% < 1 or
33b
sup (b) L1 Stability
\g(t - x)k(z)\ dx = cj < 1
We have immediately wx eL 1
=> e1? e2 e L1
if either 34a
sup T>0
r00
\k(t)g(t - T)| dt = c / < 1
^T
or 34b
sup |/C(T)| T>0
|flf(0|
dt<\
^0
(c) I? Stability If (33a) and (34a) hold, or if (33b) and (34b) hold, then for any p e (1, GO), ut e IF => eu e2 e IF. Exercise 5 Consider a linear time-varying system described by (30) and (31). Suppose that gel} and that k(i) = k0 + 2kx cos co0t9
Vf e U+
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where k0, ku and co0 are real constants. Apply the loop shifting theorem (II.6.8) (with the linear map K — kQI) so that the feedback is 2k1 cos co0 t, and the forward loop is yt = gx * ex. Suppose that giijco) is of low-pass character, say, for co>coc, \gl(jw)\ < |^fO'coc)|(coc/co)2. Discuss the L2 stability of the system and show that as a>0 becomes much larger than coc, then kx can take very large values.
7.3 A perturbational result Consider the feedback system shown in Fig. IV. 1. Our ultimate objective is to show that if the time-varying feedback gain k(-) e L1 [0, oo), then the open-loop system is LP stable if and only if the closed-loop system is LP stable, under quite broad conditions. We begin with a few preliminary results. Since all of our main results are based on an integral equation relating the closed-loop and open-loop impulse responses, we first show that the closed-loop system is uniquely solvable and that its input-output relation is of the form (18). The two lemmas that follow immediately are in the nature of " dogwork," and the reader may skip to Theorem (50) without loss of continuity. 37
Lemma Consider the feedback system shown in Fig. IV. 1, and suppose u, e, y: U+ -> Un. Suppose G(t, x) is bounded over every compact subset of U2, and that K(-) e L1nxne(U+). U.t.c, for each u e Llne(U+), there exists a unique e e L|,e((R+) satisfying the equation
38
e(t) = u(t) - K{t) f G(t, T > ( T ) dx Jo
Moreover, VT< oo, 3cT < oo such that ||erlli ^ c rll w rlliRemark Lemma (37) is stated for u belonging to L / , since Lex is the largest of the extended spaces, in the sense that Lex contains L e 2 , L™, and indeed all L / f o r 1
Equation (38) is of the form
39
e(t) = u(i) - (KGe)(t) LAwe(R+)
where KG: ->Lj e (R+). Taking truncations of (38) over [0, T], we get an equation of the form 40
eT(t) = uT(t) - (FeT)(t) l
1
where F: Ln [0, T] -• L,, [0, T] is defined by 41
(Fx)(t) = f K{t)G(t, J
o
T)JC(T)
dx,
Vf e [0, T]
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117
Note that F can be interpreted as the operator PTKGPT, restricted to the space Lnx [0, T], where PT is the projection operator defined by Eqs. (III. 1.4). We know from Eq. (II.7.4) that (40) has a unique solution er e Lnl [0, T] corresponding to each uTe L„l[Q, T] if the spectral radius of the operator FT is less than 1. Now define a family of operators (RT, TeU+), with
(RTx)(t)=
(\K(t)\
gTx(x)dx
where 42
gT=
sup
\G(t,x)\
0
We show that p(RT), the spectral radius of RT when viewed as a mapping from L1 [0, T] into itself, is zero. It is left as an exercise to the reader to verify that, for meZ+ and m > 1, the operator RTm: l}[0, T] is given by (RTmx)(t) = I K(t) | J ^ I J^ | K(s) | gT ds\
43
• ^ r x(t) dr/(/w - 1)!
Now by slight modifications of Theorem (19) and Remark (21), we have \\RTm\\=
44
sup
\ \K(t)\\
< I \K(t)\dt\
\K(s)\gTds\ gTml(m - 1)! = (bTgTTI(m
gTdtl(m-\)\ - 1)!
where 45
rT bT= \ \K(t)\ dt < 00 ^0
So clearly ||i? T w || 1/m ^ 0 as m -> oo, so that p(7?r) = lim^^,, ||^ r w ll 1/m = 0. At this point, one can show by routine majorization arguments that p(FT) < p(RT) = 0. Hence (/ + Fj)'1 is a well-defined operator on L/IX), T], and IKZ + ^r)" 1 !! < E ^ o l l F r i l = ^ r , say. Thus, (38) has a unique solution
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IR2 and K(-) e Lj,e[0, oo), then for each u(-) e L^,[0, oo), there exists a unique y e L*e[0, oo) satisfying the equation 46
y(t) = f G(t, T)W(T) JT •'O
f G(f, T)K(T)J>(T) rfr •'o
Moreover, V r < oo, 3 J r < oo such that Halloo < dT ||wrlla>Exercise 8 Prove Lemma (37) and do Exercise 7 by using Corollary (III.5.8). 47
48
Lemma Let G(t, T) be bounded over every compact subset of U2, and suppose K(-) e L1nxne(R+). U.t.c, there exists a unique solution H(-, •) of the equation H(t, T) = G(t, T) - f G(r, v)K(v)H(v, r) dv,
0 < T < t < oo
Moreover, //(*, T) is bounded over every compact subset of IR2. Proof Fix T G I R + , and consider (48) to be an equation in the unknown function H( •, T). Then by Exercise 7, we see that H(,r) is uniquely determined from (48), and that for each T < oo, there exists a finite constant dT such that 49
\\H(t, T)|| < dT sup ||G(r, T)||,
Vr e [T, T]
T
It is clear from (49) that since G(t, t) is bounded over every compact subset of R2, so is H(t, T). S: We are now in a position to state our main result concerning the nature of the input-output relation of the closed-loop system in Fig. IV. 1. 50
51
Theorem Consider the system shown in Fig. IV. 1, where w, e, y: U+ -> Un. Suppose G(t, T) is bounded over every compact set in R2 and that K( •) e L\ x ne (R+). U.t.c, for each u e LFne(R+), the equations y(t) = (G(U TMT) dx, J
e(t) = u(i) - K(t)y(t)
o
have a unique solution y e LPne(R+) given by y(t) = f //(r,
T)W(T) JT
= (Hu)(t)
where H(-, •) is the unique solution of (48). /V00/ Suppose weLJ e (R + ) for some pe [1, oo]; then ue l}ne(U+). Hence the system of equations (51) has a unique solution for e (and hence y) in
7
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119
L*e([R+). Now consider the function z = Hu. Clearly, z e LFne(U+) and hence z e I}ne(M+), and moreover (51) is satisfied with y = z. Since (51) has a unique solution, this implies that y = Hu. :: 52
Remark The thrust of Theorem (50) is in showing that (i) the feedback system in Fig. IV. 1 is solvable; (ii) the input-output relation of the closed-loop system is of the form (18); and (iii) the open-loop impulse response and the close-loop impulse response are related by (48). The next three theorems relate some properties of the open-loop impulse response with the corresponding properties of the closed-loop impulse response.
53
Definition Let F: U+2 -»UnXn be measurable and locally integrable, and suppose further that F(t, x) = 0 whenever x > t. Define
54a
fh=\\F\\b=
sup
54b
/ i = l|F|li= sup xeU+
54c
|F(*,T)|
C\F{t,x)\dt J
x
/ „ = IIFIL = sup
f\F(t,T)\
With these definitions, let Sb be the normed linear space 55 Sh = {F: \\F\\b < ex)} The normed linear spaces Sl and S^ are defined similarly. 56
Remark Consider a system whose input-output relation is rf
r00
57
y(t) =
F(t, T)M(T) dx = Jo
F(t, x)u(x) dx J0
Then (i) Fe Sb if and only if u e l}=> y e L00, and moreover, there exists a finite constant cb such that ||y || ^ < ^^||^/|| i; (ii) FE SX if and only if the system (57) is L1 stable; (iii) Fe S^ if and only if the system (57) is L00 stable. 58
Theorem Suppose K: U+-^UnXn and that KeL\U+), and suppose G: R+2 -> Rnxn and H: U+2 -> UnXn are related by (48). U.t.c, He Sb iff G e Sb.
IV
120
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Proof <= By hypothesis, G and H are related by (48), and G e Sb. Let 9b = \\G\\b be defined as in (54a). Then, from (48), we have 59
\H(t9x)\
\K(v)\
\H(v9x)\dv
Let i e ( R + be fixed for the moment. Applying the Bellman-Gronwal) inequality to (59), we get 60
\H(t9x)\
( \K(v)\ J
dv] < gb exp[gb • \\KJH,]
X
where 61
K
=
\K(v)\dv
Since the bound in (60) is independent of both t and T, it follows that H e Sb. => Equation (48) can be equivalently expressed as 62
G(t, x) = H(t9 T) + f H(t9 v)K(v)G(v9 T) dv ^X
The equivalence of (48) and (62) is most easily seen by observing that if H is obtained from G by applying a feedback of X, then G is obtained from H by applying a feedback of —K. Now (62) is of the same type as (48), except that the roles of G and H are interchanged and K is replaced by —K. However, if K e l}(U+), then so does —K. Thus, if HeSb, then so does G, by the same reasoning as above. » 63
Theorem Under the conditions of Theorem (58), H e Sb n S^ if and only if G e Sbn S^. Proof <= Suppose G e Sbn S^. Then by Theorem (58), HeSb, only remains to show that He S^. From (48), we have
64
and it
f \H(t9x)\ dx< ( \G(t9T)\ dx+ ( f | G ( f , v ) | |X(v)| \H(v9x)\ dv dx
(\K(v)\
J
0
= gx+gb(
J
\H{v,x)\dvdx
J
x
\K(v)\ ( 0
J
\H(v,r)dxdv
0
after changing the order of integration. Now define m(t)=
f \H(t9x)\ dx
7
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121
SYSTEMS
Then (64) becomes 65
m(t) < g^ + gb f | K(v) \ m(v) dv Applying the Bellman-Gronwall inequality to (65), we get
68
m(i) < do0 exp[gb f |X(v)| dv]
o
Hence ra(-) is bounded, and so He
S^.
=> This implication follows by symmetry from the proof of <=, as in Theorem (58). ss 69a Theorem Under the conditions of Theorem (58), H e Sbn S1 if and only if The proof is left as an exercise. Exercise 9 Interpret Theorems (58), (63), and (69a) in terms of the stability of the closed-loop system versus the stability of the open-loop system. Now we state the main result of this section. 69b Theorem Consider a system of the type shown in Fig. IV. 1, where u, e, y: IR+ -• W\ the forward path contains a linear time-invariant element with a transfer function k
70
»=i
71
mi
G(s) = Gb(s) + 1 1 RtjKs -
j
Pi)
= Gb(s) + Gu(s)
j=i
where Gbe L\xnn L* xn , Re^ f > 0 for / = 1, . . . , k and mt = 1 if Re pt = 0, and the feedback element is a memoryless time-varying gain multiplier with gain F(t) = F0 + Fv(t) where F0eUnXn equivalent:
and Fv (•) e L1nxn. U.t.c, the following three statements are
(I) The closed-loop system is L00 stable, and W(-, •) (the closed-loop impulse response) is bounded on U2. (II) The closed-loop system is LP stable for all/? e [1, oo], and W(-, •) is bounded on U2. 72
(III)
(i)
inf
| det[/ + F0G(s)] | > 0
Res>0
and 73
(ii)
dQt[D(pi) + FoN(Pi) + FoGb(pi)D(pi)]^0,
for
/=!,...,£
122
IV
LINEAR
SYSTEMS
where Gu(s) = N(s)D(s) \ N and D are right coprime, and f) is column proper. Proof We proceed in the following order: (II) => (I), (I) => (III), (III) => (II). (II) => (I) Obvious. (I) => (III) Suppose (I) holds. Then W(-, •)> the closed-loop impulse response, belongs to Sbn S^. So by Theorem (63), it follows that the function H1: (i, T) I—• H(t — T), where H is the inverse Laplace transform of G(I + F0Gy\ also belongs to Sb n £«,. By Remark (2611), this means that H( •) e IJ([R+). But by Theorem (4.36), this implies (III), namely, (72) and (73). (III) => (II) Let H = G(I + F0GyK By Theorem (4.36), (III) implies that He ,
Since HeL\xn and Gft eL*xn nL^° x „, the first three terms in (74) are analytic for Re s > 0 and are bounded and uniformly continuous on Re s = 0. Hence so is the last term, namely, (/ — HF0)GU. In order to show that (/ — HF0)GU has an inverse transform in L™xn, we consider an arbitrary element of (/ — HF0)GU, which is of the form r(s)(j)(s), where r is a partial fraction expansion at p{, / = 1, . . . , k, and 0 e ,Q/. Thus, we can consider one pole of r(-) at a time. Now we claim that if 0 e J / and if [m\/(s — a)m + 1](j)(s) is analytic at a, with meZ+ and Re a > 0, then and that if Re a = 0 and (a) = 0, t h e n / e L00 provided m = 0. We note that
using the binomial expansion, we express the first integral as
7
LINEAR T I M E - V A R Y I N G
SYSTEMS
123
where the last equality is a consequence of the fact that <j)(s)/(s — a) w + 1 is analytic at a if and only if (l)(a) = 0 for / = 0, . . . , m. Furthermore, if either Re a > 0 and m e Z + , o r R e a = 0 and m = 0, then the function T i-> (f — r) m e«(t-x) | s bounded for 0 < t < %, say, by y. Consequently, for all t > 0, we have Hence fe L°°, and so H e L™xn. 76
SB
Corollary Under the conditions of Theorem (69b), suppose (72) and (73) hold. Let rj and y be the solutions, respectively, of the equations
77a
y =
lib
Gu-GF0y-GFvy
t] — Gu — GF01] 00
Then, whenever u e L , we have r\{i) — y(i) -* 0 as t -• oo. Proof Let H = (I + GF 0 ) _ 1 G. Then HeL^xnnL^xn. Moreover, by Theorem (69b), y e L^. Now (77a) and (77b) can be rewritten as 78a
y =
H*u-H*Fvy
78b
rj = H * u Subtracting (78a) from (78b), we get t] - y =
H*Fvy
Since y e L„°° and Fv e L*xn, it follows that F y y e L„ . Since H eLlxnn we finally have that (rj — y)(t) = (// * ^ ^ ( 0 -* 0 as f -> oo.
L„x„, SS
Exercise 10 Suppose Fv( -) e L^°x „, but not necessarily L\ x „ . Using the small gain theorem, obtain a bound on ||F(-)lloo t n a t ensures closed-loop L°° stability. 79
Remarks 1. With regard to Theorem (69b), note that the conditions (72) and (73) are sufficient for the closed-loop system to be L00 stable, but are not necessary, strictly speaking. If (72) and (73) are violated, in which case H = G(I + FQG)'1 $ jtfnX", we can only conclude that either the closed-loop system is not L^ stable, or that closed-loop impulse response [equivalently H()] is unbounded. 2. Theorem (69b) and Exercise 10 show that the feedback can be perturbed by an arbitrarily large L^xn gain without affecting the stability status of the system, but that there is, in general, a limit on how large an L™xn perturbation can be made in the feedback without affecting stability. Now consider the discrete-time analogues of the preceding results. The theory in the discrete-time case is made much simpler by the fact that any
124
IV
LINEAR
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sequence in /„* necessarily belongs to /n°°. We state only the theorems and leave the proofs as exercises. 80
Theorem Consider the feedback system shown in Fig. IV. 1, where u, e, y: Z + -> Un; the elements G and F are represented by the relations
81
(Ge)(k) =
^G(kJ)e(i) ;=o
82
(Fy)(k) = F(k)y{k) Then the system of equations
83a
e = u — Fy
83b
y = Ge has a unique solution of the form fc-i
85
y(k) = X H(k, i)u(i) i=0
where //(•, *) is the unique solution of the equation k-l
86
H(k, i) = G(k, i) - £ G(k, l)F(l)H& i) 1=1
87
Remarks 1. As is common practice, we take empty summations to equal zero. 2. The element G is not only nonanticipative [which is equivalent to requiring that G(k, i) — 0 for / > k], it has the additional property that G(k, k) = 0 for all k e Z + . Physically, this means that there is a delay of one sampling instant before the effect of the input is felt on the output. This is the case, for example, if the subsystem G has a state representation of the form x
k+l
=
S*kxk + "kuk->
yk ~
^kXk
where x: Z+ -• Rm, and A, B, C map Z+ into an appropriate space. In analogy with the sets Sb, StJ and S^, we define 88a
S?b = {G:Z+-+UnXn:
sup \G(i,j)\ < ao} i,jeZ
+ 00
88b
£fx ={G: Z+->UnXn:sup jeZ+
£ \G(i,j)\ < 00} i=0 00
88c
STn = {G: Z+ -> K"x": sup £ \G(i,j)\ < <»} ieZ+
j=0
8
SLOWLY V A R Y I N G
SYSTEMS
125
Exercise 11 Show that ^x and ^^ are subsets of 9"h. Give a physical interpretation for each of the sets Sfh^ Sfu ¥'(X). 89
Theorem Suppose G, H: Z + -> U"Xn, that G(k, i) = H(k, i) = 0 for k > i, and that G and / / a r e related by (86). Suppose F: Z + -> MnX" belongs to l*xn. U.t.c, / / e « ^ ( r e s p . ^ 1 ? y 0 0 ) i f and only if G e ^ ( r e s p . , ^ , ^ J .
90
Theorem Consider the system of the type shown in Fig. IV. 1, where u, e, y: Z+ -• Un; the forward element is linear and time invariant with a transfer function. k
91
G(z) = z'\Gb{z) + I
mi
X Ru/(z -
j Pi) }
= z~J Gb(z) + SB00
where Gbel*xn, \pt\ > 1 for / = 1, . . . , /:, and the feedback element is a memoryless time-varying gain multiplier with gain F: Z+ -+MnXn of the form F(k) = F0 + Fv(k) nXn
where F0eM equivalent:
and Fvel*xn.
U.t.c, the following three statements are
(I) The closed-loop system is Z00 stable. (II) The closed-loop system is lp stable for all p e [1, oo]. 92
(III) (i)
inf | det[/ + F0 G(z)] \ > 0 |z|>l
and 93
(ii) detlDfa)+
Fofit(pd + Fopr1Gb(pi)B(pi)]*0,
where Gu(z) = N(z)D(z)~\ proper. 94
for i = l,...,fc
N and D are right coprime, and D is column
Remark In contrast with the continuous-time case, (92) and (93) are necessary and sufficient for closed-loop Z00 stability.
8
Slowly varying systems
It frequently happens in engineering that some system parameters vary slowly. Then a typical question arises: Given that the system is stable (in some sense) when its parameters are constant, to what extent is it true that the system will still be stable when its parameters vary slowly ? In this section we consider differential systems; furthermore, since there is only a small increase in difficulty to consider the nonlinear case rather than only the linear case, we
126
IV
LINEAR
SYSTEMS
shall treat the nonlinear case. The approach is due to J. Barman [Bar. 1]. Suppose that the system is represented by 1
x = f(x, et) where e is a positive number. We denote by |x| the I2 norm of x e Un. We assume throughout (Al) / : W x IR+ -» Wn is in C 1 and/(0,/?) = 0, for all/? > 0; (A2) / i s globally Lipschitz—there is a constant / > 0 such that \f(x,p)-f(x',p)\
V/7>0,
V^jc'eR";
(A3) there is a positive constant rj such that |0 2 /(*,/>)l
2
x=f(x,p) where p e U+ is fixed. We think of (2) as describing the "frozen" system or, more precisely, the system frozen at t — p/e. To describe stability, we introduce two definitions.
3 4
The zero solution of (1) is said to be global uniform exponentially stable iff there are positive constants c and a such that \(j)(t; t0,x)\
for all
t > t0 > 0,
VxeK"
where cf)(t; t0, x) is the solution of (1) at time t, starting from x at t0. 5 8
The zero solution of (2) is global uniform exponentially stable uniformly in p iff there are positive constants k and o such that \4>f(t,x\p)\
Vf>0,
V(x,p) e W x U +
where (j)f (t, x; p) is the solution of the frozen system (2) at time t, starting from x at t = 0. 7
Theorem Consider the systems (1) and (2), where/(•, •) satisfies assumptions (Al) to (A3). U.t.c, if the zero solution of (2) is global uniform exponentially stable uniformly in p, then for e sufficiently small, (1) is global uniform exponentially stable. In the linear case, (1) would read x = A(t)x and (2) would read x = A(p)x. Condition (6) is equivalent to requiring that all eigenvalues of A(p) are in Re s < — a for all p > 0. (Al) is automatically satisfied provided t\-+A(t) is C 1 ; (A2) is equivalent to t\-+A(t) bounded on IR+; (A3) is equivalent to \DA(t)\
8
Proof
SLOWLY V A R Y I N G
SYSTEMS
127
The solution of the frozen system (2) satisfies the integral equation
8
<j>f(T,x;p)=x+
I
J
f[(j)f(t',x;p),p]dt'
o
Differentiating with respect to the real variable p, we have VT, peU VxelR"
+,
Using (6) and (A2) in (8), we obtain 10
<£/(T,*;/0>i|*|,
for
0 < T < l/lkl^T,
V(x,p)eUn
x U+
Using (6), (A2), (A3), into (9), and invoking Bellman's lemma, we see that for some positive constants m and m\ 11
\D34>f{x,x\p)\
<m'\x\emx
for all (T, x,p) e U+ x Un x U +
Now choose P > (m + o)\2o and choose r oo
12
V(x,p)=
\4>f(z,x;p)\2lldr
as a possible Lyapunov function for the frozen system (2). Clearly, Ve C 1 and V(0, p) = 0, Vp > 0. From (6) and (10), there are constants kl and k2 such that 13 Finally, from (12),
Thus V is a Lyapunov function for the frozen system (2). We are going to show that the Lyapunov function Kof (2) is also a Lyapunov function of (1), provided £ is small enough. Consider the time derivative of V(x, st) = Jo 10/(T> x> 8t) I 2/? di along solutions of (I):
Now -00
Hence
128
IV
LINEAR
SYSTEMS
where M is finite in view of the choice of /?. Then (16) reads sM)\x\2(i
V(l)(x, et)< - ( 1 -
If s < l/M, V(l) < 0; using (13) and estimating V{i)/V, we see that the solutions of (1) are bounded by 18 where c = (&2/£i)1/2/? and a = (1 -
£ M)/(2^ 1 ).
::
Exercise State precisely the theorem for the linear case, expressing the conditions in terms of characteristics appropriate for the linear case.
9
Linearization
One of the most important and the most frequently used method of analysis and design is that of linearization; that is, a nonlinear system is replaced by an (approximate) linear system which is tangent to the given nonlinear system. This approximation is, of course, valid only locally, in the same sense that, in U2, a tangent approximates a curve only locally.
9.1
Formulation
We restrict ourselves to differential systems; we wish to give conditions under which the linearized equations about a given trajectory will give valid results on the whole ofU + . In the language of circuit theory or control theory, we are going to develop conditions under which the small-signal equivalent circuit or the variational equations give predictions for the state-space trajectory which are valid for all t > 0. Consider the system of differential equations 1
x =f(x, u, t) n
m
n
where/: U x U x R + -> U
2
7(0,0,0 = 0,
3
x(0) = 0
W>0
Equation (1) relates the state trajectory x(-) to the input u(-). Equations (2) and (3) assume that the velocity under zero input is always zero and that the system starts from the zero state at t = 0; this is just a matter of convenience. This can always be achieved by a change of coordinates. First, we impose assumptions which guarantee the existence and uniqueness of solutions.
9
4
5
LINEARIZATION
129
Assumption For every fixed x e IRn and u e IRm, f(x,u,-): IR+->lR" is regulated; for every fixed u e (Rm, t e U + ,/(*, w, 0 is locally Lipschitz in x; for every fixed l e H " , t e M + ,f(x, •, 0 is continuous. We think of the input u as consisting of a "reference" input w0 and a small perturbation w; thus, u = u0 + u We assume throughout that u0 and w are regulated functions of t, defined on U+; as a consequence, for fixed jceIR", t\-+f[x,u(i), t] is regulated. Consequently, the state trajectory can similarly be thought of as a sum of two terms
6
x = x0 + £ where we assume that x 0 is defined for all £ > 0 by
7
x 0 =/(*(, > w o, 0,
x0(0) = 0
Note that £ is defined by (6). Consider the function t i-> [x0(i), u0(t)], and in Un x [Rm x U+ the set # defined by 8
V = {[x0(t) + £, u0(t) + u, t]\
\Z\
\u\
t>0}
where £m and um are positive numbers. We think of # as a " tube " surrounding the "reference trajectory": t i-*[x0(t), u0(t)]. In order to carry out our approximations, we require an additional assumption. 9
Assumption For some £m > 0 and um > 0,/has indwell-defined, continuous, second partial derivatives with respect to x and u ,and
are finite for all appropriate values of i, k, l,j, and h. Let us expand/about the reference trajectory:
where DJdenotes the derivative of/with respect to its /th argument. Observe that D{f[x0(t), u0(t), t] is the Jacobian matrix of/evaluated along the reference trajectory; therefore, it is a known function of time mapping U+ into
IV
130
LINEAR
SYSTEMS
UnXn. Similarly, D2f[x0(t), u0(t), t] is the n x m. matrix of partial derivatives with respect to the w/s; therefore, it is a known function of time mapping U + into R"Xm. By the standard expression for the remainder of a Taylor series [Die.l, p. 190]
assumption (9) implies that 35 < oo such that 11 In other words, g is of second order uniformly in t since S does not depend on t. Calling A(t) and B(i) the derivatives in (10), we rewrite (1) as 12
| = A(t)£ + *(*>/ + <7«, u, 0,
(0)
=0
If we drop the second-order term g(£, w, 0, we have 13
£0 = A(t)£0 + B(t)u9
£0(0) = 0
Equation (12) is an exact nonlinear equation equivalent to (1). Equation (13) is an approximate linear equation', it is the linearized equation about the reference trajectory [x0(*)> u0(-)]. Intuitively, we would expect that if the linearized equation (13) is " stable," then for " small" u, Eq. (13) will give valid results. The theorem that follows will make this basic idea precise. From (13), using standard notation, we obtain 14
where
Show that (9) implies that V e > 0, 35(E) s.t.
( | { | + \u\)<2S=>\g(Z9u,t)\
< e ( | { | + |i/|),
W>0
9
9.2
LINEARIZATION
131
Main result
In terms of the above formulation, we state the following theorem. 20
Theorem Consider the system (1) subject to Assumptions (4) and (9). Let u0: R + -> Um be regulated and x0(-) be defined on IR+ by (7). Let £(•) and £0(-)be defined by (12) and (13), respectively. Assume further that (Al)
3 M < oo s.t. V f > 0
21
f (A2)
|0(*,T)|
dx<M
3N< oo s.t. V f > 0
22 (A3) V a > 0 , 3 5(e) s.t. U.t.c, if, for some e e (0, 1/M) and a 5 corresponding to it according to (23),
then, using sup norms throughout, 25
28
Comment Let us discuss the meaning of Theorem (20). Assumption (21) together with (22) are the necessary and sufficient conditions for the approximate linearized equation (13) to be bounded input and bounded state stable [Des.14]. Assumption (23) requires g to be of the second order (in |£| and | u |) uniformly in t. Turning now to the conclusions: Conclusions (25) and (26) say that if the perturbational input u is bounded according to (24), then both £0 and £, the state of the approximate linearized system and the state of the exact nonlinear system, will remain within a distance 3 of the reference trajectory. Conclusion (27) says that the ratio of the " p e a k " difference between £(f) and £0(t) to the peak input \\u\\ can be made arbitrarily small; i.e., for small inputs the response t h-» £(f) of the exact nonlinear system is arbitrarily close to the response t f-» £0(i) of the approximate linearized system over the whole half-line U + . If the right-hand side of (1) depends linearly in u, then this last conclusion can be made even sharper [see (34), in the following].
IV
132
30
Corollary in u; i.e.,
LINEAR
SYSTEMS
Consider the special case where the right-hand side of (1) is linear
31
x = / ( * , 0 + B(t)u Equation (12) becomes
32
t=A{t)£
+
B(t)u+g{Z9t)
Consequently, assumption (23) can be simplified to read (A37) V e > 0, 3 S > 0 s.t. 33 U.t.c, if (24) holds, then (25) and (26) hold, and we also have 34 35
Comment In this case, the conclusion is much sharper because it says that by taking \\u\\ small, (i.e., S small), s can be taken arbitrarily small; hence by (34), ||{ — £oll/ll£oll> the ratio of the " p e a k " difference between £(f) and £ o (0 to the peak value of £o(0> can be made arbitrarily small. This truly shows that the relative error between the nonlinear and the linearized system [i.e., £(0 — £ 0 (0] c a n be made arbitrarily small. This fact has been verified computationally [Des.5]. Proof of Theorem (20) If necessary, increase TV [from (22)] so that 1 — eM < N + sM. For purposes of the proof, assume temporarily that
36 The solution of (12) is given by 37 or, by virtue of (13),
38 From (14), taking sup norms of both sides, we obtain
from which (25) follows. Using the temporary assumption (36) in (38), we get
40
9
LINEARIZATION
133
Pick a time T in (0, oo) and consider (40) for t e [0, T]. Since A(-) is regulated (hence bounded on [0, T]), (£, t') is bounded on [0, T] x [0, T]; thus, the Bellman-Gronwall lemma applied in (40) implies that for any finite T \\€T\\ < oo. Hence, for any Tin (0, oo), we obtain from (40) Thus, using sM < 1,
Since the right-hand side is independent of T, if we let T-* oo, we see that the function Ti—> ||^ r || is monotonically increasing and bounded; hence,
i.e., (26) has been established under the temporary assumption (36). Looking back over the derivation, we see that since, over [0, oo), | £(f) | < 3 and |w(OI ^ <^ w e n e e ( i on ^y require that (36) holds for all (£, w, r) in ^ = {[x0(r) + £, w0(r) + w, f] | | £| < 3,\u\ < 3 and t > 0}. Hence conclusions (25) and (26) hold under the assumptions of the theorem. From (38)
Hence (27) is established. Proof of Corollary (30) Conclusions (25) and (26) follow as before. Start with (38), where now g depends only on £ and f, so that by (33) we have successively
from which (34) follows.
9.3
::
Discrete time case
The linearization for the discrete case follows the same pattern as the continuous case. The system is specified by the difference equation
134
IV
where/: Un xUm xI+^Un. 51
LINEAR
SYSTEMS
We assume throughout that x(0) = 0
As before, we split the input u and the response x in two parts: 52
u(k) = u0(k) + u(k),
x(k) = x0(k) + £(k)
where x 0 (') is defined by 53
x0(k + 1) =f[x0(k),
u0(k), k]
x0(0) = 0
As before, let £m > 0 and um > 0 and define 54 We further assume that 55
(a) for some £m > 0, um > 0, / has continuous second partial derivatives with respect to x and w in # ; 56 (b) for all appropriate values of /, /:, /, 7, and //,
are finite. [The sups are taken over all (x, u, k) e^.] From Taylor's expansion theorem applied to (50) about [x0(k), u0(k)], we obtain the exact nonlinear difference equation 57 The approximate linearized equation is 58 Note that (55) and (56) imply that the remainder term g in (57) satisfies the following condition: 59
For any e > 0, there is a 5(e) > 0 (independent of k) s.t.
60 whenever \£\ + \u\ <25. 61
62
Theorem Consider the system (50) and (51). Suppose that (55) and (56) hold. Suppose also that (AI) there is an M < 00 such that
NOTES A N D
(A2)
REFERENCES
135
there is a K < oo such that
and let e e (0, 1/M) and 3(e) be picked according to (59), U.t.c, if 64 then, using || • || to denote /°° norms, 66
67 Corollary 68
Iff in (50) is linear in u, i.e., x(k + 1) =f[x(k), k] + B(k)u(k)
then (67) can be sharpened to
Notes and references The difficulties associated with " pole-zero " cancellations in the multivariate case can be treated by state-variable techniques using the concepts of controllability and observability. The purely algebraic approach developed in Sections 1 to 3 is due to several researchers [Pop. 1, Ros. 1, Wan. 1, Wol. 1, Des. 12]. The material in section 4 has been developed in stages by several researchers. Desoer [Des. 2] gave a general formulation of the Nyquist criterion, and this was extended by Desoer and Wu [Des. 6] and Desoer and Callier [Des. 12, Cal. 2]. The method of pseudoright-coprime factorizations is due to Vidyasagar. The graphical test of Section 5 was developed for the case of periodically spaced impulses by Willems [Wil. 1, 5], and for the general case by Callier and Desoer [Cal. 1]. Davis [Dav. 1, 2] has obtained related results. A thorough treatment of the discrete-time case, which contains the elements of the discussion of Section 6, can be found in [Des. 9] and [Wu 2]. The calculations of norms in Section 7 appear to have a long history in the mathematical literature [Edw. 1]; for the engineering literature see for example [San. 12, Wil. 2, Wu. 1]. The idea of using the principle of uniform boundedness to prove the equivalence of the two types of L00 stability is found in Desoer and Thomasian [Des. 1]. The spectral radius calculations in Section 7 are taken from [Vid. 3], and earlier versions of the perturbational results can be found in [Che. 1] and [Vid. 2]. The method employed in Section 8 is due to Barman [Bar. 1]. Finally, the results of Section 9 are updated versions of those found in [Des. 4].
V
APPLICATIONS OF THE SMALL GAIN THEOREM
The purpose of this chapter is to illustrate the concept that once the small gain theorem [(III.2.1) and (III.3.1)] is understood and the techniques of Chapter II, Chapter IV, and the Appendices are available, it is easy to establish a large number of input-output stability results. Our emphasis is more on displaying the flexibility of the method rather than giving an exhaustive account of known results. In Section 1, we obtain an LP stability result based on the results of Chapter IV and the small gain theorem. Section 2 contains a sufficient condition for L2 stability that differs in two important respects from that of Section 1: (i) The condition of Section 2 can be verified in the frequency domain, in contrast with that of Section 1. (ii) The condition of Section 2 is less conservative than that of Section 1. In Section 3, the important concept of exponential weighting is discussed, and a criterion for L00 stability, which can be verified directly in the frequency domain, is derived. Discrete-time analogues of some of theresuits of the first two sections are given in Section 4. In Section 5, we study 136
1
CONTINUOUS-TIME
SYSTEMS-!"
STABILITY
137
slowly varying linear systems and obtain a condition that ensure LP stability when each " frozen " system is stable. Section 6 presents a nonlinear circuit example that illustrates the application of the small gain theorem. In Section 7, we show how small gain techniques can be used to prove the existence of periodic solutions to a nonautonomous nonlinear differential equation. Section 8 contains a proof of the well-known Popov criterion using the small gain approach. Finally, in Section 9 we study two approaches for obtaining conditions that ensure the instability of a nonlinear feedback system.
1
Continuous-time systems—Lp stability
We consider feedback systems of the form shown in Fig. III.l, where H1 is a linear convolution operator and H2 is a memoryless time-varying nonlinearity. The general idea is to assume that H2 can be approximated by a time-invariant linear map y2(t) = Ke2(t) and to use the small gain theorem to show that if the linear time-invariant system with Hx in the forward path and K in the feedback path is stable, then so is the original nonlinear system, provided H2 does not deviate too much from K. We consider first an LP stability result where p e [1, oo]. 1
Theorem Consider a multivariable feedback system of the form shown in Fig. III.l, where uu u2, eu e2, yu y2\ U+ -» Un. The subsystem H1 is linear and time-invariant and is represented by
where G(-) is Laplace transformable, and furthermore, where Jf, Q) e s3nXn and the pair (^T, # ) constitutes a prcf of G. The subsystem H2 is memoryless and is represented by
where 0 : W x U+ -> Un; 4> is continuous in its first argument and regulated in its second. Suppose that uuu2eLnp=>
yuy2eLnpe
U.t.c, if there exist a constant matrix K eUnXn and real constants y and /? (with ft = 0 whenever p < oo) such that
138
V
APPLICATIONS
OF THE S M A L L G A I N
THEOREM
and if 8
(a)
inf
|det[/ + KG(s)]\ > 0
Res>0
9 10
(b) det[^0) + Kjy{s)} ^ 0, (c)
whenever
Re s > 0 and det §{s) = 0
\\HK(-)U-y<\
where HK(s) 4 G(s)[I + KGis)]-1 = [I +
11
G&Ky'Gis)
Then, for eachp e [1, oo], 12
uuu2e
Lnp =>el9e29yi9y2e
Kp
and the system is LP stable. Proof The details of the proof are left as an exercise. Apply the loop shifting theorem (III.6.3). In view of (8) and (9) and Theorem (IV.4.27), the function HK(-)e jtfnXn. Assumption (10) and the small gain theorem lead readily to the conclusion (12). :: Exercise 1 Explain why it is assumed in (7) that fi = 0 if p < oo, while fi may be nonzero if p = oo. If (j> satisfies a global Lipschitz condition, we can invoke the incremental form of the small gain theorem (III.3.1). Thus, we state the following corollary. 13
Corollary
If in the assumptions of Theorem (1) we replace (6) by
14 then the conclusions of Theorem (1) can be strengthened to (a) for each wl5 u2 e Lnp, there is a unique solution el9 e2, yl9 y2e Lnp (and it can be computed iteratively); (b) the solution depends continuously on u1, u2. Remark It is important to note that our results in no way depend on the manner in which 4> varies with t. Exercise 2 When p = oo (i.e., uu u2eLnco), show that condition (7) of Theorem (1) and condition (14) of Corollary (13) can be relaxed so that they have to hold only for all o in some ball whose radius depends on 11^ H^ and ll«2ll».
2
L2 S T A B I L I T Y — C I R C L E
CRITERION
139
Exercise 3 Suppose 0 satisfies (14) for some y eU, and suppose that G(-) contains at most impulses in some neighborhood of the origin. Let G0 be the "strength" of the impulse at t = 0. Show that, if det(7 + KG0) ^ 0, then ul9 u2eLpne=>eue2, yl9 y2eLpne.
2 L2 Stability—circle criterion The results of Section 1 are based on the inequality 1
\\HK*e\\p<\\HK\\s/\\e\\p which is valid for all/?e [1, oo]. Theorem (1.1) and Corollary (1.13) are very powerful in that they provide a sufficient condition for LP stability for all values of/?. However, the application of Theorem (1) to practical situations is, in some instances, limited by the fact that the crucial small gain condition (10) involves \\HK(-)\\^. At the present time, there are no means available for either determining || HK{ •) || ^ or for obtaining an upper bound for || HK{ •) || ^ based solely on the knowledge of HK(jco), co e R + . | Therefore, in order to obtain an explicit value for ||//*(•) II .*J w e a r e obliged to actually find the inverse Laplace transform of the function HK{-) defined by (11), and then apply the definition of || • \\^. In applications such as distributed RC networks, this process is very cumbersome and usually involves some form of interpolation. In this section, we use the small gain theorem to obtain a sufficient condition for L2 stability, and we derive the so-called circle criterion. The results of this section have two main advantages over those of Section 1: (i) The small gain condition in Theorem (4) in the following is based on the induced L2 norm of the operator HK, and as a result (8) in the following is less conservative than (1.10). (ii) Condition (8) in the following can be readily verified by examining only HK(joS). Much of what follows is based on the following fact.
2 3
Fact
Suppose H e $$n x n and e e Ln2. Then \\H *e\\2< {sup Xmax[H (»*H(yco)]} 1 / 2 • ||e|| 2
The proof is based on Parseval's equality and is left as an exercise. t We can only obtain a lower bound for \\HK(-)\\^, namely,
140
V
APPLICATIONS
OF THE S M A L L G A I N
THEOREM
We first state an L2 stability result that brings out the main idea, and then we specialize to the circle criterion. For convenience in later formulations, we define D[a, b] to be the closed disk in U2 whose diameter is the line segment joining the points (a, 0) and (b, 0). 4
Theorem Consider a multivariable feedback system of the form shown in Fig. III. 1, where ul,u2,el9e29yi,y2' ^+ -+ K"; the subsystem//satisfies (1.2) and (1.3), and the subsystem H2 satisfies (1.4). Suppose uu u2,e L2 implies that yl9 y2eL2e. U.t.c, if there exist a matrix KeUnXn and a constant y e U such that
5
|0(or, t)-K
V*eR + ,
Va e Un
and if 6
(a)
inf, Res>0
7
(b) det[&(s) + KJP(S)]
^ 0, whenever
Re s > 0 and
det @(s) = 0
where HK(-) is defined by (1.11); then 9 The proof is left as an exercise. As it stands, Theorem (4) cannot be applied to G(jco) alone, since the verification of (6) still requires knowledge of G(s). However, once the condition (6) has been verified, the small gain condition (8) can be applied in the frequency domain directly, and there is no need to compute the time-domain function HK(-). Also recall (Section IV.5) that under certain conditions, the inequality (6) can be verified graphically by examining G(jco) alone. We now state the celebrated circle criterion in a form more general than the conventional one. Because of its importance, we give both the result and its proof in full. The importance of the theorem lies in the fact that the condition is readily applicable to directly available experimental data, namely, the Nyquist diagram. 10
Theorem (circle criterion) Consider a scalar system of the form shown in Fig. III.l, where uu u2, eu e2, yt, y2: U+ -* U. The subsystem H1 is represented by
11 where
2
L2 S T A B I L I T Y — C I R C L E
CRITERION
141
Re pt > 0 for / = 1, . . . , k, and gbe L1. The subsystem H2 is represented by 13
y2(t) =
14 For convenience, let 15 and assume that ^ ^ 0. U.t.c, we have that wl5 w2 e L2 => el9 e 2 , y l5 y 2 e ^2> if the pole locations and Nyquist diagram of g [i.e., the map co — i > #(yco) with indentations as required] satisfy one of the following conditions, as appropriate : (a) if 0 < a < /?, there are no restrictions on the location of the poles of g, and the Nyquist diagram of g is bounded away from the disk Z>[— 1/a, — 1//5] and encircles it in the counterclockwise direction np times, where np is the number of poles of g with positive real part; (b) if 0 = a < /?, g has no poles in the open right half-plane, and the Nyquist diagram of g must remain to the right of the vertical line of abscissa — 1/jS; i.e.,
(c) if a < 0 < /?, g has no poles in the closed right half-plane, and the Nyquist diagram of g is completely contained in the interior of the disk D[-l/oc, -1/jS]. Proof The result follows directly from the loop shifting theorem and the small gain theorem and can also be thought of as a corollary to Theorem (4). First of all, it is clear that whether we are in case (a), (b), or (c), the Nyquist diagram of g does not intersect the point (— l/£, 0) and encircles it exactly np times in the counterclockwise direction, where np is the number of poles of g in the open right half-plane. Therefore, by the graphical test [Theorem (IV.5.5)], it follows that h^s) = g(s)/[l + ^g(s)]eJ. Apply Theorem (4) with K = £/. Then the small gain condition (8) becomes
It is a simple exercise to show that for z e C , 17a
if 0 < a < /?, then p | z | < 11 + £,z\ is equivalent to: z is bounded away from the disk D[- 1/a, - 1/jS];
142
V
APPLICATIONS
OF THE S M A L L G A I N
THEOREM
if 0 = a < ft, then p\z\ < 11 + £z\ is equivalent to: Re z > — 1//?; if a < 0 < /?, then p | z | < 11 + £z | is equivalent to: z is in the interior of the disk D [ - l / a , -1/jS]. Hence the conditions of Theorem (4) are satisfied, and the conclusion follows. :: Remark As fi -> a > 0, the sector condition (14) reduces <j> to a linear characteristic of slope a; the critical disk D[— 1/a, — l/j8] shrinks to the critical point (— 1/a, 0); and the circle criterion above reduces to the graphical test of Theorem (IV.5.5). Exercise 1 In the hypothesis of Theorem (10), suppose gb in (12) belongs to s0 and not L1. Give an appropriately modified version of Theorem (10) and prove it. Exercise 2 Consider an «-input-«-ouput feedback system where Ht is specified as the zero-state response of a system whose input is ex and output yl with x = Ax + Bel9 dxd
dXn
x{0) = 0,
yx = Cx
nXd
{A e U , Be U , Ce U are constant); all eigenvalues of A have negative real parts. Call G(jco) = C(jcoI — A)~XB the transfer function of Hv Let H2 be specified by y2{i) = 4>[e2{t), t]9 where
then, for any . {Hint: Use / 2 norms in IR", and for x e Ln2, \\x\\2, = Jo Z i \xi(t)\2 dt> n o t e t n a t t n e impulse response of Hx is G(0 = C exp{At)B e L*XB, and G(0 = C55(0 + C4 exp^O^ej^V) Exercise 3 where
Consider a single-input-single-ouput system with unity feedback y{t)=\tg{t-T)k{x)e{x)dx
with
= {Ge){t)
3
where
EXPONENTIAL WEIGHTING—/.00
STABILITY
< oo. Show that
(a) for all T> 0, G is a continuous linear map from PTL2e
(b) (c)
3
143
to
PTL2e;
if y(G) < 1, then the closed-loop system is L2 stable.
Exponential weighting—L 00 stability
The results of Section 1 are quite powerful in that they provide conditions for II stability for all values of p, but the price we pay for these powerful results is that the crucial small gain condition (1.10) is of a form that cannot be verified directly in the frequency domain. In Section 2, we formulated a theorem containing a small gain condition that can be directly verified in the frequency domain, but as a result we could conclude only L2 stability. In this section, using the important technique of exponential weighting, we obtain a criterion for IT stability that is, in fact, stated in terms of the Nyquist diagram, and in return we are obliged to make a few extra assumptions. The exponential weighting technique is predicated on two facts: (1) If then, for all aeU, 2
y(t)eat = g(t)sat * e(t)eat (2) If^f(•) denotes Laplace transform, then In the above, we use/(0e a ' to mean the function th->f(t)sat, but this slight notational abuse causes no confusion and saves us from unnecessary clutter. We use sat to denote cxp(at), thus avoiding multiple usage of e.
3
Theorem Consider a scalar feedback system of the form shown in Fig. III. 1. The subsystem Hx is represented by
4 while the subsystem H2 is represented by
5 we assume that u2(t) = 0, and that 6 7
eatg(t) eL1 n L2, (p e sector [a, ],
for some a > 0 with
>0
144
V
APPLICATIONS
OF THE S M A L L G A I N
THEOREM
U.t.c, if the a-shifted Nyquist diagram of g, namely, the image co i—• g( — a +jco), satisfies the circle condition of Theorem (2.10), then ux e L00 simples e1? ^ e L00, and moreover, there exist finite constants ml and m2 such that 8a 8b Proof
The feedback system is characterized by
9
y(t) = f g(tJ
10
T)W(T) dx - f g(t - T)4>[y(x\ T] rfr •'O
o
=v(t)-
\
g{t-T)(j)[y(x\T\dx
J
o
where we have dropped the subscript " 1 " for convenience, and denote (g * u) by v. Clearly, by (6), v e L00. Now (10) can be equivalently rewritten as 11 If we define = ea'y(t)
12a
yM
12b
vw(t) = satv(t)
12c
gjt)
=
fg{t)
Equation (11) becomes 13
yjt) = vw(t) - (gjt J
- T)8«0[e-«^(t), T]
o
Since the map 14 belongs to the sector [a, /?], and since the Nyquist diagram co i-> gw(jco) = g( — a+jco) satisfies the circle criterion, we see by Theorem (2.10) that yw( •) e Le2 whenever vw( •) e L e 2 and moreover that, for some finite constant p, 15 where ||[j\v(*)Lll2 denotes the L2 norm of the truncated function However, it is easy to show that
16
[yw(')]t.
3
E X P O N E N T I A L WEIG HTI NG—Z.00
STABILITY
145
By the sector condition on the map defined in (14), we see that
Going back now to (9), we get 18
(by Schwartz' inequality) Hence (8b) is established. The proof of (8a) is left as an exercise. 19
::
Corollary Under the conditions of Theorem (3), if u(-) is bounded on U + by a decaying exponential, then so are e(-) and y(-). Proof
By assumption, there exist r, a e U + such that
20 Without loss of generality, suppose 0 < a < a. From (9) we have
Note that (i) the function t h-> satu(t) is bounded; (ii) the map o H-> 8aT0(£~aTcr, t) belongs to the sector [a, /?]; (iii) the function t \—> satg(t) satisfies (6), with a replaced by (a — a)/2. Hence by Theorem (3), saty(i) is bounded.
s:
We now present a slight generalization of Theorem (3). The generalization is based on the following lemma. 21
Lemma
Let k
22
mi
G(s) = X I RijKs - PdJ + Gb(s) = Gu(s) + Cb(s)
V
146
APPLICATIONS
OF THE S M A L L G A I N
THEOREM
where Re pt > 0, mi = 1 if Re pt = 0, and Gbe L^xnn L2nxn. Suppose F e Jnxn. U.t.c, if H = G(I + FGy1 e JnXn, then in fact HeLlnxnn L2xn. Proof
By hypothesis, H = G(I + FG) _ 1 e ^ " X n . Now i / satisfies
23 or 24
H + HFGb -Gb = (I- HF)GU Since (/ — HF)GU is analytic for Re s > 0 and bounded and uniformly continuous on Re s = 0, we can show, as in the proof of Theorem (IV.7.69), that in fact J2?_1[(/ - HF)GU] e L\xn n Ln°°xn, and in particular J^" 1 [(/ - i?F)GJ
25
= - H F G , + Gb + (I- HF)GU and all terms on the right side belong to L\xn n L^ x „, the conclusion that H E Llxnn L2xn follows. ::
26
Theorem Consider a feedback system of the form in Fig. IIl.l and suppose the subsystems H1 and H2 are represented by (4) and (5), respectively. Suppose that u2(t) = 0, and that for some a > 0
27 28 29 30 U.t.c, if the a-shifted Nyquist diagram of g satisfies the circle conditions of Theorem (2.10) (where the right half-plane is taken as {s: Re s > —a}), then ux E L00, implies el9 yt e L°° and moreover (8a) and (8b) hold. The proof is straightforward and is left as an exercise.
4
Discrete-time systems—Lp stability
In this brief section, we state what are essentially the discrete-time analogues of the results of Section 1. We do not explicitly state the results corresponding to those in Sections 2 and 3, since these are quite obvious.
5
SLOWLY V A R Y I N G
LINEAR
SYSTEMS
147
1
Theorem Consider a multivariable feedback system of the form shown in Fig. III. 1, where ut, u2, el9 e2, yi9 y2 ' ^+ -* Kn. The subsystem H1 is linear and time invariant and is represented by
2
yt = G * et where
3 Jf, <3el*xn, and (Jf, §)) constitutes a prcf of G. The subsystem H2 is memoryless and is represented by 4 U.t.c, if there exist a constant matrix iC e UnXn and real constants y and v (with v = 0 if p < oo) such that
5 6
inf | det [ J + KG(z)] | > 0
7
det[J(z) + KJT(z)] / 0,
whenever
| z\ > 1 and
det #(z) = 0
8 where 9 then for each/? e [1, oo], 10
uu u2el/=>
eue2,yuy2elnp
Exercise 1 Prove Theorem (1). Exercise 2 State and prove the discrete-time analogues of Theorems (2.3) and (3.26).
5
Slowly-varying linear systems Consider a multivariable feedback system where
1
yi(t)
= (G * ex)(t)
2
y2(t) = K(t)e2(t)
148
V
APPLICATIONS
OF THE S M A L L G A I N
THEOREM
If K(-) takes values in some subset $f of UnXn9 and if (/ + GMY^G e JnXn for all M e S, we expect intuitively that the time-varying system characterized by (1) and (2) is also stable, provided K(-) varies sufficiently slowly. In what follows, we give a precise interpretation to the phrase "sufficiently slowly." Note that if K(-) were to be "frozen" at the value K(x)9 the closed-loop transfer function of the resulting linear time-invariant system is 3
Hx{s) =[I + G{s)K{x)Y1G{s)
4
Theorem Consider a multivariable feedback system of the form shown in Fig. III. 1, where ul9 ul9 el9 el9 yl9 y2' ^+ ->t^n- Suppose the subsystem H± is represented by (1), and that the subsystem H2 is represented by (2), where K(-) takes values in some bounded subset £f of Un. Suppose that for each l e y , the function [I + G(f)M]-lG(s) belongs to stfnXn. U.t.c, if r 00 .
5
sup
| Hx(x - x')[K(x) - K{x')]| dx < 1
T6IR+ ^ 0
where Hx(s) = [/ + e(j)X(T)]- , e(j)
6 then ul9 u2el? 7
=> eu e2, yl9 y2 e L00. If sup f °° | Hx(x - x')[K(x) -
K(x')]\dx<\
then ul9 u2el} => el9 e2, yl9 y2 eO. If both (5) and (7) hold, then for each pe [1, oo], ul9 u2eLp =>el9e29 yl9 y2eLp. Proof
The closed-loop system is characterized by
8
yx(t) = - (G * Kyi)(t) - (G * Ku2)(t) + (G * 1,0(0 = - ( G * X y 1 ) ( 0 + (G *u) where
9
u(t) = w^r) -
K(t)u2(t)
Let T G IR + . Then (8) can be written as 10
yi(t)
+ [G * X W ^ ] = (G * M)(f) + {G * [K(x) -
K(t)]yi}(t)
Convolve both sides of (10) with the inverse Laplace transform of [I +
G(s)K(x)]-\
which by assumption belongs to ,s/" x "; then (10) becomes 11
yi(t)
= (Ht* u)(t) + ( Ht{x - t') [K(r) - K(z')]y(T') dx'
6
NONLINEAR
CIRCUIT
EXAMPLE
149
In particular, for t = x, (11) becomes y,{x) = (HT * «)(T) + ( Hz(x - x')[K(x) - K(x')]y(z') dx'
12
J
0
Now consider the mapping defined by (Fy)(T) =
13
CHX(X
- x')[K(x) - K(x')]y{x') dx
By the results of Section IV.7, the left side of (5) is the induced L00 norm of the mapping F, while the left side of (7) is the induced L1 norm of the mapping F. The conclusions now follow from the small gain theorem. s: 14
Remark By hypothesis, Hx() eJnXn for each x e R + . So both (5) and (7) are satisfied if K(-) varies slowly enough. The idea is to have K(-) remain substantially constant during the " memory time " of Hx( •), i.e., the time when Hx() differs appreciably from zero.
6
Nonlinear circuit example
Consider Fig. V.l. The linear time-invariant one-port Jf is connected to a nonlinear time-varying capacitor, and the parallel combination is driven by
Figure V.l
a current source u(-). Letg(-) be the impulse response of JV to a unit impulse of voltage; so Jf is characterized by i(t) = f g(t -
T)V(%) dx,
t>0
The characteristic of the nonlinear time-varying capacitor is vc =
150
4
V
APPLICATIONS
OF THE S M A L L
GAIN
THEOREM
Assertion Consider the circuit shown in Fig. V.l and described by (3). Let (a) (j>: U x U+ -> U be, for each fixed q, regulated in t; (b) for all compact intervals I cM + , cj) satisfy a global Lipschitz condition with Lipschitz constant y(I) < oo; (c) (f) belong to the sector [a, /?] with p > oc. Let
5
a 4 (a + j8)/2,
a = (j8 - a)/2
6 7 then, for any u el} u L2, g, q, ve L2, #, ^ e L00, and #(0, K0 -• 0 as t -> oo. 8
Comments (a) When weL 1 , the current source delivers a finite charge to the parallel connection of the one-port Jf and the nonlinear capacitor. (b) The time variations of the nonlinear capacitor are restricted by the sector condition and the condition that (j)(q, •) be regulated. Note that a time-varying capacitor is an active element; i.e., it may deliver energy. (c) Since g(s) is the input admittance of J/', (6) means that the parallel combination of Jf and a linear capacitor of 1/a Farad is open circuit stable. Proof As a consequence of the global Lipschitz condition on (/>, Theorem (III.5.2) can be used to prove that for any u e L1 u L2, (3) has a unique solution q : U+ -» U. With the sector condition in mind and the notations defined in (5), we set
9 We note that both in Lel and L e 2 , $ has a gain
q(t)
+ *{g
* q)(t)
= u(t)
-
\ g(t
-
T)<£[?(T), T]
A
Let 11
w(^) - [s + ocg(s)V* and
h(s) ± g(s)w(s) = g(s)[s + otg(s)]~1
Since g e stf and, by (6), s + &g(s) is bounded away from zero i n C + , w e ^ . Indeed, (11) can be rewritten as
s+1
I s+ 1 J
7
EXISTENCE
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SOLUTIONS
151
Thus, w is the product of two elements of j$9 hence is in j$. For the same reason, hesi. Consider (11) and recall that g e stf''; then (Oh->w(jao) and co h-> h(joS) e L2 since they are both 0(1/co) as | co | h-> oo. Thus, w, h e srf n L2; i.e., they have no impulses. Consequently, w, he 1} n I}. Convolving (10) with w, we obtain 13 Consider (12) as the equation of a feedback system: For any u e L1 u L2, w * w e L2; furthermore, (7) is the small gain condition using L2 norms. Hence uel)u L2 implies qsl}. By the sector condition, K 0 = [#(')> ' ] a n d ' ] a r e m ^25 therefore, by (10), 4 e L2. Thus, # e L00 and #(0 -> 0 as t-> oo (Exercise 1,B.2). These results, together with the sector condition, yield v e L00, t?(r) -» 0 as t -• oo. SS Exercise 1 State and prove the generalization to «-ports terminated by n nonlinear time-varying capacitors. Exercise 2 Prove that w and h defined by (11) are elements of L00. [Hint: Seiyi) = sw(s).]
7
Existence of periodic solutions
In this section, we illustrate how a local form of the incremental small gain theorem can sometimes be used to demonstrate the existence of a periodic solution of a nonautonomous differential equation and to obtain bounds on the amplitude of the periodic solution. We first formulate an existence and uniqueness result, which is in a form convenient for the present application. 1
Lemma Let $ be a Banach space (i.e., a complete normed linear space), and l e t / : ^ - * ^ be continuously Frechet differentiable for all x e J . Suppose there exist a n i 0 e J and a k e [0, 1 ] such that
2
whenever
xe B[x0; p/(l - k)]
where 3
4 U.t.c, there exists a unique 3c e B[x0; j8/(l — k)] such that 5
x
=f(x)
152
6
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THEOREM
Remark The lemma does not assert anything concerning the possible existence of fixed points of/outside the ball B[x0; /?/(l — k)]. Proof
Whenever x, y belong to B[x0; /?/(l — k)], we have
Since k < 1,/is a contraction on the complete metric space B[x0; /?/(l — k)]. Hence there exists a unique x e B[x0; /?/(l — k)] such that x =f(x). :: Exercise 1 Prove Lemma (1) in detail. [Hint: Consider the sequence Cx*)00, where xt =f(xi_i) for / = 1, — ] The best way to explain the technique is to work out an example and describe the steps so that the generality of the method becomes apparent. 7 8
Example Consider the Duffing equation y + a2y + by3 = a cos cot,
T = Injco
where a, b, a, and co are constants. We assume throughout that co ^ a, for otherwise even with b = 0, (8) would not have a periodic solution. We wish to discuss the existence and local uniqueness of periodic solutions of (8) when b is small (" small" nonlinearity) and when a is small ("small" drive). Since co ^ a, if b were zero, (8) would have a periodic solution of the form y0(t) = a0 cos cot. Considering it as an approximate solution of (8) and neglecting the third harmonic term [(cos x)3 = f cos x + \ cos 3.x], we obtain an equation for a0: 9
(36/4K 3 + (a2 - co2)a0 = oc Writing (8) in state form, with x(t) = [x^t), x2(i)]', we obtain
Considering the term [^(f)] 3 so that we have
as
given, (10) is of the form x = Ax + g[x(t)],
where
7
EXISTENCE OF P E R I O D I C
SOLUTIONS
153
which is of the form 11
x(0 =
W(t, T)g[x{x)] d%,
0
Exercise 2 Give an explicit expression for the matrix W(t, T) for (t, T) e [0, T] x [0, T], for the state equations (10). Let x 0 be the approximate solution of (10) corresponding to y0; so we write 12 We are going to use Lemma (1) to show that (11) has a solution close to x0(-). Note that (11) is of the form x =f(x). We use the following norm 13
||x||=max i=l,2
sup
\xt(t)\
0
It is well known that £%, the space of continuous functions 0: [0, T] -> U2 with the norm (13), is a Banach space. So let us obtain the constant j3 of (4) and a bound for/'(•*)•
where wM is a constant which depends on T and the resonant frequency a. Inequality (14) gives us /?. Estimating/'(x), we obtain
Now max|^(01 < niax|x l o (0| + max|x x (0 — x 10 (0l = a0 + \\x — x0\\ Thus, 15
\\f\x)\\<3\b\wM[a0+\\x-x0\\}2 which is of the form of (2). Using (14) and (15) to write out condition (2), we see that we must have a k e [0, 1) such that
16
154
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THEOREM
Thus, if such a k exists, there is a periodic solution x such that
where k0 is the largest solution of (16). The existence of such k is immediate in certain limiting cases: (I) If a, a, co are given constants [and hence a0 is determined by (9)], and if b is sufficiently small, then inequality (16) is easily satisfied for some k < 1. In other words, if the nonlinearity is small, a periodic solution with period Tis guaranteed to exist near x0 and satisfies the bound (17). (II) If b, a, and co are given, then for sufficiently small a, (16) is satisfied for some k < 1. Indeed, under these conditions as a decreases to zero, so does a 0 , the solution of (9), and, for a0 small, some k < 1 satisfying (16) always exists. So in this case a periodic solution with period T is guaranteed to exist near x0 if the input has a small enough amplitude. 18
Comments (I) Unfortunately, the method does not apply to autonomous systems. The first difficulty is that the absence of forcing function leaves us without any hint as to the period of the periodic solution. The second difficulty is that even if that were known, in a neighborhood of the solution, the equation similar to (18) is not a contraction. Indeed suppose, that x = Ax + g(x)
19
(where A e Un x n) has a periodic solution x of period T. Then, as before, we obtain rT 3c(r)= W(t, x)g[x(x)] dx Now x is also a periodic function with period T; it satisfies (d/di)x = Ax +g'[x(t)]$(t) as can easily be seen by differentiating the equation of x. Thus,
20
*(t) = f
W(t, T)0'[X(T)]*(T) dx
Hence the norm of the map T
z(-)h-f
W(',x)g'[x(x)]z(x)dx
J
o
is at least 1. Thus, although (19) is of the form x = / ( x ) , we observe that (20) implies that ||/'(x)ll > 1. Hence we cannot have a contraction. In order to establish existence, topological methods are required.
8
POPOV
CRITERION
155
(II) In contrast to our previous work, the contraction (incremental small gain) condition is valid only locally; consequently, the existence and uniqueness are valid only in the ball B[x0; /?/(! — k)].
8
Popov criterion
In this small gain exposition small gain
section, we prove a simple form of the Popov criterion using the theorem. In Chapter VI, Section 6, we develop a more complete of the Popov criterion. Here we use it as an illustration of the theorem.
Theorem Consider the single-input-single-output described by e1=u1-g*e2
time-invariant system
e2 = <£0i) where 4> : U -> U is continuous and belongs to the sector [0, k], gel}, and gx e stf. We assume that for any ux e L2, ex and e2 are in Le2. If for some q > 0, there is a S > 0 such that 4
Re [(1 + qj(D)g(jco)] + \/k > S > 0,
Vco>0
then whenever ut and ut e L2, eu e\, e2 e L2. Proof Without changing e1 and e2 , let us insert in front of the nonlinearity (j) the transfer function 1/(1 -f qs) and insert after the transfer function g(s) the transfer function (1 + qs); clearly, we need to replace ux by ut + qu1. Next, apply the loop transformation theorem and put a feedforward of —(k/2) around 1/(1 + qs) and 0, and a feedback of —(k/2) around (1 + qs)g(s). A compensating input of (/:/2)(w1 + qui) has to be inserted as shown in Fig. V.2. To apply the small gain theorem, we use L2 norms («£? = L2). We
Figure V.2
156
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THEOREM
claim that the gain of Hl9 yC^i), is smaller or equal to k/2. The output of Hx is yl = (/>(^i) — krjll2, and its input is rj. = et + qe\. From Fig. V.2, for any
T>o,
The first integral is nonpositive by the sector condition. If
where we noted that (j)(o) > 0, Vor e !R by virtue of the sector condition. Hence 8
yiH,) < k/2 We claim that the gain of H2 is smaller than (2/fc). For simplicity, let h(jco) 4 /*r(/co) + jhi(jto) = ( 1 + qjco)g(jco),
co e U
and define 9 Then assumption (4) becomes 10
P(jco)
>d>0
The gain y(//"2) is given by the expression
11
For simplicity, let us drop the dependence of co. Then in view of (10), we have 12
1
9
INSTABILITY
157
so that
where
Substituting back in (11), we finally have that y(H2) = (2/Jfc)(l - e)1/2 < 2/k
15
From (8) and (15), we see that y{Hx) • y(H2) < (1 - e)1/2 < 1; therefore tfi = ei + a^i e ^2- Now the L2 gain of the transfer function (1 + qs)~x is 1, so that HeJ^ < II^/iII2 < °°- Thus, el,elel?. By the sector condition, e2 = (K^) e L2. s:
9
Instability
Virtually all of the foregoing sections are devoted to obtaining conditions under which a given feedback system is stable. In this section, we obtain conditions that ensure that a given system is unstable. From a mathematical point of view, obtaining sufficient conditions for instability is much more tricky than obtaining sufficient conditions for stability. In the latter case, one is required to show that all inputs belonging to some class produce outputs belonging to the same class. As a result, sweeping general methods can be used to achieve this purpose. In the former case, however, one is obliged to show that some inputs belonging to a certain class do not produce outputs belonging to that class. Thus, one is concerned not with general statements and general proofs, but with particular situations involving particular inputs. This is the major hurdle to proving instability theorems. In this section, we give two possible interpretations to instability and derive conditions that ensure each of these types of instability. In Section 9.1, we show that under certain conditions, an L2 input to a system does not produce an L2 output, and moreover, we delineate the set of L2 inputs that do not produce L2 outputs. In Section 9.2, we study the case where an LP input produces an LP output, but the relationship between them is not causal. Consider a feedback system described by the equations 1 2
e1 = u1 — H2 e2 e2 = u2 + H1e1
158
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THEOREM
where ux and u2 are the inputs, and ex and e2 are the errors. We designate yi = N1el and y2 = H2 e2 as the outputs. Moreover, we assume that t/j, u2, e i> e2> yu J2 belong to an extended Banach Space $£e, in accordance with Chapter III. Now suppose in the interests of simplicity that u2 = 0. Then (1) and (2) can be combined into the single equation e1=u1H2 H1el There are several ways in which a system represented by (3) can be " unstable " (i) For a particular u1 e $£, (3) may not have a solution in 5£'e. (ii) For a particular ux e ££, (3) has a solution in =£?e, but this solution does not belong to $£; (iii) For each ux e if, (3) has a solution ex e ££, but the relation between ul and ex is noncausal. (Notice that we do not claim that these are all the possible ways in which instability can occur, but only some of the ways.) With regard to these three possibilities, (i) can be ruled out under relatively mild conditions (see Section III.5). Roughly speaking, the conditions under which (iii) occurs are explored in Section 9.2, while the conditions under which (ii) occurs are the subject of Section 9.1.
9.1
Orthogonal decomposition approach
In this section, we study the case where a "bounded" input produces an "unbounded" output. The approach we present below is not inherently limited to L2 instability, but evidently we can obtain readily verifiable conditions only in this case. The method of this section can be briefly outlined as follows. Given a causal linear operator G : L2 -• L e 2 , suppose G is "unstable" in the sense that the image of L2 under G is not wholly contained in L2, and let M denote the inverse image of L2 under G. Then M is a subspace of L2 and, under suitable conditions on G, is a closed subspace of L2. With regard to the feedback system of Fig. III. 1, where Hl is represented by the operator G, we show that if a small gain type of condition is satisfied, then the output yt does not belong to L2 whenever u2 = 0 and u1 e M 1 / ^ } , where M1 is the orthogonal complement of M. We begin with a lemma on the closedness of M. Throughout this section, we use || • || to denote the L2 norm. Lemma Let G : Le2 -> Le2 be linear and causal, and let M be the subspace of L2 defined by M = {xeL2 : Gx e L2}
9
INSTABILITY
159
If G satisfies the following two assumptions 6
Vr < oo, 3fiT < oo, such that
7
3yM(G) < oo, such that
||(Gx) r ||
||Gx|| < yM{G)\x\,
Vx e L2
\fx e M
2
then M is a closed-loop subspace of L . 8 9
Remarks (1) The condition (6) essentially states that if we define a mapping GT : L2[0, T] -> L2[0, T] defined by V/eL 2 [0, T],
GTf=(Gx)T
2
where x is any element of L such that xT = f then GT is continuous. Notice, however, that the family of constants (/iT) is not assumed to be bounded. (2) The assumption (7) means that G has a finite gain over the elements of M. Proof Let (x() be a sequence in M converging to x0e L2. We must show that x0el}. By hypothesis, GxteL2, V/eZ + , and moreover (Gxt) is a Cauchy sequence in L2 in view of (7). Since L2 is complete, this implies that (Gxi) converges to an element of L2, which we denote by y. By virtue of (6), we see that for each T< oo, the sequence ([Gxt]T) converges to [Gx0]T. But this means that [Gx0]T = yT, VT< oo, i.e., that Gx0 — y and hence that Gx0 el}. By the definition of the set M, this implies that x0 e M. :: The following theorem can be thought of as an instability counterpart of Theorem (III.2.1). 10
Theorem Consider a feedback system of the form shown in Fig. III.l, where Hl and H2 map Le2 into itself. Suppose that corresponding to each ul9 u2 in L2 there exist ei9 e2 in L2 satisfying the system equations
11a 11b
e1=u1H2 e2 e2 = u2 + H1e1 Suppose there exist finite constants y(H2) and p such that
12 Further, suppose Ht = G : L2 -> L2 is linear and causal, satisfies (6) and (7), and that the subspace M defined by (5) is not all of L2. U.t.c, (i) if yM (G) - y(H2) < 1 and /? = 0, then y1 $ L2 whenever u2 = 0, ux e M1 and ux ^ 0 where M1 = orthogonal complement of M; (ii) if yM(G) • y(H2) < 1, then for each w 2 eL 2 , there exists a corresponding constant m(||w2|l) s u c n that ^ $L2 whenever ux e M1 and HwJ > w(||w2||).
160
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THEOREM
Proof The hypothesis on H1 implies that M i s a closed subspace of L2. Since M is not all of L2, it follows that M1 is a nontrivial subspace of L2. We first tackle the case where yM(G) • y(H2) < 1, P = 0, and u2 = 0. Suppose by way of contradiction that ux ^ 0, ux e M 1 , and that yt = GeY e L2. Then et e M, and 13
y2 = ux - ex Since ux ^ 0, uxe M1, and ex e M, we have that <wx | ^ > = 0 , and hence ll^2ll2 = lkll 2 + lkill 2 >lkill 2
14
On the other hand, we have y2 = H2 Htel9 so that 15
\\y2\\
where 17
c=
y[H2)\\u2\+p
The inequality (16) can be rearranged as follows: 18
For u2 e L2 fixed, k and c are fixed constants, and it is routine algebra to verify that the maximum value of the right-hand side of (18) with respect to \\ex\\ is c 2 /(l -k)2. Therefore, if \\ux\\ > c2/(l - k2) 4 m(||w2||), then (18) can never be satisfied, and this contradiction establishes that ux$l}. " We close the section with an instability counterpart of the circle criterion [Theorem (2.10)]. 19 20
Theorem Consider the feedback system of Fig. III.l, where ul9 u2, eu yl9 y2: U+-+U. The subsystem Hx is represented by
e2,
9
INSTABILITY
161
where g(-) is Laplace transformable, and g(s) is of the form
with gb e L1, Re pt > 0 for all / = 1, . . . , k, and rtj belongs to C for all ij.The subsystem H2 is represented by 22
(H2e2)(t) = $[e2(t),t] where 0 : ( R x l R + - > [ R i s continuous in the first variable and regulated in the second variable, and furthermore,
23
4>e sector [a, j8],
fi
>0
Suppose that, for all compact intervals / c [ R + , the function
We show that the hypotheses of Theorem (10) are satisfied with Hx and H2 replaced by H1^ and H2^. First, since the mapping <j>^ e sector [ — p,p], H2^. satisfies (12) with y(H2^) = p and /? = 0. Next, the conditions on the Nyquist diagram of g(joj) imply that g^jco) is bounded as co varies over R, but that g^(s) has one or more poles in the open right half-plane. It is easy to verify that since g%(-) eLex, the mapping H^ satisfies (6), and it only remains to verify (7). It is clear that in the present case, the subspace M consists of func-
162
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THEOREM
tions in L2 whose Laplace transforms have zeros at the points where g^(-) has poles. Suppose / ( • ) e M; then g^(s) is analytic in Re s > 0 and g^(jco)f(jco) e L2(U). Moreover, for a l l / e M, we have by Parseval's equality that
Therefore, H^ satisfies (7) with
The present conclusion now follows from Theorem (10), since the circle criterion on the Nyquist diagram of g(jco) ensures [see Theorem (2.10)] that s: Another approach to L2 instability based on orthogonal decomposition is found in Section VI. 12.
9.2
Noncausality approach
In this subsection, we study a very special kind of "instability," namely, the case where a "bounded" input does produce a "bounded" output, but the map relating the input to the output is not causal. We first prove a general result, which we then specialize to obtain instability conditions. Theorem Let & be a Banach algebra with unit element J, and let &+ be a subalgebra of $ containing / (i.e., xe&+, y e &+ => xy e J t + , yx e J , + ) . Let x, y e &+ and let Sf be a bounded connected subset of C or [R, as appropriate. Suppose that for each a e ^ , the element x + ay is invertible [i.e., (x + a y ) - 1 e J*], and that the collection of these inverses is bounded; that is, there exists a finite constant m such that ||(x + ay)" 1 1| < m < oo, U.t.c, (x + a y ) - 1 e J*+, Va e $f, if and only if (x + ay)" 1 e @+ for some Proof => Obvious <= The statement is clearly true if y = 0; so suppose y # 0, and suppose (x + a0 y)~* e J*+ where a 0 e «9*. First of all, note that if a e £f, (x + ay)"* e J^ + , and | a — a' | < (\\y\\ • m)" 1 , then (x + a'y)" 1 exists and belongs to J t + (whether or not a' belongs to Sf). This is immediate from the observation that x + a'y = x + ay + (a' — a)y = [J + (a' — a)y(x + ay) - 1 ](x + ay) and the contraction mapping principle for Banach algebras.
9
INSTABILITY
163
Given that (x + a 0 j/) _ 1 e J*+, let a e ^ . Since £f is connected, there exists a finite sequence of points al5 a2 , . . . , ocN+1 = oc such that | at — a i + 11 < Obll * w ) - 1 for / = 0, . . . , TV. Now since |ax — a 0 | < (H^Hw)"1 and (x + (Xoyy1 e &+, it follows that (x + a ^ ) - 1 e &+ ; since |a 2 —a x | <(IMI -ra)" 1 and (x + oqy)" 1 e J* + , it follows that ( x - f a 2 j / ) _ 1 G l + . Repeating this reasoning N times, we see that (x + %+iJ 7 ) - 1 = (x + a y ) - 1 e ^ + . Since a e ^ is arbitrary, this completes the proof. :: 31
Remark The basic idea of the proof is a follows. Given that (x + a 0 >') _1 e ^ + , it can be shown that (x + ccly)~1 e J*+, provided | oq — a01 is sufficiently small. The assumption (30) that the collection of inverses (x + a y ) 1 ls bounded really means that the upper bound on | oq — a01 is independent of a0 . Now given that (x + oc0 y)~l e ^ + , in order to show that (x + ay)'1 e J + where | a — a01 exceeds the aforementioned upper bound, we construct a "chain" of numbers a1? . . . , ocN+1 = a in £f such that | a£ — a£ + 11 is suitably small. Exercise 1 Under the conditions of Theorem (29), suppose (x + a 0 - y) _ 1 G @ + . Show that (x + ay)'1 e &+ whenever
32 Exercise 2 Prove the following stronger version of Theorem (29). Let & be a Banach algebra with unit element / ; let &+ be a subalgebra containing J; and let x, y e $+. Let (oHj)jej be a finite or countable collection of points in C or U (as appropriate), such that
U.t.c, if (x + ocjy)"1 e J , + for some 7'eJ, then (x + a y ) 1
G
^ + whenever
33 The significance of the above results to the stability question is now illustrated. 34
Fact Let 1
164
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THEOREM
a unit element; is a subalgebra of J*(LP); and &+(Lp) contains the unit element. We are now in a position to state a small gain theorem for instability in the sense of noncausality. 35
Theorem Consider a system represented by (1) and (2). Suppose H1 and H2 are extensions to J^e of continuous linear mappings on $£ (we denote the restrictions of H1 and H2 to <£ also by Hx and H2). Suppose there exists a continuous linear map K on if, which can be extended to ££e, such that the operator (J + KH^1 is well defined on S£e and continuous on $£'. U.t.c, if
36 then (i) (7 + H2H1)~1 is well defined on $£e and continuous on <£\ (ii) (7 + H2 Hi)"1 is causal o (I + KH^'x is causal; (iii) Equations (1) and (2) have unique solutions for et and e2 given by 37a
ex = (7 + H2H1)~\ul
-
H2u2)
37b
e2 = (7 + H.H^1^
+ 7/lW2)
Comment In view of the linearity of H1 and 7/ 2 * ei 38
e2 = u2 + HX(I + H2 H^1^
can a
ls° be expressed as
- 7/ 2 i/ 2 )
Proof Conclusions (i) and (iii) are proved by the familiar methods of the small gain theorem. To prove conclusion (ii), consider the family of elements in J'(x) defined by 39
Fa = I + KHX + oc(H2 - K),
a e [0, 1]
By virtue of the assumption that (7 + KH^'1 exists, and in view of condition (36), we see that F " 1 e &(x) exists for all a e [0, 1]. Moreover, the collection of these inverses is bounded, since 40 Hence by Theorem (29), F~a * is causal for all oc e [0, 1 ] if and only if F~* is causal for some oc e [0, 1]. In particular, F^1 = (7 4- H2H^)~1 is causal if and only if FQ 1 = (7 + KH^'1 is causal. :: Let us now turn to convolution feedback systems. We give only a few results, and we do not always state the strongest possible results, in the interests of simplicity.
9
41
INSTABILITY
165
Definition The set «s/2 consists, by definition, of all generalized functions / ( • ) of the form 0
where tt, ft e R, fa : R ->R, and
The set jrf2 c a n ^>e made into a linear space in the obvious way. If we define
and if we let the product of two elements/and # of ^ 2 De their convolution, then j2/2 becomes a Banach algebra. One can think of stf2 as a two-sided version of the Banach algebra sf defined in Appendix D. In fact, if we define 45 then jrf is a subalgebra of ja/2 • Let j2?2 denote the set of Laplace transforms of all elements of srf2 • The conditions under which an element of j£2 is invertible are discussed in the following theorem. 46
Theorem Suppose / e s£2. Then l/fe s4\ if a n d only if
47
inf |/(5)| > 0 Res=0
Suppose / e i . Then l / / e «s?2 if and only if (47) holds, and l/fe s£ if and only if 48
inf |/(5)| > 0 Res>0
Now we turn to the question of feedback stability. 49 50
Theorem Suppose G, FeJn2Xn. only if
Then ft = G(I + FG)" 1 e i f "
if and
inf | det[J + F(s)G(s)] \ > 0 Res = 0
With the above results in hand, we can state the instability results.
166
51
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THEOREM
Theorem Consider a multivariable feedback system of the form shown in Fig. III.l. Suppose the subsystem Hl is represented by
52
yi(t)
= (G * ei)(t)
where G e stfnXn and that the subsystem H2 is represented by 53
y2(t) =
n
where <> / : U x U -+M is continuous in the first argument and regulated in the second argument. Suppose further that el9 e2, yl9 y2 e Lnep(U) whenever wl5 u2 e L/(U). U.t.c, if there exist a matrix K eUnXn and real constants y and p with /? = 0 whenever p < co such that
Moreover, the map (w1? w2) -» (^i, e2) *s causal if and only if 59
inf \det[I + KG(s)]\ > 0 Re s > 0
The proof is immediate from Theorem (35). 60
Corollary Under the assumptions and conditions of Theorem (51), suppose (56) is replaced by
Then the conclusions of Theorem (51) hold for/? = 2. 62
Theorem Consider a scalar feedback system of the form shown in Fig. III.l, where the subsystem Hl is represented by yi(t)
= (g *
ei)(t)
with g e Zi(lR), and the subsystem H2 is represented by y2(t) =
NOTES A N D
REFERENCES
167
with <> / e sector [a, /?], 0 < a < fi. Let D[— 1/a, — 1//?] be the disk in C whose diameter is the line segment connecting the points — 1/a + y'0 and — 1//? + y'0. U.t.c, if the Nyquist diagram of g(jco) does not intersect the disk D[ — 1/a, — 1//?] and encircles it in the clockwise direction at least once, then ul9 u2 eL2(U) => el9 e2, yl9 y2 e L2(U) and the map (ul9 u2) -• (el9 e2) is noncausal.
Notes and references Early versions of the small gain theorem and circle criterion were given by Sandberg [San. 4] and Zames [Zam. 3], both of whom also explored the technique of exponential weighting [San. 8], [Zam. 2]. Exponential weighting techniques are also utilized in [Ber. 1]. Stability results involving a circle criterion based on a Liapunov approach were derived by Brockett [Bro. 1, 2], and by Narendra and Goldwyn [Nar. 1]. The theorem on slowly varying linear systems is a generalization of an idea due to Sandberg [San. 12], while those on the existence of periodic solutions are due to Holtzman [Hoi. 1, 3]. The nonlinear circuit example is in the spirit of [San. 7]. Takeda and Bergen [Tak. 1] were the first to study the orthogonal decomposition approach to L2 instability, while the noncausality approach is due to Willems [Wil. 1]. Instability results have also been obtained by Brockett [Bro. 5], who used Liapunov theory, and by Davis [Dav. 1, 2], who used the concept of Fredholm operators. Finally, an application of small gain techniques to pulsewidth modulated systems can be found in [Sko. 1].
VI
0
PASSIVITY
Introduction
The thrust of this chapter is the systematic use of the concept of passivity, which plays an essential role in circuit theory and, more generally, in physics. The first section uses a very simple circuit example to motivate the formalism. The technical properties of scalar products are presented in Section 2. Then, as in Chapter III, a section is devoted to the general framework. This abstract section serves an important unifying role in that it allows us to see many problems as special cases of a general problem. The main difference is that now we work in inner-product spaces rather than normed spaces. Section 4 defines passive operators and presents many examples of passive operators which are used later. The reader need not study every example and work out every exercise because whenever one of these examples is used in later developments, a specific reference is made to it. The passivity theorems of Section 5 correspond to the small gain theorem of Chapter III. They constitute the main results of this chapter. Several forms of the Popov criterion 168
1
MOTIVATION
FROM CIRCUIT
THEORY
169
are developed in Sections 6 and 7 for the scalar case, the multivariate case, the continuous-time case, and the discrete-time case. The logarithmic variation criterion is the subject of Section 8. The interesting feature of that criterion is that if the average rate of variation of a feedback gain is small enough, then the stability of the frozen system implies that of the time-varying system. This complements the results of Chapter V, Section 5. The technique of noncausal multipliers with applications is the subject of Section 9: The general pattern of the technique is presented, and two methods of factorization are given. In the next section, it is shown that under rather general conditions, the passivity theorem and the small gain theorem are equivalent; that is to say, that if we are given a feedback system whose stability can be established by the passivity theorem, then, under some general conditions, this system can be transformed into another equivalent one whose stability can be established by the small gain theorem, and vice versa. The problem of invertibility of an operator of the form I + H is fundamental to the study of feedback systems. Passivity is used in Section 11 to obtain sufficient conditions for J + H to be a bijection. The final section presents a general instability theorem based on passivity. The technique is then applied to obtain an instability counterpart to Popov's criterion.
1
Motivation from circuit theory
Consider a one-port whose port voltage (current) is v(-) [/(•), resp.]. The power delivered to the one-port at time t is v(t)i(t), provided of course that the usual reference directions are used (see Fig. VI. 1). If $(t0) denotes the energy stored in the one-port at time t0, then it is natural to say that the one-port is passive iff
This is a natural definition since the integral evaluates the energy delivered to the one-port during the time interval [t0, t]. More generally, an n-port Jf is viewed as a relation in
170
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PASSIVITY
(Here the superscript asterisk (*) denotes, as usual, the complex conjugate transpose.) Of course, in the present case taking the complex conjugate is vacuous; however, it becomes important when voltages and currents are represented in terms of phasors. One of the earliest connections between passivity and stability is due to Youla who considered the linear case. The technique, however, carries over to the nonlinear case. We state it as a theorem. 3
Theorem Let the (possibly nonlinear) w-port Jf have the property that v9 ieL2e(U), V(#, i)eJf. Assume that for the connection shown in Fig. VI. 1 and for any input eeL 2 e (R), there is a response v, ieL2e. U.t.c, if Jf is passive, then e e Ln2 implies that ||tf||2 < \\e\\2 and ||/||2 < \\e\\2.
Figure VI.l
Proof
By definition, let (v\i)T=
v(t)*i(t)dt
Hence the passivity assumption is equivalent to 4 Now from Kirchhoff's voltage law, e = v + / and since e e L2, 5
oo > (e\e}
> (e\e}T,
VJGIR
= O 4- i\v + i}T = {v\v}T + <j)\i}T + < i » r + r >
Comment We had to make the assumption that for any eel} there is at least one response. So existence is assumed, and passivity is used to prove stability. No assumption nor conclusion can be made concerning uniqueness and/or continuous dependence.
2
2
SCALAR
PRODUCTS
171
Scalar products
Any function (denoted below by < • | • » mapping «f x j f into C, where Jf is a linear space over the field C, is called a scalar product iff it satisfies the following axioms (we use the superscript asterisk (*) to denote the complex conjugate): si.
<*|j> + / > = < * b > + <*!/>'
SII.
<x\yy =
SIII. SIV.
<x\Xy} =l(x\y} (x\x}>0ox=/=6
where 9 denotes the zero vector in 34f. Exercise 1 Prove the following consequences of the axioms: (a) The function XF-» by ||;c||; so, by definition,
<X|X>1/2
is a norm on J-f. It will be denoted below
\\x\\2 £ <x|x>
1
This norm and this scalar product are related by Schwarz's inequality: 2
\<x\y>\ >\\x\\\\y\\,
Vx, y e #>
and equality holds if and only if (x, y) are linearly dependent or one or both are zero. 3
(b)
<x + x'\y} = <x\y} + <x'\yy
4
(c)
5
Comments (a) For fixed x e Jf, y f-> <x| y> is a linear functional on J-f. (b) In most of our applications, the linear space Jf is over the field U, and the scalar product < • | • > takes real values. Then by Axiom SII, <x | y> = (y | x>, for all x, y e #P. Furthermore, the function mapping the ordered pair (x, y) e 2tf x j f into <x| y> G IR is bilinear•; that is, Vx e J-f, 3/1—> <^c 13^> is linear, and Vy G 3tf, x i-» <x| y> is linear.
6
Examples (I) Let x, y G U or C, then <x| y> = x*y, the usual multiplication and where * denotes the conjugate. (II) Let x, y G Cn with x = (£1? £ 2 , . . . , Q, y = (rjl9 rj2, . . . , >/„), then
VI
172
PASSIVITY
(III) Let x, y: U+ -• R with x, yeL2, then <x|.y> = $$x(t)y(t)dt. [If x and y are complex valued, the integrand is changed to x(t)*y(t).] (IV) Let x, y: U+ -> C" with x, y e L„2 and use * to denote the conjugate transpose, then (x\ y} = Jo x(t)*y(t) dt. (V) In the previous cases, we can also use a weighting factor and write Jo w(i)x(t)*y(t) dt, where w(-) is bounded, continuous, and w(t) > 0 for all t. Typically, w(t) = e~ 2at (a > 0). (VI) Let x, y e I2, with x(/) = [xt(i), x2(i), . . . , x„0')] anc * each xfc(/) e R
3
Formal framework
Let 2T be the set of instants of time of interest. (Typically, &~ = U+,U, Z, or Z + = { 0 , 1 , 2 , . . . } . ) Let y be a linear space with scalar product (•, •). Let 8F be a class of functions x: 3~ -• TT. Let a scalar product < • | • > be defined on a subset of #". If 2T = R + , TT = [Rw, and (•, •) is the usual scalar product in Un, we might have 1 similarly, with
2 For any t e ^~, let PT denote as before the linear operator defined on 3* by the condition that for any x e #", T e J ,
where 6 denotes the zero vector in ir. Clearly, PT2 = PT; hence PT is, for each T, a projection. We shall often use the notation T —
T
The projections PT identify linear subspaces of J*7, namely, PT& ={ge&\g Let 5
= PT / f o r s o m e / e#"} = {g e ^\g(t)
= d for t> T}
4
PASSIVE
SYSTEMS
173
Thus, $?e is the space of all x with the property that for all f e J , xT has finite norm. From Schwarz's inequality it follows that if
Similarly, let 6 The scalar product is required to fulfill the following conditions: HI. For all xe Jf e , the function r - » | | x T | | is monotonically increasing. 9 HII. For all 10 HIII. For all 8
Note that the six examples of scalar products given in Section 2 satisfy these three conditions.
4
Passive systems: definition and examples
We use in this section the framework just developed to define passive systems. We give several examples and exercises, which are important instances of passive systems. Let H: J-fe -» ffle. We say that H is passive iff 3 some constant /? s.t. 1 We say that H: j f e -• «#% is strictly passive iff 3<5 > 0 and 3£ s.t. 2 We say that H: J-fe -• 3/Fe is incrementally passive iff 3 We say that / / : ^f e -* ^f e is incrementally strictly passive iff E3<5 > 0 s.t. 4 5
Comments (a) If H is linear, then /? can be taken to be zero in (1) and (2). More importantly, when H is linear, (i) H is passive if and only if it is incrementally passive; (ii) H is strictly passive if and only if H is strictly incrementally passive. (b) It is very important to distinguish between passivity, strict passivity, etc., which are notions tested on the extended space ^fe, and the more
174
VI
PASSIVITY
common notion of positivity. In Section 8 we shall define ffl positivity by the condition:
where H: Jf -» Jf. When H is causal (nonanticipative), these two notions are identical [see Lemma (8.2)]. They are distinct when H is noncausal (see Exercise 2, Section 8). The remainder of this section displays many classes of passive systems both by demonstration through examples and by exercises. 6
Example 1 Let H: Le2 -» Le2 and be defined by
7
Hu = h* u,
where
h e $$ and
u e Le2
Note that since he jtf, H is causal. We assert that: 8
H is passive iff Re[fi(yco)] > 0, Vco e U;
9
/ / i s strictly passive iff, for some S > 0,
The proof is based on the following calculation: by (3.10) because / / is causal (Parseval) observing that
Statements (8) and (9) follow easily. Example 2 Let H be an n x « matrix whose elements belong to stf, and let the operator H be defined by Hu = / / * i/, for all ueL2e. Thus, H : L^e -»Z^ e . We assert that 10 11
H is passive iff the Hermitian matrix H(jco) + ft(j(D)* is positive semidefinite, Vco e IR. H is strictly passive iff, for some d > 0,
i.e., for all co e U, the least eigenvalue of the Hermitian matrix H(jco) + H(jco)* is > (5.
4
PASSIVE
SYSTEMS
175
Exercise 1 Prove (10) and (11) in detail. Exercise 2 (discrete-time case) Let H : I2 -> I2 and be defined by Hu = h * u, where h = (h0, hl9 h2, ...) e Z1. Let ue le2 . Show that if is passive (strictly passive) iff Re[fi(e/0)] > 0 (for some S > 0, Re[£(e/fl] > 5) for all 0 e [0, 2TT]. Exercise 3 (discrete-time case) Let H : l2e -» l2e and be defined by Hu = / / * w , where H = (H0, Hl9 H2, ...)el*xn. Let uel2e. Show that H is passive (strictly passive) iff the Hermitian matrix H(eje) + 0(eje)* is positive semidefinite (for some 8 > 0, ,l m i n [iJ(^) + i?^' 0 )*] > 5, resp.) for all 0 e [0, 2TT].
Example 3 Consider the system whose input is u and whose output is y = Hw, where w, y: U + -> [Rn. The equations of the system are
where ^ : U+ -* Rn and i? e lR"Xn. The system can be visualized as a bank of n integrators whose input is u(t) and output is £(0 [see Eq. (12)], and the integrator outputs are combined linearly to obtain the system output y(t), as required by (13). The system H has a well-defined response for any u(-) in L%e(U+). We assert that if R is a symmetric positive semidefinite matrix, then H is passive. Proof
For any ueL2e(^+)
an
d any Te U + ,
This is a line integral in Un along the curve t h-> £(t) for t e [0, T]. Since R is symmetric, the value of the integral is independent of the path; so we can evaluate it along the line segment A i—^ (1 — A)£0 + A^(T), where £0 = £(0). Thus,
And passivity follows from the positive semidefiniteness of R.
::
176
VI
PASSIVITY
Exercise 4 (a) Show that even if £0 = 0 and if R is positive definite, it still does not follow that the system defined by (12) and (13) is strictly passive. (Use an example.) (b) Show that the path integral above depends only on its initial point and its final point by observing that since R is symmetric, it follows that
Exercise 5 Let H: J^e -> 34fe. Let K denote a time-varying gain: K: J^e -> Jf e such that (Kn)(0 = k(t)u(t) where k: 3T -* U and k is a bounded function on P. Show that (a) if H is passive, then KHK is also passive; (b) if H is strictly passive and if k is bounded away from zero on ^~, then KHK is strictly passive. Example 4 Consider the system shown in Fig. VI.2; the input of H is u and
'1
H I
1U
Figure VI.2
its output is y. We have 14
where w,
For all T > 0 and all t/ e L e 2 we have
Since 0 e sector [0, oo), the last integral is nonnegative, and if we define <S>: R-*R by
16
4
PASSIVE
SYSTEMS
177
then $>(&) > 0, V(7 e U. Finally, using <X> to evaluate the first integral above, and noting that q > 0, we obtain
Hence the system is passive.
»
Exercise 6 Give an example to show that if in the system above <> / is time varying but (p e sector [0, oo), Vf e R + , then / / is not necessarily passive. Exercise 7 Suppose that in Fig. VI.2 we interchange the order of the blocks. Show that the resulting system is not passive for all <> / e sector [0, k). [Hint: Consider $(G) = a for a > 0 and (f)(a) = 0 for a < 0.] Exercise 8 Consider the «-input-«-output generalization of the system shown in Fig. VI.2. Let P, Q e Unx"; cr(0, u(t), y(t) e Mn, and 17
QCT(0 + Pa(0
18
=
u(t)
X 0 = 4>K0] n
where
178
VI
PASSIVITY
Proof
where we used the fact that
Figure VI.3
characteristic (/>(•, i) e sector (0, k) for some k e (0, oo]. Let N: L2 -> Le2 be arbitrary except that for some 3 > 0, (Nw)(0 > 6 > 0,
Vw e L e 2 ,
Vf > 0
The output y is the product of the output of N and of (f>. U.t.c, show that the system is passive. Give some additional restriction that will make the system strictly passive. Example 6 Let P be an operator from L2 ->Le2. Let K be a time-varying gain; i.e., K: L2 -+L2 with 26
(Kx)(t) = k(t)x(t), where k(-) is absolutely continuous on U+ with 0 < k 4 inf k(t) < k(t) < sup A:(0= /c < oo The class of all such operators is labeled Jf 0 ; if k > 0, it is called X. We assert that
27
if P is passive (strictly passive) and passive (strictly passive) ( \ means monotone nonincreasing); 28 if P is passive (strictly passive) and k(-) e X is /*, then PK is passive (strictly passive).
4
29
PASSIVE
SYSTEMS
179
Comment The importance of the assumption that k(-) is absolutely continuous is that k(-) is absolutely continuous if and only if (a) k(-) is defined almost everywhere, and (b) k is the indefinite integral of k(-) [McS.l]. Proof of (27) For all u e L2 and for all T > 0 T
T
30 because the first term is nonnegative and the integral is nonpositive because P is passive, k(T) > 0 for all T, and k(t) < 0 for all t. :s Proof of {2%) Mu e Le2, VT > 0, with Ku ^ v T
Note that u e Le2 ov = Kue Le2 because k(-) e JT. So we have the same integral as (30) except that k(t) is replaced by l/k(t). Hence the same calculation shows that PK is passive. s: Exercise 10 Prove the assertions for the strictly positive case, and prove related assertions for the discrete-time case. The following example is a variation on Example 6 in that/: is not assumed to be increasing, but using exponential weighting techniques, it considers the case where t h-> k(t) exp(at) is nondecreasing. Example 7 Suppose there are numbers k and k such that 0 < k < k(t) < k < oo for all t. Let k: L2 ->L2 and be defined by (26). Let g: L2 -+L2 with gx = Q * x, where g e s/. Let /? > 0 and assume that g can be continued analytically up to Re s = —f$. We assume that there is a 3 > 0 s.t. 31
Re g(j(D) > 3 > 0,
Vo> e R
and 32
Re g(jco - p) > 0,
Vco e R
U.t.c, (i) if for some a e (0, 2/?), t H-• A:(0 exp err is /, then gk is strictly passive and has finite gain; with 3 = 0, fi — o > 0, gk is passive; (ii) if for some o e (0, 2/?), t H-> /:(r) exp( — <jr) is \ , then kg is strictly passive and has finite gain.
VI
180
33
PASSIVITY
Comment If k e Jf, then, in (i), we could write the assumption as d log k(t) dt Proof
kit) k(t)
(i) Let p(t) = l/k(t); then for some numbers g and p, we have
35
0 < p < p{t) < p at
and t h->p(t)e~ is \ . Now for some <5t > 0, sufficiently small, since 2fi > o > 0, we have that [p(t) — St] cxp(-lfit) is \ . (Take the logarithm and differentiate !) Then, Vw e Le2, W > 0, with v = kt/, we have
The first integral is larger than 38 The second integral is nonnegative. Apply to it the second mean value theorem [McS.l, p. 210], noting that the bracket is positive and \ ; hence it is equal to
where T' is some suitable number in [0, T]. Now t \-> v(t) exp fit, truncated at T', is in L 2 ; hence if we call vT its Laplace transform and if we note that ept(gv)(t) = ep\g * v)(t) = (ge" * ve") we conclude, from Parseval's theorem that the second integral is equal to 40 where the last inequality follows from (32). From (36), (38), and (40), we conclude that gk is strictly passive. Also, y(gk) = llgll^K. The proof of (ii) is immediate because 41 and exp(
::
Comment The manipulation in (41) is valid under general assumptions; namely, k must be self-adjoint, y(k) < oo, and y(k _1 ) < oo.
5
5
PASSIVITY
THEOREM
181
Passivity theorem
In this section, we formulate a simplified version and a generalized version of the passivity theorem. We are guided by our circuit-theoretic intuition discussed in Section 1. We use throughout the general framework of Sections 3 and 4. We start by proving a simple version of the passivity theorem. 1
Theorem Consider the usual feedback system of Fig. III.l with u2 = 0 so that
2
ex=u1-H2
3
e2 =
e2
H^
Here H1 and H2 map 2tfe into 3tfe. We assume that for any ut e Jf, there are solutions e1, e2 in Jf e . U.t.c, if Hy is passive, H2 is strictly passive, and if ut e jf, then H^e^ e 3tf. Proof
We obtain successively by using (2) and (3)
where S2 and fi2 are constants with b2 > 0, since H2 is strictly passive. Hence by Schwarz, and noting that ut e 34f, we obtain 5 from which it follows that Ih-> H/Z^Hr is bounded; i.e., e2 = H1e1 e ffl. •• ••
Exercise 1 Suppose that, in the strict passivity condition for H2, the constant /?2 is nonnegative. Then calculate from (5) the constant k such that
6
Comments (a) This form of the passivity theorem is easy to prove, easy to apply, and easy to remember. Its circuit-theoretic interpretation makes it obvious (refer to Fig. III.2): The current sources u1 face the parallel combination of a passive «-port and a strictly passive «-port. The theorem proves that if ut e 2tf, the port voltages H1el = e2 e Jf. (b) Note that it is not claimed that et and H2 e2 e Jf. Indeed, in case of one-ports, a counterexample is easily built up by taking ux(-) to be a rectangular pulse of current, H1 a 1-henry inductor, and H2 a parallel connection of a 1-ohm resistor and a 2-henry inductor; after the pulse of current has
182
VI
PASSIVITY
become zero, there remains a constant current through the inductors! Thus, et and H2e2 are not in L2. (c) In order to be able to claim that both ex and e2 = H1e1 e 2tf, one must have a finite-gain assumption on H1 or H2 as done in Theorem (10) in the following. (d) Intuitively, one would expect that H1 and H2 need not both be passive; what should suffice is that any "activity" (circuit-theoretic term for nonpassivity) of one should be compensated by some "passivity" of the other so that the parallel connection is strictly passive. This is done precisely in Theorem (10). (e) Clearly, by permutations and combinations of the examples of passive systems of Section 4, we can construct dozens of Instability theorems. We turn now to the more general version of the passivity theorem. 10
Theorem
Consider a feedback system as shown in Fig. III. 1 and described by
12
el=u1-
H2e2
13
e2 = u2 4- H^x where H1 and H2 map Jfe into ^fe. Assume that for any ut and u2 in Jf, there are solutions ex and e2 in Jf e . Suppose that there are constants yl9 pl9
14
U.t.c, if 17
5t + e2 > 0 then , t/2 e J'f imply that eu e2, H^e^ H2 e2 e 2tf
18
Comments (a) Generality This theorem applies to continuous-time (2T = U+) or discrete-time ( ^ = 2+) systems. The systems may be single input-single output (i^ = U) or multiple input-multiple output (Y = Un) or have distributed inputs and outputs {V is an inner-product space). (b) (14) means that Hi has finite gain. (c) When s2 = 0, (17) requires that 3t> 0; then the theorem holds if // x is strictly passive with finite gain and H2 is passive. When e2 ¥^ 0, if we go back to the circuit-theoretic interpretation of Fig. 111.2 for the one-port case, we see that the "impedance operator" H1 dissipates more energy than a
5
PASSIVITY
THEOREM
183
resistor of d± ohm, and the " admittance operator " H2 dissipates more energy than a resistor of £2 ohm; thus, the condition (17) says that the net amount of energy dissipated must be positive. (d) If H2 is a bijection of Jf e onto J^ e, (16) can be restated as
thus if £ 2 >0 (16) means that H_2 Proof
is strictly passive.
For any ux, u2 e Jf, for any f e « f ,
where we used (12) and (13). In the last equality, use (14), (15), (16), and Schwarz's inequality
To eliminate H2 e2 in (21), note that H2 e2 = wx — et so that 22 Substituting (12) in (21) and replacing £2 by |£ 2 | in the right-hand side, we obtain
Using the key assumption 5X + £2 > 0, we rewrite this inequality in the form 24 where b(T) and c(T) are monotonically increasing and, as T -> oo, tend to y?mte constants since z/1? u2e.tf?. Call the constants & and c, respectively. From (24) we have 25
ItaUr < b(T) + [Z>(T)2 + c(T)] 1/2 < b + (Z>2 + c) 1/2 ,
VTe iT
So ex e Jf. By (14) the same holds for ^ = i / ^ . By (12) and (13) it follows that e2 and H2e2 are also in J-f. SS 26
Corollary Under the conditions of Theorem (10), (a) if «x = 0, then the map u2 H» // 2 ei *s passive; (b) if u2 = 0, then the map u± t-> / / ^ is passive.
184
VI
PASSIVITY
Exercise 2 Give a circuit-theoretic interpretation of Corollary (26). Proof
(a)
By (20), since ut = 0, we have Vw2 e 34?e and VT e ^",
Hence u2 H-» #2^2 *s passive; we cannot conclude that it is strictly passive because the assumptions do not exclude the case where H2 sends all inputs into the zero output. (b) The proof is similar. s: 27
Corollary If in addition to the assumptions of Theorem (10), we have Pi — Pi —Pi = 0> t n e n the maps sending (uu u2) into eu e2, H^u a n d H2 e2 are L2 stable. Proof U.t.c, the quantities b and c of Eq. (25) are homogeneous polynomials in || ux\\ and \\u2\\ of degrees one and two, respectively. Hence from (25), there is a k < 00 s.t. The remaining maps are L2 stable as a consequence of (12), (13), and (14).
In discussing the small gain theorem (III.2.1), we noted that we obtained sharper results by requiring conditions on the incremental gains rather than on the gains themselves; in fact, in the first case, we could only assert that the norms of the outputs were finite; with conditions on the incremental gains we obtained, in addition, existence, uniqueness, and continuous dependence. A similar behavior occurs when one considers incremental passivity. 30
Theorem Consider the usual feedback system described by (12) and (13), where Ht and H2 map 3tfe into 3tfe. Suppose that any uu u2 in Jf produce el9 e2EJ^e. Let Ht0 = 0; H20 = 0. U.t.c, if (a) H1 has finite incremental gain and is strictly incrementally passive, i.e., there are constants 0 < % < 00, dl > 0 such that
(b) H2 is incrementally passive, i.e.,
5
PASSIVITY
THEOREM
185
then (i) for any input ul9 u2e ^f, the outputs el9 el9 yi9 y2 e #P and are unique \ (ii) the mappings (ul9 u2) -+eu (uu u2) -*e2 are uniformly continuous on 2/e x tf; (iii) the mappings sending (ul9 u2) into eu e2, H1e1, and H2e2 are L2 stable. Proof (i) and (ii). Since by assumption H10 = 0, / / 2 0 = 0 , by putting x' = 0 in (31), (32), and (33), we obtain (14), (15), and (16) with yt =yu dt = S1 > 0, e2 = 0, and all the /Ts zero. Consequently, by Theorem (10), e u e2 > yu yi e ^- To establish uniqueness, consider the input pairs (ul9 u2), (uxf, u2') and their corresponding solutions (eu e2)9 (ex\ e2). Now, as in the proof of Theorem (10), we obtain that V T E &~
Using (31), (32), (33), and Schwarz's inequality, we obtain the inequality
Hence if wx = ut' and u2 = u2, (35) implies that ex = e±\ whence e2 — e'2 by (13). Therefore, uniqueness is established. (ii) To establish uniform continuity, we rewrite the right-hand side of (35) as
The two bracketed terms are increasing functions of T and as T -+ oo, they tend to two constants, say, 2bdl and cS1. Thus, from (35), (36), and (5X > 0, we conclude that
where b and c are homogeneous polynomials in \\ux — ux'\\ and \\u2 — u2'\\ of degrees one and two, respectively, and by elementary algebra,
38 Since Z? and c are constants and || • || r increases monotonically with T, we may drop the subscript T in (38).
186
VI
PASSIVITY
Now if Ui -+ux and u2 -• u2 in 2tf (i.e., \u{ — u/\\ -*0 for / = 1, 2, and consequently \ui — u(\T ->0 for any T), then by (38), \ex — e/|| ->0. The same holds for
6
The Popov criterion
A very important application of the passivity theorem is the derivation of the Popov criterion and its extensions. Case I 1
Theorem Let : U -• U be continuous and be in the sector [0, oo). Let g: U+ -» U, g e LX([R+), and g e stf. For the system shown in Fig. VI.4, we
Figure VI.4
assume that for any ul such that ul9 u± e L2, then el9 e2 e Le2. U.t.c, if there is a q > 0 such that 2
inf Re [(1 + gj(o)§(Jco)] = <S > 0 and if ui9 ul e L2, then (i) el9 e\9 el9 yi9 yl9 and y2eL2; (ii) el9 yl9 and y2 are continuous, belong to L00, and go to zero as t -• oo.
3
Exercise 1 Show that ul9 ule L2 imply that u1 e L00, ^ is continuous, and i(0 -»0 as £ -• oo. [Hint: t i—• w1(0wi(0 e L1; so its integral is continuous and tends to a finite constant as t -> oo, . . . .] w
Exercise 2 Show that if an input u2 were inserted at the usual location in the system of Fig. VI.4, then all the conclusions of Theorem (1) would still hold, provided that u2el}. (Hint: Use the linearity of H2 to replace u2 by an equivalent input at ux; note that if z2 = g * u2, then z 2 = g * u2, where # is the derivative of g in the distribution sense.)
6
THE POPOV
CRITERION
187
Proof Transform the system of Fig. VI.4 into the one shown in Fig. VI.5. Define H1 and H2 by H1r\1 = yl9 H2 e2 — y2 + gy2 • By Example 4 of Section 4, Hl is passive, and by Example 1 of Section 4, H2 is strictly passive, as a consequence of (2). Therefore, by Corollary (5.26), ul9 u1 el} imply that
Figure VI.5
y1 = e2 e L2. Since gel} and g e jrf, H2 has finite gain; hence y2, y2, and y\1 e L2. Going back to Fig. VI.4, we have el9 e\el}\ hence ex e L00, ex{-) is continuous, and e^-^O as t -• oo. The same holds for yx(t) indeed, 4> is continuous and the continuous function cj) maps compact sets in IR into compact sets in U. The same properties hold for y2 since y2 = g * e2
and gel}. Case II 5
:: (j) e sector [0, k].
Theorem Consider the system shown in Fig. VI.4, where now c\> e sector [0, k]. Assume that all previous assumptions hold except that (2) is replaced by: If there is a q > 0 and some d > 0 such that
6
Re[(l + qjco)g(jto)} + l/k ^ S > 0,
Vco > 0
then the same conclusions follow. 7
Graphical Interpretation If we write hold if and only if
We need only check this inequality for co > 0 because gr(jo)) and cogt(jto) are even functions. The inequality is equivalent to the following: The complex plane curve co — i > [gr(jco), cogtijco)] lies below the straight line of slope l/q, which goes through the point (d — (l/k)fi). Since S > 0 can be arbitrarily small, this means that the curve co cog^jco)] lies below and is bounded away from the straight line of slope l/q, which goes through the point (-l/fc,0). Proof
Consider Hl shown in Fig. VI.5, we have for all T> 0,
188
VI
PASSIVITY
where we used the notation of Example 4, Section 4. In other words, calling /?! the last term, we have
Now H2 is a linear operator with finite gain since gel}
and g e s/, where
Therefore, H1 and H2 satisfy the conditions of the passivity theorem (5.10), and the map ux + qux H-» (rjt, e2) is L2 stable. The remaining conclusions are obtained by the same considerations as above. :: 8
Comment The proof can also be obtained by direct calculation as follows. The system equation reads (we drop the subscript 1 for simplicity)
9a
u = e + g * cj)(e) Since ue L2 and g e sf, we obtain by differentiation
9b
u = e + g * (f>{e) Since all terms in these two equations are in Le2, we can take truncated scalar products,
Using Schwarz's inequality, the sector condition, ^ > 0, and (6), we obtain Hence (e) e L2. Then e e L2 by (9a) and e e L2 by (9b). 10
Remark It is important to note that Theorems (1), (5), and (11) assume that the memoryless nonlinearity is time invariant. If arbitrary time variations are introduced while still obeying the sector conditions (say, as in the circle criterion), one can show by examples that instability results for some chosen (f)((T, t) and some chosen inputs. Case III g includes an integrator and <j> e sector [e, k + e] for some small positive s.
11
Theorem Let 0 be continuous, <j>: U -> R, and <j> e sector [a, k + e] for some sufficiently small positive e. Let 9(t) = r +
gi(t)
6
THE P O P O V
CRITERION
189
with the constant r > 0; gx e sd9 gA e l}(U+). Assume that, for the system of Fig. VI.4, ul9u1 e Le2 implies el9 e2 e Le2. U.t.c, if there is a q > 0 such that for some 3 > 0 12 where
and if ul9 ui e L2, then el9 el9 e2el}\
el9 e2 e L00 and go to zero as t -> oo.
Proof Since r > 0, for s > 0 sufficiently small, [1 + s^(s)]~x e i . Consequently, ge(s) and ^ e (^) e $4 [Theorem (IV.4.10)]. Hence COH-^(1 + qjo^)ge{jco) represents a continuous linear operator (finite gain). Starting with the system of Fig. VI.4 by a feedback of s around g9 we change g to gE [see (13)] and by a feedforward of s around
Thus, (f)E belongs to the sector [0, k]. Thus, with ge and e satisfying these properties, we are back to the previous case. s: 14
Remark In Theorems (1), (5), and (11), the linearity of the subsystem g was never used in the proofs; what was used repeatedly was that the transfer function (1 + qs)g{s) represents a system with finite gain and the appropriate passivity properties [as in (2), (6), or (12)]. Extensions of the Popov criterion to multi-input multi-output systems are quite easy. The basic principle is illustrated by the following theorem.
15
Theorem Consider the system described by
16
ex = ux — He2
17
e2 =
18 and let there exist a scalar function 19
VI
190
PASSIVITY
Let H and DH map L2e into L 2 e .f Assume that ul9 ut e L2 imply that eu e2 e L2e. U.t.c, if there is a (5 > 0 s.t.
then uu u1e L2
imply that
e2 =
If, in addition, / / and DH map L„2 into L„2, then el9 Q^ eL„ 2 ; with Q nonsingular, eu e2e L„°° and go to zero as t -> oo. Proof in Outline Introduce the multipliers P + QD and (P + QD)'1 and the feedback and feedforward of gain K. The result is Fig. VI.6. The operator
Figure VI.6
H1 is passive since
The first term is nonnegative by the sector condition (18). The second term can be integrated
Since V(£) > 0, V^ e Un, we conclude from (22) that 7/x is passive. Now assumption (21) simply states that H2 , shown in Fig. VI.6, is strictly passive. Hence by Theorem (5.1), ut, u1 e L2 imply that e2 = (// map L,,2 into L„2, we have 7/
t Z>/y denotes the map that sends x into
(d/dt)[(Hx)(t)].
7
Now since (P + sQ)
1
DISCRETE-TIME
CASE
191
is an exponentially stable matrix transfer function,
e1=(P
+ QDy1(m
+
Ke2)eLn2
Finally, Pet + Qe\ =m+
Ke2 e L2
from which we obtain Qe\ e L2. Thus, with Q nonsingular, we have el9 e1 e Ln2 so that eA eL„°° and 1^(01 -*0 as t -> oo. The same holds for e2 because (18) implies that
where Xmin(K) denotes the least eigenvalue of K, which is positive since K is positive definite. » Exercise Write down a formal proof of Theorem (15) in the style of Comment (8).
7
Discrete-time case
The techniques required for the discrete-time case are somewhat different. For simplicity, we consider below the single-input-single-output case. Thus, e i> ei-> y\i a n d yi map Z+ into U and
Instead of having a differential operator, we have the difference operator 2
(Ve)(ri) = e(n) — e(n — 1)
(backward difference)
(Ae)(fl) = e(n + 1) — e(n)
(forward difference)
or 3 Similarly, (VGx)(n) = (Gx)(n) - (Gx)(n - 1) 4
Remark The discrete-time problem has some features that make it much simpler than the continuous-time problem. For example, if x = [x(0), x(l), x(2), ...] e I2, then necessarily x e /°° and x(n) ->0 as n -> oo.
VI
192
5
PASSIVITY
Theorem Consider a system where ul9 eu e2, )>i, a n d yi
6
ex = ux — Ge2
1
e2 = <£0i)
ma
P Z + - R and
Let 4>: U -> IR belong to the sector [a, /?] with 0 < a < /?. Let G be causal and map I2 -> /e2. Suppose that for some q > 0, there is a <5 > 0 such that
(a)
U.t.c., if a > 0 and if 3 - (qP/4(x2) > 0
9
then ut eI2 implies that eu e2el2. (b) U.t.c, if 0 G sector [0, /?] and is monotonically increasing and if G has finite gain, i.e., 3 y (G) e U s.t. 10 then u1el2 implies that eu e2 e I2. Proof
(a) We have u1=el
+ G4>{et)
Hence
and, for all N e Z + ,
By the first sector condition, the first term is nonnegative. Now
7
DISCRETE-TIME
CASE
193
Since q > 0, we obtain from (15)
17 Noting that || Vt/JI < 211^ || and using Schwarz's inequality, we obtain 18 Hence 19
hence We also have e1 el2 since HeJ 2 < ||(^i)||2/a2. (b) Now <> / belongs to the sector [0, ft], but (j) is monotonically increasing (i.e., {a1 — o-2)[(cr1) - $(o2)] ^ 0, Vcrl5 a2 e IR). As before, we have (15). The estimate in (16) is different:
Note that we used the fact that ex{— 1) = 0 . Thus, in this case, the last term in (17) drops out, and (18) becomes 23 hence, letting iV -> oo, ||0O?i)ll < 11^11(1 + 2
24 26
Remarks (I) The theorem does not require that the subsystem G be linear or time invariant. (II) The derivations could just as well have been carried out using the forward difference operator (3) rather than the backward difference operator (2). Slight modifications are required.
27
Corollary If the operator G is linear and time invariant and, in particular, is a convolution operator whose impulse response (g0, gl9 g2, ...) e / 2 and has a z transform g(z) which is bounded on the unit circle (\z\ = 1 ) , then
194
Vi
PASSIVITY
condition (8) can be checked by considering the graph of 6 \-> g(ejd) in the complex plane (here 6 varies from 0 to n). Indeed, condition (8) becomes Re[l + q(l - z~l)]g(z) + (l/j8) > 8,
for
z = ej\
O<0
Recall that if e(z) = £P[e(')] is the z transform of the sequence {e(0), e(l), e(2),...}, then z~le(z) is the z transform of the sequence {0,
8
Average logarithmic variation criterion
The average logarithmic variation criterion is a general Instability result concerning a linear time-varying system shown in Fig. VI.7(a). The interesting
Figure VI.7
feature of this criterion is that it puts a condition on the average rate of increase of the logarithm of the feedback gain k(t) and on the shifted Nyquist diagram of w. Although the technique of proof requires the construction of multipliers, the L2-stability conditions do not.
Description of the system and assumptions The two subsystems are described by 1
y1=w^ei ei=u-
y2(t) = k(t)yi(t) y2
where u, e l5 yl9 and y2 map U+ into U, and 2
w(r) = w0 (5(0 + wa(t) with w 0 e H and for some a0 e U, t \-^ wa(t) exp(cr0 i) e l}(U+);
3
k(-)eJf where Jf = {/: U+ -> U\f sup, f(t) < oo}.
4
absolutely continuous and
0 < inf, f(t) <
w(-) = J?[w] can be continued analytically so that it is an analytic function in Re s > —or for some or > 0.
8
5 6
AVERAGE
LOGARITHMIC
VARIATION
CRITERION
195
For all u e L2, et and y1 e Le2. Let (7sh e (0,
8
Theorem For the system shown in Fig. VI.7(a), let assumptions (1) to (5) hold. U.t.c, if the o"sh-shifted Nyquist diagram of w does not intersect the negative real axis (i.e., {z e C | Re z < 0, Im z = 0}) and if for some T > 0 and some r > 0
(where we use the notation [ / ( T ) ] + = max[/(t), 0]), then the system is L2 stable. The method of proof consists in two steps: first, a factorization of the gain k(-) and the introduction of a " multiplier " m, second, the use of the passivity theorem (5.10) and of Example 7, Section 4. The multiplier m is a causal convolution operator on Le2 defined by its transfer function m(-) with the property that m and m _ 1 , restricted to L2, have finite gains. We start by proving a factorization lemma for k(-). 12
Lemma If k(-) e JT and satisfies (10), then there are two functions k+(-) and k__(-) such that
13
(i)
14
(ii) k+ and k_ are bounded away from zero and bounded on IR+;
15
(iii)
16
k(t)=k+(t)k.(t)
k+(t) exp(2rr) i s / 1 fc_(0exp(-2rf)is\
The multiplier m is constructed by the following lemma.
196
18
VI
PASSIVITY
Lemma Given that w satisfies assumptions (2) and (4) with w0 > 0, and the shifted Nyquist diagram condition of Theorem (8), there exists a causal convolution operator m such that m is analytic in Re s > — ash with 11—• m(t) exp(crsh i) e <srf and m(s) -* 1 as \s\ -> oo; furthermore, for some S1, <5 2 >0
for all w e R ; finally, m^s)" 1 is analytic in Re»s> — ash and bounded in
Res>-ash. 21
Comments It follows from Lemma (18) that the operators m and m~ *, when restricted to L2, have finite gains; indeed, since <7sh > 0, (D\-+m(jco) and co i—• [mijco)]'1 are bounded on R. Exercise Show that if w0 = 0, by the loop transformation theorem, one can transform the system to one where vv0 > 0, though possibly very small. We shall prove Lemmas (12) and (18) later. Using these lemmas, we prove now Theorem (8). Proof of Theorem (8) Refer to Fig. VI.7(a) and (b). In view of the properties of k+, k_, m, and m" 1 , the system shown in (a) is L2 stable if and only if the system shown in (b) is L2 stable. Now the forward loop gain, mwk+ = gk + , is strictly passive and has finite gain. Indeed, we can show that the assumptions of Example 8, Section 4 are satisfied; consider (10), (15), and (20). Furthermore, m(s)w(s) is analytic in Re s > — ash < 0 and tends to w0 > 0 as | s \ -• oo; therefore, we have Re[ni(jco)w(jcD)] > d2 > 0, Vco e U, because the real part of an analytic function reaches its minimum on the boundary of its domain of analyticity. The feedback gain k _ m _ 1 is strictly passive. Indeed, the assumptions of Example 7, Section 4 are satisfied; see (16) and the fact that m ^ ) - 1 is analytic together with (19). Thus, the L2 stability of the system (b) follows by the passivity theorem (5.10). :: Proof of the Factorization Lemma (12) Let l(t) = log k(t). Since /: IR+ -> IR is bounded. Consider the triangular weighting function
keJf,
elsewhere Note that fi(t) = d(t) - [1(0 - l(t - T)]/T, where 1(0 is the unit step. Define 24
8
AVERAGE
LOGARITHMIC
VARIATION
CRITERION
197
and 25 By (10) we see that /+ is bounded on IR+; in fact, 26 And, since / is bounded by assumption, /_ is also bounded. Consequently, k+ = exp /+ and k_ = exp /_ are bounded away from zero and bounded on R + . Finally, by (25), k(t) = k+(t)k_(t), W > 0. This establishes (i) and (ii). Differentiate (24) to obtain 1+(T) = [l(t)]+ - f f [/(T)] + (l/T)[l(r - T) - l(t - T - T)] dx Since the first term is nonnegative and since (10) bounds the integral by 2r, we obtain equivalently k+(t)exp2rt / Similarly, from (25) we show that :: It remains to construct the causal convolution operator m. We first establish a technique to construct a causal convolution operator with prescribed phase. 30
Lemma We are given a phase characteristic oo i—• (froti00)) oo eU, which is odd, i.e., differentiate, and such that 4>0 and its derivative <$>0' are in L2(IR). U.t.c, (i) there exists a function X e l}(U) with X(t) — 0 for t < 0 and
31
Im X(joo) =
MM
(ii) there exists a function z e l}(U) with z(f) = 0 for t < 0 and 32
1 + z(joo) = exp[!(yco)] (iii) if, in addition, — n <
33
arg[l 4- z(joo)] = (j)0(joS)
, then and
1 + z(s) ^ 0,
for
Re s > 0
Exercise Show that $ 0 , $ 0 ' e L2 imply that 0 O is in L00, is uniformly continuous, and ->0 as |co| -^ oo. (Hint: Think of Riemann-Lebesgue.) Proof (i) Since 0 O is in L2, it has an inverse Fourier transform which is also in L 2 :
198
VI
PASSIVITY
and since $ 0 and 0' are in L2, (j)0el}(U). t -> 0O(O is odd, consider the even function
(See Appendix B.2.5.) Since
is even, GL*([R), and $e(jco) is real for all co. Let A(0 = 0e(O + o(0; then A G l}(U), X(t) = 0 for t < 0 and (31) holds. Also X(s) is analytic and bounded in Re s > 0. (ii) Note that 36
exp[l(yco)] = exp[0e(y'co)] • exp[# 0 (yco)] and
In the time domain 38
z(t) = X(t) + (X * A)(0 + ••• + (! * ; [ * • • • * A)(0/«! + • • • Since A e LX([R) and since ||A * A^ < |{A|[^ • ||A||! it is clear that the series in (38) converges in the L1 norm and that Hz^ < exp^/ll^) — 1. Since l}(U) is complete, z e L 1 ^ ) . Also, z(f) = 0 for t < 0, by (38). (iii) If — 7i < <po(joj) < 7i, we may take the logarithm of (36) (since we avoid crossing the branch cut on the negative real axis), and
Since X is analytic and bounded in Re s > 0, by (37), 1 4- z(s) is bounded away from zero in Re s > 0 by exp(— \\XWi). « Lemma (30) shows us how to construct, on the basis of the phase 0 O , a function 1 + z such that itself and its inverse are analytic in Re s > 0 and bounded in Re s > 0. We now use this result to construct the multiplier m. Proof of Lemma (18) (I) By assumption (2), vv is analytic in Re s > — <7sh > —ac. By assumption, w(jeo — ash) ^ 0, \fcoeU; hence log[w(yco — ash)] is analytic for all co e U. Therefore, arg w(ja> —
9
MULTIPLIER
THEORY
199
where c e R is chosen so that c$0 is continuous. In view of the properties of the Nyquist diagram and by construction, (41), we have
for all co e U. Clearly, by (41), c/>0, 0' e L 2 . (c/>0' is well defined on U except at | ao \ = Q, where it may have a jump.) By Lemma (30) there is a z(s) with the properties (33). Let 44
m(s) = From (41), (44), and the properties of z, it follows that (I) m(s) and ra(s)-1 are analytic and bounded in Re s > 0 ; (II) the functions w i—> m(jao —
9
Multiplier theory
The basic idea behind the multiplier technique is that by multiplying certain operators by appropriately chosen "multipliers," the product can be doctored to satisfy the conditions of the passivity theorem. For example, u h-> g * u is passive if and only if Re[$(y'co)] > 0 for all co; i.e., the Nyquist diagram of g lies in the right half-plane. In the Popov criterion, the multiplier 1 + qs = m(s) is introduced and, consequently, Re[(l + qjco)g(jco)] is required to be nonnegative; i.e., the curve co — i > [gXJ0^)^ ^i(jco)] must lie below the line of slope l/q through the origin. Thus, for convolution operators we can think of the multiplier as a device which makes g(s) look passive by having the phase of m{j(o)g(jaS) remain in ( — n/2, n/2). If the multiplier m(s) is rational and causal and if u — i > m * u maps L2([R) into 2 L (IR), then m may not have poles in the closed right half-plane. This last requirement severely constrains the choice of the phase. If, for this reason, we let m(s) have right-half-plane poles and if we require u h-> m * u to map l3(U) into L2([R), then in order to calculate the kernel 11—• m(t) from m(-) we have to choose the jco axis as the path of integration in the calculation of the inverse Laplace transform. Then, as a consequence, 11—> m(t) is not identically zero for t < 0; i.e., the multiplier is noncausal. Since we shall allow noncausal multipliers throughout, we start by examining some of the consequences of noncausality.
200
9.1
VI
PASSIVITY
Causal and anticausal operators
Throughout this section, let 34? be a Hilbert space over the field U with scalar product denoted by < • | • >. Consequently, (x, y) i—• <x| y} is a bilinear map. Typically, 34? will be L2(U), Ln2(U), / 2 , or I2. Let //map either 34? into itself or 3^e into itself. Recall that H is said to be causal (or, equivalently, nonanticipative) iff PTx =PTy implies that PT Hx = PT Hy, Vx, y e 34? (or 34?e), VT e F. This condition can also be written PTH = PT HPT, VT e &~. Thus, for causal operators, past outputs depend only on past inputs. Recall also that if His causal and if H: 34? -> 34?, then / / c a n be extended into a map, also denoted by //, which maps 34?e into 34?e; the extended map is defined by the relation PTHx ^ PTHPTx, Vxe34?e, V7'e ST\ Note that if xe 34?e, PTx G 3^ so that HPTx is well defined. Finally, as with the term nonlinear, we say noncausal when we mean not necessarily causal. The map H: 34? -* 34? is said to be anticausal iff (/ - PT)x = (I - PT)y implies (/ - PT)Hx = (I - P T )#(J - ^V))7, Vx, j; e Jf, VT e R; equivalently, (/ - PT)H = (/ - PT)H{I - PT\ VTe «T. Thus, for anticausal maps, future outputs depend only on future inputs. Exercise 1 Let / / : u-+h*u, where hel}(U), uelI(U). Give conditions on h so that / / is (a) causal, (b) anticausal, (c) neither causal nor anticausal. The presence of noncausal operators (as a result of introducing noncausal multipliers) creates some difficulties in the application of Fourier techniques to demonstrate passivity. The following lemma and exercise exhibit useful facts. Lemma Let / / : 3tf -> J-f and be causal. U.t.c, if and only if
where the same d can be chosen in (3) and (4). Proof
=> For any ue34?e and TeM,uTe3tf? (u\Hu)T
= (uT\(Hu)T>
and by (3.10)
= <wT | (Hur)Ty by causality = (uT\HuTy >d\\uT\\2 by (3) Hence (4) holds for the same d. <= Let ue 34? in (4); let T -• oo and use (3.9).
s:
9
MULTIPLIER
THEORY
201
The importance of the causality assumption in Lemma (2) is stressed in the following exercise. Exercise 2 Let H: L2(R) -* L2(U) with Hu~-=h*u. e x p ( - 0 + l ( - f ) exp t,teU. Show that (a) Re[h(jco)] > 1, VcoeR; (b) VxeL2([R), (x\h*x}> \\x\\2; (c) find some uel} and some TE R s.t.
Let h(t) = 3(t) + 1(f)
(u\h * */>r < 0 This exercise shows that when H is noncausal, it is very important to distinguish between condition (3) and condition (4). Of course, when H is causal, these conditions are equivalent. Recall that according to (4.2), H: 34?e -*34?e is said to be passive (strictly passive when 3 > 0) when H satisfies (4). Now let H: #e -» 3tf. When H satisfies (3) with 3 = 0 (3 > 0, resp.), we say that H is 34? positive (strictly 34? positive, resp.). We say 34? positive to emphasize the fact that 3^ positivity is tested over 34?, whereas passivity is tested over 34?e. Lemma (2) says that when H: 3tf -» Jf and is causal, H is 34? positive (strictly J-f positive) if and only if H is passive (strictly passive, resp.). Remark In the case of convolution operators, condition (3) is easily tested by Fourier methods. Condition (4) is the one required for the passivity theorem. Therefore, in certain manipulations that follow, we shall have to assure ourselves that the operators under consideration are causal. For future reference, we must define adjoint operators. Let H be linear and map 34? into 34?. The map //* defined by (x\Hy}
= ,
Vx,y e 34?
is called the adjoint of H. If H = H*, H is said to be self-adjoint. Exercise 3 Use the bilinearity of the map (x, y) H-> <X | y} to show that if H, H*: 2tf -» #P are related by (7), then H and H* are necessarily linear maps. Exercise 4 Show that J and PT are self-adjoint. Show that
Lemma Let H: 34? -* 34? be linear. U.t.c, H is anticausal if and only if its adjoint H* is causal. Proof
=> For all x, y e 34?, for all Te R, since H is anticausal < x | ( / - PT)Hy> = < x | ( J - P T )//(J - P r ) j >
VI
202
PASSIVITY
hence by (3.9) and the definition of the adjoint,
= <(/ - PT)H*(I -PT)x\
y>
Since y is arbitrary and Jf is a Hilbert space, PT H*(I - PT)x = 0, and by linearity pTH*
=PTH*PT
<= Retrace backward the steps of the proof above.
9.2
::
Pattern of the noncausai multiplier technique
Keeping in mind the difficulty caused by noncausai operators, we can outline the general technique as follows. 9
10
Assertion Let j f be a Hilbert space over the field U. Assume that there is a multiplier M, i.e., a noncausai map M: 3^ -> Jf. Suppose M can be factored so that M= M_M+,
M_ , M+ mapping Jf into 3/e
where M""1, MZ1 and M+ 1 are well-defined mappings from 2tf into J-f; 11
M_ is linear, hence M_* and (M_*) _ 1 well defined;
12
M, M+ , M_*, and their inverses have finite gain. U.t.c, the L2 stability of any one of the three feedback systems S, SM, and SM+ (shown in Fig. VI.8) implies the L2 stability of the others.
Figure VI.8
9
MULTIPLIER
203
THEORY
Exercise 5 Prove Assertion (9). [Use (12).] In most applications, H1 and H2 are causal. Since M is noncausal, MH^ and H2 M are noncausal. If M+ is also causal, the operators of the feedback system SM+ , namely, M+ / ^ ( M . * ) - 1 and M_* H2M^i, are causal. Hence in that case Lemma (2) is available to establish passivity. The following lemma establishes an important connection between the operators MH and M+ / / ( M _ * ) - 1 . 15
Lemma Let H: tf -* tf and M satisfy (10) to (12). U.t.c, the following statements are equivalent:
16
(i)
3S> Os.t. (u\MHu}>3\\u\\2,
17
(ii)
3<5'>Os.t.
where 3 and 3' are both positive or both zero. Proof
Let x and u belong to Jf and be related by x = M_*u; hence by (12)
||x||
= (M_*u\M+Hu}
= (x\M+
77(M_*)_1x>
The equivalence follows.
::
From these considerations, we can state the following theorem. 20
Theorem Consider the system S of Fig. YI.8, where H1 and H2 are causal maps from Jf into jtf. Suppose that for all uu u2 e J^, the solutions eu e2e^e. Let there be a noncausal multiplier M: satisfying (10), (11), and (12), and the condition that M+ , (M+)_1,
21
M_*, (M_*) _ 1
are causal
U.t.c, if 22
(i)
23
(ii)
24
(iii)
y(MHi) < oo for some 3 > 0
then S is L2 stable. Proof The assumptions of the passivity theorem hold for the system SM+ . Indeed,
where we used (22) and (12).
204
VI
PASSIVITY
Applying Lemma (15) to (23) and (24) and noting the causality of the operators, we conclude that
Hence SM+ is L2 stable by (5.27), and so is S by the L2 stability equivalence established in Assertion (9). :: 27
Remark Note that the operators MH1 and H2M~1 3tf positive; they may or may not be passive.
are only assumed to be
Clearly, the key step in the procedure is the factorization of the multiplier M. Before developing the factorization theory, let us consider some examples. Note that in applications it is sufficient to know that a particular multiplier M can be factored in the manner required by Assertion (9); it is not necessary to calculate the factors M_ and M+. The factorization theorem (62) (which is given at the end of this section) will give sufficient conditions under which the factors M+ and M_ exist and have the required properties.
9.3
Examples
Example 1 Consider a feedback system of the form shown in Fig. III. 1 and where 28
where 0 : IR -»M, <j> is f\ i.e., 30
[(j)(a) - 0(j8)](a ~ ^) > 0,
Va, fi e U
and for some k < oo,
Theorem Consider a feedback system with H1 and H2 satisfying (28) to (29). Suppose that there is a noncausal convolution operator M whose impulse response is of the form
32
m(t) = 1 - z(t) = 1 - X zt d(t - td - zj®
+ 00
— 00
where 33
9
MULTIPLIER
THEORY
205
34 and that there is a S > 0 such that 35 U.t.c, if either z f l (-)>0
36
and
zt > 0,
Vi
or (/>(•) is an odd function
37
then the system is L stable. 38
Remark It follows immediately from (32) to (34) that (a) M is a linear map; (b) the incremental gain of M satisfies y{M) < 2; (c) M is invertible, thus M'1 : L2(U) -»L2((R), and, in fact, K ^ - 1 )
<(i - ii^rir1).
Proof The factorization theorem (62) establishes that the multiplier M can be factored as required. As a consequence, Theorem (31) is established if it is shown that y{MH^) < oo, that the noncausal operator MHX is strictly 3tf positive [condition (23)], and that H2 M " 1 is #? positive [condition (24)]. Now (23) holds because by Parseval's theorem we have
The 3tf positivity of H2M~X is a consequence of Lemmas (40) and (46). Hence, once the factorization is justified, the proof of Theorem (31) will be complete. :: 40
Lemma If 0 : U -» U is / and belongs to some sector [0, k], k < oo, then, for all x(-) e L2(U) and all TEU,
^ -oo
If , in addition, is an odd function, then 42
206
VI
PASSIVITY
Proof Note that x e L2(U) implies that 0 W 0 1 e L2(R) and <&[*(•)] e L\U), where by definition d$)/dx =
Letting a = x(f), j8 = x(t + T), and integrating over [R, we obtain
Hence (41) follows. Now when 0 is odd, O is even, and we have, in addition to (41),
since O is even Hence (42) follows. 46
::
Lemma Let 0 : U H+ IR, 0 /*, and 0 e sector [0, k], for some k < co. Let M satisfy (32) and (33). U.t.c, if either
47
z a (0 > 0,
48
Vf e U, or
and
z, > 0, V/
0 is odd
then (24) holds. Proof For convenience, let Z: x h-> z * x, x e L2(IR), where z(0 = £ Zi8(t - tt) 4- za(t). Then M = / - Z. Now y(M) < 2 and y(M _ 1 ) < oo. Let y = Mx\ then *; e l3(U) if and only if x e L2((R). We obtain successively
9
MULTIPLIER
THEORY
207
and upon expansion the right-hand side becomes [all integrals are over (-00,
GO)]
Now by invoking Lemma (40) and either (47) or (48), we obtain
- \\z\\)jx(t)
Vi;eL2([R)
where the last inequality follows from (34) and the sector condition.
::
Example 2 Consider a system similar to that described in Theorem (31). Let H1 be defined by (28); let 4>: U -• U and have its slope restricted as follows. For some /? > 0, e > 0, and some oceU,
If the Nyquist diagram of g(jco) does not encircle the point ( — 1/a, 0) (in the sense of Section IV.5), then the present system can be reduced to the case of Theorem (31) by simple loop transformations. Step 1 Apply a feedforward of — a/ around H2 and a feedback of — ocl around H1. The new nonlinearity is / / 2 — a/; its characteristic is and belongs to sector
[0, /? — s]
The new linear subsystem is H±(l + a / ^ ) " 1 . It is a convolution operator whose impulse response is in s/ since g e srf and because the Nyquist diagram does not encircle (—1/a, 0). Ste/? 2 Apply positive feeedback of (1//?)J around the nonlinearity and a positive feedforward of (l/P)I around the linear subsystem; the resulting nonlinearity is
Note that the inverse is well defined because the nonlinear characteristic of P~1{H2 — ocl) is strictly increasing and its slope is less than (/? — e)/P < 1. The new linear subsystem is
208
VI
PASSIVITY
Finally, noting that the constant factors /? and f$ * may be canceled out without affecting the L2 stability of the system, we end up with
H2=(H2-aI)[(a
+
p)I-H2]-1
and R1 = [I + (a + P)HX](I + ocH.y1 The characteristic of H2 is strictly increasing and is in the sector [0, /?/e]. The impulse response of Ht is in s/. Hence we have the same setup as in Theorem (31).
9.4
Factorization by logarithms
Lemma Let ^f be a Hilbert space. Let M be a linear continuous map of J4? into #?. U.t.c, if M is Jf positive and M~1 is a well-defined linear continuous map of #? into «?f, then so is log M. It is convenient to think in terms of Banach algebras. The class [34f] of all linear continuous maps from the Hilbert space ^f into itself constitutes an algebra; addition and scalar multiplication of such maps are defined in the usual manner. The multiplication is taken to be the composition. If, in addition, we choose the norm of such a map as the norm induced by the norm of Jf, then we have ||/||=1
and
\\M1M2\\<\\M1\\ \\M2\\
where J is the identity map and Ml9 M2 any pair of continuous linear maps from 2tf into #t. Since tf is complete, the normed algebra [jf ] is complete; thus it is a Banach algebra. Proof Let J be the identity map on Jff;\cts e C. By definition, the resolvent of M is the map s \-> (si — My1 from some subset of C into j ^ ; the domain of the resolvent, p(M), is an open set. Since M'1 is a well-defined continuous linear map from ff into Jf, (si — M ) " 1 is well defined in some open disk B(0; r) centered on the origin. On the other hand, since M is continuous, the spectrum of M, c(M) [i.e., the set C - p(M)] is in the disk 5(0; ||M||). Since M is linear, ^f positive, and continuous, (si — M)'1 is well defined for all Re s < 0. Hence we can find a simple closed rectifiable curve T which does not touch the negative real axis (including the origin) and which encloses the spectrum of M. Let D be the domain enclosed by T. In D and on T, the function s h-> log s is analytic; hence log M: I}(U) ->L2([R) is given by the Dunford-Taylor integral [Dun.l] log M = (1/2TT/) f log s(sl
-My1
ds
ss
9
MULTIPLIER
THEORY
209
Exercise Show that (si — M) i is a well-defined continuous linear map on JUT for \s\ > \\M\\. [Hint: Consider (si - M)" 1 = (\/s)(I - M/s)~x = (\/s)^M"/s\] Let $ be the class of convolution operators M such that M: uv-+ m*u, where 00
56
m(t) = X mt S(t - tt) + ma(i) — 00
with (m^el1, ma(-) e l}(U), and t{ e U, Vz. Note that M is not causal in general. M e & maps l3(W) into L2(R), and for all u e L2([R),
J
If we denote the bracket by ||M||, it is easy to show that the function || • ||, thus defined, is norm on &. So Me J* is a linear continuous map. If M, Ne &, then the maps M + N, aM (Va e C), MN, and NM are in ^ . By Fubini's theorem, MN = NM, and ||MiV|| < ||M|| ||N||. Since the identity J e J , we conclude that 58
& is a commutative Banach algebra with a unit. Let P+ be a projection on J 1 (hence P+ P+ = P+)defined as follows: Since Mu = m * w, we have
59
Define P+ M by
60 P + M is a causal convolution operator from l3(U) into L2(IR). Let 61 P_ Mis an anticausal (or purely anticipative) convolution operator from L2(R) into L2(U). Let ^ c (^fl) be in the class of all causal (anticausal) continuous operators onL 2 (R). &c and £%a are closed under addition, multiplication (i.e., composition of functions), and multiplication by scalars. Thus, £%c and &a are subalgebras of ^ . If P+M and P_M have finite norms, then P+M and P _ M belong to @c and @a, resp. Exercise Show that if A, B e @c, then AB = BAe^c, integers m, n e1+ .
AmBn e @c for any
VI
210
PASSIVITY
Exercise Let A, B e UnXn. Show that if A and B commute (i.e., AB = BA), then exp A • exp B = exp B • exp A = txp(A -f B). Show, by an example, that if AB =£ BA, the two equalities above fail in general. 62
Factorization Theorem Let & be the Banach algebra of continuous convolution operators defined in (56). Let M be an operator in &. If M is #P positive and if M~1 is in ^ , then with ||P + log MII and II P_ log M \\ finite (a)
63
M can be factored as M = M_ M+ , with
M_ e J f l ,
M+ e @c
(b) M;1e^candMI1eJffl. Proof Since L2(lR) is a Hilbert space, M is Jf7 positive, and M _ 1 e ^ , by Lemma (55), log Mis a well-defined element of Jf; soareP+ log M,P_ log M. Let M+ ^ exp(P + log Af);
M_ = exp(P_ log M)
Since the exponential function is analytic everywhere in C, M + and M_ -are continuous convolution operators in &c and ^ a , resp. The same holds for their inverses because 64
M'1 = e x p ( - P + log M);
M I 1 - e x p ( - P _ log M)
Now since M+ , M_ e ^ , and ^ is a commutative algebra, 65
M_ M+ = exp(P+ log M) • exp(P_ log M) = exp[(P + + P_) log M] = exp(log M) = M Similarly, we have M+ M_ = M. Finally, M+ 1 e Mc because P+ log M e &c and 00
67
Remark The preceding proof requires & to be a commutative algebra. Therefore, this factorization technique fails (except for special cases) in the case of convolution operators from Ln2(U) into Ln2(U), because their kernels are matrices, and the product of two matrices is, in general, not commutative. For this reason, this factorization technique fails in the multi-input-multi-output case.
9
9.5
MULTIPLIER
THEORY
211
Factorization by projection
The factorization theorem in the following does not require that the Banach algebra be commutative. 68
Factorization Theorem Let J* be a Banach algebra with identity I. Let 11 + be a projection on £8 i.e., I l + : J* -> J> is linear and 11+ 11+ = I1+). Let the projection n _ be defined by
69 where J denotes the identity map on J*. It is assumed that 70
l|n+||
and
||I1_|| < 1
Let ^ + and ^ _ be the ranges of 11+ and Il_, resp. It is assumed that $+ and J*_ are subalgebras (i.e., are closed under multiplication). Let M e l U.t.c, if there exists a Z e J such that 71
\\Z\\ < 1 and
72
M =I - Z then (i) M can be factored as follows
73
M = I - Z = M_M+,
with
M_ , M+ e J*
where M_ and M+ are invertible in J*;
where / © ^ + (/ © ^ _ ) denotes the subspace of all elements in ^ of the form a/ + M, where a e R , A f e ^ + ( ^ _ , resp.). 76
Comments (I) The Banach algebra $ is not assumed to be commutative. This is important for the multivariable case. (II) From (71) and (72), we have ||Af|| < 1 + ||Z|| < 2. If some given element N is to be factored and has a norm larger or equal to 2, then a scale factor k > 1 may be introduced, and the given element N may be written as N = k(I - Z). (Ill) There are two extremely important conditions, namely, (70) [because projections in normed spaces do not necessarily obey (70)] and the condition that J*+ and J*_ be subalgebras. The proof of the factorization theorem (68) uses the following lemma.
212
79
VI
PASSIVITY
Lemma LetCQ^ 00 , ( P ^ 0 0 , and (N^ °° be sequences in ^ , J*+ , and J*_ , resp. Suppose that for \r\
converge. [Consequently, Q, P, A/" are analytic functions of r in B(0; rc), taking values in &, I © ^ + , and J © ^ _ , resp.] U.t.c, if these functions are related by Q = PN, then the sequence (Qfc)j °° c= J* determines uniquely the sequences (P*),00 c J*+ and ( A ^ 0 0 c <#_ . Proof In Q = PiV, replace each function by its power series; then using a well-known theorem of analytic functions, we equate coefficients and obtain
whence
and
Qk=Pk + Nk + kfpk.iNi,
£ = 1,2,...
whence Pk = U+(Qk-Y
Pk-iNd,
£ = 1,2,...
1=1
and a similar relation for Nk. Thus, the Pfc's and Nks are successively determined uniquely in terms of the Qks. :: Proof of the Factorization Theorem (68) Step I
Consider for \r\ < 1, the following equations
84
P = / + ( P ) ^ / + rn+(ZP)
85
N =f-(N)
= / + rU_(NZ)
Note that to simplify notations, we wrote P for P(r) and N for N(r). The functions/+ and/_ are contractions on £8\ indeed, for any Pl9 P 2 e &, 86
\\MPt) -f+(P2)\\
= \\rTV + [Z{P1 -P2)]\\ < \r\ \\Z\\ \\Pt -P2\\
9
MULTIPLIER
THEORY
213
where in the first step we used the linearity of 11+ and the distributive law in £8\ in the second step we used (70). Since \r\ • ||Z|| < 1,/+ is a contraction on £%. A similar proof holds for/_ . Thus, P and N can be calculated by iteration starting with P0 = N0 = I; thus, again using linearity and the distributive law, 00
00
8
with Pk + 1^ Il+(ZPk)9
# * + ! = n_(ATfcZ), k = 0, 1, 2, . . . . Also, for all £, . The solutions P and JV are analytic functions in P(0; ||Z|| _ 1 ). Also
88
| | P | | < ( 1 - |H lizil)" 1 and similarly for ||iV||. Step II
From Eq. (84) and (85), recalling that 11+ + EL = J, we obtain
89
(/ - rZ)P = I -
rU_(ZP)
90
N(I -rZ)
rU+(NZ)
=I-
Since by assumption \\Z\\ < 1, for \r\ < ||Z|| _ 1 , (I — rZ)" 1 e J* and is given the power series £ o Zkrk convergent in P(0; \\Z\\ ~*). Now, using (70) and (88),
the same bound holds for ||rII + (iVZ)||. Therefore, the elements I — r n _ ( Z P ) and / - rU+(NZ) of @ are invertible in J* for \r\ < \\Z\\_1/2. Their inverses are given by power series convergent in P(0; ||Z|| - 1 /2). The inverses take values in / © J L and / © J* + . Hence we have from (89) and (90) 91 Since, in these two equalities, the five power series involved have, as common domain of convergence, the disk P(0; \\Z\\~x/2)7 the uniqueness guaranteed by Lemma (79) gives and 92 Now in (92), (J — rZ)" 1 , P, and TV are represented by power series convergent for |r| < ||Z|| _ 1 ; hence setting r = 1, we obtain 93 with M_ = iV"1, M+ = P - 1 . All the properties of M_ and M+ have been established in the course of the derivation. ::
214
VI
PASSIVITY
94 Application Consider now the application of this factorization theorem to the following special case: Let s#2 be the Banach algebra defined in Section V. 9.2; namely, stf2 consists of generalized functions / ( • ) of the form 00
where tteU with *. < 0 ( = 0, > 0 resp.) for / < 0 ( = 0, > 0 resp.);
Let s/%x n denote the set of n x n matrices whose elements belong to sf2 • If F(-) of the form OO
96
F(t)=
I i=
Ft5(t-t,)
+ FM
— oo
is an element of s$\x ", we define its norm by ^
97 where | • | is any induced norm on Un x n. Finally, we define the product of two elements in $0n2*n to be their convolution. Under these conditions, one can easily verify that s/2xn is a Banach algebra, that / • 3(t) is the identity element of stfn2xn, and that stfn2xn is not a commutative algebra unless n = 1. Let stfnxn be the subset of jtf2xn consisting of elements F(-) that satisfy 98
Ft = 0 V/ < 0,
Fa(t) = 0 W < 0
Then s4nxn is a subalgebra of $4n2xn and contains the identity element. Similarly, let $4nxn be the subset of stfn2xn consisting of elements F(-) that satisfy 99
Ft = 0 V/ > 0, nxn
Fa(t) = 0 W > 0
Then stf is also a subalgebra of srfn2xn, and it too contains the identity element. If we think of stfn2xn as the set consisting of the impulse responses of stable «-input-«-output systems, then stfnxn consists of all causal impulse responses in stfn2xn, while s$nxn consists of all anticausal impulse responses in s/tt2xn. Next, let n + : st™"->s/n2xn and n _ : $4n2xn -* srf\xn be the mappings defined by ( n + F)(i) = £ « 0 Fs 8{t - tt) + Fa + (t) and ( n _ 70(0 = L ^ - c o + Fa„(t) Fid{t-t^
10
PASSIVITY
VERSUS
SMALL GAIN
THEOREMS
215
where
It is routine to verify that 17+ maps J ^ * " into s#nxn, that IT+2 = 11 + (so that TI+ is indeed a projection), and that stfnxn is the range of U+ . Similarly, n _ maps stf\xn i n to sfn_xn, XlJ = IT_ , and s/n_xn is the direct sum of the identity operator and the range of II _ . Also, in view of the definition (97) of the norm on s/^", it is clear that
so that
Thus all the hypotheses needed to apply the factorization theorem (68) are satisfied. Hence we have the following 101
Theorem Let Me stf\xn, and suppose M = 1 - Z where \\Z\\^2 < 1. Then M can be expressed in the form
102
M=M_*M+ where M_ , M+ are invertible in srfn2xn,
103 104 Remark In factoring by logarithms, as in (62), the projections P+ and P_ could have norms larger than 1, but the Banach algebra had to be commutative. In (101) the projections II + and II _ must have norms less than or equal to 1, but the Banach algebra need not be commutative.
10
Relation between the passivity theorem and the small gain theorem
We have derived a form of the Popov criterion by applying the small gain theorem (see V.8), and also by applying the passivity theorem (see VI.6.1). This suggests that these two theorems are related. It should be stressed that they are not equivalent. Indeed, referring to the general formulation of Sections II 1.1 and 111.2, we see that the small gain theorem requires only a normed space structure ( L / , L„2, L„°°, . . . ) , whereas the passivity theorem
216
VI
PASSIVITY
requires an inner-product space structure (usually, a Hilbert space, with L2 as the prototype). We shall see below that further assumptions are required in order that the " equivalence " hold. The intuitive idea behind the relation between passivity and small gain is supplied by linear circuit theory. Indeed, consider a linear time-invariant «-port and let it be described by its impedance matrix Z. If v and i represent the Laplace transforms of its port voltages and currents, we have 1
v=Zi The «-port may also be described by its scattering matrix S, which relates the incident wave a to the reflected (scattered) wave 5:
2
b — Sa where, using the standard normalization of the 1-ohm transmission line,
3
v = b + a,
i=a— b
Hence 5 = (Z-/)(/+Z)-1
4
It is also physically obvious that if the linear time-invariant ft-port is passive (hence Z is a passive operator; positive semidefinite, as they say in circuit theory), then the reflected wave b carries an amount of energy no larger than the incident wave a, and conversely. Thus, 5
Z is passive <=> y(S) < 1 This suggestion from linear time-invariant n-povt theory extends to the general framework of Section 3. It is made precise in the following lemma.
7
Lemma Let H: Jfe -> Jf e. Assume that
8 is well defined. Defined: 9
eby S
= (H-I)(I
+ H)-1
U.t.c, (a)
H is passive with the constant /? = 0, i.e.,
10 if and only if S has a gain < 1, i.e., 11 (b) 12
H has a finite gain y with the additive constant /? = 0, i.e.,
10
PASSIVITY
VERSUS
SMALL GAIN
THEOREMS
217
and H is strictly passive, i.e., there is a 5 > 0 s.t. 13 if and only if S has a gain smaller than 1, i.e., for some y' e (0, 1), 14 Exercise 1 Since, in (4), Z is a linear operator, show that S — (I 4- Z ) - 1 (Z — /). [Hint: Calculate the difference between this expression and that of (4); exhibit the places where linearity is used.] Show, by example, that in the nonlinear case, (H - /)(/ + H)~l ^ (/ + H)~\H - I). 15
Comment The lemma establishes that if (8) holds, then whenever the "impedance" operator H is passive, then the corresponding "scattering" operator S has a gain < 1 , and conversely; also whenever the "impedance" operator H has finite gain and is strictly passive, then the corresponding "scattering" operator S has a gain smaller than 1. In contrast to the usual derivations found in circuit theory texts, the "impedance" operator and the "scattering" operator are neither assumed to be linear nor time invariant. Before proving lemma (7), let us establish a few identities. Given any y e 2tfe, define x e Jf e by
16 As a consequence, 17
Sy = (H-I)x
and
y = (I + H)x
Hence 18 and 19 Thus, for all ye34fe, for all T e
> y'\\y\\T2 = y'[\\Hx\\T2 + ||x|| r 2 + <2x|//x> r ]
218
VI
PASSIVITY
Since y' e (0, 1), we obtain where y" = y'/(4 — 2y') > 0. So (14) implies that H is strictly passive in the sense of (13). Eliminate y from (14) by (18) and (19) and rearrange terms to obtain yf\\Hx\\T2 < (4 - 2y')\\Hx\\T \\x\\T -
y'\\x\\T2
The finite-gain condition (12) follows if we choose y = (4 — 2y')/y'.
::
Exercise 2 Suppose H is given as a black box; devise block diagrams for S. (Hint: S = 2H(I + H)'1 - I;S = (H - I)QI) • [ / + ( / / - / ) ( i / ) _ 1 ; watch out that in the nonlinear case, H • (21) ^ 2IH.) Exercise 3 To appreciate the importance of assumption (8), construct an example where H: Jf e -» 3tfe has finite gain and is strictly passive but where I+H does not have an inverse. (Hint: Take &~ = {0}, 2tf = U; recall Chapter I.) Exercise 4 Show that if (8) holds and if H is incrementally passive, then S has an incremental gain < 1. To exhibit the relation between the small gain theorem and the passivity theorem, we consider a system $f described by u1=e1+
H2 e2
u2=e2-
H1el
where ut, u2 , eA, e2 e 3tfe and Hx, H2 map 2tfe into itself. We apply successively the loop transformation theorem (II.I.6.3). These transformations are described as follows: Forward subsystem H\ Apply feedback with gain — 1;
Feedback subsystem Hi Apply feedforward with gain — 1:
Multiply output by 2:
Multiply input by \\
Apply feedforward with gain — 1:
Apply feedback with gain — 1:
11
INVERTIBILITY
OF
I+H
219
It can be checked that in the process, the inputs have been changed to ux — u2 and ux + u2, respectively. Thus, we obtain the system S? shown in Fig. VI.9 The loop transformation theorem (111.6.3) yields the conclusion that, provided
Figure VI.9
(I + H^'1 and (I + H2)~1 are well-defined maps from 34?e into Jf e , the system 9 described by (24) is $f stable if and only if & is stable. We make now two observations: (a) If H1 satisfies (10) (is passive) and if H2 satisfies (12) (finite gain) and (13) (strictly passive), then, by Lemma (7), y{Sx) < 1 and y(S2) < (1 — y') < 1. So by the small gain theorem (III.2.1), 9 is Jf stable; so is Sf by virtue of the loop transformations. (b) If y(St) < 1 and y(S2) < 1 - / < 1, then H1 satisfies (10) and H2 satisfies (12) and (13); therefore, by the passivity theorem (VI.5.10), Sf is Jf stable; so is & by virtue of the loop transformations. Note that we have not shown that the small gain theorem and the passivity theorem are equivalent; i.e., whenever the ^f stability of a system can be established by one of the theorems, it can be established by the other. In the first place, the passivity theorem requires an inner-product space structure, whereas the small gain theorem requires only a normed space structure. It would seem that (/ + Hx)~l and (/ + H2YX be well defined maps from J-fe into Jf e is also required. However, if (/ + H^'1 and (/ + H2)~1 are interpreted as the inverse relation ofI+Hl and I + H2, then the conclusion still holds. In the next section, we consider the problem of finding conditions under which I + H has an inverse.
11
Invertibility of
I+H
Throughout our study of feedback systems, we encounter equations of the form u = (I + H)e If J + H is known to be invertible in some space, say, Jf, the conclusion follows that for any u e Jf, the equation has a unique solution in Jf. Note also that in order to legitimately apply the loop transformation theorem, we
VI
220
PASSIVITY
need to know that a map of the form (/ + KH^) is invertible. We give below two theorems that guarantee the invertibility of such maps. Both theorems require an inner-product space structure. 1
2 3
Theorem Let 2tf be a Hilbert space. Let F: ^fe^J^e. We assume that F is incrementally passive and has finite incremental gain; more precisely, there is a 3 > 0 and a y < oo s.t. %\\x-x'\\T2, Vx,x'eJfe, \\Fx — Fx'\\T < y\\x — x'\\T
(Fx-Fx'\x-x'}T>
V J e 3~
Under these conditions, (i) F'\s a bijection from 3tfe onto Jf c ; equivalently, F _ 1 is a well-defined map from tffe onto tfe\ (ii) F'1 has a finite incremental gain: yCF 1 ) ^ l/?j (iii) F " 1 satisfies the condition 4
< F - 1 z - F~xz'\z - z'}T > 5\\F~lz -
F-^'WT2,
VZ, Z'
e jfe,
VTe F
Exercise 1 Show that (2) and (3) imply that y > 5. Exercise 2 Show that (3) implies that F is nonanticipative, and (4) implies that F'1 is nonanticipative. We could express the idea of (3) by saying that for all f e « f , the causal map F: $?e -> ffle is globally, uniformly (in T), Lipschitz continuous in PT Jf e , with the Lipschitz constant y. Similarly, conclusion (ii) says that for all Te^T, the causal map F"1: Jfe->Jfe is globally, uniformly, Lipschitz continuous in PTJife, with the Lipschitz constant 1/5. Proof (i) From Exercise 1, y > 5. Pick any z e <#?e and a nonzero scalar a; then the equation z = Fx is equivalent to 5
x = az + x — OLFX = f(x,
z)
If we set a equal to a 0 = 5/y2, we claim that/defined in (5) is a contraction. Indeed, Vre«T, Vx, x' e J^e 6
||/(x, z) - / ( * ' , z)\\T2 = ||x - x' - a(Fx - 7V)|| r 2 = \\x — x'\\T2 — 2a<x — x'\Fx — Fx'}T + a2\\Fx-Fx'\\T2 < l l * - * ' l l r 2 [ l -2ocS + (x2y2] where we used (2) and (3). Minimizing the right-hand side of (6) with respect to a, we see that the minimum k02 = 1 — (52/y2) is reached when a =S/y2.
11
IIMVERTIBILITY
OF
I+H
221
Thus, k0, the contraction constant, is <1 and independent of T. Consequently, given any z e 3tfe, there is one and only one x e J^e such that z = Fx; equivalently, F _ 1 is a well-defined function mapping Jfe onto Jf\ (ii) From (2) \\Fx - Fx'\\T > S\\x - x'\\T, with z = Fx and x = F~1z, we have ||F _ 1 z - F _ V | | T < (l/<5)||z - z'|| T ,
7
Finally, (iii) is immediate from (2), by replacing x and x' by F~xz and F ~ V , resp. s: 8
Corollary Let F be a nonanticipative linear map from ^fe -• 3tfe which has finite gain. U.t.c, if Fis strictly passive, then F _ 1 is a nonanticipative, linear, strictly passive map from 3tfe into Jf e s.t.
Vze,?f e ,
\fTeF
By linearity, assumptions (2) and (3) reduce to those of the corollary. •• ••
The following theorem shows that the requirement of finite incremental gain in Theorem (1) can be replaced by a continuity requirement. 10
Theorem Let J-f be a Hilbert space and PT 3tfe be complete for all T e &~. Let F map &?e into J f e . For each 7'e &', let PTF be a continuous map of PTJ^e into PTM?e. U.t.c, if F i s strictly incrementally passive, i.e., there is a 3 > 0 s.t.
11
(Fx - Fx'\x - x'}T>d\\x
- x'\\T2,
Vx,i'e/e,
VFG^
then (a) F is a bijection of Jfe onto 34?e, and (b) F _ 1 has an incremental gain smaller or equal to <5_1. 12
Comments In feedback systems, Theorem (10) is used as follows. Suppose that H: 34?e^3*?e and that, for all T, PTH: PT3tfe -+PT3tfe is continuous. U.t.c, (a) if H is incrementally passive, then (/ + H) is a bijection of Jf e onto 3tfe, and (I + H)'1 has an incremental gain smaller or equal to 1; (b) if H is strictly incrementally passive (with constant SH), then (/ + H) is a bijection of Jf e onto J-fe , and (J + H)~l has an incremental gain smaller or equal to (1 + 3H)~l.
222
VI
PASSIVITY
Exercise 3 Give an example of a map H: Jf e -> Jf\ s.t. H has finite incremental gain, H: PT3tfe ->PT3fe is continuous, and is strictly passive (and not incrementally passive), but which does not have an inverse. [Hint: Take Jf e = R9 2T = {0} so H is given by its graph in U2. Choose H so that for some y e R, (I + H)x = y has several solutions; hence (J + / / ) _ 1 can at best be a relation, not a function. Use Chapter I.] Proof
From (11) and Schwarz's inequality,
Hence we have 12a which shows that Fis injective (one to one) on J-fe. Therefore, F defined function on &(F), the range of F, and, from (12a),
1
is a well-
13 Thus, F - 1 has finite incremental gain on ^ ( F ) . It is easy to show that the continuity of F and (13) imply that ${PTF) is closed (see Exercise 4, in the following.) It remains to show that 01{F) = J"fe. To simplify the calculations below, let S = 1 in (11) and define H: Jfe - Jf e by 14
Fx^(I
+ H)x
or
F^I+H
Thus, H is continuous on PT J4fe, and (11) with S = 1 becomes 15
r>0,
V;c,x'eJf e ,
VFG^
i.e., / / is incrementally passive. Let S: &(F) -> ffle be defined by 16
S £ (H - I)(I + Hy1 so if y and x are related by y = Fx, we have
17
y=Fx<=>j; = x + Hx o Sy = Hx — x Since H is incrementally passive, then S: 31(F) -> J*fe has an incremental gain < 1; see Exercise 4, Section 10. Thus, we have
20
\\Sy - Sy'\\T2 <\\y - / | | r 2 ,
Vy, / e ^ ( F ) ,
VFe^
Now it is well known that [Min. 1] a map Fr*S from the closed set <%(PTF) c Pr 2tfe into PT 34?e and which satisfies a global Lipschitz condition like (20) an can be extended to a map S: PT^e -*PT^\ d still preserve (20). Define the map B: J^e-^^fe defined by B: y i-> i(y - Sy). Now on ^(F),
11
21
INVERTIBILITY
OF
I+H
223
B = ±(J - S) = ±[J - (if - /)(/ + J/)" 1 ] = i[(/ + / / ) - ( # - /)](/ + i / ) - 1 = (I + H)~1 =F~X In other words, FB = I on &(F). We are going to show that B is the inverse Fon all of «#%, i.e., that m{F) = jf c . Pick any j / e ^ and define
22
3c 4 By We claim that Fx = y. Equivalently, FBy = y for any y e 3tfe\ i.e., 5 is the inverse of F o n all of J-fe. The proof is by contradiction. Suppose that for the chosen y e 3tfe there is some l e J such that
23
||j;-Fx||r>0 Since H is continuous on PT jf?e, there is an r\ > 0 such that
24
| JC - x\\T > Y\ => ||//x - / / x | | r < (i)||j; - Fx|| r Choosex 0 = x + X(y — Fx)with/i > 0 and small enough so that ||x0 — x|| r < rj. Given that particular x0, define y0 by y0 = Fx0 =x0 + Hx0 Therefore, we have y0 e 01{F) and By0 = x 0 . We obtain successively, using (21) and expanding the scalar products,
25
(By0 - By) | (By0 - By)}T
where the last inequality follows by (20). Now by By0 = x 0 and by (22), (25) becomes 27
(y0 - y - (x 0 - x) | x 0 - x> T > 0 N o w on the one hand, y 0 = Fx0 = (J + H)x0, and by the choice of x 0 , x0 — x = X(y — Fx) = X(y — x — Hx) (27), then becomes, after noting that X > 0 (Hx0 — y +x\y — x — Hx}T > 0 Adding and subtracting Hx and noting that Fx = x + Hx, we obtain
-Fx)\y-
Fx}T > 0
or equivalently, 30
(Hx-Hx\y-Fx}T>
\\y - Fx\\T2
224
VI
PASSIVITY
Finally, by Schwarz's inequality and (24), (30) becomes
This shows that the supposition (23) leads to a contradiction. Hence B 4 i ( i - S); tfe -»tfe is the inverse of F = I + H on tfe and ^ ( F ) = tfe. Thus, F = (I + H) is a bijection of Jf e onto 2tfe. Part (b) of the theorem follows from the above and (13). :: Exercise 4 Show that PTF continuous and (13) imply that &(PTF) is a closed set in PTJ^e. [Hint: Consider a convergent sequence in &(PTF); use (13) and the continuity of PTF].
12
Instability theorems
In this section, we derive an instability counterpart to the Popov criterion of Section 6. As in Section Y.9.1, the present results are based on an orthogonal decomposition of L2, and we make use of some of the theorems from Section Y.9.1. We begin with a general result that illustrates the method. 1
Theorem Consider a feedback system represented by the equations
2
ex = ul — He2
3
e2 = u2 + Gex n
where el9 e2, ul9 u2: U+ ->U ; G, H: L%e(U+) -^L 2 e (R + ), and suppose that corresponding to each ul9 u2 in Ln2(U+), there exists a unique pair (el9 e2) in L2e([R+) x Lle(U+) such that (2) and (3) are satisfied. Suppose further that 4
G is causal and linear, and satisfies assumptions (V,9.6) and (V.9.7)
5
He =
0=>e=0
There is some S e U such that 6 where M is defined by Eq. (V.9.5), and eelR such that 7 U.t.c, if M is not all of L2, and if 8
(5 + s > 0 2
then yx = Gex $ L whenever ux = 0, u2 # 0, and u2 e M1.
12
INSTABILITY
THEOREMS
225
Proof [As in Theorem (V.9.10) if M is not all of L2, then M1 is a nontrivial subspace of L2.] Let ux = 0, u2 ¥" 0, and u2e M1. Suppose by way of contradiction that Ge1 e L2. Then this implies that e1 e M. Now we have
since ux = 0 and u2 e M1. But, on the other hand, we get from (6) and (7) that 10
ie^Ge^r
+ <e2\He2}T > S\\eir\\2 + s\\(He2)T\\2 = (6 + s)\\elT\\2
Since S + e > 0, (9) and (10) can be simultaneously satisfied only if et = 0. But this implies, on the one hand, that Get = 0 and, on the other hand, that He2 = 0, whence e2 = 0 by (5). Finally, since ex = Ge1 + u2, the conditions e = i 0> e2 = 0? a n d w2 7^ 0 lead to a contradiction. Hence Ge1 $ L2. Next, we consider the instability counterpart to Theorem (6.5). 11
Theorem Consider a system described by (2) and (3), where uu u2, eu e2: R+ -> U. The subsystem G is linear and causal and is represented by
12
(Ge)(t) =
(g*e)(t)±y(t)
where 13
g(-) is absolutely continuous. [This implies that g(') e L e \ that is g measurable except possibly for an impulse at t = 0, and that gT(-)e srf for all T < oo.] Further, g(-) is Laplace transformable; g(-) can be analytically continued to the region {s: Re s > 0}; and g(-) is meromorphic in this region. Also, we assume that
14
the maps co *->
#0'CD)
and co v-+jojg(joS) are bounded on R.
The subsystem H is represented by 15
(He)(t) = $[e(t)] where : R -> R is continuous and (f> e sector [0, /:]. Suppose finally that corresponding to each wl5 w2 in L2, there exists a pair (e l9 e2) in L e 2 x L e 2 satisfying (2) and (3). U.t.c, if g(-) has at least one pole in the region {s: Re s > 0}, if there exist real numbers q > 0 and r > 0 such that
16
Re[(1 + jqoj)g(jco)] + l/k > r > 0, and if
17 then the system under study is L2 unstable.
226
Proof
VI
PASSIVITY
If uY = 0, the feedback system can be modified as shown in Fig. VI. 10
Figure VI.10
With this transformation, the system equations can be rewritten as
18
z(t) = [(g + qg) * ^](r) + u2(0 = i>(0 + " 2 (0
19 20 Let H1 and H2 be the mappings defined by Hl: et H-> W and 7/ 2 : Z K — ^ . Then we know from Example 6 that the operator H2 satisfies (7) with s = l/k. The operator Hx is a convolution operator with kernel hx = g + qg. In view of the assumptions on g and #, we see that the operator H1 satisfies conditions (V.9.6) and (V.9.7). Hence the preimage of L2 under Hl9 say Mu is a closed subspace of L2. Also, since g has a singularity in the region {s: Re s > 0}, we see that M\ is a proper subspace of L2. Now it is easy to verify by ParsevaPs equality that if *(•) and [hx * x(-)] belong to L2, then 21 Making use of the causality of H1 and invoking Lemma (9.2), we conclude from (21) that
So if (16) holds, then the mapping H1 satisfies (6) with 3 = r — l/k. Finally, we have 3 4- & = r > 0. Thus, we see from Theorem (1) that there exists a u2 e L2 such that the corresponding y $1}. For this w2, let w2 = «^ _ 1 {[V0 + ^)]^ 2 (^)}- Then w2 e ^ 2 but the corresponding y <£ L2, and hence the system is L2 unstable. s:
Notes and references Theorem (1.3) was established in Youla et al. [You.l] for the linear case. The formal framework of Section 3 is to be found in Sandberg [San.8], Zames [Zam.3], and Willems [Wil.5]. In Section 4, Examples 1, 2, and 3 are
NOTES
A N D REFERENCES
227
well known in the circuit theory literature. Example 5 is based on an observation of Cho and Narendra [Cho.l]. Examples 6 and 7 can be found in Sundareshan and Thathachar [Sun.l ]. The passivity theorems were developed by Sandberg [San.8], Zames [Zam.3], and Cho and Narendra [Cho.l]. The first treatment of the Popov criterion, which considered distributed systems, is found in Desoer [Des.3], and its generalization to the multivariable case is due to Lee and Desoer [Lee.l]. The discrete-time case development follows [Kri.l]. Theorem (8.8) is a sharpened version of the well-known paper of Zames and Freedman [Zam.6]; the improvement is due to Ramarajan and Thathachar [Ram.l]. The multiplier theory has been applied by many researchers. The idea of using noncausal multipliers is due to O'Shea [O'Sh.l]. The factorization technique using logarithms is originally due to Zames and Falb [Zam.5] and is found in its present form in Sundareshan and Thathachar [Sun.3]. The example involving monotone nonlinearities is due to Zames and Falb [Zam.5]. The second factorization technique can be found in Willems [Wil.5]. The application to multivariable systems using L2 norms is due to Vidyasagar. The relationship between the passivity theorem and the small gain theorem pervades the circuit theory literature in the linear case. The development of Section 10 is based on Anderson [And.4]. Theorem (11.1) is based on Sandberg [San.ll], and Theorem (11.10) on Browder [Bro.6]. The extension theorem used in the proof of (10.10) is in Minty [Min.l]. The instability results of Section 12 are due to Takeda and Bergen [Tak.l].
A
INTEGRALS AND SERIES
A.1
Regulated functions
In system theory there is one particularly useful class of piecewise continuous functions, namely, the regulated functions. We define them below and state some of their properties. 1
Let (is,|| • ||) be a complete normed linear space, i.e., a Banach space. Let I be an interval in U. The mapping/: I -* E is said to be a step function iff there exists a finite partition of i, i.e., k
I = {j[Xi,xi+1]
with
Xi<xi
+ l9
Vf,
i= 0
such that / is constant on each open interval (xi9 xi+1). x0= - o o , xk= oo.) 2
(Note: If I = R,
Let I be an interval in U and E be a Banach space. A function/: J -> is is said to be regulated iff, for all x e I,/has one-sided limits. More precisely, let the closure of I be [a, /?]; then, for all x e / distinct from /?, for h -> 0, 229
APPENDIX
230
3
\imf(x
A
+ A) 4 / ( * + )
INTEGRALS
exists
AND
SERIES
(hence \\f(x +)|| is finite)
h>0
and for all x' el distinct from a, for h-+0, 4
\imf(x'
— h) = f{x' —)
exists
(hence \\fix — )|| is finite)
h>0
5
Theorem A function / : J -> E is regulated if and only if, over all compact subintervals of /, it is the limit of a uniformly convergent sequence of step functions. (Recall that an interval is compact if and only if it is closed and bounded.) It is immediate that iff: I -> E is continuous, then it is regulated; also if / : / -* R is monotonic, then it is regulated.
6
Corollary Iff is regulated, then it can only have a denumerable number of discontinuities. Example: / : (0, 1) -> R where 11-» sgn[sin(s1/r)].
7
Corollary If / : J -> E is regulated, then / is bounded on every compact subinterval of I.
8
Corollary Let f: / -* Et be regulated functions, where each Et is a Banach space, / = 1, 2, . . . , n. Let g: f|"= 1Ei-* E (where F is a Banach space) be continuous, then the composite function
t^g[f1(t),f2(t),...,L(t)] is regulated from J to F. An immediate consequence of Corollary (8) is that if/and g are regulated functions from J to M. so are t \-+f(t) + git), t y^f(t)g{t\ and t t->f(t)/g(t), where for the last one it is required that git) ^ 0, W e I. Exercise 1 Give examples to show the following: Suppose that/: J -» R and 0:/(J)->R.U.t.c, (i) iff and g are regulated, then g °fis not necessarily regulated; (ii) if / is continuous and g regulated, then g of is not necessarily regulated. 9
10
The indefinite integral of a regulated function / is easily defined. Let / : [a, b] -» E, then, by definition,
A.1
REGULATED
FUNCTIONS
231
where the / v 's are step functions, which converge uniformly to / over (a, i). 11
Clearly, (/> has a derivative everywhere, except at the points of discontinuity of/, but at such points, say t', <j> has a well-defined right derivative and left derivative:
12
and
13
A function / : [a, b] -» R is said to be absolutely continuous on [a, b] iff, for any s > 0, there is a (5 > 0 such that
£ = 1
for every finite collection of subintervals (<xf, j8f) of [a, b] with
14
Special Cases (I) By choosing the collection above to consist of one subinterval, we see that if/is absolutely continuous on [a, b], then it is uniformly continuous on [a, b], (II) If / o b e y s a Lipschitz condition on [a, b], then it is absolutely continuous on [a, b]. (III) If/is absolutely continuous on [a, b]9 then it is of bounded variation on [a, b]. (IV) If / : [a, b]-+U is the integral of a regulated function, then / is absolutely continuous. (V) There are classical examples of monotonically increasing functions that are continuous but not absolutely continuous. The importance of absolutely continuous functions lies in the following theorem [McS.l, p. 208].
1 5 Theorem If/: [a, b] -> U is absolutely continuous, then (a) its derivative f\x) exists and is finite almost everywhere, and (b) if we set/(A;) =f'ix), where/' is defined, and zero elsewhere, then we have
f(x)=ff(Odt+f(a),
Vxe[a,b]
Conversely, if/ is of the form
/ ( * ) = fg(Odl; +/(*), J
Vxe[a,b]
a
where g is integrable on [a, b], then / is absolutely continuous on [a, b].
APPENDIX
232
A.2
A
INTEGRALS
AND
SERIES
Integrals
We state in the following a number of properties of the Lebesgue integral that are used in the main development. For readers acquainted with Lebesgue integration, this will be a review. For readers who have not gone through a detailed development of the Lebesgue integral, they need only know that (1) The Lebesgue integral is an extension of the usual Riemann integral; the class of Lebesgue integrable functions includes the class of Riemann integrable functions (which includes continuous functions, piecewise-continuous functions, regulated functions etc.). (2) The analytical properties of Lebesgue integrable functions are simpler than those of Riemann integrable functions. The logic of the main development will not fail if such readers take the properties in the following as axiomatic. One of the most important analytical properties of the Lebesgue integral is the following theorem. 1
Dominated Convergence Theorem [Ros. 1, p. 76] Let g be integrable on an interval J finite or infinite. Let {/n}0°° De a sequence of functions on / such that Vw, \fn{x)\
2
Holder's Inequality [Roy. 1, p. 95] Let/, g: U-+U. Let p and q be nonnegative extended real numbers (i.e., 1
(b) When p = 2, the Holder inequality becomes the Schwartz inequality. 3
Minkowski Inequality
[Roy. 1, p.5] For p e [1, oo], /, g e LP implies that
4
Fact Let / : U -> U and 1 < p < oo. If fe LP and if / i s uniformly continuous on IR, then \f(t)\ ->0 as \t\ -> oo. (Note t h a t / e L 0 0 =>/is uniformly continuous.)
5
Fact If/: IR -> IR and fe L1 n L00, then for pe[l,ao],fe
LP.
A.2
6
INTEGRALS
233
Fact Let / : [a, b] -* U, where oo < a < b < oo. If for some pe[l, feLp[a, b], then feL1 [a, b].
oo],
Give an example to show that the converse is not true. [McS.l, p. 227] If/: U -> U a n d / e L1, then
7
Fact
8
Tonelli's Theorem [McS.l, p. 145] Let m and n be positive integers. Let / : Um x IR" -» R be locally integrable on Rm x DT. If (a) for all x e Rm, except for a set E of measure zero, the function y * |/(*> J7) I is integrable over IRn, (b) the iterated integral l-
then (i) (x, y) *->f(x, y)
is integrable over
Um x Un
(ii) 9
Remarks (1) The absolute value sign in assumption (b) is essential; see counterexample [McS.l, p. 144]. (2) Roughly speaking, the theorem says that if one iterated integral of the absolute value of/exists, then it is equal to the double integral o f / By symmetry, the double integral is equal to each of the iterated integrals. (3) In applications, one establishes via Tonelli's theorem the existence of the double integral by studying one of the iterated integrals, i.e., by establishing inequality (b).
Indefinite integrals 10
Theorem Let I be an interval (finite or infinite) of M. A function F: I -> U is the indefinite integral of some integrable function/if and only if F is absolutely continuous. Also, F ==/a.e., on L Exercise 1 If fe L1, then
exists (and is finite).
234
APPENDIX
A.3
A
INTEGRALS AND
SERIES
Series
In our work we use sequences n i—• xn, where x „ e i o r Un or Rnx", etc. So it is simpler to think in terms of sequences which take values in a normed space E. The norm in E is denoted by | • |. In most cases, E will be complete, i.e., a Banach space. Given a sequence (xn)o (i.e., the function n H-» xn, from Z + to E), we define the nth partial sum sn by 1 sn = x0 + xt + • • • + xn 2
We say that the series Zo xn converges to s iff lim sn = s. (This is convergence in terms of the norm of E.)
3
For the case where the normed space E is complete, we say that the series YJS xn converges absolutely if and only if the series of positive real numbers Zo \xn\ converges. To discuss the convolution of sequences, we need a theorem analogous to Tonelli's. Instead of functions of two variables, we have double sequences.
4
Theorem Let (m, n) — i > amn be a double sequence in E (i.e., a function mapping Z + x Z+ into E). If oo
Z | amn |
converges for
n = 0, 1, 2, . . .
m= 0
and
then 00
0 0 / 0 0
\
0 0 / 0 0
\
The two expressions on the right are called repeated series. 5
Theorem (Holder's inequality for sequences) Let x = (x1? x2, .. .)> J = (Ji? ^2» • • •) be sequences in E, i.e., x, j ; : Z + -• .E. Let/?, ^r e [1, oo] with
p'1 +q~x = 1. Ifxelpandyelq,
then
(i.e., the scalar product xTy = Zo xkyk converges absolutely in £), and 00
6
X \xkyk\
< \\x\\p\\y\\q
k= 0
For convolution of sequences, see Appendix C. 4.
B
FOURIER
TRANSFORMS
We collect in the following some basic results that are frequently used in our work.
B.1 L Theory 1
Theorem If/: U-+R a n d / e L1, and iff(jco) j
(a) (JO n->/0"co) = jt™f(t)e~
4
t
dt, then
dt is uniformly continuous on U,
(b)
(c) (d) m a x | / ( » | < ll/Hx coeK
(e) /(*) =
(1/2TT)
(Riemann-Lebesgue)
J? „, f(j(jo)ej0it dco
almost everywhere in U.
(f) if, in addition, / is of bounded variation in a neighborhood of t, then 2
* ( / ( , + 0) -f{t
- 0)] = (1/2*) lim f +°J(Jo))e"" dco 235
APPENDIX B
236
B.2 1
FOURIER TRANSFORMS
L2 Theory
Theorem Let / : R -> R and fe L2. If
2
N = 1, 2, . . . then (a) as (b) (!l/ll 2 ) 2 = (l/27i)(||/|| 2 ) 2 (c) as
3
Remarks (a) The convergence in (a) and (c) is convergence with respect to the L2 norm. It is not pointwise convergence (e.g., the Gibbs phenomenon). (b) The Fourier transform maps L2 onto L2, and the L2 norms o f / a n d / are equal except for the factor of l/(2n)1/2. (c) fe L1 ¥>fe L1 (e.g.,/(f) = 1 for \t\ < 1 and zero elsewhere). (d) feL2*feL°. (e) feL2j>fel}.
4
Parseval Theorem If/ g e L2, then (a)
f*g=f§ r
(b)
f J
(c) 5
+ oo
- + 00
f(t)g(t) dt = (1/2*)
f*(jco)g(jco) dco J
-00
-00
II/U2 = [(l/27t) 1/2 ]||/|| 2
Theorem (Sandberg) (a) I f / a n d / e L 2 , t h e n / e L 1 . (b) If / a n d / e L2, t h e n / e L1. Proo/ f,feL2
imply t h a t / e L2 and j(of(ja>) e L2. Thus,
::
B.3
LAPLACE
TRANSFORM
237
Exercise 1 If / a n d / e L 2 , show that fe L00 and \f(t) | -> 0 as t -• oo. Consider the integral of//, — ) Exercise 2 I f / e L 2 and f/e L 2 , t h e n / e L 1 .
B.3
Laplace transform
The Laplace transform has simpler properties than the Fourier transform because it considers only functions that are identically zero for t < 0. Except for immediate consequences of the Fourier theory, the most interesting results are Theorems (4) and (5). 1
If/: U+ -• R is locally integrable, its Laplace transform is defined by r00 f(t)e~stdt, s = a+ja> o where the domain of definition of / depends on the convergence of the integral. f(s)=
J
We use .£?(/) to denote the Laplace transform of/. 2
Fact If the integral converges absolutely for s = s0, then s \-^f(s) is analytic for all s such that Re s > Re s0.
3
Fact If feL1(U+) Re s > 0.
[or L2(U+) or L°°(R+)], then s\-+f(s)
is analytic in
In the case of distributions (generalized functions) and of L2 functions, one can give a necessary and sufficient condition as follows. 4
Theorem [Sch.l] Let / b e an analytic function defined on some half-plane Re s >
5
Theorem A function / is the Laplace transform of a function / : U+ -> U w i t h / e L2 if and only if (a) / is analytic for a > 0, (b) for some M < oo (which depends on / ) , ^ + 00
\f(a+jcD)\2
dco<M,
for all
a>0
C
CONVOLUTION
C.1
Introduction
Given two functions u and h mapping M+ into U, we may compute (under some conditions) their convolution u * h which is defined by the integral 1
(u * h)(t) = f h(t - T)W(T) dx = y(t),
for
^> 0
The main reasons for using the convolution as a mathematical representation for linear time-invariant systems are the following: (a) Under very mild conditions, L. Schwartz has proved that any linear time-invariant system is represented by a convolution with a kernel, which is a distribution (generalized function); (b) It allows consideration under the same framework of lumpedparameter systems, distributed systems, and systems with transportation lags [Sch.l, p. 162]. 238
C.2
2
CONVOLUTION
OF
FUNCTIONS
239
The simplest case is where /?(•) is the impulse response of a linear, timeinvariant, nonanticipative system whose transfer function h(s) is a rational function of s with \h(s)\ -•O as \s\ -• oo. Note that u.t.c, (A) h(s) has all its poles in the open left half-plane if and only if he I}. If (A) holds, then (i) h decays exponentially, i.e., for some hm < oo, and some a > 0, \h(t)\ (h)
h(t) = h(o+)d(t) + /?1(0
where hx{-) decays exponentially; /z(0 + ) is possibly zero. Indeed, /?(() + ) = l i m , ^ sh(s). (iii) uel}=>yel}nL?, y e L1; furthermore, j is continuous and 2 y(t) ->0asf-+oo.we L = > j / e L 2 n L00, j e L 2 ; furthermore, j is continuous and j ( 0 -• 0 as t -» oo. (iv) For 1
C.2
Convolution of functions
L1 Theory 1
Theorem
Let u, v, w: U+->U. If u, v, we L1, then
2 3 4
w*w = w*w
and
(u * v) * w = u * (i; * w)
Comment The linear space L1, together with the convolution as a product, constitutes an algebra, because the addition, scalar multiplication, and multiplication obey all the required axioms (see Appendix D). Furthermore, it is a Banach algebra, because \\u * w\\x < \\u\\t • \\w\\t and because L1 is complete. In view of (3), it is a commutative Banach algebra which, however, does not have a unit. Exercise 1 Discuss the system-theoretic implications of Theorem (1). Consider the zero-state response of interconnections of linear time-invariant subsystems whose impulse response is in L1.
APPENDIX
240
C
CONVOLUTION
Proof The key tool in the derivation is Tonelli's theorem. All integrals below are onlR + ; so we omit the limits on integral signs. Let us bound 5
\(w*
u)(t)|
=
f w(t -
T)W(T) dx
< f | w(t -
T)I
|W(T)| dx
where the last integral may take the value oo for some t. Anyway, let us calculate ||w * wld: 6
j\(w* u)(t)| dt < jU\w(t
- x)| |u(x)| rfr] A
Consider the other iterated integral
8
=j\u(x)\[j\w(t-x)\
9
=NlilNli
dt\dx
The inequality (6) is immediate; however, at this stage the right-hand side might be infinite. (7) is simply the iterated integral in reverse order, which is then evaluated by (8) and found to be finite in (9). ByTonelli's theorem, since the iterated integral (7) is finite, so is the integral (6). Therefore, conclusion (2) is established. Conclusion (3) follows from (2) and Tonelli's theorem. :: 10
Corollary Let W: U+ -* UnXn and u: U+ -> Un. If all elements of W and of u are in L1, then
11 where 12 with | u(t) | designating some Un norm of u(t), and | W(t) \ the corresponding induced matrix norm in UnXn. 13
Remark
If w, w e L1, then
(a) w * u need not be bounded (on U+) nor continuous, (b) w * u need not go to zero as t -* oo, (c) w * u = wu is bounded, uniformly continuous, and goes to zero as 0) -> oo, (d) w * u need not be in L1.
C.2
CONVOLUTION
OF
FUNCTIONS
241
Exercise 2 L e t / : M+ -> U a n d / : t \-+ (1 + 0 _ 1 (0~ 1 / 2 - L e t #: ff3+ -> [R with g(f) = / ( l - 0 for f e (0, 1) and zero elsewhere. Show that f*g$ L°°(R+). Let (/)(0 = £^° =2 2~w/(f —ft).Show that g * 0 does not go to zero as t -> oo and that VT > 0, 3f' > T such that | (g * )(*') \ > T.
Lp Theory 14
Theorem
Let w, w: U+ -> (R and/? e [1, oo]. If we II and w eL 1 , then
15 Proof
Consider 1 < p < oo; then we have |(w*w)(0l
<J|H^-T)|
1 / P
|
|W(T)|
• |w(r-T)|17^T
The first factor is in II [by Theorem (1)], and the second is in 13; hence by Holder's inequality,
Note that the last factor is bounded by || w|| \lq. Taking II norms of both sides, ||w*i/|| p <|| W || 1 ^{J[J|H'(f-T)| \u(r)\> dr] dt}1" where we used again Theorem (1). For p = 1, (15) becomes the known inequality (2). For p = oo, (15) is immediate since we can pull the essential supremum of u out of the convolution integral. :: 16
Remarks (I) Inequality (15) of Theorem (14) extends to the matrix-vector case as in Corollary (10). (II) In Theorem (14), w * u need not be bounded, nor continuous nor tend to zero as t -> oo. (III) If w e 1} and u e L2, then we have a sharper bound (II.6.7) ||w * u\\2 < max| w(jco)\ \\u\\2 (O
Furthermore, it can be shown that [Tit. 1, Boc.l] w * u = wu (IV) Inequality (15) is sharp only forp = l,oo, because then \\wWt is the induced norm of the map u — i > w * u from IT into IT (II.6.3).
242
APPENDIX
C
CONVOLUTION
Exercise 3 If w e L1 and weL 00 , then (a) w * u is uniformly continuous. [Hint: Use Fact (A.2.4).] (b) w * u need not go to zero as t-* oo. (Hint: Think of some step response!) Exercise 4 If w e L2 and ue L2, then (a) w * w = wu e L1; (b) w * w G L00, w * u is uniformly continuous, and (w * w)(0 -• 0 as f-> 00.
Exercise 5 Let us impose more restrictions on w. Let w el} n L°; hence w e L p for 1
C.3
Convolution of a measure and a function
Models of physical systems involve often impulse responses which include impulses. So instead of considering its impulse response w, we think in terms of its step response fi. For example, if w consists of a sum of a locally integrate function wx and impulses 1
w{i) = w,(t) + f; ockS(t - tk),
0
o
then the corresponding \i defined by 2
pit) = f' w(f) d? includes steps of size ak at tk9 k = 0, 1, . . . . The upper limit is designated f + because we require that \i be continuous from the right. The function [i defined by (2) is a special case of a measure on U + . In particular, the measure of the
C.4
CONVOLUTION
OF
SEQUENCES
243
interval (a, b] is given by fi(b) — 11(a). In the standard manner [Roy.l], the integral of the measurable function/: U+ -> U is defined and denoted by
Let where the supremum is taken over all/: IR+ -* U which are continuous, have compact support and such that sup r > 0 |/(0| < 1The measure \i is said to be bounded iff \\i\ < oo. Exercise 1 In case /i is defined by (2) in terms of w given by (1), show that „00
00
J
0
0
The convolution of a function g: U+ -> U with the measure fi is denoted by (fi * g) and is defined by
For example, for the measure fi defined by (1) and (2),
where k is the largest integer such that tj < t. The following result can be proven (Die.2). 6
Theorem (a)
Let \i be a bounded measure on U + , If p e [1, oo ] and g e LP(U+), then /x e g is defined almost everywhere
and 7 (b) If g is continuous, then \i * g is defined everywhere and is continuous. (c) If g is continuous and -»0 as t -» oo, then so does /* * #.
C.4
Convolution of sequences
Let a = (a0, au a2, ...) and b = (b0, b1, b2, ...) be sequences of real numbers, i.e., functions from Z+ to R. The convolution of a and Z? is defined to be the sequence 1
a*b
= (y0, yu y2, ...)
244
APPENDIX
C
CONVOLUTION
where n
2
y„ = 2 > „ - A ,
n = 0,1,2,...
fc = 0
It is a fact that the convolution of two I1 sequences is an I1 sequence. More precisely, we have the following theorem. 3
Theorem (Apostol)
If a and b are I1 sequences and if
then a * b e ll
5 and, with yn defined by (2),
Since Holder's inequality holds for sequences, we have the following theorem. 7
Theorem Let a and b be sequences and p e [1, oo]. If a e lp and b e I1, then
8 9
Remark The conclusions (5) and (8) still hold if a and b are sequences that take values in UnXn or in any normed space, provided that in the latter case, the product of two elements is defined and that \\ab\\ < \\a\\ \\b\\ for all a, b. Indeed, careful examination of the definitions and of the proofs shows that, with the general definition of convergence in normed spaces (A.2.2) and the properties of the norm, every step of the proof is valid.
D
ALGEBRAS
The purpose of this appendix is to introduce certain general definitions and facts concerning algebras, to discuss in detail the algebras $0 and and, finally, to state same useful inversion theorems.
D.1
Algebras
1
A linear space is, by definition, a set V of elements together with a field IK and two operations; an addition (i.e., a map V x V-> V) is defined on V, and it is associative, commutative, has a zero vector 6, and every v e V has an inverse -v (under addition); a scalar multiplication (i.e., a map V x K-*V) is defined, and it is associative, distributes with respect to addition, and lv = v, Vv e V.
2
If in addition to scalar multiplication we have a multiplication, we obtain an algebra. More precisely, we say that A is an algebra over the field IK iff 245
246
APPENDIX
D
ALGEBRAS
(1) A is a linear space over IK (hence an addition and a scalar multiplication are defined in A); (2) a multiplication, i.e., a map (x, y) \-> xy from A xA-+A,is defined and is associative (xy)z = x(yz) = xyz,
Vx, y, z e A
(3) multiplication distributes over addition x(y + z) = xy + xz
(y + z)x = yx + zx,
Vx, y, ze A
(4) multiplication and scalar multiplication commute (ax)(py) = (ocP)xy,
Va, £ e K,
Vx, j e v4
Notes (a) Another way of expressing conditions (3) and (4) is to say that multiplication is a bilinear map from A x A into A (over the field IK). (b) If the multiplication is commutative, the algebra is said to be commutative. (c) An algebra A need not have a unit element, i.e., an element e such that ex = xe = x, Vx e A. (d) If, in addition, A is a Banach space over IK (i.e., a complete normed linear space over K) and if \\xy\\ < ||x|| \\y\\, Vx, y e A, then A is said to be a Banach algebra. If A has a unit element e, the norm is usually scaled so that
INI = 1. Examples 1. The set of all n x n matrices with elements in IK with the usual operations is an example of a noncommutative algebra with a unit element (the identity matrix). 2. Ditto for the class of all continuous linear maps from E into E, where E is any normed linear space. (Multiplication here is the composition of linear maps.) 3. Let A = I}(U) with the convolution taken to be the product. This is a commutative algebra without unit. [The unit is d(t), which is not an element ofl}(M).] 4 Let s& have elements of the form
where
D.1
ALGEBRAS
247
On stf, addition is defined pointwise, and scalar multiplication by real numbers is defined in the usual manner. The product of two elements/, g e s/ is defined to be their convolution. More precisely, let g be given by (5) and / be given by
Then the convolution/* g is the function
Exercise 1 Use the completeness of L1 and the properties of s/ to show that the second and third terms of (7) are in L1. Finally, prove that if/ g e s/, t h e n / * g = g * / e «s/. By Exercise 1, disclosed under this multiplication, so that s/ is a commutative algebra (over IR) with a unit element, namely, S(t). Define a norm on s/ by 0
Then it is easy to show that
9
\\f*g\h<\\f\U\g\L and
10 Thus, with the norm (8), s/ is a normed algebra which can be shown to be complete [Hil.l]. In other words, s/ is a commutative Banach algebra with a unit. Exercise 2 L e t / e LP((R+), where 1
Exercise 3 L e t / e L 2 (R + ) and w e i . Show t h a t / * w e L 2 and
Generalize to the case where fe Ln2(U+) and w e i " x " . [Use Theorem (II.6.7).]
248
APPENDIX
D
ALGEBRAS
Extensions 11
Let a e IR. Define the algebra s${c) as consisting of elements of the form (5) where now we require that
12 and
The norm in S0(G) is taken to be (abusing notations!) 14 Exercise 4 Show that s&(6) is closed under the convolution defined by (7). Exercise 5 Show that if g e «s/(
Definition Let a e IR and stfn{o) be the set of all ordered n tuples g = (#i, Qi, • • • > 0„) s u c n t n a t 0; e ^(o") for / = 1, 2, . . . , n. Similarly, let <stfnXn(a) be the set of all n x n matrices G = {g{J) (/, j\= I, 2, . . . , n) such that all 9ij G ^ O ) Exercise 6 Show that s$n x "(a) is a noncommutative Banach algebra with a unit element. Exercise 7 Let HestfnXn be viewed as defining a map of Ln°° into itself, namely, H: u\-> H * u. Pick in U" the /°° norm, namely, | x(t) | = maxf | xf(/) |. Show that the induced norm of H is
where h^ is the (/, j) element of H.
D.2 1
Ideals
Given a commutative algebra A, I is said to be an ideal of A iff (a) (b)
/ is a linear space a A;
D.3
2
INVERSES
IN sJ
249
The algebra A and the empty set 0 are ideals of A; they are said to be trivial ideals of A. An ideal / of A is said to be a proper ideal of A iff / ^ 0 and / =£ A; i.e., / is a nonempty proper subset of A. Example 1 Consider the commutative algebra of all n x n diagonal matrices with elements in U. Show that I = set of all diagonal matrices X with xlx = 0 is an ideal. Example 2 I?(IR+) is an ideal of the algebra stf defined above. Example 3 Let A = {fe l!(IR); convolution product}. Show that I = {fe A | f(jcoi)0 = 0,
for some fixed a)0}
is an ideal of A. Example 4 Consider the algebra s/ of Example 4 above. Show that I = {fe stf:f(s0) = 0 for some fixed s 0 with Re s0 > 0} is an ideal of stf. 3
Theorem Let A be a commutative algebra with unit element e\ let / be a proper ideal of A. If x e I, then x has no inverse in A; equivalently, xy ^ e,
Vy e A
Proof (by contradiction) Suppose that some xe I has an inverse x~l; thus x x _ 1 = e. Now for any y e A, x~1y e A and, since / is an ideal, x(x~xy) e I. On the other hand, by associativity, x(x~~ly) = (xx~x)y = ey — y Thus, the product of x with every element of A will cover A. This contradicts the assumption that I was proper. ::
D.3
Inverses in
The main result is due to Hille and Phillips. 1
Theorem
2
Let g e jrf; then g has an inverse in stf if and only if inf \g(s)\ > 0 Res>0
Exercise 1 Suppose that g e L1 c ^ . Show that (2) is never satisfied. (Hint: Use Riemann-Lebesgue.) Proof
=> Call x the inverse of g: # * x = d. Since g, x e <stf, g(s)x(s) = 1
V^ with
Re s > 0
APPENDIX
250
D
ALGEBRAS
Now x e ^ ; s o sup Res > 0 |x(X)| < ||x||^ < GO. Hence \
W with
Re^>0
<= For a complete proof, refer to Hille and Phillips.
"
It is illuminating to observe that if there were an s0 e C with ReCs*0) > 0 such that g(s0) = 0, then g would be in a proper ideal of j?/, hence would not have an inverse in s#. Therefore, to guarantee that g have an inverse in s/9 we must have | g(s) \ > 0, Vs e C with Re s > 0. Similarly, if in Re s > 0, there were a sequence of points sk with \sk\ -> oo with | ^ ( ^ ) | -^ oo, we would contradict the boundedness in Re s > 0 of the Laplace transform of g ~1. Corollary
Let G e j / n X " ; then G has an inverse in &?nXn, if and only if inf|det 6 ( j ) | > 0 Res > 0
Proof Follows immediately from the fact that /, ^ A e i = $£{stf) implies that s i-+f(s)g(s) e s$, and from Cramer's rule. :: Exercise 2 Extend Theorem (1) and Corollary (3) to stf(a) and
jtfnXn(a).
Discrete-time case There is a discrete-time analogue to Theorem (1) above. The setup is as follows. Consider all one-sided sequences, in particular those in l1 = {g = (9o>9i,92> • • • ) k ; e ^ > Z o ) \gt\ - IIgII < oo}. The linear space/ 1 becomes a commutative algebra with a unit if one uses convolution of sequences as the multiplication. By Theorem (C.4.3), I1 is closed under this multiplication. Exercise 3 Verify that with the multiplication, I1 is an algebra over U, according to definition (1.2). If g el1, define OO
0
It is easy to prove the following theorem. Theorem If g e I1, then (a) z h-> g(z) is an analytic function of z for | z \ > 1; (b) 6 i—• g(sjd) is uniformly continuous on [0, In]; (c) \g(z)\ <\\9l V|z| > 1 ; (d) as \z\ ->oo, g(z)->g0. The analogue to Theorem (1) is Wiener's theorem (Wie.l).
D.3
6 7 8
INVERSES
IN sf
251
Theorem Let g = (g0, gl9 g2 , • •.) be a (one-sided) sequence in I1 with g0 ^ 0. U.t.c, g has an inverse in I1 if and only if the curve 6 h-> g(eje),
the image in C of the interval [0, 27r],
does not go through the origin and does not encircle it. The theorem can be stated as: Let gel1; then g has an inverse in I1 if and only if i n f ^ ^ \g(z)\ > 0. The equivalence between (6) and (8) is immediate in view of (d) of Theorem (5) and the principle of the argument. Exercise 4 Instead of the algebra ll, consider the algebra ll(p). Let 0 < p e U; then l1(p)±{g = (g0, 9l9 . . . ) | ^ e R , £ o \gk\p~k< co}. Check that it is a commutative algebra with a unit and write down the inversion theorem corresponding to (6).
E 1
B E L L M A N - G R O N W A L L LEMMA
Lemma
Let
(i) / , g, k; U+ -* U and locally integrable; (ii) g>0, k>0; (iii) geLe*>,; (iv) gk is locally integrable on K + . U.t.c, if w: M+ -> R satisfies 2
w(0 < / ( / ) + g(t) \ k(x)u(x) dx,
V/eR +
then
3 4 5
i/(0
V/e(R +
Let k: U+ -> IR locally integrable on U+ and /: > 0; if w(0 < c +
f £(T)W(T) 252
rfr,
W > 0
APPENDIX E
BELLMAN-GRONWALL
LEMMA
253
then
Proof (The standard proof uses differentiations and an integrating factor k(t) exp[— Jo k(x)g(x) dx].) Let us prove it directly. Consider the integral equation, 7
y(t) = / ( 0 + 5(0 f' k(x)y{x) dx, •'o
y(0) = / ( 0 )
Its solution is 8
y(t) = / ( 0 +
dx
Subtracting (2) and (7) and defining £, we obtain 9
« 0 = y(t) - u(t) > g{t) f' *(t)b(T) - «(T)] dx which is of the form Hence
Now
10 Let 11 Note that ra(-) is continuous and m{t) > 0, for / > 0; so the bracket in (10) is no larger than [m(t)]n~l because k(x)g{x) > 0, Vr by (ii). Hence (10) becomes
254
APPENDIX
E
B ELLM A N - G RO N WALL
LEMMA
For any fixed t, as n -> oo, we obtain 12
C(0 > 0 where we used (iii). Finally, assertion (3) follows from (12), (9), and (8). For the discrete case, we have the following lemma.
13
Let (wk)0°°, (/Jo00*
Lemma
14
hk>0, U.t.c, if
15
£ = 0,1,2,... 0
then
where Proof
when i = k — 1. By direct substitution in (15), using (14), we see that
The proof of (16) is easily obtained by induction. Note that (a) if for some constant hM,hi
(b) if for some constant fM ,f
V/, then
W, then
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Wu.l
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Wu, M. Y., and Desoer, C. A., //-stability (1
INDEX
A «< 246 jtf(o\ 248 j/nxn(o), 248 .p/ 2 , 165 Algebra, 245 commutative, 246 Algebra V, 246 inverse in, 249 Algebra sJ (o-), 248 Algebra/ 1 , 250 inverse in, 251 Anticausal, 200
C
B ys, 209 •&+, 211
^ _ , 211 38a, 209 ^ c , 209
Banach algebra, 22, 239, 246 Banach spaces, 17 Bellman-Gronwall discrete time inequality, 254 Bellman-Gronwall inequality, 252 Bellman-Gronwall lemma, 252-254 Boundedness, 46
C, 11 C, 58 Causal, see Causality Causality, 38 Characteristic, 2 Characteristic equation, 78 Chattering, 49 Circle criterion, 139 Circuit interpretation of feedback system, 37 motivation for passivity, 169 nonlinear, 149 261
INDEX
262
Closed-loop characteristic, 4 derivative of, 7 Closed-loop transfer function exponentially stable, 67 poles and zeros of, 60 Continuous dependence, 5, 44, 184 Continuous linear maps, 22 Contraction mapping theorem, 28, 44, 47,48, 151 Convex function, 29 directional derivative of, 29 set, 18 Convolution definition, 238 feedback systems, 82flf. of functions, 239ff. properties of, 239ff. of sequences, 243 Coprime left, 63 pseudo-right, 87, 105 right, 63 Cross-correlation bounds, 205
pseudo-right coprime, 88, 105 of rational matrices, 62ff. of time-varying gains, 195 of transfer functions, 85, 87, 105 Feedback formula, 54 Feedback system, 37 Finite-gain property, 5, see also Theorems small gain Function absolutely continuous, 231 almost periodic, 96 generalized, 246fT. integration of, 230ff. regulated, 229 G Gain, 41, see also Theorems, small gain incremental, 44 loop, 43 Graphical test, 92, 107 H Holder inequality, 232, 234
D I Divisors common left, 62 common right, 62 greatest common divisor extraction, 74 greatest common left, 62 greatest common right, 62 E Eigenvalues closed loop, 81 Encirclement, 98, 107 Equivalence, 13 Equivalence classes, 16 Existence, 5, 44, 47 of inverse, 219, 249 of periodic solution, 151
Ideal, 248 proper, 249 Incremental passivity application to invertibility, 220, 221 definitions, 173 theorem, 184 Instability, 157fT., 224fT. relation to stability, 53 Instability theorems, 159ff., 164fT., 224 Integrals, 232 double, 233 iterated, 233 Invertibility o f / 4 / / , 219 relation to ideals, 249 L
F Factorization of operators, see Operators, factorization of
Z 1 ,/ 2 ,/ 0 0 , 11 Ll,L2, L00, 12 Lp, 16 LP [a, b], 17
263
INDEX
L„P(M + ) , 18
(E, E\ 20 .(E, E), 22 ^-stability, 52 Linearization, 128, 133 Lipschitz condition, 50 Locally integrable, 16 Logarithmic variation criterion, 194 Loop gain, 43, see also Return difference matrix transformation, 50 M Matrix factorization of, 63ff. polynomial, 58 proper, 58 strictly proper, 58 transfer function exponentially stable, 59 poles, 71, 103 zeros, 71, 103 unimodular, 63 Measure, 242 bounded, 243 examples of, 33 of a matrix, 30 properties of, 31 Minkowski inequality, 232 Multipliers factorization of, 208 for logarithmic variation criterion, 196, 198 pattern of noncausal, 202 with prescribed phase, construction of, 197 N Nonanticipative, 38 Noncausality, 53, 162 Norm, 11 compatible, 23 of convolution maps, 23, 25, 239fT. equivalent, 13 Euclidean, 11 induced, 20
of inverse, 12 of matrix, 13 operator, 21 of sequences, 15 Normed space, 11 O Operators factorization of in commutative algebras, 210 using L 2 -induced norms, 214 using logarithms, 210 in noncommutative algebras, 211 by projections, 211 logarithms of, 210 P P+,P-, PT, 38
209
II+, n_,211 Passivity definitions, 173 examples, 174-180 instability theorems, 224 motivations from circuit theory, 169 with Popov-type multipliers, 176, 177 relationship to positivity, 200 to small gain, 215 stability theorems, 181, 182 with time-varying gains, 178, 179 Perturbation effect on output, 54-55, 123 effect on stability, 116ff. Poles closed loop vs. open loop, 69 of / + EG(s), 71, 72 of/-I- EG(z), 103 Popov's criterion for stability for discrete-time case, 191 for multivariable case, 189 for single-input case, 186ff. by small gain methods, 155 Popov's criterion for instability, 224 Positivity, 174 relationship to passivity, 174, 200
I N DEX
264
R U[s], 57 Unxn[s], 58 U (s), 57 [R n x n 0), 58 Ray, 18 Relation, 6 Return difference matrix, 70, 102 Riemann-Lebesgue lemma, 235 S Scattering operators, 216 Schwartz inequality, 232 Sector, 2 Sector conditions, 2, 5 equivalence, 2, 3 Series, 234 convergence, 234 Set absorbing, 18 balanced, 18 convex, 18 equilibre, 18 Slope relations, 7 Slowly varying systems, 125, 147, 194 Space Banach, 17 extended, 38 normed, 11 Spectral radius of integral operators, 117 of linear map, 29 of matrix, 27 Stability exponential, 67 global uniform, 126 J2\ 52 L 00 , 108, 115, 121, 143 L p , 112, 115, 121, 137, 147 L 2 , 139 L 1 , 115, 121
l\ 125 Systems autonomous, 154 linear convolution, 82ff. with rational transfer functions, 57 linearization of, 128, 133 linear time-varying, 108ff. T Theorems boundedness, 46 circle, 139 dominated convergence, 232 existence and uniqueness, 48 incremental passivity, 184 incremental small gain, 44 loop transformation, 50 Parseval's 236 passivity, 181, 182 Popov, see Popov criterion small gain, 40 relation to passivity theorem, 215 Tonelli, 233 Transfer function factorization of, 85, 87, 105 poles, 69-72, 103 zeros, 70-72, 103 Truncation, 38 U Uniqueness, 5, 34, 44, 118, 184 Z z-transforms, lOOff. Zeros of I+F G(s), 71, 72 of I+FG(z), 103 of a matrix, 70
