BLOCK DIAGRAM REPRESENTATION OPEN LOOP PLANT INPUT
x(t ) X (s)
G (s )
OUTPUT
y (t ) Y ( s)
Gc R
Y ( s) = G ( s) × ...
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BLOCK DIAGRAM REPRESENTATION OPEN LOOP PLANT INPUT
x(t ) X (s)
G (s )
OUTPUT
y (t ) Y ( s)
Gc R
Y ( s) = G ( s) × X ( s) Y (s) G ( s) = X (s)
CONTROLLER
GP
X
PLANT
Error, E = R − Y = 0 is desired
Y
X = RGC Y = XGP ⇒ Y = RGC GP Now , E = R − RGC GP = 0 ⇒ R(1 − GC GP ) = 0 Q R ≠ 0, 1 − GC GP = 0 1 GC = GP
CLOSED LOOP CONTROLLER
R
E ─
C
PLANT
U
G1
MEASUREMENT
H
Y = UG1 U = EC E = R − YH
Y CG1 = R 1+ CG1H
Denominator : 1 + CG1 H = 1 + GH { G
Y
1 + GH → Characteristic polynomial 1 + GH = 0 → Characteristic equation (CE) Y G = R 1 + GH If CE is known, we can determine, • order of system • eigen values • stability • response characteristics
EXAMPLE Applying Conservation of Angular Momentum h& = τ − mgl sin θ
h = ml θ&
(
h& = ml 2θ&&
⇒
2
h = Iθ&
ml 2θ&& + mgl sin θ = τ 1 mgl & & θ + 2 sin θ = 2 τ ml ml
τ l
θ
)
m
ml 2θ&& = τ − mgl sin θ Need to linearize • Operating point • Static equilibrium point
θ = θ (t )
g 1 & & θ + sin θ = 2 τ l ml
τ = τ (t ) θ& = θ&& = 0
Static Equilibrium,
θ 0 ,τ 0
l
1 sin θ 0 = τ = τ 0 0 2 gml gml Assume τ 0 = 0 ⇒ ⇒
180o
sin θ 0 = 0
θ 0 = 0 or 180
© 2005 P. S. Shiakolas
o
o
0o
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
LINEARIZATION Æ Taylor’s Series Expansion
(
g 1 & & θ + sin θ − 2 τ = 0 = f θ&&, g , l , m,τ , θ l ml
(
)
)
∂f && && ∂f ∂f ( ( − θ0 + − l0 ) f − f0 = g − g0 ) + l θ 424 3 ∂l 123 ∂θ&& 0 ∂g 0 1 0 Δg =0 Δl =0 ∂f ∂f ∂f (1 ( ( + m − m0 ) + τ −τ 0 ) + θ − θ0 ) 3 ∂m ∂m 0 424 ∂θ 0 0 Δm = 0 ⎛ 1 ⎞ ⎛g ⎞ & & Δf = 0 = 1Δθ + ⎜⎜ − ⎟⎟ Δτ + ⎜ cos θ ⎟ Δθ ⎝l ⎠0 ⎝ ml 2 ⎠0 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
⎛ 1 ⎞ ⎛g ⎞ & & Δθ + ⎜ cos θ ⎟ Δθ = ⎜⎜ ⎟⎟ Δτ ⎠0 ⎝l ⎝ ml 2 ⎠ 0 If the equilibrium point is θ = 0o ⇒ 1 ⎛g ⎞ & & Δθ + ⎜ cos 0 ⎟Δθ = Δτ ⎝l ⎠ ml 2 g 1 & & Δθ + Δθ = Δτ l ml 2
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
If the equilibrium point is θ = 180o ⇒ 1 ⎞ ⎛g & & Δτ Δθ + ⎜ cos180 ⎟Δθ = ⎠ ⎝l ml 2 1 g & & Δθ − Δθ = Δτ l ml 2
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
LAPLACE TRANSFORM (Assume Zero ICs) g 1 & & Δθ + Δθ = Δτ ⇒ Laplace Transform 2 l ml
g 1 ΔΘ( s ) s + ΔΘ( s ) = ΔΓ l ml 2 2
1 ⎛ 2 g⎞ ⎜ s + ⎟ΔΘ = 2 ΔΓ l⎠ ⎝ ml ⎛ ⎞ ΔΘ 1 ⎜ 1 ⎟ ⎜ ⎟ = ΔΓ ml 2 ⎜ s 2 + g ⎟ ⎜ ⎟ l ⎠ ⎝ © 2005 P. S. Shiakolas
PLANT
ΔΘ
ΔΓ
or
⎛ ⎞ 1 ⎜ 1 ⎟ ⎜ ⎟ΔΓ ΔΘ = ml 2 ⎜⎜ s 2 + g ⎟⎟ l ⎠ ⎝
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
g s + = 0 ⇒ characteristic equation l g Im eigen values ⇒ s = ± i l 2
C.E. for θ 0 = 180o ⇒
Unstable region
g 2 g s − =0 ⇒ s=± l l If the eigen values have • positive real parts, then the system is unstable • negative real parts, then the system is stable • zero real part, then the system is marginally stable © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Re
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
g 1 & & Δθ + Δθ = Δτ l ml 2
State space ⇒ representation
Set of first order DFQs z1 = Δθ z 2 = z&1 = Δθ& ⇒ z&1 = z 2 Number of DFQs = Order of system g 1 z&2 + z1 = Δτ 2 l ml g 1 z&2 = − z1 + Δτ 2 l ml © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
z&1 = 0 z1 + 1z2 + 0Δτ g 1 Δτ z&2 = − z1 + 0 z2 + 2 l ml x& = Ax + Bu where ⎛ z&1 ⎞ ⎡ 0 ⎜ ⎟=⎢ g ⎜ z& ⎟ ⎢− ⎝ 2⎠ ⎣ l Output : © 2005 P. S. Shiakolas
A = coefficient matrix − square B = order of the system
1⎤ ⎛ z ⎞ ⎛ 0 ⎞ 1 ⎥ ⎜ ⎟ + ⎜ 1 ⎟Δτ 0⎥ ⎜ z ⎟ ⎜⎜ 2 ⎟⎟ ⎦ ⎝ 2 ⎠ ⎝ ml ⎠ y = Cx + Du ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C.E. ⇒ det (SI − A) = 0 ⎡s det ⎢ ⎢0 ⎢⎣ s det g l
0 − g s − l
0
1⎤ ⎥=0 0⎥ ⎥⎦
−1 s
=0
g s + =0 l 2
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
PROJECT PENDULUM
g
mc
θ 0 = 180o
F
l
x
θ
© 2005 P. S. Shiakolas
mr
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
MAGLEV
electro magnet
g
sensor
object mass, m L
R
Magnet
V
I
Fmagnet
mass
Fmagnet = f (k , I , x )
weight © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
• Derive system dynamic equations • Linearize, if needed • Derive state space representation • Find transfer function • Find characteristic equation
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
LAPLACE TRANSFORMS Final Value Theorem (FVT)
f (∞) ← lim s F ( s ) s →0 Condition: All the poles of F(s) other than s = 0 must have negative non - zero real parts. Poles: Are points at which the f(t) or its derivatives → infinity Zeros: Are points at which the f(t) equals zero © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
( s + 2 )(s + 10 ) G ( s) = s(s + 1)(s + 7 )(s + 17 )2 2 zeros, 5 poles Im
Re -17
© 2005 P. S. Shiakolas
-10
-7
-2 -1 0
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
TRANSFER FUNCTIONS (T.F.) L[output ] T.F. = G ( s ) = L[input ] Zero I.C.'s Y ( s ) bm s m + bm −1s m −1 + ... + b1s + b0 = , n≥m X (s) s n + an −1s n −1 + ... + a1s + a0 n → Order of the system • Applicable only for Linear Time Invariant (LTI) Systems • Independent of input (magnitude and time) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
BLOCK DIAGRAMS Output
Input
a1 a2
Description of component or subsystem
Σ
Branching point
b1
─
B
A
Signal
a3
Summing JCT: a1 + a2 − a3 = b1 © 2005 P. S. Shiakolas
C
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
D E
R
─
C
P
G H
PRINCIPLE OF SUPERPOSITION (LTI)
C = C R, D = 0 + R
─
CR ─
P
G H
D P
G H
GP CR = R 1 + GPH © 2005 P. S. Shiakolas
C D, R = 0
P CD = D 1 + PHG
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CD
GP P C= R+ D 1 + GPH 1 + GPH 1 + GPH → Characteristic polynomial
STATE SPACE REPRESENTATION System of linear DFQ’s Æ System of First Order linear DFQ’s
x
k m
f (t )
b © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
m&x& + bx& + kx = f (t )
2 nd order linear DFQ ⇒ 2 first order linear DFQ in state space State space variables : z1, z2 z1 = x
z2 = z&1 = x& z&2 = &z&1 = &x& m z&2 + bz2 + kz1 = f 1 z&2 = (− kz1 − bz2 + f ) m & z&1 = z2
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
x& = Ax + Bu y = Cx + Du Y X
( x, y ) 2 variables l 2 = x 2 + y 2 Constraint Equation
l
θ
Only one variable for system state x& = A x + Bu L ⇒ sIX − x0 = AX − BU
(sI − A)X = BU + x0 X = (sI − A)−1{BU + x0 } © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
y = C x + Du L ⇒ Y = CX + DU
[
]
Y = C (sI − A)−1{BU + x0 } + DU
{
}
Y = C (sI − A)−1 B + D U + C (sI − A)−1 x0
Denominator of the above eq. sI − A Therefore, sI − A is the Characteristic Polynomial & sI − A = 0 is the Characteristic Equation of the system. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
• Time Step for Numerical Integration • Dynamics Æ Differential Equations ¾ Linearize ¾ State Space • Block Diagram Æ State Space form
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
INTEGRATOR
x&
1
x1
x
s
x2
1 x1 = x2 s
x1 = x&2 State variables at output of integrators
1 © 2005 P. S. Shiakolas
f ( s)
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Example: U
10 s ( s + 2)
─
Z
1 s +1
U
1 s
─ X3
© 2005 P. S. Shiakolas
X2
10 s ( s + 2)
1 s +1 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
X1
Z
Y FF CE : = U 1 + FF × FB 10 1 Denominator : 1 + =0 s ( s + 2) s + 1
(s 2 + 2s )(s + 1) + 10 = 0 ← 3rd Order
10 X2 = X 1 ⇒ 10 x2 = x&1 + 2 x1 s+2 1 X1 x1 = x&3 + x3 = X3 ⇒ s +1 1 (U − X 3 ) = X 2 ⇒ u − x3 = x&2 s © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
x&1 = −2 x1 + 10 x2 + 0 x3 + 0u x&2 = 0 x1 + 0 x2 − 1x3 + 1u x&3 = 1x1 + 0 x2 − 1x3 + 0u ⎛ x&1 ⎞ ⎡− 2 ⎜ ⎟ ⎢ ⎜ x& ⎟ = ⎢ 0 ⎜ 2⎟ ⎢ ⎜ x& ⎟ ⎢ 1 ⎝ 3⎠ ⎣
10 0 0
0 ⎤ ⎛ x1 ⎞ ⎡0⎤ ⎥⎜ ⎟ ⎢ ⎥ − 1⎥ ⎜ x2 ⎟ + ⎢1⎥ u ⎥⎜ ⎟ ⎢ ⎥ 1 ⎥⎦ ⎜⎝ x3 ⎟⎠ ⎢⎣0⎥⎦
Z = 1x1 + 0 x2 + 0 x3 + 0u Y = [1 © 2005 P. S. Shiakolas
0
0] x + 0u ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Transfer Function has zeros or derivatives of input
Y f1 ( s ) = U f 2 (s) X s+a = U s 2 + 2ξωn s + ωn2 &x& + 2ξωn x& + ωn2 x = u& + au Converting into state space z1 = x z2 = x& = z&1 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
z&1 = z2 z&2 = −ωn2 z1 − 2ξωn z2 + au + u&
U
X
TF
X X Z TF = = U ZU
U
© 2005 P. S. Shiakolas
1 DEN
Z
NUM
X
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
State space representation w.r.t. variable z
X s+a = U s 2 + 2ξωn s + ωn2 From the block diagram, Z X 1 = = s+a & 2 2 U s + 2ξωn s + ωn Z Z ⇒ &z& + 2ξωn z& + ωn2 z = u U Converting into state space, φ =z φ = φ& = z& 1
© 2005 P. S. Shiakolas
2
2
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
φ&1 = 0φ1 + 1φ2 + 0u φ&2 = −ωn2φ1 − 2ξωnφ2 − 1u ⎛ φ&1 ⎞ ⎡ 0 ⎜ ⎟=⎢ ⎜ φ& ⎟ ⎢− ω 2 ⎝ 2⎠ ⎣ n
⎤ ⎛ φ1 ⎞ ⎛ 0 ⎞ ⎥⎜ ⎟+ ⎜ ⎟u − 2ξωn ⎥⎦ ⎜⎝ φ2 ⎟⎠ ⎜⎝ − 1⎟⎠ 1
X ⇒ x = z& + az Z x = aφ1 + 1φ2 + 0u © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
FEEDBACK CONTROL SYSTEM CHARACTERISTICS
STATE SPACE
BLOCK DIAGRAM
TRANSFER FUNCTION
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
PLANT
CONTROLLER
R
E ─
C
U
Y
G1
MEASUREMENT
H
F = EG
R
Φ
X GP = R 1 + GPH
─
CE :1 + GPH = 0
Find Φ such that the block diagrams are equivalent.
© 2005 P. S. Shiakolas
Unity Feedback
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
X
EFFECTS OF FEEDBACK • Affects • closed loop behavior • stability • bandwidth • overall gain • Reduces sensitivity of the output to disturbances and system/plant parameter changes. • Improve the steady state error • Ease control of transient response © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
GENERAL PROPERTIES OF F.B.C.S. T R
E ─
P
G H
GP P X= R+ T 1 + GPH 1 + GPH CE :1 + GPH = 0 E = Reference Signal − Output Signal E = R − XH © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
X
DEP
SENSITIVITY
L2 L1
Output , C or X C = f (G, P, H , T , R ) Linearize ⇒ ∂C ∂C ∂C ΔG + ΔP + ΔH ΔC = ∂H 0 ∂G 0 ∂P 0 ∂C ∂C + ΔR + ΔT ∂R 0 ∂T 0 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
IND
OPEN LOOP
H =0
CLOSED LOOP
GP P R+ T C= 1 + GPH 1 + GPH
C = GPR + PT SENSITIVITY TO DISTURBANCES ΔR = 0 ; ΔT ≠ 0 ΔG = ΔP = ΔH = 0 P ⎛ ⎞ ∂C ΔC = ⎜ ⎟ΔT ∂C = ΔT 1 + GPH ⎠ ⎝ ∂T 0
ΔC = PΔT © 2005 P. S. Shiakolas
P ΔC ΔT 1 + GPH if R = 0, = P C T 1 + GPH ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
SENSITIVITY TO SENSOR GAIN
GP C= R 1 + GPH − GPGPR ∂C ⇒ ΔC = ΔH 2 ∂H (1 + GPH ) ΔC ⇒ Normalize C ΔC − G 2 P 2 R (1 + GPH )2 = ΔH C GP 1 + GPH R © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ΔC − GP = ΔH C 1 + GPH if 1 + GPH >> 0 ⇒ ΔC ΔH GP ≅− ΔH ≅ − C GPH H ΔC ΔH ΔC ΔH ≅ − ⇒ Abs Value = C H C H One − to − one mapping
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
EXAMPLE
G=
R unit step
ωn2
s(s + 2ζω n )
─
C
G
ζ = 0.6 & ωn = 5 rad/sec Find the time domain performance specifications.
π −β Rise time → t r = ωd t r = 0.554 sec
ωd = ωn 1 − ζ 2 ωd = 4 rad/sec β = atan
1− ζ
ζ
β = 0.927 rad © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
2
π Peak time → t r = = 0.785 sec ωd Settling time → 2% t s = 5% t s =
4
ζω n 3
ζω n −
Max overshoot → M p = e
© 2005 P. S. Shiakolas
= 1.333 sec = 1.000 sec ζ
1−ζ
2
π
= 0.095
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
EXAMPLE k G= s(s + 1)
R
C
G
─
H
H = 1 + kh s
k
Determine k & k h s.t.
1 s (s + 1)
max. overshoot = 0.2 units peak time = 1.0 sec R → unit step Obtain tr , t s © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
−
Mp =e
ζ 1−ζ
2
π
= 0.456
π tp = ⇒ ωd = 3.14 ωd ωd = ωn 1 − ζ 2 ⇒ ωn = 3.53 CE : 1 + GH = 0 k (1 + kh s ) = 0 1+ s (s + 1) s 2 + (1 + kkh )s + k = 0 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Compare with standard 2 nd order s 2 + 2ζω n s + ωn2 = 0
ωn2 = k ⇒ k = 12.5 2ζω n − 1 2ζω n = 1 + kkh ⇒ k h = = 0.178 k
π −β tr = = 0.65 sec ωd t s = (2% )
4
ζω n
© 2005 P. S. Shiakolas
= 2.48 sec ; (5% )
3
ζω n
= 1.86 sec
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
EXAMPLE
Negative Feedback C.L.
k (s + 2 ) GH = s (s − 2 )
a) k such that ζ of C.L. poles = 0.707 b) k such that C.L. poles are on Im. axis 1 + GH = 0 ⇒ s 2 + (k − 2 )s + 2k = 0
ωn2 = 2k ⇒ ωn = 2k 2ζω n = k − 2 ⇒ 2ζ 2k = k − 2 a)
ζ = 0.707 ⇒ 2(0.707 ) 2k = k − 2
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
k 2 − 8k + 4 = 0
Im
Re
b) C.L. poles on Im − axis
Re − part of eigen value = 0 s1,2 = ±ωj ⇒ (s − s1 )(s − s2 ) = 0 s 2 + ω 2 = 0 ⇒ 2ζω n = 0 = k − 2 ⇒k = 2 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Im
constant ζ − line
k =2
k = 0.53
jω d
β
Re
− ζω n
⎛ ω 1− ζ 2 ⎛ ωd ⎞ ⎟⎟ = atan⎜ n β = atan⎜⎜ ⎜⎜ ζω ζω n ⎝ n⎠ ⎝ © 2005 P. S. Shiakolas
⎞ ⎛ 1− ζ 2 ⎟ = atan⎜ ⎟⎟ ⎜⎜ ζ ⎠ ⎝
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
⎞ ⎟ ⎟⎟ ⎠
STANDARD CONTROLLERS R
─
GC
GP H
ONE TERM CONTROLLER
Gc = k H =kf
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C
TWO TERM CONTROLLERS a) PI: Proportional Integral ⎛ 1 ⎞ kc ⎛ τ i s + 1 ⎞ ⎟⎟ ⎟⎟ = ⎜⎜ Gc = kc ⎜⎜1 + ⎝ τis ⎠ τi ⎝ τis ⎠ b) PD: Proportional Derivative
Gc = k p + τs = kc (τ d s + 1)
THREE TERM CONTROLLERS PID: Proportional Integral Derivative
⎛ 1 ⎞ ⎟⎟ Gc = kc ⎜⎜1 + τ d s + τis ⎠ ⎝ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
R
─
GC
GP H
1 Gp = s (τs + 1) No Controller
1 C τ = R s2 + 1 s + 1
τ
© 2005 P. S. Shiakolas
τ
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C
PD Controller
PD ⇒ Gc = kc (1 + τ d s )
kc (1 + τ d s )
GcG p C = = R 1 + GcG p 2 ⎛ 1 + kcτ d s +⎜ ⎝ τ PI Controller
τ
⎞ kc ⎟s + τ ⎠
⎛ 1 ⎞ ⎟⎟ PI ⇒ Gc = kc ⎜⎜1 + ⎝ τis ⎠
kc (1 + τ i s )
( τ iτ ) C = R s 3 + 1 s 2 + kc s + kc τ
© 2005 P. S. Shiakolas
τ
τ iτ
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Two Term – Velocity Feedback
H = 1+ k f s
R
─
kc
KC
K cG p C τ = = R 1 + K cG p 2 ⎛⎜ 1 + kc k f s +⎜ ⎝ τ R
─
─
kc τs + 1
GP H
⎞ kc ⎟⎟ s + τ ⎠ GP
kf H © 2005 P. S. Shiakolas
C
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C
STEADY STATE ERROR TABLE Type 0 1
© 2005 P. S. Shiakolas
Step A 1+ k p 0
INPUT Ramp Parabolic
∞ A kv
∞ ∞
2
0
0
A ka
3
0
0
0
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
EXAMPLE Design a controller that would satisfy :
⋅ ess for ramp imput < 10% of input ⋅ ζ of dominant poles ≥ 0.707 ⋅ settling time (2% ) ≤ 3 sec
k1 G= s (s + 2 ) H = 1 + k2 s R=
A s2
© 2005 P. S. Shiakolas
R ─
G H
C k1 = R s 2 + (2 + k1k 2 )s + k1 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C
Type of system = Type 1 GH =
k1
s1 (s + 2 )
(1 + k2 s )
Assume FVT applicable ess = lim sE ( s ) = lim s(1 − TF )R s →0
s →0
⎛ ⎞ + + s k k 2 1 2 ⎟ = lim A⎜ s → 0 ⎜⎝ s 2 + (2 + k1k 2 )s + k1 ⎟⎠ A(2 + k1k 2 ) 10 ess = < A k1 100 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CE : s + (2 + k1k 2 )s + k1 = 0 2
∴ ωn2 = k1 & 2ζω n = 2 + k1k 2
ωn = k1 2ζω n 2 + k1k 2 < 0.1 ⇒ < 0.1 k1 k1 2(0.707)ωn Let ζ = 0.707 ⇒ < 0.1 k1 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
2 ωn > 14.14 ⇒ k1 = ωn = 199.94 ≈ 200
2ζω n = 2 + k1k 2 ⇒ k 2 = 0.09 Check third requirement 4 t s (2% ) = = = 0.4 < 3 sec ζω n (0.707)(14.14) 4
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
STABILITY
STABLE t
Im
NEUTRAL Re
t
UNSTABLE t © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
RUTH-HURWITZ CRITERION CE : an s n + an −1s n −1 + an − 2 s n − 2 + ... + a1s + a0 = 0 TESTS FOR APPLYING R.H.
⋅ a0 ≠ 0 ; All ai Real ⋅ If any of the coefficients are zero or change sign, then system is unstable
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ROUTH TABLE
Power
Coefficients
sn
an
an − 2
an − 4
s n −1
an −1 an −3
an − 5
sn−2
b1
b2
s n −3
c1
c2
b3
M s0 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
− 1 an b1 = an −1 an −1
an − 2
− 1 an b2 = an −1 an −1
an − 4
an − 3
an − 5
− 1 an −1 c1 = b1 b1
an − 3
− 1 an −1 c2 = b1 b1
an − 5
© 2005 P. S. Shiakolas
b2
b3 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
RH States – The number of poles of CE with positive real parts (unstable poles) is equal to the number of sign changes in the first column of the Routh array. Case 1 2 nd Order System : a2 s 2 + a1s + a0 = 0
s
2
1
a2
a0
s
a1
0
s0
b1 = a0
− 1 a2 b1 = a1 a1
a0 0
−1 (− a1a0 ) b1 = a1 b1 = a0
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
3
2
CE : a3 s + a2 s + a1s + a0 = 0 a3 a1 3 1 − s a3 a1 b1 = a a0 2 a2 2 s a2 a0 −1 (a0a3 − a1a2 ) b1 = s1 b1 b2 a2 s0
c1 = a0
− 1 a2 c1 = b1 b1 © 2005 P. S. Shiakolas
a0 0
1 (a1a2 − a0a3 ) = a2
= a0
For stable system, b1 > 0 ⇒ a1a2 > a0 a3
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CE : s 3 + s 2 + 2 s + 24 = 0 2 −1 1 3 s 1 2 b1 = = −22 1 1 24 s2 1 24 24 −1 1 = 24 s1 b1 = −22 c1 = b 0 1 b1
s0
c1 = 24
System is unstable as there is a sign change in the first column
# of unstable poles = # of sign changes in the first column
Eigen values : s1,2 = +1 ± j 7 & s3 = −3 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CL feedback system with negative feedback
Gp =
1 a3s 3 + a2 s 2 + a1s + a0
Gc = kc
H =1
• Will this system be stable or unstable? • Is there a limiting value for kc for stability?
CE : 1 + Gc G p = 0 a3s 3 + a2 s 2 + a1s + a0 + kc = 0 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Applicable ? All coefficients > 0
a0 + kc > 0 ⇒ kc > − a0 a1a2 − a3a0 − a3kc b1 = >0 a2 c1 = a0 + kc ⇒ kc > − a0 a1a2 − a3a0 − a3kc a3a0 − a1a2 > ⇒ kc < a3 a2 a2 a1a2 − a3a0 − a0 < kc < a3 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CE : s 4 + ks 3 + s 2 + s + 1 = 0 • Determine the range of k for stability 4
1
1
s3
k
1
s
s2
b1
s1
c1
s0
d1
b2
1
−1 1 b1 = k k
1
−1 1 b2 = k k
1
1 = (k − 1) 1 k =1
0
k2 c1 = 1 − k −1 d1 = 1
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
1 b1 = (k − 1) > 0 ⇒ k > 1 k k2 k −1− k 2 c1 = 1 − >0 >0 ⇒ k −1 k −1 negative − 1 + k (1 − k ) >0 k −1 positive The above range of inequality is false and therefore there is no range of k that yields us a stable system. Hence, system is unstable. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Case 2 Zero in first column while all other elements are non-zero 5
4
3
2
CE: s + 2 s + 2 s + 4 s + 11s + 10 = 0
s5
1
2
11
s4
2
4
10
s
3
s2 1
s s
0
b1 c1 d1 e1
© 2005 P. S. Shiakolas
b2
ε −
12
ε
−1 1 b1 = 2 2
2
−1 1 b2 = 2 2
11
−1 1 c1 = b1 b1 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
=0
4
10 11 b2
=6
Case 3 Whole Row full of zeros 3
2
CE: s + s + 2 s + 2 = 0 ⋅ s1 row is all zeros
s3
1
2
s2
1
2
s1
b1 = 0
s0
c1
⋅ Form the auxillary polynomial
1s 2 + 2 s 0 ⇒ s 2 + 2 Eigen values of s 2 + 2 is ± 2 j which are symmetric about the real axis.
Differentiate auxillary polynomial ⇒ 2 s © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CE : s 4 + s 3 + 3s 2 + 2 s + 2 = 0
s4 s
3
1
3
k
2
s2
b1
b2
s1
c1 = 2
s0
d1
© 2005 P. S. Shiakolas
2
b1 = 1 b2 = 2 c1 = 0 Q = b1s 2 + b2 = s 2 + 2 dQ = 2s ds − 1 b1 d1 = c1 c1
b2 0
= b2
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
5
4
3
2
CE: s + 2 s + 24 s + 48s − 25s − 50 = 0 s5
1
24
− 25
b1 = b2 = 0
− 50
Q = 2 s 4 + 48s 2 − 50
4
2
48
s3
b1
b2
s2
c1
∴b1 = 8 & b2 = 96
s1
d1
c1 = 24 & c2 = 50
s0
e1
d1 = 112.7 & e1 = −50
s
dQ = 8s 3 + 96 s ds
∴ System is unstable © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CE : s 3 + 2 s 2 + 4 s + k = 0 • Determine the range of k for stability
s3
1
4
s2
2
k
1
s s
0
b1 c1
−1 1 b1 = 2 2
4
8−k = >0 2 k
−1 2 c1 = b1 b1
k
=k >0
0
8−k > 0 ⇒ k < 8 0
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
R ─
Gp =
Gp
C
k s ( s 2 + s + 1)( s + 2)
Find k for stability
Apply RH for CE : s 4 + 4 = 0
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
R ─
Gc
Gp
C
Gc = k
Steady State Error
Gp =
Type = 0; Step Input (unity)
1 ess = 1+ k p
2 s 3 + 4 s 2 + 5s + 2
k p = lim GcG p = k s →0
1 ess = 1+ k 1 1 ess < 2% input ⇒ < ⇒ k > 49 1 + k 50 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
• Determine the range of k for stability
CE: s 3 + 4 s 2 + 5s + 2 + 2k = 0 2 + 2 k > 0 ⇒ k > −1 18 − 2k 3 b1 = >0 s 1 5 4 c1 = 2 + 2k
s2
4
s1
b1
s0
18 − 2k >0 ⇒ k <9 4
c1
For stability, − 1 < k < 9 which
2 + 2k
contradicts k value from steady state error. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
p - controller cannot be used. Investigate PI controller. Gc =
k p s + kI s
(
)
CE : s 4 + 4 s 3 + 5s 2 + 2 + 2k p s + 2k I = 0 2 + 2k p > 0 ⇒ k p > −1 kI > 0
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
s4
1
s3
4 2 + 2k p
s2
b1
s1
c1
s0
d1
© 2005 P. S. Shiakolas
5
b2
2k I
b1 =
18 − 2k p 4
>0
⇒ kp < 9 b2 = 2k I d1 = 2k I > 0
[(
)(
)
4 c1 = 1 + k p 9 − k p − 8k I 18 − 2k p
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
]
ROOT LOCUS (RL) • Graphical technique that will allow us to plot the path or locus of the eigen values as one parameter changes (0 to ∞) • RL has the ability to • Determine closed loop system behavior from open loop conditions • Determine the effects of one parameter qualitatively
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Complex Algebra Review
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ROOT LOCUS (RL) • Graphical technique that will allow us to plot the path or locus of the eigen values as one parameter changes (0 to ∞) • RL has the ability to • Determine closed loop system behavior from open loop conditions • Determine the effects of one parameter qualitatively
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Complex Algebra Review
A = α + β j = A e jθ
Im
⎛β ⎞ θ = atan2⎜ ⎟ = ∠A ⎝α ⎠
A
θ
A = α2 + β2
A1 e jθ1
A1 A1 j (θ1 ±θ 2 ) = = e j θ A2 A2 e 2 A2 A1 ± A2 = (α1 + β1 j ) ± (α 2 + β 2 j ) = (α1 ± α 2 ) + (β1 ± β 2 ) j © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Re
Background on RL
CLTF
C G = R 1 + GH
R ─
CE : 1 + GH = 0
G H
GH: Ratio of polynomials in s Where s: complex quantity
s = σ ± jω 1 + GH s =σ ± jω = 0 GH s =σ ± jω = −1 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C
Magnitude GH s =σ ± jω = − 1 Angle
∠GH s =σ ± jω = ∠(− 1 + 0 j ) = ±180(2n + 1) n = 0,1, 2...
EXAMPLE Im
k GH = s +α GH =
k
(σ + α ) + jω
© 2005 P. S. Shiakolas
−α
s = σ ± jω
θ
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Re
GH =
k
(σ + α )2 + ω 2
∠GH = θ = −atan
ω σ +α
= ±180(2n + 1)
k +1 = 0 GH + 1 = 0 ⇒ s +α
s + (k + α ) = 0 ⇒ s = −(k + α )
k =∞ x ∞ © 2005 P. S. Shiakolas
Im
k =0 x −α ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Re
Properties of RL
GH = −1 where G = k g kg
Ng Dg
kf
Nf Df
Ng Dg
& H =kf
= −1 ⇒ k = k g k f = −
Nf Df
Ng N f Dg D f
Dg D f = 0 ⎫ ⎪ if k = 0 ⇒ ⎬ Open loop poles N g N f = ∞ ⎪⎭ Dg D f = ∞ ⎫ ⎪ if k = ∞ ⇒ ⎬ Open loop zeros N g N f = ∞ ⎪⎭ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
OLTF :
CLTF :
k
Ng N f Dg D f
C = R
kg
Ng
kg
Ng
Dg Dg = Dg D f + kN g N f Ng N f 1+ k Dg D f Dg D f
(
k → 0 : CL poles approaches OL poles k → ∞ : CL poles approaches OL zeros
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
)
EXAMPLE
k GH = s +1
0≤k ≤∞
CE : s + (1 + k ) = 0 OL poles : s = −1 OL zeros : ∞
k =∞ x ∞ © 2005 P. S. Shiakolas
Im
k =0 x -1 ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Re
EXAMPLE
k (0.4 s + 1) OL pole : s = −1 GH = s +1 OL zero : s = −2.5 Im
TP2
o
TP3
TP1
φ TP5
x
α
-1
Re
TP4
∠GH = ∠k + ∠0.4 s + 1 − ∠s + 1 ∠GH = 0 + φ − α = ±180(2n + 1) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CE : (1 + 0.4k )s + 1 + k = 0 1+ k s=− 1 + 0.4k k = 0 ⇒ s = −1 k = ∞ ⇒ s = −2.5 k = 5 ⇒ s = −2 101 k = 100 ⇒ s = − 41 1001 k =0 ⇒ s=− 401 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
GH s = −1.5 = 1 k (0.4 s + 1) =1 s + 1 s = −1.5 − 0.8k = 1 k = 1.25
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ROOT LOCUS (Cont…) Starts at O.L. poles k varies from 0 to ∞ Ends at O.L. zeros.
1 + kGH = 0 where kGH is OLTF k OLTF: GH = s (s + 1)
(
)
OL Poles :
s = −1 OL Zeros : none
CE : s 2 + s + k = 0 1 1 − 1 ± 1 − 4k 1 − 4k s1,2 = =− ± 2 2 2 © 2005 P. S. Shiakolas
s=0
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
∞
x
θ2
-1
-1/2
Im
x
θ1 0
−∞ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Re
k
CL poles
0
0
−1
1 4
1 − 2
1 − 2
1 2
1 1 − + j 2 2
10
1 39 − + j 2 2
1000 © 2005 P. S. Shiakolas
1 3999 − + j 2 2
1 1 − − j 2 2 1 39 − − j 2 2 1 3999 − − j 2 2
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Identify controller gain k to yield eigen value at 1 s = − ±3j 2 GH for all s = − 1 = 1 k =1 s (s + 1) s = − 1 + 3 j 2
find k .
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
1 0 < k < , the eigen values are on the the Re - axis 4 i.e. they behave as the first order system.
(s + α )(s + β ), ζ
> 1.
1 k = , the eigen values are on the Re - axis 4 & repeating. (s + γ )2 , ζ = 1. 1 k > , the eigen values are complex conjugates. 4
ζ < 1. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Im
s 2 + 2ζω n s + ωn2 = 0
x
-1
− ζω n
ωd
β
x
-1/2
0
ωd ωn 1 − ζ 2 ⎛⎜ 1 − ζ 2 tan β = = =⎜ ζω n ωnζ ⎜ ζ ⎝
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Re
⎞ ⎟ ⎟⎟ ⎠
EXAMPLE ζ = 0.6 (desired damping)
1 − 0.6 2 tan β = = 1.333 ⇒ β = 53.1o 0.6 1 − ζω n = − ⇒ ωn = 0.833 2
ωd = ωn 1 − ζ 2 = 0.6664 Point of interest ⇒ Desired eigen value 1 Find k to yield s = − ± 0.6664 j 2 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
RULES FOR DRAWING ROOT LOCUS Example k OLTF : GH = (s + 1)(s + 2)(s + 3)
• Find OL poles : 3 (−1, − 2, − 3) OL zeros : 0 • Draw OL poles & zeros • Start OL poles (k = 0 ) End OL zeros (k → ∞ ) • # Loci = 3 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
• Symmetry about Re − axis • n → # OL poles = 3 m → # OL zeros = 0 • # asymptotes = na = n − m = 3 (loci going to ∞) • Angle of asymptotes with the Re − axis ± 180(2q + 1) θq = n−m
± 180(2q + 1) = = ± 60(2q + 1) 3
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
q
θq
0
± 60
1 ± 180 2 ± 300
• Intersection of asymptotes with Re − axis
(Centroid of
poles & zeros )
∑ poles − ∑ zeros σa = n−m
( - 1 - 2 - 3) − 0 = = −2 3
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
• Loci on Re − axis : Loci lie on Re - axis in regions where there is an odd number of poles and zeros (on Re - axis) to the right of it. • Intersection of RL with Im - axis will yield limits of gain for stability − Routh - Hurwitz − On Im - axis s = jw Substitute s = jw in CE CE : s 3 + 6 s 2 + 11s + 6 + k = 0 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
⇒ ( jw)3 + 6( jw)2 + 11( jw) + 6 + k = 0
( ) Im ⇒ (− ω 3 + 11ω ) j = 0 ω (11 − ω 2 ) = 0 ⇒ ω = 0
Re ⇒ − 6ω 2 + 6 + k = 0
ω = ± 11
( )
2
− 6 11 + 6 + k = 0 ⇒ k = 60 • Breakin/Breakaway point CE :1 + kGH = 0 = f ( s ) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
A + kB = 0 = f ( s )
( A = A( s), B = B( s) )
A k =− B dk d ⎛ A⎞ min/ max ⇒ =− ⎜ ⎟=0 ds ds ⎝ B ⎠ The value of s is breakin/away pts if k for that s is positive.
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
RULES FOR DRAWING ROOT LOCUS (Cont..)
A k=− B
A = (s + 1)(s + 2)(s + 3)
B =1
dk d ⎛⎜ s 3 + 6 s 2 + 11s + 6 ⎞⎟ =0 =− ⎟ 1 ds ds ⎜⎝ ⎠
(
)
= − 3s 2 + 12 s + 11 = 0 s1 = −2.6 & s2 = −1.4 1. Evaluate k ⇒ breakin/away. Reject if k < 0 2. Examine the Re − axis © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ROOT LOCUS Root Locus 5 4
Constant ζ line
3
IMAGINARY AXIS Imaginary Axis
ω = 11, k = 60
2
β
1 0 -1
s = −1.4 k = 0.384
-2 -3 -4 -5 -8
-7
-6
-5
-4
-3
-2
-1
Real Axis
REAL AXIS © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
0
1
Example GH =
10(s + α ) 2
s + 5s + 6
0<α < ∞
N (s) k = GH D( s) CE : 1 + GH = 0 s 2 + 5s + 6 + 10 s + 10α = 0
(s 2 + 15s + 6)+ 10α = 0 1+ © 2005 P. S. Shiakolas
10α s 2 + 15s + 6
=0
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
n=2 ; m=0 na = n − m = 2 ± 180(2q + 1) = ±90(2q + 1) ≡ ±90, ± 270 θa = n−m ∑P−∑Z σa = = −4.5 n−m Breakin/away point s 2 + 15s + 6 dα α= ⇒ = −(2 s + 15) ds 10 s = −7.5 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Root Locus
6
Imaginary Axis
4 2 0 -2 -4 -6 -15
-10
-5 Real Axis
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
0
Example
GH =
n=3 m=3
k (s + 3)(s + 1 ± 3 j )
s (s + 1)(s + 2 )3 (s + 4 )(s + 5 ± 2 j )
(0,−1, −2,−2,−2,−4,−5 ± 2 j ) (− 3,−1 ± 3 j )
na = n − m = 5 ± 180(2q + 1) θa = = ±36(2q + 1) ≡ ±36, ± 108, ± 180 n−m ∑P−∑Z = −3.2 σa = n−m © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Root Locus
15
Imaginary Axis
10 5 0 -5 -10 -15 -20
© 2005 P. S. Shiakolas
-15
-10
-5 Real Axis
0
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
5
10
Example
Gp =
s −3
H =1
s 2 + 5s + 6
Gc = 1 (proportional) Is the system stable or unstable? OLTF :
k (s − 3) 2
s + 5s + 6
assuming Gc = k
How to obtain a quadratic type response?
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Root Locus 1
Imaginary Axis
0.5
0
-0.5
-1 -14
-12
© 2005 P. S. Shiakolas
-10
-8
-6 -4 Real Axis
-2
0
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
2
4
STANDARD CONTROLLERS P: k
No OLP/OLZ
⎛τis +1⎞ ⎟⎟ PI : k ⎜⎜ ⎝ τis ⎠
⎧ 1 OLZ ⎨1 OLP (origin) ⎩
PD : k (τ d s + 1)
1 OLZ
⎛ τ dτ i s 2 + τ i s + 1 ⎞ ⎟ PID : k ⎜ ⎜ ⎟ τ s i ⎝ ⎠
⎧ 2 OLZ ⎨1 OLP (origin) ⎩
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Im
PI x
o
x
x
o
Re Im
0 < k < klimit
x
o
x
Im
o
x
© 2005 P. S. Shiakolas
x
x
o
Re
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Re
PD
Im
Im
ζ constant
x x
x
o
o
o
x
Re
Re Im
Im x x
x
o
o
o
x
o
Re
o
Re
Re Im
0 < k < klimit © 2005 P. S. Shiakolas
o
x
x
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
PID Im
Im
x
x
x
o
Re
x
x
o
Marginally Stable → unstable
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Re
Example
( s + 2 )(s + 7 ) GH = kc (s − 1)(s + 1)(s + 10)(s + 3) Root Locus
m=2
na = n − m = 2
15
± 180(2q + 1) θa = n−m ∑P−∑Z σa = n−m
20
Imaginary axis
n=4
Root Locus 25
10 5 0 -5
-10 -15 -20 -25 -10
© 2005 P. S. Shiakolas
-8
-6
-4
Real axis
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
-2
0
2
Example R ─
─
A
1/s
C
k
A 1 + Ak
20 A= (s + 1)(s + 4)
• Draw Root Locus • Determine the value of k such that the damping ratio of the dominant pole C.L. poles = 0.4 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
A 1 C CLTF : = 1 + Ak s R 1+ A 1 1 + Ak s
CE :
⎛ A 1+ ⎜ ⎝ 1 + Ak
1⎞ ⎟=0 s⎠
(s3 + 5s 2 + 4s + 20)+ 20ks = 0 1+ K
© 2005 P. S. Shiakolas
s 3
2
s + 5s + 4 s + 20
=0
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
OLTF for RL :
s K (s + 2 j )(s − 2 j )(s + 5) Root Locus 15
n=3 m =1 na = n − m = 2
θ a = ±90(2q + 1) σ a = −2.5 © 2005 P. S. Shiakolas
Imaginary Axis
10 5 0 -5 -10 -15 -5
-4
-3
-2 Real Axis
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
-1
0
Desired eigen value is the intersection of the RL plot with constant ζ - line. s1 = −1.05 ± 2.41 j
s = −2.9
s2 = −2.16 ± 4.96 j
s = −0.68
K ( ) s=s = 1 1 for s1 : K = 8.98 or k = 0.45 for s2 : K = 28.26 or k = 1.43
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
INTRODUCTION TO MATLAB MATLAB commands used • rlocus • rltool Use MATLAB help for further information.
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
EXAMPLE ELECTROMAGNETIC RELAY CONTROL (Refer Handout for figure)
Θ( s ) = OLTF : G ( s ) = s 3 + 16s 2 + 44 s − 160 Vin ( s ) 1
Controller R
─
k
Vin
C
Plant: Electromagnetic Relay (3rd Order) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CE : s 3 + 16s 2 + 44s − 160 + k = 0 Stable or not? and why? RH : k − 160 > 0
Root Locus 20
n=3 na = 3
θ a = ±60(2q + 1) σ a = −5.33
10 Imaginary Axis
m=0
15
5 0 -5 -10 -15 -20 -30
© 2005 P. S. Shiakolas
-25
-20
-15
-10 -5 Real Axis
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
0
5
10
At intersection with Im − Axis s = jω CE : ( jω )3 + 16( jω )2 + 44( jω ) − 160 + k = 0 − jω 3 − 16ω 2 + 44 jω − 160 + k = 0
(−16ω 2 −160 + k )+ (− ω 3 + 44ω )j = 0 ω (− ω 2 + 44) = 0 ⇒ ω = 0 or ω = 44 − 16ω 2 − 160 + k = 0 ⇒ − 16(44 ) − 160 + k = 0 k = 864 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
RH : Let α = k − 160 s 3 + 16 s 2 + 44s + α = 0
1 1 b1 = − 16 16
1 = − (α − 704) α 16
44
s3
1
44
s2
1
α
s1
b1
k < 864
s0
c1
c1 = α = k − 160 > 0 ⇒ k > 160
b1 > 0 ⇒ α < 704 ⇒ k − 160 < 704
160 < k < 864
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
FREQUENCY RESPONSE METHODS Frequency Response: It is the steady state response of a system due to a sinusoidal input for a stable or marginally stable system. Why use frequency response methods? For LTI system, the response is of the same form as the input. If input is sinusoidal (w, A), output will also be sinusoidal (w, B, φ). © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
INPUT/OUTPUT RELATIONSHIP FOR F.R. y (t ) x(t ) STABLE G (s ) L.T.I. Y (s) X (s)
L Let : x(t ) = X sin ωt ⎯⎯→ X ( s ) =
Xs s2 + ω 2
p( s) p(s) = G ( s) = q ( s ) (s + s1 )(s + s2 )......(s + sn ) Y ( s) = G (s) X ( s) p(s) p(s) Xs = X (s) = (s + s1 )(s + s2 )......(s + sn ) s 2 + ω 2 q(s) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Partial Fraction Expansion bn b1 b2 a a Y ( s) = + + ... + + + s + s1 s + s2 s + s n s + jω s − j ω a = conjugate of a L- 1 ⎯⎯ ⎯→ y (t ) = b1e − s1t + b2e − s2t + ... + bn e − sn t + ae − jωt + a e jωt 1444442444443 = 0 for steady state (t ≥ 4τ ) yss (t ) = ae − jωt + a e jωt © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Residue theorem ⇒ a & a a = Y ( s ) ( s + jω ) s = − j ω
ωX (s + j ω ) s = − j ω a = G( s) (s + jω )(s − jω ) ωX a = G ( − jω ) − 2 jω
ωX & a = G ( jω ) 2 jω
G ( j ω ) = G ( jω ) e j φ
φ = ∠G ( jω )
G ( − j ω ) = G ( jω ) e − j φ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
a=
G ( jω ) Xe − jφ
y ss (t ) =
−2j
& a=
X G ( j ω ) e − j φ e − jω t −2j
G ( jω ) Xe jφ 2j +
X G ( j ω ) e jφ e jω t 2j
⎧⎪ e − j (φ +ωt ) e j (φ +ωt ) ⎫⎪ + y ss (t ) = X G ( jω ) ⎨ ⎬ 2 j ⎪⎭ ⎪⎩ − 2 j e iα − e − iα We know , Euler' s identity sin α = 2j © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
y ss (t ) = X G ( jω ) sin(ωt + φ )
φ > 0, phase lead φ < 0, phase lag EXAMPLE
x(t )
G (s )
k G(s) = τ s +1
y (t )
x(t ) = X sin ωt
y ss (t ) = X G ( jω ) sin(ωt + φ ) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
k k ⇒ G ( jω ) = G ( jω ) = 1 + τωj 1 + τ 2ω 2
φ = ∠G ( jω ) = ∠k − ∠(1 + τωj ) ⎛ τω ⎞ = −atan⎜ ⎟ ⎝ 1 ⎠ yss (t ) = X
© 2005 P. S. Shiakolas
k 1 + τ 2ω 2
sin (ωt − atan (τω ) )
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
OBSERVATIONS :
τω << 1 (ω very small) Magnitude of s.s. is almost kX Phase of s.s. is very small
τω >> 1 (ω very large) Magnitude of s.s. is approximately 1
ω
Phase of s.s. is approximately − 90o
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
BODE PLOT Amplitude & Phase Linear
MAGNITUDE
Logarithmic w
Linear
PHASE
Logarithmic winput © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Bode Plot: It is a graphical representation of the steady state output due to a sinusoidal input for a range of values of the input frequency. Standard Representation for Magnitude of G(jw) is logarithmic and is given as
20 log10 G ( jω ) db (decibels)
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
How to Draw a Bode Plot
In General, a T.F. G ( s) =
k g (s + a )
2 2 ( ( ) + + + 2 s s b s ζω n s ωn )
Normalizing, ⎛1 ⎞ k g a⎜ s + 1⎟ a ⎝ ⎠ G ( s) = 2 ⎛ ⎞ ⎛ ⎞ 2ζ ⎛1 ⎞ 2 ⎜⎜ s ⎟ ⎟ 1 s b⎜ s + 1⎟ ωn ⎜ ⎜ s + + ⎟ ⎟ b ω ω ⎝ ⎠ n ⎝⎝ n ⎠ ⎠ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
s = jω k (τ aωj + 1) G ( jω ) = ⎛ ⎛ jω ⎞ 2 2ζω ⎞ ⎟⎟ + jω (τ bωj + 1)⎜⎜ ⎜⎜ j + 1⎟⎟ ωn ⎠ ωn ⎝ ⎝ ⎠ G ( jω ) =
k (1 + τ aωj ) ⎛ ⎛ ω ⎞ 2 2ζω ωj (1 + τ bωj ) ⎜⎜1 − ⎜⎜ ⎟⎟ + ωn ⎠ ωn ⎝ ⎝
Plot Magnitude in db © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
⎞ j ⎟⎟ ⎠
20 log G ( jω ) = 20 log k + 20 log (1 + τ aωj )
− 20 log ωj − 20 log (1 + τ bωj )
⎛ ⎛ ω ⎞ 2 2ζω ⎟⎟ + − 20 log ⎜⎜1 − ⎜⎜ ωn ⎜ ⎝ ωn ⎠ ⎝
(
⎞ j ⎟⎟ ⎟ ⎠
20 log G ( jω ) = 20 log k + 20 log 1 + (τ aω )2 − 20 log ω − 20 log 2
© 2005 P. S. Shiakolas
)
(1 + (τ bω )2 )
⎛⎧ 2 ⎫2 2⎞ ⎜ ⎪ ⎛ ω ⎞ ⎪ ⎛ 2ζω ⎞ ⎟ ⎟⎟ ⎟ ⎟⎟ ⎬ + ⎜⎜ − 20 log⎜ ⎨1 − ⎜⎜ ⎜ ⎪⎩ ⎝ ωn ⎠ ⎪⎭ ⎝ ωn ⎠ ⎟ ⎝ ⎠ ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Phase
∠G ( jω ) = ∠k + ∠(1 + τ aωj ) − ∠jω ⎛ ⎛ ω ⎞ 2 2ζω ⎟⎟ + − ∠(1 + τ bωj ) − ∠⎜⎜1 − ⎜⎜ ωn ⎜ ⎝ ωn ⎠ ⎝ ⎛ τ aω ⎞ ⎛ω ⎞ = φ + a tan⎜ ⎟ − atan⎜ ⎟ ⎝0⎠ ⎝ 1 ⎠ ⎛ 2ζω ⎞ ⎜ ⎟ ωn ⎟ ⎛ τ bω ⎞ ⎜ − a tan⎜ ⎟ − a tan⎜ 2⎟ ⎝ 1 ⎠ ⎜⎜ 1 − ⎛⎜ ω ⎞⎟ ⎟⎟ ⎝ ⎝ ωn ⎠ ⎠ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
⎞ j ⎟⎟ ⎟ ⎠
Constant
20 log k ⇒ constant in db
∠k = 0
Linear Axis
| | db
0.1
1
10
ω
o 0.1 © 2005 P. S. Shiakolas
1
10
ω
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Pole at Zero or Origin 1 1 → Pole at origin s jω
1 = −20 log jω = −20 log ω 2 20log jω = −20 log ω db
ω 1 ∠ = −atan = −90o 0 jω ω
© 2005 P. S. Shiakolas
(Constant )
0.1
20 log ω − 20
− 20 log ω 20
1 10
0 20
0 − 20
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
40 Passes Through ω=1
20
| | db
0.1
1
10
100
ω (log)
-20
Slope -20 db/dec
-40
⎛⎜ 1 ⎞⎟ 2 ⎛⎜ 1 ⎞⎟ ⎝ jω ⎠ ⎝ jω ⎠
Slope -40 db/dec
o 0.1 -90 -180 © 2005 P. S. Shiakolas
1
10
100
⎛⎜ 1 ⎞⎟ ⎝ jω ⎠ 2 ⎛⎜ 1 ⎞⎟ ⎝ jω ⎠
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ω (log)
Dual Pole at Origin
20 log −
1
ω
2
= 20 log
1
ω
2
( )
= −20 log ω
2 2
= −20 log ω 2 = 2(− 20 log ω ) ⎛ 0 ⎞ ∠− = ∠1 − ∠ − ω = 0 − atan⎜⎜ ⎟⎟ ω2 ⎝ −ω2 ⎠ 1
2
( )
= −180o = 2 90o
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
⎛ ⎛ 1 ⎞⎞ ⎟⎟ ⎟⎟ ⇒ N ⎜⎜ − 20 log⎜⎜ SN ⎝ jω ⎠ ⎠ ⎝ 1
where N → slope − 20 × N db/dec ⇒ N (− 90 ) S ± N ⇒ Mag passes through ω = 1 Slope ± N × 20 db/dec ⇒ Phase ± N × 90o © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Simple Pole or Zero
1 1 + τωj
(1 ± τωj )
±N
Break Frequency Corner Frequency
1
τ
1 2 2 20 log = −20 log 1 + τ ω 1 + τωj
(
= −10 log 1 + τ 2ω 2 when ω =
1
τ
© 2005 P. S. Shiakolas
)
⇒ τω = 1⇒ τ 2ω 2 = 1 = −10 log(1 + 1) = −10 log 2 = −3.01 db ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ωτ << 1 42 4 31
ω << 1τ ωτ = 1 ωτ >>1
⇒ Mag ≈ − 10 log 1 = 0 db ∠ ≈ 0o ⇒ Mag ≈ − 3.01 db ∠ ≈ −45o
(
⇒ Mag ≈ − 10 log τ 2ω 2 ≈ − 20 log(τω )
)
≈ − 20 db/dec ∠ ≈ −90o © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
0.1
τ
| | db
1
τ
10 1
τ
100 1
τ
ω (log)
-20 Slope -20 db/dec
-40
0.1
o
τ
1
τ
10 1
τ
100 1
τ
-45
-90
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ω (log)
(1 + τωj )± N Mag : ωτ << 1 ⇒ 0 db
ωτ = 1
⇒ 3.01 db (± N )
ωτ >> 1 ⇒ 20 × (± N ) db/dec Phase : ωτ << 1 ⇒ 0o
ωτ = 1
⇒ 45 × (± N )
ωτ >> 1 ⇒ 90 × (± N )
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Quadratic Poles & Zeros Quadratic Poles 1 G ( s) = s 2 + 2ζω n s + ωn2
G ( s) =
1
1
ωn2 ⎛ s ⎞ 2 s ⎜⎜ ⎟⎟ + 2ζ +1 ωn ⎝ ωn ⎠
G ( jω ) =
1 2
⎛ω ⎞ ω ⎟⎟ + 2ζ 1 − ⎜⎜ j ωn ⎝ ωn ⎠
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
2
⎛ω ⎞ ω ⎟⎟ + 2ζ j Mag : 2 − log G ( jω ) = −20 log 1 − ⎜⎜ ωn ⎝ ωn ⎠ ⎡⎧ 2 ⎫2 2⎤ 1 ω⎫ ⎥ ⎢⎪ ⎛ ω ⎞ ⎪ ⎧ ⎟⎟ ⎬ + ⎨2ζ = −20 × log ⎢⎨1 − ⎜⎜ ⎬ ⎥ 2 ωn ⎠ ⎪ ⎩ ωn ⎭ ⎝ ⎪ ⎢⎩ ⎥ ⎭ ⎣ ⎦
ω Let =u ωn
(
)
⎡ 2⎤ 2 2 = −10 log ⎢ 1 − u + (2ζu ) ⎥ ⎦ ⎣ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
⎛ 2ζu ⎞ Phase ∠G ( jω ) = −atan⎜⎜ ⎟ 2⎟ ⎝1− u ⎠ if u << 1, Mag ≈ −10 log 1 = 0 Phase ≈ 0o if u >> 1, Mag ≈ −10 log u 4 = −40 log u slope = −40 db/dec ⎛ 2ζu ⎞ ⎛ 1 ⎞ ⎟ ≈ −atan⎜ Phase ≈ -atan⎜⎜ ⎟ 2⎟ ⎝ −u ⎠ ⎝ −u ⎠ ≈ −180o © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
if u = 1, Mag = −10 log(2ζ )2 = −20 log(2ζ ) ⎛ 2ζ Phase = −atan⎜ ⎝ 0 | | db -20
0.1u
u =1
⎞ o 90 = − ⎟ ⎠ 10u
-40
o
ω (log) Slope -40 db/dec
0.1u
u =1
10u
ω (log)
-90 -180 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Max Amplitude occurs at ωr (resonant frequency)
ωr = ωn 1 − 2ζ 2
ζ < 0.707
M Pω r = G ( jωr ) = ⎧⎨2ζ 1 − ζ 2 ⎫⎬ ⎩ ⎭
(
−1
⎛ Mag ⎞ ⎟⎟ × G ( jω ) × sin ωt + 0o Output = ⎜⎜ ⎝ Input ⎠
© 2005 P. S. Shiakolas
)
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
First Order Simple Pole at ω = 0.1 ⇒ 0 db
τ
20 log G ( jω ) = 0 db ⇒ G ( jω ) = 1 ⎛ mag ⎞ ⎟⎟ × 1× sin(ωt + 0o ) output = ⎜⎜ ⎝ input ⎠ at ω = 100
τ
20 log G ( jω ) = −40 db ⇒ G ( jω ) = 0.01 ∠ ≈ −90
o
© 2005 P. S. Shiakolas
⎛ mag ⎞ ⎟⎟ × 0.01× sin(ωt − 90o ) output = ⎜⎜ ⎝ input ⎠ ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Example
G ( s) =
10 ( s + 3) s ( s + 2) ( s 2 + s + 2)
Normalize ⎛1 ⎞ 10 × 3 ⎜ s + 1⎟ ⎝3 ⎠ G (s) = 2 ⎞ ⎛ 1 ⎛1 ⎞ ⎜⎛ s ⎞ s × 2⎜ s + 1⎟ × 2 ⎜ ⎟ + s + 1⎟ ⎟ 2 ⎝2 ⎠ ⎜⎝ ⎝ 2 ⎠ ⎠ ⎛ jω ⎞ 7.5 ⎜ + 1⎟ 3 ⎝ ⎠ G ( jω ) = 2 ⎛ jω ⎞⎟ ⎛ jω ⎞ ⎜ ⎛ jω ⎞ jω ⎜ +1 + 1⎟ ⎜ ⎟ + ⎟ 2 ⎝ 2 ⎠ ⎜⎝ ⎝ 2 ⎠ ⎠ © 2005 P. S. Shiakolas ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
FACTORS Break Freq.
Slopes db/dec
Angle deg.
7.5
j
ω 3
+1
( jω )
−1
⎛ ω ⎞ ⎜ j + 1⎟ ⎝ 2 ⎠
−1
(Quad )−1
*
3
1
2
2
Low Freq.
*
0
-20
0
0
High Freq.
*
20
-20
-20
-40
Low Freq.
0
0
-90
0
0
-45
-90
-90
-180
Break Freq. High Freq.
© 2005 P. S. Shiakolas
45 0
90
-90
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Final Slope (ω >> 1) = − 60 db/dec Final Phase (ω >> 1) = −270o
Constant : 20 log(Constant) db = 20 log(7.5) = 17.5 db Quadratic : Find ωr , ζ , M pω r from formulae. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Discussion/review from last example
ω1
| | db -20 -40
© 2005 P. S. Shiakolas
Simple Zero Slope = 20 db/dec
ω2
Constant
ωr = M p
ω3
Simple Pole Slope = -20 db/dec
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ω (log)
Quadratic Slope = -40 db/dec
Magnitude
ω1
| | db
ω2
ω3
-20 -40
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ω (log)
Phase 90 45
ω (log)
o -45 -90 -135 -180 -270
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
MINIMUM PHASE & NON-MINIMUM PHASE SYSTEMS • Poles & Zeros in left-hand s-plane Æ Minimum Phase • Poles or Zeros in right-hand s-plane Æ Non-Minimum Phase 1 + j ωT 1 − j ωT G1 ( jω ) = G2 ( jω ) = 1 + jωT1 1 + jωT1
⎛ ωT ⎞ ⎛ ωT1 ⎞ ∠G1 = atan⎜ ⎟ − atan⎜ ⎟ ⎝ 1 ⎠ ⎝ 1 ⎠ ⎛ − ωT ⎞ ⎛ ωT1 ⎞ ∠G2 = atan⎜ ⎟ − atan⎜ ⎟ ⎝ 1 ⎠ ⎝ 1 ⎠ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
1 1 > T T1 o o
1 T1
1 T1
1 T
1 T
CUT-OFF FREQUENCY OR BANDWIDTH Cut-off frequency is the frequency at which the magnitude of the closed loop frequency is at 3 db below the zero frequency value. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
| | db
ωb
0
ω (log)
-3 db Bandwidth
| | db
ω (log)
0 -3 db Bandwidth
Find the attenuation factor at -3 db magnitude. (Find real # corresponding to magnitude of -3 db) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
e − jω L G ( jω ) = 1 + j ωT
L = 0.5 T = 1.0
Magnitude (db) = 20 log G ( jω ) = 20 log e − jωL − 20 log 1 + jωT Phase ∠ = ∠e − jωL − ∠1 + jωT ⎛ ωT ⎞ = − ωL − a tan⎜ ⎟ { ⎝ 1 ⎠ 57.3ωL deg
φ = −ωL or − 57.3Lω © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
STABILITY ANALYSIS IN FREQUENCY DOMAIN USING BODE PLOTS
ωcg
| | db 0 o -180
ω (log) Positive Gain Margin (G.M.)
Positive Phase Margin (P.M.)
© 2005 P. S. Shiakolas
ωcp
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ω (log)
Phase Margin: It is the amount of additional phase lag at ωcg required to bring the system to the verge of instability. γ = 180 + φ ω cg
( ) if γ < 0 (φ < −180o ) , System is unstable if γ > 0 φ > −180o , System is stable
Gain Margin: It is the reciprocal of the magnitude |G(jω)| at ωcp. GM indicates how much the gain can be increased before the system becomes unstable OR how much it must be decreased for the system to become stable. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
kg =
1 G ( jω ) ω
cp
k g db = 20 log k g = −20 log G ( jω ) ω
cp
if k g (db) > 0, system is stable if k g (db) < 0, system is unstable For satisfactory performance and to guard against variations in performance of system components,
30o < PM < 60o GM > 6 db © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Example
k G (s) = s (s + 1)(s + 5)
Find k for stability and draw BP for k = 1,10,100 Root Locus 10
n=3
k = 30
m=0
ω= 5
na = n − m = 3
θ a = ±60(2q + 1) −6 σa = = −2 3
Imaginary Axis
5
0
-0.472 -5
-10 -14
© 2005 P. S. Shiakolas
-12
-10
-8
-6 -4 Real Axis
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
-2
0
2
4
STATIC ERROR CONSTANTS Slope of magnitude curve at low frequencies
G( s) = Type 0
k (T1s + 1)(T2 s + 1)....
s N (Ta s + 1)(Tb s + 1)....
(s = jω )
Static position
At low frequencies, the magnitude of G(jω) equals to k or kp
lim G ( jω ) = k = k p
ω →0
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
kp db
| | db
-20 db/dec -40 db/dec
ω (log)
Type 1 Static Velocity
k (s + 2 ) G( s) = s (s + 5)(s + 8)
| | db
k ( jω + 2 ) G ( jω ) = jω ( jω + 5)( jω + 8) © 2005 P. S. Shiakolas
1
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ω (log)
At low frequencies, kv , ω << 1 | | G ( jω ) = jω db kv = ω1
-20 db/dec 20logkv 0
ω1 ω (log)
ω =1
Type 2 Static Acceleration At low frequencies, the slope is -40 db/dec
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
G ( jω ) =
20 log
ka
( jω )2
ka
( jω )2 ka
( jω ) ka
ω
2
© 2005 P. S. Shiakolas
= 20 log 1
=1
=1
, ω << 1
-40 db/dec
| | db
20logka 0
ω (log) ω =1
⇒ ωa = k a
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ωa = k a
COMPENSATION • Series/Cascade R
• Feedback
─
• Output/Load • Input
R
Gp
Gc
H Gc
─
Gp H
P, PI, PD, PID, VF © 2005 P. S. Shiakolas
C
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C
LEAD / LAG COMPENSATORS
s +α G( s) = k s+β Lead:
Relates to phase lead
Lag:
Relates to phase lag
Lead-Lag:
Low frequencies – Lag High frequencies – Lead
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Lead: - Improves transient response - Small Effect if ess - Accentuates high frequency noise effects Lag: - Increases transient response time - Appreciable improvement in ess steady state accuracy - Suppresses high frequency noise effects Lead-Lag: - Combines characteristics of both (diff. freq. range) - Increases system order by 2. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
LEAD COMPENSATORS 1 s+ 1 1 s+ T T G (s) = = β s+ 1 ⎛1⎞1 βT s + ⎜⎜ ⎟⎟ ⎝β ⎠ T
β <1 1 s + ⎛1⎞ T = ⎜⎜ ⎟⎟ ⎝ β ⎠ s + α 1T LEAD Pole to the left of zero © 2005 P. S. Shiakolas
α >1 1 −α T
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
1 − T
| | db 0
LEAD
s+ 1 T G ( s ) = kc s+ 1 αT
90 45 o 0 - 45 - 90
0 <α <1 STATIC / DC GAIN
kcα
α min = 0.07 Maximum Phase Lead = 60o © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Find Maximum Phase Lead Max. value occurs at ωMEAN, geometric mean between pole and zero Is the halfway distance on the log plot
1 1 1 ωm = zp = = T αT T α T ωj + 1 − α T ωj + 1 G ( jω ) = kcα αTωj + 1 − αTωj + 1 G ( jω ) =
( 1 + αT 2ω 2 + jTω (1 − α )) (αTω )2 + 1
© 2005 P. S. Shiakolas
kcα
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
1−α φmax ω = = atan 2 α T α 1
1−α sin φmax = 1+ α HOW TO DESIGN A LEAD COMPENSATOR BASED ON FREQUENCY APPROACH R ─
© 2005 P. S. Shiakolas
Gp
C
4 Gp = s (s + 2 )
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Design a Compensator such that • static velocity error const. kv=20 sec-1 • phase margin at least 50o • gain margin at least 10 db
s+ 1 Ts + 1 T = kcα G ( s ) = kc αTs + 1 s+ 1 αT
0 <α <1
OLTF compensated system
4 Ts + 1 GcG p = kcα αTs + 1 s (s + 2 ) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Let kcα = k 1 Ts + 1 GcG p = 4k (αTs + 1) s(s + 2) 1 Let 4k = G1 s (s + 2 ) Ts + 1 GcG p = G1 (αTs + 1)
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
4k kv = 20
kv = lim sGcG p s →0
Ts + 1 1 4k = lim s s (s + 2) s → 0 αTs + 1 kv = 2k ⇒ k = 10 = kcα © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
k = 10 1 Bode Plot : 4k s (s + 2 ) Examine new Bode Plot with k = 10 kv = 20 sec-1 (See Bode Plot: Draw a line extension of the low frequency – 20 db/dec part of the magnitude plot and find its intersection with the 0 db line) Find the new phase and gain margins. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
GM = ∞ PM = 180 − 162 = 18o Requirement for PM ≥ 50° Therefore, need to add phase (without changing k) Amount of phase needed = 50 – 18 =32° • Need a LEAD compensator • Lead compensator will shift gain crossover frequency to the right, thus reducing the PM. So, require the compensator to provide additional phase of (say) 5°. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Therefore, the total phase lead needed, 32° + 5° = 37° Determine the frequency at which the magnitude of 1 the uncompensated system G1 equals − 20 log
α
Select this frequency to be the new gain crossover frequency. • Determine the attenuation factor α 1−α sin φ = Solve for α 1+ α © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
• Identify frequency at which max phase lead occurs 1 ωmax = T α The frequency corresponding to 40° PM is about 0.7 rad/sec. The new gain crossover frequency must be chosen near this frequency. 1 ( zero of lag compensator) ω= T Lets select this to be 0.1 rad/sec (not far below 0.7 ?? small modification on phase plot) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
G1 ( jω ) = −20 log
1
α
1 = −20 log = −6.2 db 0.24 Refer Bode Plot and find ω for G1 ( jω ) = −6.2db
ω = 9 rad/sec Determine the corner frequencies for L.C.
zero
1 ωz = T
© 2005 P. S. Shiakolas
&
pole
1 ωp = αT
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ωcg = 9 =
1 T α
1 = 4.41 & T
⇒ T = 0.23 1 = 18.37 αT
L.C.
s+ 1 s + 4.41 T Gc = kc = kc s + 18.37 s+ 1 αT 0.23s + 1 or Gc = kcα 0.054 s + 1 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
10 k = kcα ⇒ kc = = = 41.7 α 0.24 k
s + 4.41 Gc ( s ) = 41.7 s + 18.37 1 G p ( s) = 4 s (s + 2 )
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
EXAMPLE: Position Control System Preamplifier
R
Shaft Motor Amplifier & Load velocity
k
─ Compensator
1/s 100 s + 100
C
1 s + 36
Design a lead compensator to yield a 20 % overshoot kv =40 sec-1, peak time Tp = 0.1 sec Solution: Closed loop bandwidth to meet the speed of response requirement imposed by Tp=0.1 sec © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ω BW =
π Tp 1 − ζ 2
1 − 2ζ 2 + 4ζ 4 − 4ζ 2 + 2
ζ is found from overshoot requirement, for 20%
ζ = 0.456 ω BW = 46.59 rad • Meeting kv requirement, need to increase the constant gain, gain = 1440 • Plot the uncompensated gain bode plot. Find PM and GM. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
• 20% overshoot requires
φm = tan
−1
2ζ − 2ζ 2 + 1 + 4ζ 4
= 48.15°
⎛ s + 25.27 ⎞ G ( s ) = 2.38⎜ ⎟ ⎝ s + 60.17 ⎠
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
LAG COMPENSATION
1 + s Ts + 1 T = kc Gc ( s ) = kc β βTs + 1 s+ 1 βT | | 0 zero : − 1 db T pole : − 1
βT
o 0
1 − T © 2005 P. S. Shiakolas
β >1
−α
1 T
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Example
1 Gp = s(s + 1)(0.5s + 1)
R ─
Gp
• Static velocity error constant kv = 5 sec-1 • PM at least 40° • GM at least 10 db Plot BP Solution
ω g = 0.75 rad GM = 9.5 db ωcp = 1.4 rad PM = 32° kv = 1sec −1 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C
Ts + 1 Gc ( s ) = kc β βTs + 1
β >1
G1 ( s ) = kG p ( s ) = adjust gain k to meet kv req. kv = lim sG ( s ) G p ( s ) s →0
Ts + 1 1 = lim s k s → 0 β Ts + 1 s (s + 1)(0.5s + 1) kv = k = 5 Plot new BP including k = 5 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ωcg = 1.8
GM = −4.4 db
ωcp = 1.4
GM = −13°
Satisfying kv requirement made the system unstable Desired phase is " padded" by about 10° - 12° ⇒ Required phase is 52° ⇒ Phase angle for uncompensated system is - 128° at ω = 0.5 rad/sec ⇒ Select this ω as the new gain crossover frequency Mag plot = 0 db at ω = 0.5 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
⇒ The attentuation to bring the magnitude plot to 0 db 1 ⎛ attenuation to bring ⎞ ⎟⎟ 20 log = ⎜⎜ β ⎝ mag. plot to 0 db ⎠ 1 20 log = −20 db β
β = 10 5 k ⇒ k = kc β ⇒ kc = = = 0.5 β 10 1 ⇒ Plot of lag compensator ω = =1 βT © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ω zero = 0.1 ω pole = 0.01 kc = 0.5 s+ 1 Ts + 1 T = kc ⇒ Gc ( s ) = kc β βTs + 1 s+ 1 βT s + 0.1 = 0.5 s + 0.01
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
PHYSICAL REALIZATION OF CONTROLLERS & COMPENSATORS Ro Ri Vo
Vi
⎛ Ro ⎞ Vo ( s ) = −⎜⎜ ⎟⎟ Vi ⎝ Ri ⎠ Proportional Gain © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Zo Zi Vi
Vo
Z: Impedence
Zo Vo ( s ) = − Vi Zi R2 Vi
R1
C2
Vo
Zo Vo ( s ) = − Vi Zi © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
R1
R2
R − 2
R
C
1
C
R
Gain Int. Diff. PI
R C
PD
− RC s
− RCs
C
R
R2
⎞ ⎛s+ 1 R ⎜ R2C ⎟ − 2⎜ ⎟ R1 ⎜ s ⎟ ⎠ ⎝ ⎞ − R2C ⎛⎜ s + 1 R1C ⎟⎠ ⎝
R1 © 2005 P. S. Shiakolas
R1
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C1
R2
PID
C2
R1
τi 1 ⎡ ⎤ ⎛ R2 C1 ⎞ R1C2 ⎥ ⎢ ⎟⎟ + R2C1s + − ⎜⎜ + ⎢⎝ R1 C2 ⎠ s ⎥ ⎢⎣ ⎥⎦ τ d
kp
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
LAG/ LEAD
⎛ s+ 1 ⎞ C1 ⎜ R1C1 ⎟ − ⎜ ⎟ 1 C2 ⎜ s + ⎟ R C 2 2⎠ ⎝
© 2005 P. S. Shiakolas
C1
C2
R1
R2
LAG LEAD
R2C2 > R1C1 R1C1 > R2C2
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Passive Components LAG Vi
R1
Vo
R2 C
s+ 1
Vo ( s ) R2 = Vi ( s ) R1 + R2 s + 1
© 2005 P. S. Shiakolas
R2C
(R1 + R2 )C
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
LEAD
C Vi
R2 V o
R1
s+ 1
Vo ( s ) = Vi ( s ) s + 1
© 2005 P. S. Shiakolas
R1C1 + 1 R1C R2C
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
LAG-LEAD
C1 Vi
Vo ( s ) = Vi ( s )
s 2 + ⎛⎜ ⎝
R1
R2 V o C2
⎛s + 1 ⎞⎛ s + 1 ⎞ ⎜ ⎟⎜ ⎟ R C R C 1 1⎠⎝ 2 2⎠ ⎝ ⎞s+ 1 1 + 1 + 1 R1C1 R2C2 R2C1 ⎟⎠ R1R2C1C2
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Example: Implementation of PID ( s + 55.91)(s + 0.5) Gc ( s ) = s
27.95 Gc ( s ) = s + 56.41 + s From Realization table, ⎛ R2 C1 ⎞ ⎜⎜ ⎟⎟ = 56.41 + ⎝ R1 C2 ⎠
1 = 27.95 R1C2
R2C1 = 1 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
3 equations, 4 unknowns. Let C2 = 0.1μF ⇒ C1 = 5.59 μF R1 = 357.73 kΩ
R2 = 178.86 kΩ R2=179 kΩ
Vi
C1= 5.6 μF
C2= 0.1 μF
R1= 358 kΩ
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Vo
Determination of OLTF from Experimental Data Sensor
Shaker Table
Sensor Æ measures displacement Shaker table Æ generates sinusoidal signal of certain Amplitude & frequency
• Turn displacement into dB • Plot Bode Plot •
Determine if sys is Min/Non-Min phase examine high frequencies slope ~ -60dB/dec Æ relative order -3 phase = -270° Æ relative order -3 Æ Minimum phase
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
•
Identify quadratic terms Æ establish natural frequency for quadratics Æ at φ = −180° ω = 2 rad/sec
max. amplitude (ω = 2 ) ~ 14 db ⇒ establish ωn = 2 M p = 14 db ⇒ 20 log M p = 14 db ⇒ M p = 5.02 Mp =
© 2005 P. S. Shiakolas
1 2ζ 1 − ζ 2
⇒ ζ = 0.1
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
• Determine corner frequencies Æ examine slope changes. Corner freq at w = 0.5 for first order numerator.
j
ω ωcorner
+ 1 ⇒ ωcorner = 0.5
• Low Frequencies Establish double pole at origin
ka = 1
© 2005 P. S. Shiakolas
The extension of -40 dB/dec (low freq). Slope intersects the 0 dB line at w = 1.
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
• Set-up T.F. Æ Relative Order = -3
G ( jω ) =
=
© 2005 P. S. Shiakolas
⎛ ω ⎞ ⎜⎜ j + 1⎟⎟ ⎝ ωc ⎠ ⎛ ⎛ ω ⎞2 ⎞ ω ( jω )2 ⎜⎜ ⎜⎜ j ⎟⎟ + 2ζ j + 1⎟⎟ ωn ωn ⎠ ⎝ ⎝ ⎠ ⎛ ω ⎞ j + 1 ⎜ ⎟ ⎝ 0.5 ⎠ 2 ⎞ ⎛ ω 2⎜⎛ ω ⎞ ( jω ) ⎜ ⎜ j ⎟ + 2(0.1) j + 1⎟⎟ 2⎠ 2 ⎝ ⎠ ⎝ ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
G (s) =
© 2005 P. S. Shiakolas
(
8(s + 0.5)
s 2 s 2 + 0.4 s + 4
)
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
TUNING RULES FOR PID CONTR. Ziegler-Nichols Rules First Method: • Obtain unit step response of system Æ if unit step response looks like an S-shaped curve C(t) Inflection point Tangent at inflection point
L
t Tunit step
© 2005 P. S. Shiakolas
Plant
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Use L & T to define the gains of PID controller
⎞ ⎛ 1 Gc ( s ) = k p ⎜⎜1 + + τ d s ⎟⎟ ⎠ ⎝ τis Type of controller
kp
τi
τd
P
T/L
-
-
PI
0.9 T/L
L/0.3
0
PID
1.2 T/L
2L
0.5L
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
PID Controller T⎛ 1 ⎞ + 0.5 Ls ⎟ Gc ( s ) = 1.2 ⎜1 + L ⎝ 2 Ls ⎠ 1⎞ ⎛ ⎜s + ⎟ L⎠ = 0.6T ⎝ s
2
Im
1 − L © 2005 P. S. Shiakolas
0
Re
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Second Method: Controller
r(t)
Plant
c(t)
─
• Assume that (τ i = ∞ & τ d = 0 ) • Only proportional kp • Start increasing kp from 0 Æ critical value kcr s.t. the output sustains oscillations © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Pcr
C(t)
t
Type of controller
kp
τi
τd
P
0.5kcr
∞
0 0 0.125Pcr
PI
0.45kcr
1 Pcr 1.2
PID
0.6kcr
0.5Pcr
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
PID Controller ⎛ ⎞ 1 + 0.125 Pcr s ⎟⎟ Gc ( s ) = 0.6kcr ⎜⎜1 + ⎝ 0.5Pcr s ⎠ ⎛ 4 ⎜⎜ s + Pcr ⎝ = 0.075kcr Pcr s
© 2005 P. S. Shiakolas
⎞ ⎟⎟ ⎠
2
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Application of Z-N: r(t) ─
Gc
Gp
c(t)
Desired ~ 25% overshoot 1 G p (s) = s (s + 1)(s + 5) Use 2nd Z-N Æ start with kp kp C = R s (s + 1)(s + 5) + k p
RH : s 3 + 6 s 2 + 5s + k p = 0 ⇒ kcr = 30 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Identify Pcr (critical period)
Period =
2π
ω
CE : s 3 + 6 s 2 + 5s + 30 = 0 2π CE ⇒ s = jω ⇒ ω = 5 ⇒ Pcr = = 2.81 5 Use table to find k p ,τ i ,τ d k p = 0.6, kcr = 18, τ i = 0.5, Pcr = 1.41,
τ d = 0.125, Pcr = 0.35 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Gc
( s + 1.42)2 ( s ) = 6.32 s
Experiment with double zero at s = −0.65 s = −1.4235
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
ANALYSIS OF CONTROL SYSTEM IN STATE SPACE REPRESENTATION • Assume that system is representation in terms of nfirst order dfq’s • Preferred representations – Canonical Forms Controllable, Observable, Diagonal, Jordan • Controllable – important when discussing poleplacement approach for control system design • MATLAB: ctrb(a,b) obsv(a,c) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Example:
Controllable:
Observable:
Y (s) s+3 = U ( s ) s 2 + 3s + 1
⎛ x&1 ⎞ ⎡ 0 ⎜⎜ ⎟⎟ = ⎢ ⎝ x&2 ⎠ ⎣− 2
⎛ x1 ⎞ y = [3 1]⎜⎜ ⎟⎟ + 0 ⎝ x2 ⎠ ⎛ x&1 ⎞ ⎡0 − 2⎤ ⎛ x1 ⎞ ⎛ 3 ⎞ ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ u ⎜⎜ ⎟⎟ = ⎢ ⎥ ⎝ x&2 ⎠ ⎣1 − 3⎦ ⎝ x2 ⎠ ⎝ 1 ⎠ y = [0
© 2005 P. S. Shiakolas
1 ⎤ ⎛ x1 ⎞ ⎛ 0 ⎞ ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ u ⎥ − 3⎦ ⎝ x2 ⎠ ⎝ 1 ⎠
⎛ x1 ⎞ 1]⎜⎜ ⎟⎟ + 0 ⎝ x2 ⎠
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Controllability: A system is said to be controllable at time to if it is possible by means of an unconstrained control vector to transfer the system from an initial state x(to) to any other state in a finite time interval. x(to) Æ x(tf) Observability: A system is said to be observable at time to if, with the system at state x(to) it is possible to determine this state from the observation of the output over a finite time interval. © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Observability:
⎡ C ⎤ ⎢ CA ⎥ ⎢ ⎥ Q = ⎢ CA2 ⎥ ⎢ ⎥ ⎢ M ⎥ ⎢⎣CAn −1 ⎥⎦
⇒ Find det Q ⇒ if Q is rank n
Controllability:
[
Pc = B
AB
A2 B
L
An −1B
]
if det Pc ≠ 0 ⇒ system is controllable © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Example:
⎛ x&1 ⎞ ⎡ − 3 ⎜⎜ ⎟⎟ = ⎢ ⎝ x&2 ⎠ ⎣− 2 y = [1
1 ⎤ ⎛ x1 ⎞ ⎛ 0 ⎞ ⎜⎜ ⎟⎟ + ⎜⎜ ⎟⎟ u ⎥ − 1.5⎦ ⎝ x2 ⎠ ⎝ 1 ⎠
⎛ x1 ⎞ 1]⎜⎜ ⎟⎟ ⎝ x2 ⎠
Controllability: Pc = [B
⎛1 AB ] = ⎜⎜ ⎝4
1⎞ ⎟⎟ 4⎠
⇒ det Pc = 0 X1( s) s + 2.5 = U ( s ) (s + 2.5)(s − 1) © 2005 P. S. Shiakolas
pole/zero cancellation
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Observability:
⎡1 Q=⎢ ⎣− 5
1⎤ ⇒ det Q = 2.5 + 5 = 7.5 ≠ 0 ⎥ 2.5⎦
System observable.
© 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Controllability / Observability and Pole / Zero Cancellation
⎛1 x& = ⎜⎜ ⎝0
0⎞ ⎛ b1 ⎞ ⎟⎟ x + ⎜⎜ ⎟⎟ u 2⎠ ⎝ b2 ⎠
y = (c1
c2 ) x
y = c(sI − A)−1 Bu y (b1c1 + b2c2 ) s − (2b1c1 + b2c2 ) = (s − 1)(s − 2) u b2c2 s − b2c2 b2c2 (s − 1) Let b1 = 0 ⇒ TF = = (s − 1)(s − 2) (s − 1)(s − 2) © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
Let b2 = 0 ⇒ Controllability c1 = 0
[B
AB ]
c2 = 0
⎡ b1 ⎢ ⎣b2
⎛1 ⎜⎜ ⎝0
⎡ b1 ⎢b ⎣ 2
b1 ⎤ ⇒ 2b1b2 − b1b2 = b1b2 ⎥ 2b2 ⎦
0⎞ ⎟⎟ 2⎠
⎛ b1 ⎞⎤ ⎜⎜ ⎟⎟⎥ ⎝ b2 ⎠⎦
Observability
C1 ⎡ ⎡C ⎤ ⎢ ⎢CA⎥ = ⎢(C ⎣ ⎦ ⎢ 1 C2 ) ⎣ © 2005 P. S. Shiakolas
⎛1 ⎜⎜ ⎝0
C2
⎤ ⎡C1 ⎥ 0⎞ = ⎢ ⎟⎟⎥ ⎣C1 2 ⎠⎥⎦
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
C2 ⎤ 2C2 ⎥⎦
det = 2C1C2 − C1C2 = C1C2 Pole Placement Define the location of the closed loop poles Æ Estimate the controller gains to yield the defined C.L. poles. All system states are successfully measured. If a state is non-measurable or non-available, then we can implement an observer to estimate the state. x& = Ax + Bu y = Cx + Du
Assume Controllability
Select control signal u = − kx © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
x& = Ax − Bkx
x& = ( A − Bk )x
( A− Bk )t x(t ) = e x(0)
Stability and transient performance determined from eigen values of (A-Bk)
Generate the CE for (A-Bk); Generate the CE for desired pole locations. Equate the two. Evaluate k through direct substitution. Low order system, n = 3
k = (k1
k2
k3 )
Desired eigen values are μ1, μ 2 , μ3 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
CE : (s − μ1 )(s − μ 2 )(s − μ3 ) = 0 CE from A − Bk sI − ( A − Bk ) = 0 Equate the two CEs Ackerman’s Formula or Principle
x& = Ax + Bu ⎫ ⎬ x& = ( A − Bk )x u = −kx ⎭ k = [0
0
K
[
Controllab ility matrix
© 2005 P. S. Shiakolas
]
−1
1] B AB K A B φ ( A) 14444244443 n −1
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
φ ( A) : State transition matrix φ ( A) = An + α1 An −1 + α 2 An − 2 + K + α n −1 A + αI CE of new system with k sI − ( A − Bk ) = s n + a1s n −1 + a2 s n − 2 + K + an Example
⎡0 A = ⎢⎢ 0 ⎢⎣− 1
1 0 −5
0⎤ 1 ⎥⎥ − 6⎥⎦
⎛ 0⎞ ⎜ ⎟ B = ⎜ 0⎟ ⎜1⎟ ⎝ ⎠
Desired C.L. poles : s = −2 ± 4 j © 2005 P. S. Shiakolas
s = −10. Find k .
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
[
Check for controllability B
AB
A2 B
]
Find det = −1 ⇒ state controllable Let k = [k1
k2
k3 ]
CE of new system ( A − Bk ) sI − ( A − Bk ) = s 3 + (6 + k3 )s 2 + (5 + k 2 )s + 1 + k1 = 0 CE based on desired eigen value loc. s 3 + 14 s 2 + 60 s + 200 = 0 Equate coefficients ⇒ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
6 + k3 = 14 ⇒ k3 = 8 5 + k 2 = 60 ⇒ k 2 = 55 1 + k1 = 200 ⇒ k1 = 199 Ackerman's k = [0
0
[
1] B
AB
]
A2 B φ ( A)
⎡ φ ( A) = A + 14 A + 60 A + 200 I = ⎢ ⎣ 3
2
After all algebra ⇒ k = [199
© 2005 P. S. Shiakolas
55
8]
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
⎤ ⎥ ⎦
OBSERVER x& = Ax + Bu y = Cx Observer x&ˆ = Axˆ + Bu + L( y − Cxˆ ) as t → ∞, xˆ → x x(to ) is not known xˆ (to ) estimate Observer error, e(t ) = x(t ) − xˆ (t ) We want the observer to produce a result s.t. e(t) → 0 as t → ∞ © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
If the system is completely observable Æ we can find L so that the tracking error e(t) is asymptotically stable e& = x& − xˆ&
= Ax + Bu − Axˆ − Bu − L( y − Cxˆ ) = A( x − xˆ ) − L( y − Cxˆ ) e& = Ax − Axˆ − Ly + LCxˆ Roots or eigen values = A( x − xˆ ) − LC ( x − xˆ ) should be in the negative s-plane &e = ( A − LC )( x − xˆ ) e& = ( A − LC ) e
det sI − ( A − LC ) = 0 © 2005 P. S. Shiakolas
ME 5303 Classical Methods of Control Systems – Analysis & Synthesis
