The Handbook of Formulas and Tables for Signal Processing. “Laplace Transforms”

Poularikas A. D. “Laplace Transforms” The Handbook of Formulas and Tables for Signal Processing. Ed. Alexander D. Poularikas Boca Raton: CRC Press LLC,1999

2 Laplace Transforms 2.1 2.2 2.3 2.4

Definitions and Laplace Transform Formulae Properties Inverse Laplace Transforms Relationship Between Fourier Integrals of Causal Functions and One-Sided Laplace Transforms 2.5 Table of Laplace Transforms 2.2 Table of Laplace Operations 2.3 Table of Laplace Transforms References Appendix 1 Examples • Inversion in the Complex Plane • Complex Integration and the Bilateral Laplace Transform

2.1 Definitions and Laplace Transform Formulae 2.1.1 One-Sided Laplace Transform ∞

F( s) =

∫ f (t ) e

− st

dt

s = σ + jω

0

f (t) = piecewise continuous and of exponential order

2.1.2 One-Sided Inverse Laplace Transform f (t ) =

1 2πj

σ+ j ∞

∫ F( s) e

st

ds

σ− j ∞

where the integration is within the regions of convergence. The region of convergence is half-plane σ < Re{s}.

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2.1.3 Two-Sided Laplace Transform ∞

F( s) =

∫

f (t ) e − st dt

s = σ + jω

−∞

f (t) = piecewise continuous and of exponential order

2.1.4 Two-Sided Inverse Laplace Transform f (t ) =

1 2πj

σ+ j ∞

∫ F( s) e

σ− j ∞

where the integration is within the regions of convergence which is a vertical strip σ1 < Re{s} < σ2.

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st

ds

2.2 Properties 2.2.1 Properties of the Laplace Transform (one sided) TABLE 2.1 Laplace Transform Properties 1. 2.

Linearity L {K1 f1 (t ) ± K 2 f 2 (t )} = L {K1 f1 (t )} ± L {K 2 f2 (t )} = K1 F1 (s) ± K 2 F2 (s) Time derivative

3.

d  L  f (t ) = sF(s) − f (0 + )  dt  Higher time derivative  dn  L  n f (t ) = s n F(s) − s n −1 f (0 + ) − s n − 2 f (1) (0 + ) − L − f ( n −1) (0 + )  dt  where f (i)(0+), i = 1,2,…,n – 1 is the ith derivative of f (·) at t = 0+.  Integral with zero initial condition L  

4.

∫

t

0

 F(s) f (ξ) dξ  = s 

 F(s) f ( −1) (0 + ) + f (ξ) dξ  = where f ( −1) (0 + ) = lim t→0 + s s −∞ 

5.

 Integral with initial conditions L  

6.

Multiplication by exponential L {e ± at f (t )} = F(s m a)

7.

Multiplication by t L {t f (t )} = −

8.

Time shifting L { f (t ± λ ) u(t ± λ )} = e ± sλ F(s)

9.

  t  Scaling L  f  = aF( as) ;   a

∫

t

d F(s) ; ds

L { f (ta)} =

L {t n f (t )} = ( −1) n

1  s F a  a

∫

t

−∞

f (ξ) dξ

dn F(s) ds s

a>0

 t ∆ Time convolution L  f1 (t − τ) f2 ( τ) dτ  L { f1 (t ) ∗ f2 (t )} = F1 (s) F2 (s)  0  Frequency convolution

∫

10. 11.

L { f1 (t ) f 2 (t )} = 1 2 πj

∫

x + j∞

x − j∞

F1 ( z ) F2 (s − z ) dz = 1 {F1 (s) ∗ F2 (s)} 2 πj

where z = x + jy, and where x must be greater than the abscissa of absolute convergence for f1(t) over the path of integration. 12.

Initial value lim f (t ) = lim sF(s) provided that this limit exists.

13.

Final value lim f (t ) = lim sF(s) provided that sF(s) is analytic on the jω axis and in the right half of the s plane

14.

 f (t )  Division by t L  =  t 

t→0 +

t →∞

s→∞

s→ 0 +

∞

∫ F(s ′) ds ′ s

∫e L { f (t )} = T

f (t) periodic

15.

− st

f (t ) dt

0

1 − e − sT

f (t ) = f (t + T )

2.2.2 Methods of Finding the Laplace Transform 1. 2. 3. 4.

Direct method by solving (2.1.1). Expand f (t) in power series if such an expansion exists. Differentiation with respect to a parameter. Use of tables.

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2.3 Inverse Laplace Transforms 2.3.1 Properties 1. Linearity L−1 {c1 F1 (s) ± c2 F2 (s)} = c1 f1 (t ) ± c2 f2 (t ) 2. Shifting L−1 {F(s − a)} = e at f (t ) 3. Time shifting L−1 {e − as F(s)} = f (t − a)

t>a

4. Scaling property L−1 {F(as)} = 1 f t a a

()

a>0

5. Derivatives L−1 {F ( n ) (s)} = ( −1) n t n f (t )

F ( n ) ( s) =

d n F( s) ds n

6. Multiplication by s L−1 {sF(s) − f (0 + )} = L {sF(s)} − f (0 + ) L {1} = f (1) (t ) + f (0)δ(t )  F( s)  = 7. Division by s L−1    s 

∫

t

f (t ′) dt ′

0

8. Convolution L−1 {F(s) H (s)} =

∫ F(u)H(t − u) du = F(s) ∗ H(s) t

0

2.3.2 Methods of Finding Inverse Laplace Transforms 1. Partial fraction method: Any rational function P(s)/Q(s) where P(s) and Q(s) are polynomials, with the degree of P(s) less than that of Q(s), can be written as the sum of rational functions, known as partial fractions, having the form A/(as + b)r, (As + B)/(as2 + bs + c)r, r = 1,2,… . 2. Expand F(s) in inverse powers of s if such an expansion exists. 3. Differentiation with respect to a parameter. 4. Combination of the above methods. 5. Use of tables. 6. Complex inversion (see Appendix 1).

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2.4 Relationship Between Fourier Integrals of Causal Functions and One-Sided Laplace Transforms 2.4.1 F (ω) from F (s) F (ω ) =

∞

∫e

− jωt

t≥0 t<0

 f (t ) f (t ) =   0

f (t ) dt

0

a) The region of convergence of F(s) contains the jω axis in its interior, σ < 0 (see 2.1.2) F(ω ) = F(s) s= jω b) If the axis jω is outside the region of convergence of F(s), σ > 0, then F(ω) does not exist; the function f (t) has no Fourier transform. c) Let σ = 0, F(s) is analytic for s > 0, and has one singular point on the jω axis, hence, F(s) = 1 1 and there we obtain or F(s) = L{e jω ot u(t )}. But F{e jω ot u(t )} = πδ(ω − ω o ) + jω − jω o s − jω o the correspondence F( s) =

1 s − jω o

F(ω ) = F(s) s= jω = πδ(ω − ω o ) + F(s) s= jω

Also F( s) =

1 (s − jω o ) n

F (ω ) =

πj n−1 ( n−1) (ω − ω o ) + F(s) s= jω δ (n − 1)!

δ(n–1)(·) = the (n – 1)th derivative. d) F(s) has n simple poles jω1, jω2,…, jωn and no other singularities in the half plane Re s ≥ 0. F(s) n

takes the form F(s) = G(s) +

∑ s − jω an

n =1

where G(s) is free of singularities for Re s ≥ 0. The n

correspondence is n

F(ω ) = G(s) s= jω +

∑ n =1

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an s − jω n

n

s = jω

+π

∑ a δ (ω − ω ) n

n =1

n

2.5 Table of Laplace Transforms TABLE 2.2 Table of Laplace Operations F(s)

f (t)

∞

1

∫e

− st

f (t ) dt

f(t)

0

2 3

AF(s) + BG(s) sF(s) – f(+0)

Af(t) + Bg(t) f ′(t)

4

s n F(s) − s n −1 f ( +0) − s n − 2 f (1) ( +0) − L − f ( n −1) ( +0)

5

1 F(s) s

∫

1 F(s) s2

∫∫

6 7

f (n)(t) t

t

τ

0

0

∫

F1(s)F2(s)

t

0

8 9

–F′(s) (–1)nF(n)(s)

f (λ ) dλ dτ

f1 (t − τ) f 2 ( τ) dτ = f1 ∗ f2

tf(t) tn f(t)

∞

10

f ( τ) dτ

0

∫ F( x) dx

1 f (t ) t

s

eat f(t) f(t – b), where f(t) = 0; t < 0

11 12

F(s – a) e–bsF(s)

13

F(cs)

1 t f c  c

14

F(cs – b)

1 ( bt ) / c  t  e f  c c

∫e a

15

− st

f (t ) dt

0

f(t + a) = f(t) periodic signal

1 − e − as

∫e a

16

− st

f (t ) dt

0

f(t + a) = –f(t)

1 + e − as

17

F(s) 1 − e − as

18

F(s) coth

19

p (s) , q (s) = (s − a1 )(s − a2 )L(s − a m ) q (s)

∑ q ′( a ) e

p (s) φ(s) = q ( s) ( s − a) r

e at

20

f1(t), the half-wave rectification of f(t) in No. 16. as 2

.

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f2(t), the full-wave rectification of f(t) in No. 16. m

p (an )

an t

n

1

r

φ ( r − n ) ( a)

∑ (r − n)! n =1

t n −1 +L (n − 1)!

TABLE 2.3 Table of Laplace Transforms F(s) 1

sn

2

s

3

1

f(t) δ(n)(t) dδ(t ) dt δ(t)

4

1 s

1

5

1 s2

t

6

1 (n = 1, 2,L) sn

7

1 s

8

s–3/2

9

nth derivative of the delta function

s −[ n +(1 / 2 )] (n = 1, 2,L)

t n−1 (n − 1)! 1 πt t π

2

2 n t n −(1 / 2 ) 1 ⋅ 3 ⋅ 5L(2n − 1) π

10

Γ(k ) ( k ≥ 0) sk

tk–1

11

1 s−a

eat

12

1 ( s − a) 2

teat

13

1 (n = 1, 2,L) ( s − a) n

1 t n −1e at (n − 1)!

14

Γ(k ) ( k ≥ 0) ( s − a) k

t k −1e at

15

1 (s − a)(s − b)

1 (e at − e bt ) ( a − b)

16

s (s − a)(s − b)

1 ( ae at − be bt ) ( a − b)

17

1 (s − a)(s − b)(s − c)

−

18

1 ( s + a)

19

1 s ( s + a)

1 (1 − e − at ) a

20

1 s 2 ( s + a)

1 − at (e + at − 1) a2

21

1 s ( s + a)

1 1 at 2 1 − at  −t+ − e  2  2 a a a 

22

1 (s + a)(s + b)

1 (e − at − e − bt ) (b − a)

23

1 s (s + a)(s + b)

 1  1 (be − at − ae − bt ) 1 + ab  ( a − b) 

3

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(b − c) e at + (c − a) e bt + ( a − b) e ct ( a − b)(b − c)(c − a)

e–at valid for complex a

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)  1  2 − bt 2 − at  ( a − b) ( a e − b e ) + abt − a − b  

24

1 s 2 (s + a)(s + b)

1 ( ab) 2

25

1 s 3 (s + a)(s + b)

( a + b) 1  a3 − b3 1 1  b − at a  + t2 − t+ e − 2 e − bt    ( ab)  ( ab) 2 ( a − b) 2 ( a − b)  a 2 ab b

26

1 (s + a)(s + b)(s + c)

1 1 1 e − at + e − bt + e − ct (b − a)(c − a) ( a − b)(c − b) ( a − c)(b − c)

27

1 s (s + a)(s + b)(s + c)

1 1 1 1 − e − at − e − bt − e − ct abc a(b − a)(c − a) b( a − b)(c − b) c ( a − c)(b − c)

28

1 s 2 (s + a)(s + b)(s + c)

1  ab(ct − 1) − ac − bc + 2 e − at 2  ( abc ) − a ( b a )(c − a)   1 1  + e − bt + 2 e − ct  b 2 ( a − b)(c − b) c ( a − c)(b − c) 

29

1 s 3 (s + a)(s + b)(s + c)

1 2 ab + ac + bc  1 2  ( abc) 3 [(ab + ac + bc) − abc( a + b + c)] − ( abc) 2 t + 2 abc t   1 1 1  − ε − ct e − at − 3 e − bt − 3  a 3 (b − a)(c − a) b ( a − b)(c − b) c ( a − c)(b − c) 

30

1 s2 + a2

1 sin at a

31

s s2 + a2

32

1 s2 − a2

33

s s2 − a2

34

1 s (s 2 + a 2 )

1 (1 − cos at ) a2

35

1 s 2 (s 2 + a 2 )

1 ( at − sin at ) a3

36

1 (s 2 + a 2 ) 2

1 (sin at − at cos at ) 2a 3

37

s (s 2 + a 2 ) 2

t sin at 2a

38

s2 (s + a 2 ) 2

1 (sin at + at cos at ) 2a

39

s2 − a2 (s 2 + a 2 ) 2

40

s (a 2 ≠ b 2 ) (s 2 + a 2 )(s 2 + b 2 )

cos at − cos bt b2 − a2

41

1 ( s − a) 2 + b 2

1 at e sin bt b

42

s−a ( s − a) 2 + b 2

43

1 [(s + a) 2 + b 2 ]n

2

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cos at 1 sinh at a cosh at

t cos at

eat cos bt − e − at 4 n −1 b 2 n

n

r  2n − r − 1 r −1 d [cos(bt )]  ( −2t ) n −1  dt r

∑  r =1

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

44

f(t) r  − at  n  2n − r − 1 r −1 d  ne−1 2 n  [a cos(bt ) + b sin(bt )]   ( −2t ) dt r  4 b  r =1  n − 1    n −1 r    2n − r − 2 r −1 d  − 2b [sin(bt )] r  ( −2t ) r 1 n − dt     r =1

∑

s [(s + a) 2 + b 2 ]n

∑

 at 3 at 3  e − at − e ( at ) / 2  cos − 3 sin 2 2  

45

3a 2 s3 + a3

46

4a3 s 4 + 4a 4

47

s s 4 + 4a 4

1 (sin at sinh at ) 2a 2

48

1 s4 − a4

1 (sinh at − sin at ) 2a 3

49

s s4 − a4

1 (cosh at − cos at ) 2a 2

50

8a 3s 2 (s 2 + a 2 ) 3

51

1  s − 1 s s 

sin at cosh at – cos at sinh at

(1 + a2t2) sin at – cos at

n

Ln ( t ) =

e t d n n −t (t e ) n! dt n

[Ln(t) is the Laguerre polynomial of degree n] 52

1 ( s + a) n

t ( n −1) e − at (n − 1)!

53

1 s ( s + a) 2

1 [1 − e − at − ate − at ] a2

54

1 s 2 ( s + a) 2

1 [at − 2 + ate − at + 2e − at ] a3

55

1 s ( s + a) 3

 1  1 2 2  1− a t + at + 1 e − at   a 3   2 

56

1 (s + a)(s + b) 2

1 {e − at + [(a − b) t − 1] e − bt } ( a − b) 2

57

1 s (s + a)(s + b) 2

 1 1 1 a − 2b  − bt − e − at −  t+ 2 e ab 2 a ( a − b) 2 b ( a − b) 2   b( a − b )

58

1 s 2 (s + a)(s + b) 2

1 1  1 1 2 ( a − b) − b  − bt   e − at + 2 t − − 2 +  2 t+ 3 e a 2 ( a − b) 2 ab  b ( a − b) 2  a b   b ( a − b)

59

1 (s + a)(s + b)(s + c) 2

  1 t + 2c −2 a − b 2  e − ct  ( c − a ) ( c − b ) ( c − b )( c − a )     1 1 − at − bt  + (b − a)(c − a) 2 e + ( a − b)(c − b) 2 e 

60

1 (s + a)(s 2 + ω 2 )

1 1  ω e − at + sin(ωt − φ); φ = tan −1 2 2  a a2 + ω 2 ω a +ω

61

1 s (s + a)(s 2 + ω 2 )

1 1 1 a 1  − sin ωt + 2 cos ωt + e − at  aω 2 a 2 + ω 2  ω ω a

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where n is a positive integer

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)

62

1 s 2 (s + a)(s 2 + ω 2 )

1 1  1 − at  aω 2 t − a 2 ω 2 + a 2 ( a 2 + ω 2 ) e   1 a  + cos(ωt + φ); φ = tan −1  3 2 2  ω + ω a ω 

63

1 [(s + a) 2 + ω 2 ]2

1 − at e [sin ωt − ωt cos ωt ] 2ω 3

64

1 s2 − a2

1 sinh at a

65

1 s 2 (s 2 − a 2 )

1 1 sinh at − 2 t a3 a

66

1 s 3 (s 2 − a 2 )

1 1 (cosh at − 1) − 2 t 2 2a a4

67

1 s3 + a3

a  1  − at 3 3  t at − 3 sin at   e − e 2  cos 3a 2  2 2   

68

1 s 4 + 4a 4

1 (sin at cosh at − cos at sinh at ) 4a 3

69

1 s4 − a4

1 (sinh at − sin at ) 2a 3

70

1 [(s + a) 2 − ω 2 ]

1 − at e sinh ω t ω

71

s+a s[(s + b) 2 + ω 2 ]

 a 1 ( a − b) 2 + ω 2 − bt  2 − + e sin (ωt + φ); 2 b + b2 + ω 2 ω ω    −1  ω  −1  ω   φ = tan   + tan  b a − b 

72

s+a s 2 [(s + b) 2 + ω 2 ]

 ( a − b) 2 + ω 2 − bt  2 1 2 [1 + at ] − 2 2 ab 2 2 + e sin (ωt + φ) (b + ω ) ω (b 2 + ω 2 ) b + ω   ω −1  ω  + 2 tan −1  φ = tan    b a − b 

73

s+a (s + c)[(s + b) 2 + ω 2 ]

 a−c 1 ( a − b) 2 + ω 2 − bt  e − ct + e sin (ωt + φ) 2 2 ω c − b + ω ( ) (c − b ) 2 + ω 2     ω  −1  ω  − tan −1  φ = tan  a − b  c − b 

74

s+a s (s + c)[(s + b) 2 + ω 2 ]

a (c − a )  − ct  c (b 2 + ω 2 ) + c[(b − c) 2 + ω 2 ] e   ( a − b) 2 + ω 2 − bt  1 e sin(ωt + φ)  − 2 2 ω b 2 + ω 2 (b − c) + ω    ω  ω   ω  φ = tan −1 + tan −1 − tan −1   b  a − b  c − b 

75

s+a s 2 ( s + b) 3

a b − 3a  3 a − b a − b 2 2 a − b  − bt t+ + 4 + t + t e b3 b4 2b 2 b 3   b

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TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)

76

s+a (s + c)(s + b) 3

a − c − ct  a − b 2 c−a a − c  − bt e + t + t+ e 2 c − b c − b) 3  (b − c) 3 ( ) ( c − b ( ) 2 

77

s2 (s + a)(s + b)(s + c)

a2 b2 c2 e − at + e − bt + e − ct (b − a)(c − a) ( a − b)(c − b) ( a − c)(b − c)

78

s2 (s + a)(s + b) 2

 b2 a2 b 2 − 2 ab  − bt e − at +  t+ e 2 (b − a) ( a − b) 2   ( a − b)

79

s2 ( s + a) 3

 a 2 2  − at t e 2 − 2 at + 2  

80

s2 (s + a)(s 2 + ω 2 )

a2 e − at − (a 2 + ω 2 )

s2 ( s + a) 2 ( s 2 + ω 2 )

 a 2 a ω 2  − at ω t− 2 e − 2 sin(ωt + φ);  2 2 ( a + ω 2 ) 2  (a + ω 2 )  ( a + ω )   −1  ω   φ = −2 tan   a 

82

s2 (s + a)(s + b)(s 2 + ω 2 )

 a2 b2 − at − bt  (b − a)(a 2 + ω 2 ) e + ( a − b)(b 2 + ω 2 ) e    ω ω  ω  − sin(ω t + φ); φ = − tan −1 + tan −1  2 2 2 2  a  b    a b ( + ω )( + ω ) 

83

s2 (s + a )(s 2 + ω 2 )

−

84

s2 (s + ω 2 ) 2

1 (sin ωt + ωt cos ωt ) 2ω

ω a +ω 2

2

sin(ωt + φ); φ = tan −1

ω  a

2

81

2

2

2

a ω sin( at ) − 2 sin(ωt ) (ω 2 − a 2 ) (a − ω 2 )

s (s + a)[(s + b) 2 + ω 2 ]

 a2 1 (b 2 − ω 2 ) 2 + 4 b 2 ω 2 − bt  e − at + e sin(ωt + φ) 2 2 ( a − b ) + ( a − b) 2 + ω 2 ω ω     ω  −1  −2 b ω  − tan −1  φ = tan  b2 − ω 2   a − b 

86

s2 (s + a) 2 [(s + b) 2 + ω 2 ]

  a[(b − a) 2 + ω 2 ] + a 2 (b − a)  − at a2 te − at − 2   e 2 2 [(b − a) 2 + ω 2 ]2    ( a − b) + ω  2 2 2 2 2  (b − ω ) + 4 b ω − bt  e sin(ωt + φ)  + ω[(a − b) 2 + ω 2 ]    −2b ω   ω   − 2 tan −1 φ = tan −1 2   b − ω2   a − b 

87

s2 + a s ( s + b)

b 2 + a − bt a a e + t− 2 b b2 b

88

s2 + a s 3 ( s + b)

a 2 a 1 t − 2 t + 3 [b 2 + a − ( a + b 2 ) e − bt ] b b 2b

89

s2 + a s (s + b)(s + c)

a (b 2 + a) − bt (c 2 + a) − ct + e − e bc b(b − c) c( b − c )

90

s2 + a s (s + b)(s + c)

b 2 + a − bt c 2 + a − ct a a (b + c) e + 2 e + t− bc b (c − b ) c (b − c) b2c2

2

85

2

2

©1999 CRC Press LLC

2

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)

91

s2 + a (s + b)(s + c)(s + d )

b2 + a c2 + a d2 + a e − bt + e − ct + e − dt (c − b)(d − b) (b − c)(d − c) (b − d )(c − d )

92

s2 + a s(s + b)(s + c)(s + d )

a b2 + a c2 + a d2 + a e − bt + e − ct + e − dt + bcd b(b − c)(d − b) c(b − c)(c − d ) d (b − d )(d − c)

93

s +a s 2 (s + b)(s + c)(s + d )

 a a b2 + a − bt  bcd t − b 2 c 2 d 2 (bc + cd + db) + b 2 (b − c)(b − d ) e    c2 + a d2 + a − ct − dt  + c 2 (c − b)(c − d ) e + d 2 ( d − b)(d − c) e 

94

s2 + a (s 2 + ω 2 ) 2

1 1 ( a + ω 2 )sin ωt − ( a − ω 2 ) t cos ωt 2ω 3 2ω 2

95

s2 − ω2 (s 2 + ω 2 ) 2

96

s2 + a s (s 2 + ω 2 ) 2

a a (a − ω 2 ) t sin ωt − 4 cos ωt − 4 ω ω 2ω 3

97

s ( s + a) (s + b)(s + c) 2

b 2 − ab − bt  c 2 − ac c 2 − 2 bc + ab  − ct e + t+ e 2 b − c (c − b ) (b − c) 2  

98

s ( s + a) (s + b)(s + c)(s + d ) 2

 b 2 − ab c 2 − ac d 2 − ad − bt − ct − dt  (c − b)(d − b) 2 e + (b − c)(d − c) 2 e + (b − d )(c − d ) te    a (bc − d 2 ) + d ( db + dc − 2bc) − dt e  + ( b − d ) 2 (c − d ) 2 

99

s 2 + a1 s + ao s 2 ( s + b)

b 2 − a1 b + ao − bt ao a b−a e + t+ 1 2 o b2 b b

100

s 2 + a1 s + ao s 3 ( s + b)

a1 b − b 2 − ao − bt ao 2 a1 b − ao b 2 − a1 b + ao e + t + t+ 3 2 b b3 b 2b

101

s 2 + a1 s + ao s (s + b)(s + c)

ao b 2 − a1 b + ao − bt c 2 − a1c + ao − ct + e + e bc b(b − c) c( c − b )

102

s 2 + a1 s + ao s 2 (s + b)(s + c)

ao a bc − ao (b + c) b 2 − a1 b + ao − bt c 2 − a1c + ao − ct + t+ 1 e + e b2c2 b 2 (c − b ) c 2 (b − c) bc

103

s 2 + a1 s + ao (s + b)(s + c)(s + d )

b 2 − a1 b + ao − bt c 2 − a1 c + ao − ct d 2 − a1 d + ao − dt e + e + e (c − b)(d − b) (b − c)(d − c) (b − d )(c − d )

104

s 2 + a1 s + ao s (s + b)(s + c)(s + d )

ao b 2 − a1 b + ao − bt c 2 − a1c + ao − ct d 2 − a1 d + ao − dt − e − e − e bcd b(c − b)(d − b) c(b − c)(d − c) d (b − d )(c − d )

105

s 2 + a1 s + ao s( s + b ) 2

ao b 2 − a1 b + ao − bt b 2 − ao − bt − te + e b2 b2 b

106

s 2 + a1 s + ao s 2 ( s + b) 2

ao a b − 2a b 2 − a1 b + ao − bt 2 ao − a1 b − bt t+ 1 3 o + tε + e 2 b b b2 b3

107

s 2 + a1 s + ao (s + b)(s + c) 2

b 2 − a1 b + ao − bt c 2 − a1c + ao − ct c 2 − 2bc + a1 b − ao − ct e + te + e (c − b ) 2 (b − c) 2 (b − c)

s3 (s + b)(s + c)(s + d ) 2

 d3 b3 c3 e − bt + e − ct + t e − dt  2 2 − − − − b c d b c b d c ( )( ) ( )( ) − − ( )( ) d b c d    d 2 [d 2 − 2 d (b + c) + 3bc] − dt e  + ( b − d ) 2 (c − d ) 2 

2

108

©1999 CRC Press LLC

t cosωt

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)

109

s3 (s + b)(s + c)(s + d )(s + f ) 2

 b3 c3 − bt − ct  (b − c)(d − b)( f − b) 2 e + (c − b)(d − c)( f − c) 2 e   d3 f3  + e − dt + t e − ft  ( f − b)(c − f )( d − f ) ( d − b)(c − d )( f − d ) 2    3f 2   +  (b − f )(c − f )( d − f )    f 3 [(b − f )(c − f ) + (b − f )(d − f ) + (c − f )(d − f )]  − dt  ε  + ( b − f ) 2 (c − f ) 2 ( d − f ) 2  

110

s3 ( s + b) ( s + c) 2

−

s3 (s + d )(s + b) 2 (s + c) 2

 d3 b3 − e − dt + t e − bt 2 2 2 b d c d c b − − − ( ) ( ) ( ) (b − d )     b 3 (c + 2 d − 3b)  − bt c3 3b 2 e + t e − ct +  + 2 3 2  ( b − c ) 2 (c − d )  (c − b ) ( d − b ) (c − b ) ( d − b )      c 3 (b + 2 d − 3c)  − ct 3c 2 e +  + 3 2  2   (b − c) ( d − c) (b − c) ( d − c) 

s3 (s + b)(s + c)(s 2 + ω 2 )

 b3 c3  e − bt + e − ct 2 2 ( − )( + ω ) ( − )( b c b c b c2 + ω2 )    ω2 sin(ωt + φ)  − 2 (b + ω 2 )(c 2 + ω 2 )     c  ω φ = tan −1 − tan −1     b ω 

s3 (s + b)(s + c)(s + d )(s 2 + ω 2 )

 b3 c3 − bt − ct  (b − c)(d − b)(b 2 + ω 2 ) e + (c − b)(d − c)(c 2 + ω 2 ) e   d3  + e − dt  ( d − b)(c − d )(d 2 + ω 2 )   ω2  − cos(ωt − φ)  2 2 2 (b + ω )(c + ω 2 )(d 2 + ω 2 )     ω  ω  ω + tan −1 + tan −1 φ = tan −1   b  c d 

111

112

113

2

3

114

s ( s + b) 2 ( s 2 + ω 2 )

115

s3 s + 4ω 4

116

s3 s − ω4

4

4

©1999 CRC Press LLC

b3 b 2 (3c − b) − bt c3 c 2 (3b − c) − ct t e − bt + e − t e − ct + e 2 3 2 (c − b ) (c − b ) (b − c) (b − c) 3

 b3 b 2 (b 2 + 3ω 2 ) − bt ω2 − bt − b 2 + ω 2 t e + (b 2 + ω 2 ) 2 e − (b 2 + ω 2 ) sin(ωt + φ)    −1  b  −1  ω   φ = tan  ω  − tan  b   cos(ωt) cosh(ωt) 1 2

[cosh(ωt ) + cos(ωt )]

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)

s + a2 s + a1 s + ao s 2 (s + b)(s + c)

 ao ao (b + c) − a1 bc − b 3 + a2 b 2 − a1 b + ao − bt + e  t− b2c2 b 2 (c − b )  bc   −c 3 + a2 c 2 − a1c + ao − ct e  + c 2 (b − c) 

s 3 + a2 s 2 + a1 s + ao s (s + b)(s + c)(s + d )

 ao − b 3 + a2 b 2 − a1 b + ao − bt − c 3 + a2 c 2 − a1c + ao − ct − e − e  b (c − b)(d − b) c (b − c)(d − c)  bcd   − d 3 + a2 d 2 − a1 d + ao − dt e  − d (b − d )(c − d ) 

119

s 3 + a2 s 2 + a1 s + ao s 2 (s + b)(s + c)(s + d )

 ao a (bc + bd + cd )  − b 3 + a2 b 2 − a1 b + ao − bt  a t+ 1 − o ε  + b2c2d 2 b 2 (c − b)(d − b)  bcd   bcd   − c 3 + a2 c 2 − a1c + ao − ct − d 3 + a2 d 2 − a1 d + ao − dt e e +  + c 2 (b − c)(d − c) d 2 (b − d )(c − d ) 

120

s + a2 s + a1 s + ao (s + b)(s + c)(s + d )(s + f )

 − b 3 + a2 b 2 − a1 b + ao − bt − c 3 + a2 c 2 − a1c + ao − ct e + e  (b − c)(d − c)( f − c)  (c − b)(d − b)( f − b)   − d 3 + a2 d 2 − a1 d + ao − dt − f 3 + a2 f 2 − a1 f + ao − ft e e +  + (b − d )(c − d )( f − d ) (b − f )(c − f )(d − f ) 

s 3 + a2 s 2 + a1 s + ao s (s + b)(s + c)(s + d )(s + f )

 ao − b 3 + a2 b 2 − a1 b + ao − bt −c 3 + a2 c 2 − a1c + ao − ct − e − e  c (b − c)(d − c)( f − c)  bcdf b (c − b)(d − b)( f − b)   − d 3 + a2 d 2 − a1 d + ao − dt − f 3 + a2 f 2 − a1 f + ao − ft e e −  − f (b − f )(c − f )( d − f ) d (b − d )(c − d )( f − d ) 

s 3 + a2 s 2 + a1 s + ao (s + b)(s + c)(s + d )(s + f )(s + g)

 − b 3 + a2 b 2 − a1 b + ao −c 3 + a2 c 2 − a1c + ao e − bt + e − ct  (b − c)(d − c)( f − c)( g − c)  (c − b)(d − b)( f − b)( g − b)  − d 3 + a2 d 2 − a1 d + ao − f 3 + a2 f 2 − a1 f + ao  e − ft e − dt +  + (b − f )(c − f )(d − f )( g − f ) (b − d )(c − d )( f − d )( g − d )   − g 3 + a2 g 2 − a1 g + ao  − gt  + (b − g)(c − g)(d − g)( f − g) e 

s 3 + a2 s 2 + a1 s + ao (s + b)(s + c)(s + d ) 2

 − b 3 + a2 b 2 − a1 b + ao − bt −c 3 + a2 c 2 − a1c + ao − ct e + e  (c − b)(d − b) 2 (b − c)(d − c) 2   − d 3 + a2 d 2 − a1 d + ao − dt  te  + (b − d )(c − d )    ao (2 d − b − c) + a1 (bc − d 2 )  2 2  + + a2 d ( db + dc − 2bc) + d ( d − 2 db − 2 dc + 3bc) e − dt 2  ( b − d ) ( c − d )2 

s 3 + a2 s 2 + a1 s + ao s (s + b)(s + c)(s + d ) 2

 ao − b 3 + a2 b 2 − a1 b + ao − bt − c 3 + a 2c 2 − a 1c + a o − ct − e − e  2 b (c − b)(d − b) 2 c (b − c)(d − c) 2  bcd  − d 3 + a2 d 2 − a1 d + a o − dt 3d 2 − 2 a2 d + a1 − dt  te − e  − d(b − d )(c − d ) d (b − d )(c − d )   ( − d 3 + a2 d 2 − a1 d + ao )[(b − d )(c − d ) − d (b − d ) − d (c − d )] − dt  e  − d 2 ( b − d ) 2 (c − d ) 2 

3

117

118

2

3

121

122

123

124

2

©1999 CRC Press LLC

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

125

s 3 + a2 s 2 + a1 s + ao (s + b)(s + c)(s + d )(s + f ) 2

126

s ( s − a) 3 / 2

127

s−a − s−b

128

1 s +a

f(t)  − b 3 + a2 b 2 − a1 b + ao − bt −c 3 + a2 c 2 − a1c + ao − ct e + e  2 (b − c)(d − c)( f − c) 2  (c − b)(d − b)( f − b)  − d 3 + a2 d 2 − a1 d + ao − dt − f 3 + a2 f 2 − a1 f + ao − ft  te e +  + (b − d )(c − d )( f − d ) 2 (b − f )(c − f )(d − f )    ( − f 3 + a2 f 2 − a1 f + ao )[(b − f )(c − f )  2  + 3 f − 2 a2 f + a1 e − ft − + (b − f )(d − f ) + (c − f )( d − f )] e − ft  ( b − f ) 2 (c − f ) 2 ( d − f ) 2 (b − f )(c − f )(d − f )  1 at e (1 + 2 at ) πt 1 2 πt3

(e bt − e at )

2 1 − ae a t erfc ( a t ) πt

129

s s − a2

2 1 + ae a t erf ( a t ) πt

130

s s + a2

2 1 − 2a e − a t πt π

131

1 s (s − a 2 )

1 a 2t e erf ( a t ) a

132

1 s (s + a 2 )

2 2 e −a t a π

133

b2 − a2 (s − a 2 )(b + s )

134

1 s ( s + a)

135

1 ( s + a) s + b

136

b2 − a2 s (s − a 2 )( s + b)

∫

a t

∫

a t

e λ dλ 2

0

e λ dπ 2

0

2

2

e a t [b − a erf ( a t )] − be b t erfc (b t ) 2

e a t erfc ( a t ) 1 e − at erf ( b − a t ) b−a 2 b 2  e a t  erf ( a t ) − 1 + e b t erfc (b t ) a 

137

(1 − s) s n +(1 / 2 )

n!   (2n)! πt H 2 n ( t )   n   − x2  x2 d   H n (t ) = Hermite polynomial = e dx n (e )   

138

(1 − s) n s n +(3 / 2 )

−

n

n! H 2 n +1 ( t ) π (2n + 1)!

139

s + 2a −1 s

ae − at [ I1 ( at ) + I o ( at )]    [ I n (t ) = j − n J n ( jt ) where J n is Bessel's function of the first kind]

140

1 s+a s+b

e − (1 / 2 )( a + b )t I o

©1999 CRC Press LLC

 a−b  t  2 

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)

141

Γ(k ) ( k ≥ 0) ( s + a) k ( s + b) k

142

1 ( s + a )1 / 2 ( s + b ) 3 / 2

143

s + 2a − s s + 2a + s

145

( s + a + s ) −2 v s s+a

( s 2 + a 2 − s) v

150

(s − s 2 − a 2 ) v

( k > 0)

2

s −a 2

155

156

157

158

159

160

(v > −1)

s2 + a2

( s 2 + a 2 − s) k

154

( k > 0) (v > −1)

2

1

( k > 0)

(s 2 − a 2 ) k 1 s s +1

k − (1 / 2 )( a + b ) t  a − b  e Ik t  2  t

a v J v ( at ) π  t  Γ(k )  2 a 

k − (1 / 2 )

J k −(1 / 2 ) ( at )

ka k J ( at ) t k a v I v ( at ) π  t  Γ(k )  2 a 

k − (1 / 2 )

I k −(1 / 2 ) ( at )

∆ erf ( t ); erf ( y) the error function = 2 π

1

1

y

−u2

du

o

J1 ( at ) ; J1 is the Bessel function of 1st kind, 1st order at

s2 + a2 + s 1

N J N ( at ) ; N = 1, 2, 3,L, J N is the Bessel function of 1st kind, N th order aN t

[ s 2 + a 2 + s] N 1 s [ s + a + s] 2

∫e

J o ( at ) ; Bessel function of 1st kind, zero order

s2 + a2

2

N

1 s 2 + a 2 ( s 2 + a 2 + s) 1 s + a [ s 2 + a 2 + s] N 2

 a−b  t  2 

Jo(at)

149

153

e − (1 / 2 )( a + b ) t I k −(1 / 2 )

1 − (1 / 2 )( at )  1  e Iv at 2  av

1

1 (s + a 2 ) k

152

( k > 0)

s2 + a2

148

151

k − (1 / 2 )

1 − at e I1 ( at ) t

( a − b) k ( s + a + s + b )2k

147

 t   a − b

  a−b   a − b  t e − (1 / 2 )( a + b ) t  I o t + I1 t    2   2 

144

146

π

2

1 s − a2 2

e − ks s

©1999 CRC Press LLC

N aN

∫

t

o

J N ( au) du ; N = 1, 2, 3,L, J N is the Bessel function of 1st kind, N th order u

1 J ( at ); J1 is the Bessel function of 1st kind, 1st order a 1 1 J ( at ) ; N = 1, 2, 3,L, J N is the Bessel function of 1st kind, N th order aN N I o ( at ) ; I o is the modified Bessel function of 1st kind, zero order 0 Sk (t ) =  1

when 0 < t < k when t > k

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t) when 0 < t < k when t > k

161

e − ks s2

0  t − k

162

e − ks (µ > 0 ) sµ

0  µ −1  (t − k )  Γ(µ )

163

1 − e − ks s

1  0

164

1 + coth 1 ks 1 2 = 2s s (1 − e − ks )

n S( k , t ) =  

165

1 s (e

+ ks

− a)

when 0 < t < k when t > k

when 0 < t < k when t > k when (n − 1) k < t < n k (n = 1, 2,L)

when 0 < t < k 0  Sk (t ) = 1 + a + a 2 + L + a n −1  when nk < t < (n + 1)k (n = 1, 2,L) 

166

1 tanh ks s

 M (2 k, t ) = ( −1) n −1   when 2 k (n − 1) < t < 2nk   (n = 1, 2,L) 

167

1 s (1 + e − ks )

1 1 1 − ( −1) n  2 M (k, t ) + 2 = 2   when (n − 1)k < t < nk 

168

1 tanh ks s2

 H (2 k, t )  

169

1 s sinh ks

2 S(2 k, t + k ) − 2 = 2(n − 1)   when (2n − 3) k < t < (2n − 1) k (t > 0)

170

1 s cosh ks

 M (2 k, t + 3k ) + 1 = 1 + ( −1)n    when (2n − 3) k < t < (2n − 1) k (t > 0)

171

1 coth ks s

2 S (2 k, t ) − 1 = 2n − 1   when 2 k (n − 1) < t < 2 kn

172

k πs coth 2k s2 + k 2

sin kt

173

1 (s 2 + 1)(1 − e −πs )

sin t  0

174

1 −k / s e s

J o (2 kt )

[ H (2 k, t ) = k + (r − k )( −1) n where t = 2 kn + r ;

when (2n − 2)π < t < (2n − 1) π when (2n − 1)π < t < 2nπ

175

1 −k / s e s

1 cos 2 kt πt

176

1 k/s e s

1 cosh 2 kt πt

177

1 −k / s e s3/ 2

1 sin 2 kt πk

178

1 s

3/ 2

ek / s

©1999 CRC Press LLC

0 ≤ r ≤ 2 k; n = 0,1, 2,L]

1 sinh 2 kt πk

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)

1 −k / s e (µ > 0 ) sµ

t  k

( µ −1) / 2

179

1 k/s e (µ > 0 ) sµ

t  k

( µ −1) / 2

180

181

e −k

 k2  exp  −   4t  2 πt 3

182

1 −k e s

 k  erfc   2 t

( k ≥ 0)

s

 k2  1 exp  −   4t  πt

183

1 −k e s

184

s −3 / 2 e − k

185

ae− k s ( k ≥ 0) s (a + s )

e − k s( s + a ) s ( s + a) e− k

188

s2 + a2

(s + a 2 ) e −k

s2 − a2

(s 2 − a 2 ) e −k(

s2 + a2 − s )

(s + a 2 ) 2

191

e

192

e −k

− ks

−e

( k ≥ 0)

− k s2 + a2

s2 + a2

− e − ks

av e− k s − a v 2 (s + a 2 )  s 2 + a 2 + s   2

193

2

 k   k2  t exp  −  − k erfc   π  4t  2 t

−e ak e a e ak e a

2

2

t

t

  k  k  erfc  a t +  + erfc    2 t 2 t  k  erfc  a t +   2 t when 0 < t < k when t > k

0  − (1 / 2 ) at 1 Io ( 2 a t 2 − k 2 ) e

2

189

190

( k ≥ 0)

s

e −k s ( k ≥ 0) s (a + s )

186

187

( k ≥ 0)

s

I µ −1 (2 kt )

k

( k > 0)

s

J µ −1 (2 kt )

0  2 2  J o (a t − k )

when 0 < t < k when t > k

0  2 2  I o (a t − k )

when 0 < t < k when t > k

J o ( a t 2 + 2 kt ) 0  ak  J (a t 2 − k 2 )  t 2 − k 2 1

when 0 < t < k

0  ak  I (a t 2 − k 2 )  t 2 − k 2 1

when 0 < t < k when t > k

2

(v > −1)

0  t−k  t + k

( )

when 0 < t < k (1 / 2 ) v

J v (a t − k ) 2

194

1 log s s

Γ ′ (1) − log t

195

1 log s (k > 0) sk

 Γ ′(k ) log t  t k −1   2  [Γ(k )] Γ(k ) 

196

log s ( a > 0) s−a

e at [log a − Ei ( − at )]

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when t > k

2

when t > k

[Γ ′ (1) = − 0.5772]

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)

197

log s s2 + 1

cos t Si (t ) − sin t Ci (t )

198

s log s s2 + 1

− sin t Si (t ) − cos t Ci (t )

199

1 log (1 + ks) (k > 0) s

 t − Ei −  k

200

log

201

1 log (1 + k 2 s 2 ) s

t − 2Ci    k

202

1 log (s 2 + a 2 ) (a > 0) s

2 log a − 2Ci ( at )

203

1 log (s 2 + a 2 ) (a > 0) s2

2 [at log a + sin at − at Ci ( at )] a

204

log

s2 + a2 s2

2 (1 − cos at ) t

205

log

s2 − a2 s2

2 (1 − cosh at ) t

206

arctan

207

1 k arctan s s

208

ek

209

1 k 2s2 e erfc (ks) (k > 0) s

erf

210

e ks erfc( ks ) (k > 0)

k π t (t + k )

2 2

s

s−a s−b

k s

erfc (ks) (k > 0)

211

1 erfc ( ks ) s

212

1 ks e erfc ( ks ) (k > 0) s

213

214

 k  erf    s  k  1 k2/ s e erfc    s s

1 bt (e − e at ) t

1 sin kt t Si(kt)  t2  1 exp  − 2  k π  4k   t   2k 

when 0 < t < k when t > k

0  −1 / 2 ( πt ) 1 π (t + k )

1 sin (2 k t ) πt 1 e −2 k πt

t

215

− e as Ei ( − as)

1 ; ( a > 0) t+a

216

1 + se as Ei ( − as) a

1 ; ( a > 0) (t + a ) 2

217

π  − Si (s) cos s + Ci (s)sin s  2 

1 t2 +1

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TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t)

218

Ko(ks)

0  2 2 −1 / 2 (t − k )

219

K o (k s )

 k2  1 exp  −  2t  4t 

220

1 ks e K1 (ks) s

1 k

221

1 K1 ( k s ) s

222

1 k/s  k e Ko  s s

when 0 < t < k when t > k

t (t + 2 k )

 k2  1 exp  −  k  4t  2 K o (2 2 kt ) πt

223

πe–ksIo(ks)

[t (2 k − t )]−1 / 2  0

224

e − ks I1 (ks)

k −t    πk t (2 k − t ) 0

when 0 < t < 2 k when t > 2 k when 0 < t < 2 k when t > 2 k

∞

2

∑ u [t − (2k + 1) a] k =0

225

1 s sinh ( as)

∞

2

∑ (−1) u (t − 2k − 1) k

k =0

226

1 s cosh s

∞

u (t ) + 2

∑ (−1) u (t − ak) k

k =1

227

square wave

1  as  tanh  2 s

∞

∑ u (t − ak) k =0

228

1  as  1 + coth 2 2s 

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stepped function

[ Kn (t ) is Bessel function of the second kind of imaginary argument]

TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t) ∞

mt − ma

∑ u (t − ka) k =1

229

saw − tooth function

m ma  as  − coth −1  s2 2s  2

∞  1 t + 2 ( −1) k (t − ka) ⋅ u (t − ka) a   k =1

∑

230

triangular wave

1  as  tanh  2 s2

∞

∑ (−1) u (t − k) k

k =0

231

1 s (1 + e − s )

∞



π 

π

∑ sin a  t − k a   ⋅ u  t − k a  k =0

232

half − wave rectification of sine wave

a (s + a )(1 − e 2

2

−πs a

)

∞



π 





π

[sin (at )] ⋅ u (t ) + 2∑ sin a  t − k a   ⋅ u  t − k a  k =1

233

  a  πs   (s 2 + a 2 )  coth  2 a   

full − wave rectification of sine wave

u (t − a ) 234

1 − as e s

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TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t) u (t − a ) − u (t − b )

235

1 − as (e − e − bs ) s

m ⋅ (t − a ) ⋅ u ( t − a ) 236

m − as e s2

mt ⋅ u (t − a) or 237

 ma + m  e − as  s s 2 

[ma + m (t − a)] ⋅ u (t − a)

(t − a ) 2 ⋅ u (t − a ) 238

2 − as e s3

t 2 ⋅ u (t − a ) 239

 2 2 a a 2  − as  3 + 2 + e s s  s

mt ⋅ u (t ) − m (t − a) ⋅ u (t − a) 240

m m − as − e s2 s2

mt − 2 m (t − a) ⋅ u (t − a) + m (t − 2 a) ⋅ u (t − 2 a) 241

m 2 m − as m −2 as − e + 2e s2 s2 s

mt − [ ma + m (t − a)] ⋅ u (t − a) 242

m  ma m  − as − + 2 e s2  s s 

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TABLE 2.3 Table of Laplace Transforms (continued) F(s)

f(t) 0.5 t 2 for 0 ≤ t < 1 1 − 0.5 (t − 2) 2 for 0 ≤ t < 2

243

1 for 2 ≤ t

(1 − e − s ) 2 s3

0.5 t 2 for 0 ≤ t < 1 0.75 − (t − 1.5) 2 for 1 ≤ t < 2 0.5 (t − 3) 2 for 2 ≤ t < 3 244

 (1 − e )    s   −s

3

b + (e ba − 1) s ( s − b) 245

0 for 3 < t

(e bt − 1) ⋅ u (t ) − (e bt − 1) ⋅ u (t − a) + Ke − b ( t − a ) ⋅ u (t − a) where K = (e ba − 1)

b   s + ba  1 − 1  e − as e −   + ( − )  s b s s b   

References W. H. Beyer, CRC Standard Mathematical Tables, 2nd Ed., CRC Press, Boca Raton, FL, 1982. R. V. Churchill, Modern Operational Mathematics in Engineering, McGraw-Hill Book Co., New York, NY, 1944. W. Magnus, F. Oberhettinger, and F. G. Tricom, Tables of Integral Transforms, Vol. I, McGraw-Hill Book Co., New York, NY, 1954. P. A. McCollum and B. F. Brown, Laplace Transform Tables and Theorems, Holt Rinehart and Winston, New York, NY, 1965.

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Appendix 1 Examples 1.1

Laplace Transformations

Example 2.1 (Inversion) The inverse of

s2 + a is found by partial expansion s 2 ( s + b)

s2 + a A B c = + + ; s ( s + b) s s 2 s + b 2

B=

s2 + a a = , s + b s=0 b

C=

s2 + a b2 + a = . 2 s b2 s =− b

Hence s2 + a A a 1 b2 + a 1 = + + . s 2 ( s + b) s b s 2 b2 s + b a Set any value of s, e.g., s = 1, and solve for A = − 2 . b Hence  s2 + a  a −1  1  a −1  1  b 2 + a −1  1  a a b 2 + a − bt = − + + L−1  2 L u t t e . ( )   = − 2 L  + L  2+ b b2 b2 b b2 s + b s b s   s ( s + b)  Example 2.2 (Differential equation) To solve y′ + by = e–t with y(0) = 1 we take the Laplace transform of both sides. Hence we obtain sY(s) – y(0) + bY(s) = s+11 or Y (s) = s +1b + ( s +1)(1 s + b ) . The inverse transform is y(t ) = e − bt + L−1{−11+ b s1+1 + 1−1b s +1b} = e − bt + b1−1 e − t + 1−1b e − bt = 21−−bb e − bt − 1−1b e − t

1.2

Inversion in the Complex Plane

When the Laplace transform F(s) is known, the function of time can be found by (2.1.2), which is rewritten f (t ) = L−1{F(s)} =

1 2πj

∫

σ+ j ∞

σ− j ∞

F(s) e st ds

This equation applies equally well to both the two-sided and the one-sided transforms. The path of integration is restricted to values of σ for which the direct transform formula converges. In fact, for the two-sided Laplace transform, the region of convergence must be specified in order to determine uniquely the inverse transform. That is, for the two-sided transform, the regions of convergence for functions of time that are zero for t > 0, zero for t < 0, or in neither category, must be distinguished. For the one-sided transform, the region of convergence is given by σ, where σ is the abscissa of absolute convergence. The path of integration is usually taken as shown in Figure 2.1 and consists of the straight line ABC displaced to the right of the origin by σ and extending in the limit from –j∞ to +j∞ with connecting semicircles. The evaluation of the integral usually proceeds by using the Cauchy integral theorem (see Chapter 20), which specifies that

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FIGURE 2.1 The Path of Integraton in the s Plane

f (t ) =

1 lim 2 πj R→∞

∫ F(s) e Γ1

st

ds

= ∑ [ residues of F(s) e st at the singularities to the left of ABC ] for t > 0 As we shall find, the contribution to the integral around the circular path with R → ∞ is zero, leaving the desired integral along path ABC, and f (t ) =

1 lim 2 πj R→∞

∫ F(s) e Γ2

st

ds

.

= − ∑ [ residues of F(s) e at the singularities to the right of ABC ] for t < 0 st

Example 2.3 Use the inversion integral to find f(t) for the function F( s) =

1 s2 + ω2

Note that the inverse of the above formula is sinωt/ω. Solution The inversion integral is written in a form that shows the poles of the integrand f (t ) =

1 2πj

∫

e st ds (s + jω )(s − jω )

The path chosen is Γ1 in Figure 2.1. Evaluate the residues   e st e st e jωt Res (s − jω ) 2 = = 2  (s + ω )  s= jω (s + jω ) s= jω 2 jω    e st e st e − jωt Res (s + jω ) 2 = = 2  (s + ω )  s=− jω (s − jω ) s=− jω − 2 jω 

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Therefore, f (t ) = ∑ Res = Example 2.4

{

e jωt − e − jωt sin ωt . = − 2 jω ω

}

Find L−1 1/ s . Solution The function F(s) = 1/ s is a double-valued function because of the square root operation. That is, if s is represented in polar form by re jθ, then re j(θ+2π) is a second acceptable representation, and s = re j (θ+2 π ) = − re jθ , thus showing two different values for s. But a double-valued function is not analytic and requires a special procedure in its solution. The procedure is to make the function analytic by restricting the angle of s to the range –π < θ < π and by excluding the point s = 0. This is done by constructing a branch cut along the negative real axis, as shown in Figure 2.2. The end of the branch cut, which is the origin in this case, is called a branch point. Since a branch cut can never be crossed, this essentially ensures that F(s) is single-valued. Now, however, the inversion integral becomes, for t > 0, 1 R→∞ 2 πj

f (t ) = lim =−

1   2 πj 

∫

F(s) e st ds =

GAB

1 2 πj

∫

σ + j∞

σ − j∞

F(s) e st ds

∫ +∫ +∫ +∫ +∫ +∫ +∫ BC

Γ2

l−

γ

l+

Γ3

FG

  

which does not include any singularity.

FIGURE 2.2 The Integration Contour for L −1{1 / s }

First we will show that for t > 0 the integrals over the contours BC and CD vanish as R → ∞, from which ∫ Γ2 = ∫ Γ3 = ∫ BC = ∫ FG = 0 . Note from Figure 2.2 that β = cos–1(σ/R) so that the integral over the arc BC is, since e jθ = 1, I ≤

∫

BC

e σt e jωt jRe jθ dθ = e σt R1 / 2 R1 / 2 e jθ / 2

= e σt R1 / 2 sin −1

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σ R

∫

π/2

β

π σ dθ = e σt R1 / 2  − cos −1  2 R

But for small arguments sin–1(σ/R) = σ/R, and in the limit as R → ∞, I → 0. By a similar approach, we find that the integral over CD is zero. Thus the integrals over the contours Γ2 and Γ3 are also zero as R → ∞. For evaluating the integral over γ, let s = rejθ = r(cosθ + jsinθ) and

∫

γ

∫

F(s) e st ds =

−π

π

e r (cos θ+ j sin θ ) jr e jθ dθ r e jθ / 2

= 0 as r → 0 The remaining integrals are written f (t ) = − Along path l–, let s = –u;

∫ F( s) e l−

st

ds +

∫ F( s) e l+

s = j u , and ds = –du, where u and

∫

l−

Along path l+, s = –u;

1  2πj 

F(s) e st ds = −

∫

0

∞

st

 ds 

u are real positive quantities. Then

e − ut 1 du = j j u

∫

∞

0

e − ut du u

s = − j u ( not + j u ), and ds = –du. Then

∫

l+

F(s) e st ds = −

∫

∞

0

e − ut 1 du = j −j u

∫

∞

0

e − ut du j u

Combine these results to find f (t ) = −

1 2 2 πj  j

∫

 1 u −1 / 2 e − ut du =  π

∞

0

∞

∫u

−1 / 2 − ut

e

du

0

which is a standard form integral listed in most handbooks of mathematical tables, with the result f (t ) =

1 π

π 1 = t πt

t>0.

Example 2.5 Find the inverse Laplace transform of the given function with an infinite number of poles. F( s) =

1 s (1 + e − s )

Solution The integrand in the inversion integral est/s(1 + e–s) possesses simple poles at s = 0 and s = jnπ, n = ±1, ± 3, + L (odd values) These are illustrated in Figure 2.3. This means that the function est/s(1 + e–s) is analytic in the s plane except at the simple poles at s = 0 and s = jnπ. Hence, the integral is specified in terms of the residues in the various poles. We thus have:

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For s = 0  se st  1 Res =  = −s  + s e ( 1 )   s=0 2 For s = jnπ  (s − jnπ) e st  0 Res =  =  −s  s (1 + e )  s= jnπ 0

FIGURE 2.3 Illustrating Example 2.5, the Laplace Inversion for the Case of Infinitely Many Poles

The problem we now face in this evaluation is that  n ( s)  0 Res = (s − a) =  d ( s)  s = a 0  where the roots of d(s) are such that s = a cannot be factored. However, we have discussed such a situation in Chapter 20 for complex variables, and we have the following result d[d (s)] d ( s) − d ( a) d (s) = lim = lim since d (a) = 0. s→ a s − a ds s= a s→a s−a Combine this expression with the above equation to obtain

 n(s)  n(s) Re s (s − a) = .  d d ( s)  s = a  [d (s)] ds s=a

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Therefore, we proceed as follows:     e st e jnπt = Res =  (n = odd )  a jnπ  s (1 + e − s )   s= jnπ  ds We obtain, by adding all of the residues, ∞

f (t ) =

e jnπt 1 + 2 n=−∞ jnπ

∑

(n = odd )

This can be rewritten as follows f (t ) =

=

 e − j 3πt e − jπt e jπt e j 3πt 1  + L + + + + + L − j 3π − jπ jπ j 3π 2   1 + 2

∞

∑ n =1

2 j sin nπt jnπ

(n = odd )

which we write, finally

f (t ) =

1 2 + 2 π

∞

∑ k =1

sin(2 k − 1) πt 2k − 1

As a second approach to a solution to this problem, we will show the details in carrying out the contour integration for this problem. We choose the path shown in Figure 2.3 that includes semicircular hooks around each pole, the vertical connecting line from hook to hook, and the semicircular path as R → ∞. Thus we have f (t ) =

1 2 πj

∫

se st ds s (1 + e − s )

   1  = + +∑ − ∑ Res  2 πj { BCA Vertical connecting lines Hooks  144244 3 123 I3 I2  I1 

∫ ∫

∫

We consider the several integrals: Integral I1. By setting s = re jθ and taking into consideration that cos = –cosθ for θ > π/2, the integral I1 → 0 as r → ∞. Integral I2. Along the Y-axis, s = jy and I2 = j

∫

∞

−∞ r→0

e jyt dy jy (1 + e − jy )

Note that the integrated is an odd function, whence I2 = 0.

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Integral I3. Consider a typical hook at s = jnπ. Since  (s − jnπ) e st  0 lim  = , r→0 s (1 + e − s )  0 s→ jnπ  this expression is evaluated and yields e jnπt/jnπ. Thus, for all poles, I3 =

1 2 πj

∫

π/2

−π/2 r→0 s→ jnπ

e st ds s (1 + e − s )

 ∞    ∞ sin nπt  jπ  e jnπt 1  1  1 2 = + =  +   2 πj n=−∞ jnπ 2  2  2 π n=1 n  n odd n odd   

∑

∑

Finally, the residues enclosed within the contour are ∞

Res

e st e jnπt 1 2 1 = + = + −s s (1 + e ) 2 n=−∞ jnπ 2 π

∑

n odd

∞

∑

n =1 n odd

sin nπt n

which is seen to be twice the value around the hooks. Then when all terms are included

f (t ) =

1.3

1 2 + 2 π

∞

sin nπt 1 2 = + n 2 π

∑ n =1 n odd

∞

∑ k =1

sin(2 k − 1) πt . 2k − 1

Complex Integration and the Bilateral Laplace Transform

We have discussed the fact that the region of absolute convergence of the unilateral Laplace transform is the region to the left of the abscissa of convergence. This is not true for the bilateral Laplace transform: the region of convergence must be specified to invert a function F(s) obtained using the bilateral Laplace transform. This requirement is necessary because different time signals might have the same Laplace transform but different regions of absolute convergence. To establish the region of convergence, we write the bilateral transform in the form F2 (s) =

∫

∞

0

e − st f (t ) dt +

∫

0

e − st f (t ) dt

−∞

σt If the function f(t) is of exponential order (e 1 ), the region of convergence for t > 0 is Re{s} > σ1 If the function f(t) for t < 0 is of exponential order exp(σ2t), then the region of convergence is Re{s} < σ2. Hence, the function F2(s) exists and is analytic in the vertical strip defined by

σ1 < Re {s} < σ2 Provided, of course, that σ1 < σ2. If σ1 > σ2, no region of convergence would exist and the inversion process could not be performed. This region of convergence is shown in Figure 2.4.

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FIGURE 2.4 Region of Convergence for the Bilateral Laplace Transform

Example 2.6 Find the bilateral Laplace transform of the signals f(t) = e–at u(t) and f(t) = –e–at u(–t) and specify their regions of convergence. Solution Using the basic definition of the transform, we obtain

a.

F2 (s) =

∫

∞

e − at u (t )e − st dt =

−∞

∞

∫e

− ( s + a )t

0

dt =

1 s+a

and its region of convergence is Re {s} > –a For the second signal

b.

F2 (s) =

∫

∞

−∞

− e − at u ( −t )e − st dt = −

∫

0

e − ( s + a )t dt =

−∞

1 s+a

and its region of convergence is Re {s} < –a Clearly, the knowledge of the region of convergence is necessary to find the time function unambiguously. Example 2.7 Find the function, if its Laplace transform is given by F2 (s) =

1 (s − 4)(s + 1)(s + 2)

− 2 < Re{s} < −1

Solution The region of convergence and the paths of integration are shown in Figure 2.5.

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FIGURE 2.5 Illustrating Example 2.7

For t > 0, we close the contour to the left, we obtain f (t ) =

3e st 1 = e −2 t (s − 4)(s + 1) s=−1 2

t>0

For t < 0, the contour closes to the right, and now f (t ) =

e 4t 3e st 3e st 3 + = − e −t + (s − 4)(s + 2) s=−1 (s + 1)(s + 2) s= 4 5 10

t<0

These examples confirm that we must know the region of convergence to find the inverse transform.

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