1. CHAPTER 4
The Laplace Transform

2. Contents 4.1 Definition of the Laplace Transform
4.2 The Inverse Transform and Transforms of Derivatives
4.3 Translation Theorems
4.4 Additional Operational Properties
4.5 The Dirac Delta Function

3. 4.1 Definition of Laplace Transform
Basic Definition If f(t) is defined for t  0, then the improper integral (1)
If f(t) is defined for t  0, then (2) is said to be the Laplace Transform of f.
DEFINITION 4.1
Laplace Transform

4. Example 1
Evaluate L{1}
Solution: Here we keep that the bounds of integral are 0 and  in mind. From the definition , s> Since e-st  0 as t , for s > 0.

5. Example 2
Evaluate L{t}
Solution
, s>0

6. Example 3
Evaluate L{e-3t}
Solution

7. Example 4
Evaluate L{sin 2t}
Solution

8. Example 4 (2)
Laplace transform of sin 2t ↓

9. L.T. is Linear
We can easily verify that (3)

10. (a) (b) (c) (d) (e) (f) (g) Transform of Some Basic Functions
THEOREM 4.1
Transform of Some Basic Functions

11. A function f(t) is said to be of exponential order,
DEFINITION 4.2
A function f(t) is said to be of exponential order,
if there exists constants c>0, M > 0, and T > 0, such
That |f(t)|  Mect for all t > T. See Fig 4.2, 4.3.
Exponential Order

12. Fig 4.2

13. Examples See Fig 4.3

14. Fig 4.4
A function such as is not of exponential order, see Fig 4.4

15. If f(t) is piecewise continuous on [0, ) and of
THEOREM 4.2
If f(t) is piecewise continuous on [0, ) and of
exponential order, then L{f(t)} exists for s > c.
Sufficient Conditions for Existence

16. Fig 4.1

17. Example 5
Find L{f(t)} for
Solution

18. 4.2 If F(s)=L(f(t)), then f(t) is the inverse Laplace transform of F(s) and f(t)=L(F(s))
THEOREM 4.3
(a)
(b) (c)
(d) (e)
(f) (g)
Some Inverse Transform

19. Example 1
Find the inverse transform of (a) (b)
Solution (a) (b)

20. L -1 is also linear
We can easily verify that (1)

21. Example 2
Find
Solution (2)

22. Example 3: Partial Fraction
Find
Solution Using partial fractions Then (3) If we set s = 1, 2, −4, then

23. Example 3 (2)
(4) Thus (5)

24. Uniqueness of L -1
Suppose that the functions f(t) and g(t) satisfy the hypotheses of Theorem 4.2, so that their Laplace transform F(s) and G(s) both exist. If F(s)=G(s) for all s>c (for some c), then f(t)=g(t) whenever on [0,
+ ) both f and g are continuous.

25. Transform of Derivatives
(7) (8)

26. If are continuous on [0, ) and are of
THEOREM 4.4
If are continuous on [0, ) and are of
Exponential order and if f(n)(t) is piecewise-continuous
On [0, ), then where
Transform of a Derivative

27. Solving Linear ODEs
Then (9) (10)

28. We have (11) where

30. Example 4: Solving IVP
Solve
Solution (12) (13)

31. Example 4 (2)
We can find A = 8, B = −2, C = 6 Thus

32. Example 5
Solve
Solution (14) Thus

33. 4.3 Translation Theorems If f is piecewise continuous on [0, ) and of
exponential order, then lims L{f} = 0.
Behavior of F(s) as s → 
Proof

34. F(s) by s-a If L{f} = F(s) and a is any real number, then
THEOREM 4.6
If L{f} = F(s) and a is any real number, then
L{eatf(t)} = F(s – a), See Fig 4.10.
Translation on the s-axis
Proof L{eatf(t)} =  e-steatf(t)dt =  e-(s-a)tf(t)dt = F(s – a): replacing all s in
F(s) by s-a

35. Fig 4.10

36. Example 1
Find the L.T. of (a) (b)
Solution (a) (b)

37. Inverse Form of Theorem 4.6
(1) where

38. Parttial Fraction To perform the inverse transform of R(s)=P(s)/Q(s):
Rule 1: Linear Factor Partial Fractions
Rule 2: Quadratic Factor Partial Fractions

39. Example 2 Find the inverse L.T. of (a) (b)
Solution (a) we have A = 2, B = (2)

40. Example 2 (2)
And (3) From (3), we have (4)

41. Example 2 (3)
(b) (5) (6) (7)

42. Example 3
Solve
Solution

43. Example 3 (2)
(8)

44. Example 4
Solve
Solution

45. Example 4 (2)

46. The Unit Step Function U(t – a) defined for is Unit Step Function
DEFINITION 4.3
The Unit Step Function U(t – a) defined for is
Unit Step Function
See Fig 4.11.

47. Fig 4.11

48. Fig Fig 4.13
Fig 4.12 shows the graph of (2t – 3)U(t – 1). Considering Fig 4.13, it is the same as f(t) = 2 – 3U(t – 2) + U(t – 3),

49. Also a function of the type. (9) is the same as
Also a function of the type (9) is the same as (10) Similarly, a function of the type (11) can be written as (12)

50. Example 5 Express in terms of U(t). See Fig 4.14.
Solution From (9) and (10), with a = 5, g(t) = 20t, h(t) = 0 f(t) = 20t – 20tU(t – 5)

51. Fig 4.14

52. Consider the function (13) See Fig 4.15.

53. Fig 4.15

54. If F(s) = L{f}, and a > 0, then L{f(t – a)U(t – a)} = e-asF(s)
THEOREM 4.7
If F(s) = L{f}, and a > 0, then L{f(t – a)U(t – a)} = e-asF(s)
Second Translation Theorem
Proof

55. Let v = t – a, dv = dt, then If f(t) = 1, then f(t – a) = 1, F(s) = 1/s, (14) eg: The L.T. of Fig 4.13 is

56. Inverse Form of Theorem 4.7
(15)

57. Example 6
Find the inverse L.T. of (a) (b)
Solution (a) then
(b) then

58. Alternative Form of Theorem 4.7
Since , then The above can be solved. However, we try another approach. Let u = t – a, That is, (16)

59. Example 7
Find
Solution With g(t) = cos t, a = , then g(t + ) = cos(t + )= −cos t By (16),

60. Example 8
Solve
Solution We find f(t) = 3 cos t U(t −), then (17)

61. Example 8 (2)
It follows from (15) with a = , then Thus (18) See Fig 4.16

62. Fig 4.16

63. 4.4 Additional Operational Properties
Multiplying a Function by tn that is, Similarly,

64. If F(s) = L{f(t)} and n = 1, 2, 3, …, then
THEOREM 4.8
If F(s) = L{f(t)} and n = 1, 2, 3, …, then
Derivatives of Transform

65. Example 1 Find L{t sin kt}
Solution With f(t) = sin kt, F(s) = k/(s2 + k2), then

66. Different approaches
Theorem 4.6:
Theorem 4.8:

67. Example 2
Solve
Solution or From example 1, Thus

68. Convolution
A special product of f * g is defined by and is called the convolution of f and g.
Note: f * g = g * f

69. IF f(t) and g(t) are piecewise continuous on [0, ) and
THEOREM 4.9
IF f(t) and g(t) are piecewise continuous on [0, ) and
of exponential order, then
Convolution Theorem
Proof

70. Holding  fixed, let t =  + , dt = d The integrating area is the shaded region in Fig Changing the order of integration:

71. Fig 4.32

72. Example 3
Find
Solution Original statement = L{et * sin t}

73. Inverse Transform of Theorem 4.9
L-1{F(s)G(s)} = f * g (4) Look at the table in Appendix III, (5)

74. Example 4
Find
Solution Let then

75. Example 4 (2)
Now recall that sin A sin B = (1/2) [cos (A – B) – cos (A+B)] If we set A = k, B = k(t − ), then

76. Transform of an Integral
When g(t) = 1, G(s) = 1/s, then

77. Examples:

78. Periodic Function f(t + T) = f(t): periodic with period T
If f(t) is a periodic function with period T, then
THEOREM 4.10
Transform of a Periodic Function

79. Proof Use change of variable

80. Example 7 Find the L. T. of the function in Fig 4.35.
Solution We find T = 2 and From Theorem 4.10,

81. Fig 4.35

82. Example 8
The DE (13) Find i(t) where i(0) = 0, E(t) is as shown in Fig 4.35.
Solution or (14) Because and

83. Assuming R=1ohm, L=1 henry
Then i(t) is described as follows and see Fig 4.36: (15)

84. Fig 4.36

85. 4.5 The Dirac Delta Function
Unit Impulse See Fig 4.43(a). Its function is defined by (1) where a > 0, t0 > 0.
For a small value of a, a(t – t0) is a constant function of large magnitude. The behavior of a(t – t0) as a  0, is called unit impulse, since it has the property See Fig 4.43(b).

86. Fig 4.43

87. The Dirac Delta Function
This function is defined by (t – t0) = lima0 a(t – t0) (2) The two important properties: (1) (2) , x > t0

88. Proof The Laplace Transform is
THEOREM 4.11
For t0 > 0,
Transform of the Dirac Delta Function
Proof The Laplace Transform is

89. When a  0, (4) is 0/0. Use the L’Hopital’s rule, then (4) becomes 1 as a  0. Thus , Now when t0 = 0, we have

90. Example 1
Solve subject to (a) y(0) = 1, y’(0) = 0 (b) y(0) = 0, y’(0) = 0
Solution (a) s2Y – s + Y = 4e-2s Thus y(t) = cos t + 4 sin(t – 2)U(t – 2) Since sin(t – 2) = sin t, then (5) See Fig 4.44.

91. Fig 4.44

92. Example 1 (2)
(b) Thus y(t) = 4 sin(t – 2)U(t – 2) and (6)

93. Fig 4.45