1. Lec 4. the inverse Laplace Transform

2. The Inverse Laplace Transform
Suppose F(s) has the general form of
Finding inverse Laplace transform of F(s) involves two steps:
Decompose F(s) into simple terms using partial fraction expansion.
Find the inverse of each term by matching entries in Laplace Transform Table.
𝐹(𝑠 = 𝑁(𝑠)......numerator polynomial 𝐷(𝑠)...denominator polynomial 2

3. The Inverse Laplace Transform
Example 1
Find the inverse Laplace transform of
Solution:
𝐹(𝑠 = 3 𝑠 − 5 𝑠 𝑠 2 +4
𝑓(𝑡) =𝐿 −1  3 𝑠  − 𝐿 −1  5 𝑠+1  +𝐿 −1  6 𝑠 2 +4  3−5 𝑒 −𝑡 +3sin(2t)𝑢(𝑡),t≥0 3

4. Partial Fraction Expansion
Distinct Real Roots of D(s)
𝐹(𝑠 = 96(𝑠+5)(𝑠+12 𝑠(𝑠+8)(𝑠+6
s1= 0, s2= -8
s3= -6

5. 1) Distinct Real Roots
𝐹(𝑠 = 96(𝑠+5)(𝑠+12 𝑠(𝑠+8)(𝑠+6 ≡ 𝐾 1 𝑠 + 𝐾 2 𝑠+8 + 𝐾 3 𝑠+6
To find K1: multiply both sides by s and evaluates both sides at s=0
To find K2: multiply both sides by s+8 and evaluates both sides at s=-8
To find K3: multiply both sides by s+6 and evaluates both sides at s=-6

6. Find K1
96(𝑠+5)(𝑠+12 𝑠+8)(𝑠+6 ∣ 𝑠=0 ≡ 𝐾 1 + 𝐾 2 𝑠 𝑠+8 ∣ 𝑠=0 + 𝐾 3 𝑠 𝑠+6 ∣ 𝑠=0
∴𝐾 1 = 96(5)(12 8)(6 =120

7. Find K2
96(𝑠+5)(𝑠+12 𝑠(𝑠+6 ∣ 𝑠=−8 ≡ 𝐾 1 (𝑠+8 𝑠(𝑠+6 ∣ 𝑠=−8 +𝐾 𝐾 3 (𝑠+8 𝑠(𝑠+6 ∣ 𝑠=−8
∴𝐾 2 = 96 −3 )(4 −8 ) −2 =−72

8. Find K3
96(𝑠+5)(𝑠+12 𝑠(𝑠+8 ∣ 𝑠=−6 ≡ 𝐾 1 (𝑠+6 𝑠(𝑠+8 ∣ 𝑠=−6 + 𝐾 2 (𝑠+6 𝑠(𝑠+8 ∣ 𝑠=−6 + 𝐾 3
∴𝐾 3 = 96 −1 )(6 −6 )(2 =48

9. Inverse Laplace of F(s)
𝐹(𝑠 = 120 𝑠 − 72 𝑠 𝑠+6
𝐿 − 𝑠 − 72 𝑠 𝑠+6 𝑓(𝑡 = 120−72 𝑒 −8t +48 𝑒 −6t  𝑢(𝑡)

10. Ex.

11. 2) Distinct Complex Roots
S2 = -3+j4
S3 = -3-j4

12. Partial Fraction Expansion
Complex roots appears in conjugate pairs.

13. Find K1
𝐾 1 = 100(𝑠+3 𝑠 2 +6s+25 ∣ 𝑠=−6 = 100 − =−12

14. Find K2 and K2* Coefficients associated with conjugate
𝐾 2 = 100(𝑠+3 𝑠+6)(𝑠+3+𝑗4 ∣ 𝑠=−3+𝑗4 = 100(𝑗4 3+𝑗4)(𝑗8 =6−𝑗8=10 𝑒 −𝑗53.13°
Coefficients associated with conjugate
pairs are themselves conjugates.
𝐾 2 =6+𝑗8=10 𝑒 𝑗53.13

15. Inverse Laplace of F(s)

16. Inverse Laplace of F(s)
𝐿 −1 −12 𝑠 𝑒 −𝑗53.13° 𝑠+3−𝑗 𝑒 𝑗53.13° 𝑠+3+𝑗4 = −12 𝑒 −6t +10 𝑒 −𝑗53.13° 𝑒 − 3−𝑗4)𝑡 +10 𝑒 𝑗53.13° 𝑒 − 3+𝑗4)𝑡 )𝑢(𝑡)

17. Useful Transform Pairs
1) 𝐾 𝑠+𝑎 ⇔ 𝐾𝑒 −𝑎𝑡 𝑢(𝑡
2) 𝐾 𝑠+𝑎 ) 2 ⇔ 𝐾𝑡𝑒 −𝑎𝑡 𝑢(𝑡
3) 𝐾 𝑠+𝛼−𝑗𝛽 + 𝐾 𝑠+𝛼+𝑗𝛽 ⇔2∣𝐾∣ 𝑒 −𝛼𝑡 cos(𝛽𝑡+𝜃)𝑢(𝑡
4) 𝐾 𝑠+𝛼−𝑗𝛽 ) 𝐾 𝑠+𝛼+𝑗𝛽 ) 2 ⇔2t∣𝐾∣ 𝑒 −𝛼𝑡 cos(𝛽𝑡+𝜃)𝑢(𝑡

18. Ex.

19. Operational Transform

20. Operational Transforms
Indicate how mathematical operations performed on either f(t) or F(s) are converted into the opposite domain.
The operations of primary interest are:
Multiplying by a constant
Addition/subtraction
Differentiation
Integration
Translation in the time domain
Translation in the frequency domain
Scale changing

21. Multiplication by a constant
OPERATION
f(t)
F(s)
Multiplication by a constant
Addition/Subtra ction
First derivative (time)
Second derivative (time)
𝐾𝑓(𝑡
𝐾𝐹(𝑠
𝑓 1 (𝑡) +𝑓 2 (𝑡 − 𝑓 3 (𝑡 +⋯
𝐹 1 (𝑠) +𝐹 2 (𝑠 − 𝐹 3 (𝑠 +⋯
𝑑𝑓(𝑡)𝑑𝑡
𝑠𝐹(𝑠 −𝑓 ( 0 −
𝑑 2 (𝑡) 𝑑𝑡 2
𝑠 2 𝐹(𝑠 −𝑠𝑓 ( 0 − −𝑑𝑓 ( 0 − )𝑑𝑡

22. Translation in frequency
OPERATION
f(t)
F(s)
n th derivative (time)
Time integral
Translation in time
Translation in frequency
𝑠 𝑛 𝐹(𝑠 − 𝑠 𝑛−1 𝑓( 0 − − 𝑠 𝑛−2 𝑑𝑓( 0 − )𝑑𝑡 − 𝑠 𝑛−3 𝑑𝑓 2 ( 0 − )𝑑𝑡−⋯− 𝑑𝑓 𝑛−1 ( 0 − ) 𝑑𝑡 𝑛−1
𝑑 𝑛 (𝑡) 𝑑𝑡 𝑛
0 𝑡 𝑓(𝑥)𝑑𝑥
𝐹(𝑠)𝑠
𝑓(𝑡−𝑎)𝑢(𝑡−𝑎), 𝑎>0
𝑒 −𝑎𝑠 𝐹(𝑠
𝑒 −𝑎𝑡 𝑓(𝑡
𝐹(𝑠+𝑎

23. OPERATION f(t) F(s) Scale changing First derivative (s)
n th derivative
s integral
1𝑎𝐹 𝑠𝑎
𝑓(𝑎𝑡),𝑎>0
𝑡𝑓(𝑡
−𝑑𝐹(𝑠)𝑑𝑠
𝑡 𝑛 𝑓(𝑡
−1 ) 𝑛 𝑑 𝑛 𝐹(𝑠) 𝑑𝑠 𝑛
𝑠 ∞ 𝐹(𝑢)𝑑𝑢
𝑓(𝑡)𝑡

24. Translation in time domain
If we start with any function:
we can represent the same function translated in time by the constant a, as:
In frequency domain:
𝑓(𝑡)𝑢(𝑡
𝑓(𝑡−𝑎)𝑢(𝑡−𝑎
𝑓(𝑡−𝑎)𝑢(𝑡−𝑎) =𝑒 −𝑎𝑠 𝐹(𝑠

25. Ex:
𝐿 𝑡𝑢(𝑡 = 1s 2
𝐿 𝑡−𝑎)𝑢(𝑡−𝑎 =𝑒 −𝑎𝑠 𝑠 2

26. Translation in frequency domain
Translation in the frequency domain is defined as:
𝐿 𝑒 −𝑎𝑡 𝑓(𝑡 =𝐹(𝑠+𝑎

27. Ex:
𝐿 cos𝜔𝑡 = 𝑠 𝑠 2 +𝜔 2
𝐿 𝑒 −𝑎𝑡 cos𝜔𝑡 = 𝑠+𝑎 𝑠+𝑎 ) 2 +𝜔 2

28. Ex:
𝐿 cos𝑡 = 𝑠 𝑠 2 +1
𝐿 cos𝜔𝑡 = 1 𝜔 𝑠 𝜔 𝑠 𝜔 ) = 𝑠 𝑠 2 +𝜔 2

29. APPLICATION

30. Problem
Assumed no initial energy is stored in the circuit at the instant when the switch is opened.
Find the time domain expression for v(t) when t≥0.

31. Integrodifferential Equation
A single node voltage equation:
𝑎lg 𝐼 𝑖𝑛 =𝑎lg 𝐼 𝑜𝑢𝑡 ⇒𝐾𝐶𝐿 𝑣(𝑡 𝑅 + 1 𝐿 0 𝑡 𝑣(𝑡 𝑑𝑡+𝐶 𝑑𝑣(𝑡 𝑑𝑡 =𝐼 𝑑𝑐 𝑢(𝑡)

32. s-domain transformation
𝑣(𝑡 𝑅 + 1 𝐿 0 𝑡 𝑣(𝑥 𝑑𝑥+𝐶 𝑑𝑣(𝑡 𝑑𝑡 =𝐼 𝑑𝑐 𝑢(𝑡 =0
𝑉(𝑠)  1 𝑅 + 1 𝑠𝐿 +𝑠𝐶 = 𝐼 𝑑𝑐 𝑠

33. 𝑉(𝑠 = 𝐼 𝑑𝑐 𝐶 𝑠 𝑅𝐶 )𝑠+( 1 𝐿𝐶
𝑣(𝑡) =𝐿 −1 𝑉(𝑠

34. Ex Obtain the Laplace transform for the function below: 1 2 3 t h(t) 41 2 3 t
h(t) 4 5

35. Find the expression of f(t):
Expression for the ramp function with slope, m =2 and period, T=2:
For a periodic ramp function, we can write:
𝑓 1 (𝑡 =2t
𝑓 1 (𝑡 =2t 𝑢(𝑡 −𝑢 (𝑡−1

36. Different time occurred:
Expanding:
𝑓 1 (𝑡 =2t 𝑢(𝑡 −𝑢 (𝑡−1 =2𝑡𝑢(𝑡 −2 𝑡𝑢(𝑡−1)
Different time occurred:
t=0 and t=1

37. Equal time shift:
𝑓 1 (𝑡 =2t 𝑢(𝑡 −𝑢 (𝑡−1 =2𝑡𝑢(𝑡 −2 𝑡𝑢(𝑡−1) =2𝑡𝑢(𝑡 −2 (𝑡−1+1)𝑢(𝑡−1) 𝑓 1 (𝑡 =2 𝑡𝑢(𝑡 −2 (𝑡−1)𝑢(𝑡−1 −2u (𝑡−1)

38. Inverse Laplace using translation in time property:

39. Time periodicity property:
𝑓(𝑡)=𝑓(𝑡+𝑛𝑇)⇔𝐹(𝑠 = 𝐹 1 (𝑠 1− 𝑒 −𝑇𝑠
𝐹(𝑠 = 𝐹 1 (𝑠 1− 𝑒 −𝑇𝑠 = 2 𝑠 2 (1− 𝑒 −2s 1− 𝑒 −𝑠 − 𝑠𝑒 −𝑠