Lec 4. the inverse Laplace Transform

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

The Inverse Laplace Transform Example 1 Find the inverse Laplace transform of Solution: 𝐹(𝑠 = 3 𝑠 − 5 𝑠+1 + 6 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

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

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

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

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

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

Inverse Laplace of F(s) 𝐹(𝑠 = 120 𝑠 − 72 𝑠+8 + 48 𝑠+6 𝐿 −1 120 𝑠 − 72 𝑠+8 + 48 𝑠+6 𝑓(𝑡 = 120−72 𝑒 −8t +48 𝑒 −6t  𝑢(𝑡)

Ex.

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

Partial Fraction Expansion Complex roots appears in conjugate pairs.

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

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

Inverse Laplace of F(s)

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

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 ) 2 + 𝐾 𝑠+𝛼+𝑗𝛽 ) 2 ⇔2t∣𝐾∣ 𝑒 −𝛼𝑡 cos(𝛽𝑡+𝜃)𝑢(𝑡

Ex.

Operational Transform

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

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

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

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

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

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

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

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

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

APPLICATION

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.

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)

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

𝑉(𝑠 = 𝐼 𝑑𝑐 𝐶 𝑠 2 + 1 𝑅𝐶 )𝑠+( 1 𝐿𝐶 𝑣(𝑡) =𝐿 −1 𝑉(𝑠

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

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

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

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)

Inverse Laplace using translation in time property:

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