Signals & Systems Questions and Answers – Inverse Laplace Transform

This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Inverse Laplace Transform”.

1. Find the inverse Laplace transform for $$\frac{1}{(s+1)^2}$$.
a) tet u(t)
b) te-t u(t)
c) tu(t)
d) et u(t)

Explanation: Given X(s) = $$\frac{1}{(s+1)^2}$$
x(t) = L-1 [X(s)] = $$L^{-1} \Big[\frac{1}{(s+1)^2}\Big] = e^{-t} L^{-1} [\frac{1}{s^2}]$$ = e-t tu(t) = te-t u(t).

2. Find the inverse Laplace transform for $$\frac{1}{(s+1)^2+1}$$.
a) te-t u(t)
b) e-t sin⁡t u(t)
c) e-t cos⁡t u(t)
d) e-t u(t)

Explanation: Given X(s) = $$\frac{1}{(s+1)^2+1}$$
x(t) = L-1 [X(s)] = $$L^{-1} \Big[\frac{1}{(s+1)^2+1}\Big] = e^{-t} L^{-1} \Big[\frac{1}{s^2+1}\Big]$$ = e-t sin⁡t u(t).

3. Find the inverse Laplace transform for $$\frac{s}{(s+2)^2}$$.
a) te-t u(t)
b) e-t sin⁡t u(t)
c) e-2t (1-2t)u(t)
d) e2t (1-2t)u(t)

Explanation: Given X(s) = $$\frac{s}{(s+2)^2}$$
x(t) = L-1 [X(s)] = $$L^{-1} \Big[\frac{s}{(s+2)^2}\Big] = L^{-1} \Big[\frac{s+2}{(s+2)^2} – \frac{2}{(s+2)^2}\Big] = L^{-1} \Big[\frac{1}{s+2}\Big] – 2L^{-1} \Big[\frac{1}{(s+2)^2}\Big]$$
= e-2t – 2e-2t L-1 $$[\frac{1}{s^2}]$$ = e-2t (1-2t)u(t).

4. Find the inverse Laplace transform for $$\frac{s}{(s+2)^2+1}$$.
a) [2e-2t cos⁡t + e-2t sin⁡t]u(t)
b) [e-2t cos⁡t + 2e-2t sin⁡t]u(t)
c) [2e-2t cos⁡t – e-2t sin⁡t]u(t)
d) [e-2t cos⁡t – 2e-2t sin⁡t]u(t)

Explanation: Given X(s) = $$\frac{s}{(s+2)^2+1}$$
x(t) = L-1 [X(s)] = $$L^{-1} \Big[\frac{s}{(s+2)^2+1}\Big] = L^{-1} \Big[\frac{s+2}{(s+2)^2+1} – \frac{2}{(s+2)^2+1}\Big]$$
$$= L^{-1} \Big[\frac{s+2}{(s+2)^2+1}\Big] – 2L^{-1} \Big[\frac{1}{(s+2)^2+1}\Big] = e^{-2t} L^{-1} \Big[\frac{s}{s^2+1}\Big] – 2e^{-2t} L^{-1} \Big[\frac{1}{s^2+1}\Big]$$
= [e-2t cos⁡t – 2e-2t sin⁡t]u(t).

5. Find the inverse Laplace transform of X(s) = $$\frac{s}{s^2 a^2+b^2}$$.
a) $$\frac{1}{a^2} \,cos⁡(\frac{a}{b})t$$
b) $$\frac{1}{a^2} \,cos⁡(\frac{b}{a})t$$
c) $$\frac{1}{a^2} \,sin⁡(\frac{b}{a})t$$
d) $$\frac{1}{a^2} \,sin⁡(\frac{a}{b})t$$

Explanation: Given X(s) = $$\frac{s}{s^2 a^2+b^2} = \frac{1}{a^2} \Big[\frac{s}{s^2+(b/a)^2}\Big]$$
We know that L-1 $$\left(\frac{s}{s^2+ω^2}\right)$$ = cos⁡ωt
∴x(t) = L-1 [X(s)] = $$\frac{1}{a^2} L^{-1} \Big[\frac{s}{s^2+(b/a)^2}\Big] = \frac{1}{a^2} \,cos⁡(\frac{b}{a})t$$.
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6. Find the inverse Laplace transform of X(s) = $$\frac{s}{(s^2+a^2)^2}$$.
a) $$\frac{1}{a}$$ t sin⁡at
b) $$\frac{1}{2a}$$ t sin⁡at
c) $$\frac{1}{a}$$ t cos⁡at
d) $$\frac{1}{2a}$$ t cos⁡at

Explanation: Given X(s) = $$\frac{s}{(s^2+a^2)^2}$$
x(t) = L-1 [X(s)] = $$L^{-1} \Big[\frac{s}{(s^2+a^2)^2}\Big] = \frac{1}{2a} \Big[-\frac{d}{ds} \Big\{\frac{a}{s^2+a^2}\Big\}\Big]$$
= $$\frac{1}{2a} tL^{-1} \Big[\frac{a}{s^2+a^2}\Big] = \frac{1}{2a}t sin⁡at$$.

7. If F1 (s) = $$\frac{1}{s+2}$$ and F2 (s) = $$\frac{1}{s+3}$$, find the inverse Laplace transform of F(s) = F1 (s) F2 (s).
a) [e-2t + e-3t]u(t)
b) [e-2t – e-3t]u(t)
c) [e2t + e3t]u(t)
d) [e2t + e-3t]u(t)

Explanation: Given F1 (s) = $$\frac{1}{s+2}$$ and F2 (s) = $$\frac{1}{s+3}$$.
F(s) = F1 (s) F2 (s) = $$\left(\frac{1}{s+2}\right)\left(\frac{1}{s+3}\right) = \frac{1}{s+2} – \frac{1}{s+3}$$
Applying inverse Laplace transform, we get
f(t) = [e-2t – e-3t]u(t).

8. Find the inverse Laplace transform for X(s) = $$\frac{s}{2s^2-8}$$.
a) cosh⁡2t
b) $$\frac{1}{2}$$ cosh⁡2t
c) sinh⁡2t
d) $$\frac{1}{2}$$ sinh⁡2t

Explanation: Given X(s) = $$\frac{s}{2s^2-8} = \frac{1}{2} \frac{s}{(s^2-2^2)}$$
We know that cosh⁡ωt = $$\frac{s}{(s^2-w^2)}$$
∴x(t)=L-1 [X(s)] = $$\frac{1}{2} L^{-1} \left(\frac{s}{(s^2-2^2)}\right) = \frac{1}{2} cosh⁡2t.$$

9. Find the inverse Laplace transform for X(s) = $$ln ⁡(\frac{s+a}{s+b})$$.
a) $$\frac{e^{-at} – e^{-bt}}{t}$$
b) $$\frac{e^{-bt} – e^{-at}}{t}$$
c) $$\frac{e^{-at} + e^{-bt}}{t}$$
d) $$\frac{e^{bt} + e^{-at}}{t}$$

Explanation: Given X(s) = $$ln ⁡(\frac{s+a}{s+b})$$
x(t) = L-1 [X(s)] = L-1 $$\Big[ln (\frac{s+a}{s+b})\Big]$$
L[x(t)] = $$ln ⁡(\frac{s+a}{s+b})$$ = ln⁡(s+a)-ln⁡(s+b)
L[tx(t)] = –$$\frac{d}{ds}$$ [ln⁡(s+a)-ln⁡(s+b)] = $$\frac{-1}{s+a} + \frac{1}{s+b} = \frac{1}{s+b} – \frac{1}{s+a}$$
tx(t) = $$L^{-1}(\frac{1}{s+b} – \frac{1}{s+a})$$ = e-bt – e-at
x(t) = $$\frac{e^{-bt} – e^{-at}}{t}$$.

10. Find the inverse Laplace transform for the function X(s) = $$\frac{2s-1}{s^2+4s+8}$$.
a) e-2t cos⁡2t u(t) – e-2t sin⁡2t u(t)
b) 2e-2t cos⁡2t u(t) – $$\frac{5}{2}$$ e-2t sin⁡2t u(t)
c) 2e-2t cos⁡2t u(t) – e-2t sin⁡2t u(t)
d) e-2t cos⁡2t u(t) – $$\frac{5}{2}$$ e-2t sin⁡2t u(t)

Explanation: Given function X(s) = $$\frac{2s-1}{s^2+4s+8} = \frac{2s-1}{(s+2)^2+2^2} = \frac{2(s+2)-5}{(s+2)^2+2^2}$$
= $$\frac{2(s+2)}{(s+2)^2+2^2} – \frac{5}{2} \frac{2}{(s+2)^2+2^2}$$
Applying inverse Laplace transform, we get
x(t) = 2e-2t cos⁡2t u(t) – $$\frac{5}{2}$$ e-2t sin⁡2t u(t).

11. Find the inverse Laplace transform for the function X(s) = $$\frac{1+e^{-2s}}{3s^2+2s}$$.
a) e-(2/3)t u(t) – u(t) + e-(2/3)(t-2) u(t-2)-u(t-2)
b) e-(2/3)t u(t) + e-(2/3)(t-2) u(t-2)
c) e-(2/3)(t-2) u(t-2) – u(t-2)
d) e-(2/3)t u(t) – u(t)

Explanation: Given function X(s) = $$\frac{1+e^{-2s}}{3s^2+2s}$$
x(t) = L-1 [X(s)] = $$L^{-1} \Big[\frac{1+e^{-2s}}{3s^2+2s}\Big] = L^{-1} \Big[\frac{1}{3s^2+2s}\Big] + L^{-1} \Big[\frac{e^{-2s}}{3s^2+2s}\Big]$$
$$L^{-1} \Big[\frac{1}{3s^2+2s}\Big] = L^{-1} \Big\{\frac{1}{3s[s+(2/3)]}\Big\} = L^{-1} \Big\{\frac{-1}{s} + \frac{1}{[s+(2/3)]}\Big\}$$ = e-(2/3)t u(t) – u(t)
$$L^{-1} \Big[\frac{e^{-2s}}{3s^2+2s}\Big] = L^{-1} \left(\frac{1}{3s^2+2s}\right)_{t=t-2}$$ = e-(2/3)(t-2) u(t-2)-u(t-2)
∴x(t) = e-(2/3)t u(t) – u(t) + e-(2/3)(t-2) u(t-2)-u(t-2).

12. Given x(t)=e-t u(t). Find the inverse Laplace transform of e-3s X(2s).
a) $$\frac{1}{2}$$ e-(t-3)/2 u(t+3)
b) $$\frac{1}{2}$$ e-(t-3)/2 u(t-3)
c) $$\frac{1}{2}$$ e(t-3)/2 u(t-3)
d) $$\frac{1}{2}$$ e(t-3)/2 u(t+3)

Explanation: Given x(t) = e-t u(t)
X(s) = L[x(t)] = L[e-t u(t)] = $$\frac{1}{s+1}$$
X(2s) = $$\frac{1}{2s+1} = \frac{1/2}{s+(1/2)}$$
L-1 [X(2s)] = $$L^{-1} [\frac{1/2}{s+(1/2)}] = 1\frac{1}{2}$$ e-t/2 u(t)
L-1 [e-3s X(2s)] = L-1 [X(2s)]t=t-3 = $$\frac{1}{2}$$ e-(t-3)/2 u(t-3)
∴L-1 [e-3s X(2s)] = $$\frac{1}{2}$$ e-(t-3)/2 u(t-3) if x(t) = e-t u(t).

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