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)
View Answer
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 sint u(t)
c) e-t cost u(t)
d) e-t u(t)
View Answer
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 sint u(t).
3. Find the inverse Laplace transform for \(\frac{s}{(s+2)^2}\).
a) te-t u(t)
b) e-t sint u(t)
c) e-2t (1-2t)u(t)
d) e2t (1-2t)u(t)
View Answer
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 cost + e-2t sint]u(t)
b) [e-2t cost + 2e-2t sint]u(t)
c) [2e-2t cost – e-2t sint]u(t)
d) [e-2t cost – 2e-2t sint]u(t)
View Answer
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 cost – 2e-2t sint]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) \(\frac{1}{a^2} \,sin(\frac{a}{b})t\)
View Answer
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 \(\left(\frac{s}{s^2+ω^2}\right)\) = cosωt
∴x(t) = L-1 [X(s)] = \(\frac{1}{a^2} L^{-1} \Big[\frac{s}{s^2+(b/a)^2}\Big] = \frac{1}{a^2} \,cos(\frac{b}{a})t\).
6. Find the inverse Laplace transform of X(s) = \(\frac{s}{(s^2+a^2)^2}\).
a) \(\frac{1}{a}\) t sinat
b) \(\frac{1}{2a}\) t sinat
c) \(\frac{1}{a}\) t cosat
d) \(\frac{1}{2a}\) t cosat
View Answer
Explanation: Given X(s) = \(\frac{s}{(s^2+a^2)^2}\)
x(t) = L-1 [X(s)] = \(L^{-1} \Big[\frac{s}{(s^2+a^2)^2}\Big] = \frac{1}{2a} \Big[-\frac{d}{ds} \Big\{\frac{a}{s^2+a^2}\Big\}\Big] \)
= \(\frac{1}{2a} tL^{-1} \Big[\frac{a}{s^2+a^2}\Big] = \frac{1}{2a}t sinat\).
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)
View Answer
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) cosh2t
b) \(\frac{1}{2}\) cosh2t
c) sinh2t
d) \(\frac{1}{2}\) sinh2t
View Answer
Explanation: Given X(s) = \(\frac{s}{2s^2-8} = \frac{1}{2} \frac{s}{(s^2-2^2)}\)
We know that coshωt = \(\frac{s}{(s^2-w^2)}\)
∴x(t)=L-1 [X(s)] = \(\frac{1}{2} L^{-1} \left(\frac{s}{(s^2-2^2)}\right) = \frac{1}{2} cosh2t.\)
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) \(\frac{e^{bt} + e^{-at}}{t}\)
View Answer
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 cos2t u(t) – e-2t sin2t u(t)
b) 2e-2t cos2t u(t) – \(\frac{5}{2}\) e-2t sin2t u(t)
c) 2e-2t cos2t u(t) – e-2t sin2t u(t)
d) e-2t cos2t u(t) – \(\frac{5}{2}\) e-2t sin2t u(t)
View Answer
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 cos2t u(t) – \(\frac{5}{2}\) e-2t sin2t 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)
View Answer
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)
View Answer
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).
Sanfoundry Global Education & Learning Series – Signals & Systems.
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