This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “The Laplace Transform”.
1. The necessary condition for convergence of the Laplace transform is the absolute integrability of f(t)e-σt.
a) True
b) False
View Answer
Explanation: The necessary condition for convergence of the Laplace transform is the absolute integrability of f(t)e-σt.Mathematically, this can be stated as
\(\int_{-∞}^∞|f(t) e^{-σt}|\)dt<∞
Laplace transform exists only for signals which satisfy the above equation in the given region.
2. Find the Laplace transform of e-at u(t) and its ROC.
a) \(\frac{1}{s-a}\), Re{s}>-a
b) \(\frac{1}{s}\), Re{s}>a
c) \(\frac{1}{s×a}\), Re{s}>a
d) \(\frac{1}{s+a}\), Re{s}>-a
View Answer
Explanation: Laplace transform, L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt\)
L{e-at) u(t)} = \(\int_{-∞}^∞ e^{-at} u(t) e^{-st} \,dt = \int_0^∞ e^{-at} e^{-st} \,dt = \frac{1}{s+a}\) when (s+a)>0
(σ+a)>0
σ>-a
ROC is Re{s}>-a.
3. Find the Laplace transform of δ(t).
a) 1
b) 0
c) ∞
d) 2
View Answer
Explanation: Laplace transform, L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt\)
L{δ(t)} = \(\int_{-∞}^∞ δ(t) e^{-st} \,dt\)
[x(t)δ(t) = x(0)δ(t)]
= \(\int_{-∞}^∞ δ(t)dt\)
= 1.
4. Find the Laplace transform of u(t) and its ROC.
a) \(\frac{1}{s}\), σ<0
b) \(\frac{1}{s}\), σ>0
c) \(\frac{1}{s-1}\), σ=0
d) \(\frac{1}{1-s}\), σ≤0
View Answer
Explanation: Laplace transform, L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt\)
L{u(t)} = \(\int_{-∞}^∞ u(t) e^{-st} \,dt = \int_0^∞ e^{-st} \,dt = \frac{1}{s}\) when s>0 i.e,σ>0.
5. Find the ROC of x(t) = e-2t u(t) + e-3t u(t).
a) σ>2
b) σ>3
c) σ>-3
d) σ>-2
View Answer
Explanation: Given x(t) = e-2t u(t) + e-3t u(t)
Laplace transform, L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt\)
X(s) = \(\frac{1}{s+2} + \frac{1}{s+3}\)
ROC is {σ > -2}∩{σ > -3}
Hence, the ROC is σ > -2.
6. Find the Laplace transform of cosωt u(t).
a) \(\frac{s}{s^2+ω^2}\)
b) \(\frac{s}{s^2-ω^2}\)
c) \(\frac{ω}{s^2+ω^2}\)
d) \(\frac{ω}{s^2-ω^2}\)
View Answer
Explanation: Laplace transform, L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt\)
X(s) = L{cosωt u(t)} = \(L[\frac{e^{jωt} + e^{-jωt}}{2} \,u(t)] = \frac{1}{2} L[e^{jωt} \,u(t)] + \frac{1}{2} L[e^{jωt} \,u(t)]\)
X(s) = \(\frac{1}{2} (\frac{1}{s-jω}) + \frac{1}{2} (\frac{1}{s+jω}) = \frac{s}{s^2+ω^2}\).
7. Find the Laplace transform of e-at sinωt u(t).
a) \(\frac{s+a}{(s+a)^2-ω^2}\)
b) \(\frac{ω}{(s+a)^2-ω^2}\)
c) \(\frac{s+a}{(s+a)^2+ω^2}\)
d) \(\frac{ω}{(s+a)^2+ω^2}\)
View Answer
Explanation: Laplace transform, L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt\)
L{x(t)} = X(s) = \(L{e^{-at} \frac{e^{jωt}-e^{-jωt}}{2j} \,u(t)} = \frac{1}{2j} L[e^{-(a-jω)t} \,u(t)] – \frac{1}{2j} L[e^{-(a+jω)t} \,u(t)]\)
X(s) = \(\frac{1}{2j} [\frac{1}{s+(a-jω)} – \frac{1}{s+(a+jω)}] = \frac{1}{2j} [\frac{2jω}{(s+a)^2+ω^2}] = \frac{ω}{(s+a)^2+ω^2}\)
e^-at sinωt u(t) \(\underleftrightarrow{LT} \frac{ω}{(s+a)^2+ω^2}\);ROC Re(s)>-a.
8. Find the Laplace transform of the signal x(t)=et sin2t for t≤0.
a) \(\frac{2}{(s-1)^2+2^2}\)
b) \(-\frac{2}{(s-1)^2+2^2}\)
c) \(\frac{2}{(s+1)^2+2^2}\)
d) \(-\frac{2}{(s+1)^2+2^2}\)
View Answer
Explanation: Given x(t) = et sin2t for t≤0
∴ x(t) = et sin2t u(-t)
L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt = \int_{-∞}^∞ e^t \,sin2t \,u(-t) e^{-st} \,dt\)
= \(\int_{-∞}^0 \left(e^{j2t} – e^{-j2t}{2j}\right) = \frac{1}{2j} \int_{-∞}^0 [e^{(1-s+j2)t} – e^{(1-s-j2)t}]dt\)
= \(\frac{1}{2j} \left(\frac{1}{1-s+j2}-\frac{1}{1-s-j2}\right)\)
=\(-\frac{2}{(s-1)^2+2^2}\).
9. Find the Laplace transform of the signal x(t)=te-2|t|.
a) \(-\frac{1}{(s-2)^2} + \frac{1}{(s+2)^2}\)
b) \(\frac{1}{(s-2)^2} + \frac{1}{(s+2)^2}\)
c) \(\frac{1}{(s-2)^2} – \frac{1}{(s+2)^2}\)
d) \(-\frac{1}{(s-2)^2} – \frac{1}{(s+2)^2}\)
View Answer
Explanation: Given x(t)=te-2|t|
L{x(t)} = X(s) = \(\int_{-∞}^∞ x(t) e^{-st} \,dt = \int_{-∞}^∞ te^{-2|t|} e^{st} \,dt \)
=\(\int_{-∞}^0 te^{2t} e^{-st} \,dt + \int_0^∞ te^{-2t} e^{-st} \,dt = -\frac{1}{(s-2)^2} + \frac{1}{(s+2)^2}\).
10. Find the Laplace transform of (cos2t)3 u(t).
a) \(\frac{s(s^2+28)}{(s^2+36)(s^2+4)}\)
b) \(\frac{s(s^2+36)}{(s^2+28)(s^2+4)}\)
c) \(\frac{s(s^2+4)}{(s^2+36)(s^2+28)}\)
d) \(\frac{s}{(s^2+36)(s^2+4)}\)
View Answer
Explanation: Given x(t)=(cos2t)3 u(t) = \(\frac{[cos6t+3cos2t]}{4}\) u(t)
X(s) = L{x(t)} = \(L[\frac{(cos6t+3cos2t)}{4} \,u(t)] = \frac{1}{4}\) {L[cos6t u(t)]+3L[cos2t u(t)]}
= \(\frac{1}{4} \left(\frac{s}{s^2+(6)^2} + 3 \frac{s}{s^2+(2)^2}\right) = \frac{s(s^2+28)}{(s^2+36)(s^2+4)}\).
11. Find the Laplace transform of [1 +sin 2t cos 2t]u(t).
a) \(\frac{s^2+2s+16}{s(s^2-4^2)}\)
b) \(\frac{s^2+2s+16}{s(s^2+4^2)}\)
c) \(\frac{s^2+2s+16}{(s^2+4^2)}\)
d) \(\frac{s^2+2s+16}{s}\)
View Answer
Explanation: Given x(t)=[1 + sin 2t cos 2t]u(t) = (1 + \(\frac{1}{2} \,sin4t\))u(t)
L{x(t)} = X(s) = L[u(t) + \(\frac{1}{2}\) sin4t u(t)] = L[u(t)] + \(\frac{1}{2}\) L[sin4t u(t)] = \(\frac{1}{s} + \frac{1}{2} \frac{4}{(s^2+4^2)} = \frac{s^2+2s+16}{s(s^2+4^2)}\).
Sanfoundry Global Education & Learning Series – Signals & Systems.
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