# Signals & Systems Questions and Answers – The Laplace Transform

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

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

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

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

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

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.
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6. Find the Laplace transform of cos⁡ωt u(t).
a) $$\frac{s}{s^2+ω^2}$$
b) $$\frac{s}{s^2-ω^2}$$
c) $$\frac{ω}{s^2+ω^2}$$
d) $$\frac{ω}{s^2-ω^2}$$

Explanation: Laplace transform, L{x(t)} = X(s) = $$\int_{-∞}^∞ x(t) e^{-st} \,dt$$
X(s) = L{cos⁡ωt 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 sin⁡ωt 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) $$\frac{ω}{(s+a)^2+ω^2}$$

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 sin⁡ωt 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 sin⁡2t 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) $$-\frac{2}{(s+1)^2+2^2}$$

Explanation: Given x(t) = et sin⁡2t for t≤0
∴ x(t) = et sin⁡2t u(-t)
L{x(t)} = X(s) = $$\int_{-∞}^∞ x(t) e^{-st} \,dt = \int_{-∞}^∞ e^t \,sin⁡2t \,u(-t) e^{-st} \,dt$$
= $$\int_{-∞}^0 \left(e^{j2t} – e^{-j2t}{2j}\right) = \frac{1}{2j} \int_{-∞}^0 [e^{(1-s+j2)t} – e^{(1-s-j2)t}]dt$$
= $$\frac{1}{2j} \left(\frac{1}{1-s+j2}-\frac{1}{1-s-j2}\right)$$
=$$-\frac{2}{(s-1)^2+2^2}$$.

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) $$-\frac{1}{(s-2)^2} – \frac{1}{(s+2)^2}$$

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 (cos⁡2t)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) $$\frac{s}{(s^2+36)(s^2+4)}$$

Explanation: Given x(t)=(cos⁡2t)3 u(t) = $$\frac{[cos⁡6t+3cos⁡2t]}{4}$$ u(t)
X(s) = L{x(t)} = $$L[\frac{(cos⁡6t+3cos⁡2t)}{4} \,u(t)] = \frac{1}{4}$$ {L[cos⁡6t u(t)]+3L[cos⁡2t 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) $$\frac{s^2+2s+16}{s}$$

Explanation: Given x(t)=[1 + sin ⁡2t cos ⁡2t]u(t) = (1 + $$\frac{1}{2} \,sin⁡4t$$)u(t)
L{x(t)} = X(s) = L[u(t) + $$\frac{1}{2}$$ sin⁡4t u(t)] = L[u(t)] + $$\frac{1}{2}$$ L[sin⁡4t 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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