250+ TOP MCQs on The Laplace Transform and Answers

Signals & Systems Multiple Choice Questions 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
Answer: A
Clarification: 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
Answer: D
Clarification: 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
Answer: A
Clarification: 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
Answer: B
Clarification: 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
Answer: D
Clarification: 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 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})
Answer: A
Clarification: 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})
Answer: D
Clarification: 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)=e^t 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})
Answer: B
Clarification: Given x(t) = e^t sin⁡2t for t≤0
∴ x(t) = e^t 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})
Answer: A
Clarification: 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)})
Answer: A
Clarification: 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})
Answer: B
Clarification: 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)}).