Engineering Mathematics Questions and Answers – Laplace Transform by Properties – 3

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This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Laplace Transform by Properties – 3”.

1. Time domain function of \(\frac{s}{a^2+s^2}\) is given by?
a) Cos(at)
b) Sin(at)
c) Cos(at)Sin(at)
d) Sin(t)
View Answer

Answer: a
Explanation: L[Cos(at)] = \(\frac{s}{a^2+s^2}\)
L-1 \([\frac{s}{a^2+s^2}]\) = Cos(at).
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2. Inverse Laplace transform of \(\frac{1}{(s+1)(s-1)(s+2)}\) is?
a) –12 et + 16 e-t + 13 e2t
b) –12 e-t + 16 et + 13 e-2t
c) 12 e-t16 et13 e-2
d) –12 e-t + 16 e-t + 13 e-2
View Answer

Answer: b
Explanation:
Given, \(F(s)=\frac{1}{(s+1)(s-1)(s+2)}=\frac{-1}{2(s+1)} +\frac{1}{6(s-1)}+\frac{1}{3(s+2)}\)
Hence, inverse laplace transform is \(f(t)=-\frac{1}{2} e^{-t}+\frac{1}{6} e^t+\frac{1}{3} e^{-2}\)

3. Inverse laplace transform of \(\frac{1}{(s-1)^2 (s+5)}\) is?
a) 16 e – t136 et + 136 e-5t
b) 16 ett – 136 et + 136 e-5t
c) 16 e-tt2136 e-t + 136 e5t
d) 16 e-t t-136 e-t + 136 e5t
View Answer

Answer: a
Explanation:
Given, \(F(s)=\frac{1}{(s-1)^2 (s+5)}=\frac{1}{(s-1)} \left [\frac{1}{(s-1)(s+5)}\right ]\)
=\(\frac{1}{(s-1)} \left [\frac{1}{6(s-1)}-\frac{1}{6(s+5)}\right ]\)
=\(\frac{1}{6} \left [\frac{1}{(s-1)^2}-\frac{1}{(s-1)(s+5)}\right ]\)
=\(\frac{1}{6} \left [
\frac{1}{(s-1)^2} – \frac{1}{6} \left [\frac{1}{(s-1)} – \frac{1}{(s+5)}\right ]\right ]\)
=\(\frac{1}{6(s-1)^2}-\frac{1}{36(s-1)}+\frac{1}{36(s+5)}\)
Inverse Laplace transform is \(f(t)=\frac{1}{6} e^t t-\frac{1}{36} e^t+\frac{1}{36} e^{-5t}\)
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4. Find the inverse laplace transform of \(\frac{1}{(s^2+1)(s – 1)(s + 5)}\).
a) 112 et113 Cos(-t) – 112 Sin(-t) – 1156 e-5t
b) 112 e-t113 Cos(t) – 112 Sin(t) – 1156 e5t
c) 112 et113 Cos(t) – 112 Sin(t) – 1156 e-5t
d) 112 et + 113 Cos(t) + 112 Sin(t) + 1156 e-5t
View Answer

Answer: c
Explanation:
Given , F(s)=\(\frac{1}{(s^2+1)(s-1)(s+5)}\)
F(s)=\(\frac{1}{6(s^2+1)}\left [\frac{1}{s-1}-\frac{1}{s+5}\right ]=\frac{1}{6(s^2+1)(s-1)}-\frac{1}{6(s^2+1)(s+5)}\)
=\(\frac{1}{6} \left [\frac{1}{2*(s – 1)}-\frac{1}{2} \frac{s+1}{(s^2+ 1)}\right ]-\frac{1}{6}\left [\frac{1}{26*(s + 5)}-\frac{1}{26} \frac{s-5}{(s^2+1)}\right ]\)
=\(\frac{1}{12(s – 1)}-\frac{1}{26} \frac{2s+3}{(s^2+ 1)}-\frac{1}{156(s + 5)}\)
=\(\frac{1}{12(s – 1)}-\frac{1}{13} \frac{s}{(s^2+ 1)}-\frac{1}{12} \frac{1}{(s^2+ 1)}-\frac{1}{156(s + 5)}\)
=\(\frac{1}{12} e^t-\frac{1}{13} Cos(t)-\frac{1}{12} Sin(t)-\frac{1}{156}e^{-5t}\)

5. Find the inverse laplace transform of \(\frac{s}{(s^2+ 4)^2}\).
a) 14 sin(2t)
b) t24 sin(2t)
c) t4 sin(2t)
d) t4 sin(2t2)
View Answer

Answer: c
Explanation:
Given, \(Y(s)=\frac{s}{(s^2+ 4)^2}\)
Inverse Laplace transform of \(\frac{1}{s^2+4}\)=sin⁡(2t)
Now, \(\frac{d}{ds} (\frac{1}{s^2+4})\)=-tsin(2t)
Inverse lapalce of \(\frac{-2s}{(s^2+4)^2}=-\frac{t}{2} sin(2t)\)
Inverse lapalce of \(\frac{s}{(s^2+4)^2}=\frac{t}{4} sin(2t)\)
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6. Final value theorem states that _________
a) x(0)=\(\lim_{x\rightarrow ∞} sX(s)\)
b) x(∞)=\(\lim_{x\rightarrow ∞} sX(s)\)
c) x(0)=\(\lim_{x\rightarrow 0} sX(s)\)
d) x(∞)=\(\lim_{x\rightarrow 0} ⁡sX(s)\)
View Answer

Answer: d
Explanation: Final value theorem states that
x(∞)=\(\lim_{x\rightarrow 0} ⁡sX(s)\)

7. Initial value theorem states that ___________
a) x(0)=\(\lim_{x\rightarrow ∞} sX(s)\)
b) x(∞)=\(\lim_{x\rightarrow ∞} sX(s)\)
c) x(0)=\(\lim_{x\rightarrow 0} sX(s)\)
d) x(∞)=\(\lim_{x\rightarrow 0} ⁡sX(s)\)
View Answer

Answer: a
Explanation: Initial value theorem states that
x(0)=\(\lim_{x\rightarrow ∞} sX(s)\)
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8. Find the value of x(∞) if \(X(s)=\frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}\).
a) 5
b) 4
c) 1220
d) 2
View Answer

Answer: c
Explanation:
Given, \(X(s)=\frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}\)
Hence, \(sX(s)=\frac{2s^3+5s^2+12}{s^3+4s^2+14s+20}\)
Hence, by final value theorem,
\(x(∞)=\lim_{x\rightarrow 0} ⁡sX(s)=\frac{12}{20}\)

9. Find the value of x(0) if \(X(s)=\frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}\).
a) 5
b) 4
c) 12
d) 2
View Answer

Answer: d
Explanation:
Given, \(X(s)=\frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}\)
Hence, \(sX(s)=\frac{2s^3+5s^2+12}{s^3+4s^2+14s+20}\)
Hence, by initial value theorem,
\(x(0)=\lim_{x\rightarrow \infty} ⁡sX(s)=2\)
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10. Find the inverse lapace of \(\frac{(s+1)}{[(s+1)^2+4][(s+1)^2+1]}\).
a) 13 et [Cos(t) – Cos(2t)].
b) 13 e-t [Cos(t) + Cos(2t)].
c) 13 et [Cos(t) + Cos(2t)].
d) 13 e-t [Cos(t) – Cos(2t)].
View Answer

Answer: d
Explanation:
Given, \(Y(s)=\frac{(s+1)}{[(s+1)^2+4][(s+1)^2+1]}\)
=\(\frac{s+1}{3(s^2+ 2*s + 2)}-\frac{s+1}{3(s^2+ 2*s + 5)}\)
=\(\frac{s+1}{3[(s+1)^2+1]}-\frac{s+1}{3[(s+1^2+4)]}\)
=\(\frac{1}{3} [e^{-t} Cos(t)]-\frac{1}{3}[e^{-t} Cos(2t)]\)
=\(\frac{1}{3} e^{-t} [Cos(t)-Cos(2t)]\)

11. Find the inverse laplace transform of \(Y(s)=\frac{2s}{1-s^2}e^{-s}\).
a) -e-t + 1 + et – 1
b) -e-t + 1 – et + 1
c) -e-t + 1 + et + 1
d) -e-t + 1 – et – 1
View Answer

Answer: d
Explanation: Given,
Y(s)=\(\frac{2s}{1-s^2}e^{-s}\)
Let,G(s)=\(\frac{2s}{1-s^2}=-\frac{1}{s – 1}-\frac{1}{s + 1}\)
hence,g(t)=\(-e^{-t} – e^t\)
Since,Y(s)=\(e^{-s} G(s)=>y(t)=g(t-1)\)
hence,y(t)=\(-e^{-t+1}-e^{t-1}\)

12. Find the inverse laplace transform of \(\frac{1}{s(s-1)(s^2+1)}\).
a) 12 e-t + 12 Sin(-t) – 12 Cos(-t)
b) 12 et + 12 Sin(t) – 12 Cos(t)
c) 12 et + 12 Sin(t) + 12 Cos(t)
d) 12 et12 Sin(t) – 12 Cos(t)
View Answer

Answer: b
Explanation: We know that,
Given, Y(s)=\(\frac{1}{s(s-1)(s^2+1)}\)
Let, G(s)=\(\frac{1}{(s-1)(s^2+1)}=\frac{1}{2(s^2-1)}-\frac{s+1}{2(s^2+1)}=\frac{1}{2*(s-1)}-\frac{s}{2(s^2+1)}-\frac{1}{2(s^2+1)}\)
Now, g(t)=\(\frac{1}{2}e^t-\frac{1}{2}cos(t)-\frac{1}{2}cos(t)\)
Now, Y(s)=\(\frac{1}{2}G(s)=>y(t)=\int_0^t g(t)dt=\frac{1}{2}e^t+\frac{1}{2}sin(t)-\frac{1}{2}cos(t)\)

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

To practice all areas of Engineering Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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