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 \(f(t)=\frac{1}{6} e^t t-\frac{1}{36} e^t+\frac{1}{36} e^{-5t}\)

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)=\(\lim_{x\rightarrow ∞} sX(s)\)

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.

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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 is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn