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

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)

Explanation: L[Cos(at)] = $$\frac{s}{a^2+s^2}$$
L-1 $$[\frac{s}{a^2+s^2}]$$ = Cos(at).

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

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 $$f(t)=-\frac{1}{2} e^{-t}+\frac{1}{6} e^t+\frac{1}{3} e^{-2t}$$

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

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

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)

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 $$\frac{s}{(s^2+4)^2}=\frac{t}{4} sin(2t)$$

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(∞)=$$\lim_{x\rightarrow 0} ⁡sX(s)$$

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(∞)=$$\lim_{x\rightarrow 0} ⁡sX(s)$$

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

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

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,
$$x(0)=\lim_{x\rightarrow \infty} ⁡sX(s)=2$$

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)].

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

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)

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)=$$\frac{1}{2}G(s)=>y(t)=\int_0^t g(t)dt=\frac{1}{2}e^t+\frac{1}{2}sin(t)-\frac{1}{2}cos(t)$$

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