Laplace Transform by Properties Questions and Answers - Sanfoundry PDF

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3/31/2021

Laplace Transform by Properties Questions and Answers - Sanfoundry

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

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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)] = \(\frac{s}{a^2+s^2}\)

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L \([\frac{s}{a^2+s^2}]\) = Cos(at).

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2. Inverse Laplace transform of \(\frac{1}{(s+1)(s-1)(s+2)}\) is? a) –1⁄2 et + 1 ⁄6 e-t + 1 ⁄3 e2t b) –1 ⁄2 e-t + 1 ⁄6 et + 1 3⁄ e-2t c) 1 ⁄2 e-t – 1 ⁄6 et – 1 ⁄3 e-2 d) –1 ⁄2 e-t + 1 ⁄6 e-t + 1 3⁄ 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? 1

–t

1

t

1

-t 2

1

-t

1

-5t

a) ⁄ 6 e – ⁄36 e + ⁄36 e 1 t 1 t 1 -5t b) ⁄6 e t – ⁄36 e + ⁄36 e 1

5t

c) ⁄6 e t – ⁄36 e + ⁄36 e 1 -t 1 -t 1 5t d) ⁄6 e t- ⁄36 e + ⁄36 e View Answer Answer: a

Explanation: Given, \(F(s)=\frac{1}{(s-1)^2 (s+5)}=\frac{1}{(s-1)} \left [\frac{1}{(s-1)(s+5)}\right ]\) =\(\frac{1}{(s-1)} \left [\frac{1}{6(s-1)}-\frac{1}{6(s+5)}\right ]\) =\(\frac{1}{6} \left [\frac{1}{(s-1)^2}-\frac{1}{(s-1)(s+5)}\right ]\)

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=\(\frac{1}{6} \left [

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4. Find the inverse laplace transform of \(\frac{1}{(s^2+1)(s – 1)(s + 5)}\). a) 1⁄ 12 et – 1⁄13 Cos(-t) – 1⁄12 Sin(-t) – 1⁄156 e-5t b) 1⁄12 e-t – 1 ⁄13 Cos(t) – 1 ⁄12 Sin(t) – 1 ⁄156 e5t c) 1 ⁄12 et – 1 ⁄13 Cos(t) – 1 ⁄12 Sin(t) – 1 156 ⁄ e-5t d) 1⁄12 et + 1 ⁄13 Cos(t) + 1 ⁄12 Sin(t) + 1 ⁄156 e-5t View Answer Answer: c Explanation: Given , F(s)=\(\frac{1}{(s^2+1)(s-1)(s+5)}\) F(s)=\(\frac{1}{6(s^2+1)}\left {6(s^2+1)(s+5)}\)

[\frac{1}{s-1}-\frac{1}{s+5}\right

]=\frac{1}{6(s^2+1)(s-1)}-\frac{1}

=\(\frac{1}{6} \left [\frac{1}{2*(s – 1)}-\frac{1}{2} \frac{s+1}{(s^2+ 1)}\right ]-\frac{1}{6}\left [\frac{1} {26*(s + 5)}-\frac{1}{26} \frac{s-5}{(s^2+1)}\right ]\) =\(\frac{1}{12(s – 1)}-\frac{1}{26} \frac{2s+3}{(s^2+ 1)}-\frac{1}{156(s + 5)}\) =\(\frac{1}{12(s – 1)}-\frac{1}{13} \frac{s}{(s^2+ 1)}-\frac{1}{12} \frac{1}{(s^2+ 1)}-\frac{1}{156(s + 5)}\) =\(\frac{1}{12} e^t-\frac{1}{13} Cos(t)-\frac{1}{12} Sin(t)-\frac{1}{156}e^{-5t}\) 5. Find the inverse laplace transform of \(\frac{s}{(s^2+ 4)^2}\). 1

a) ⁄ 4 sin(2t) 2 b) t ⁄4 sin(2t) t

c) ⁄4 sin(2t) t 2 d) ⁄4 sin(2t ) 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 \(\frac{s}{(s^2+4)^2}=\frac{t}{4} sin(2t)\) advertisement

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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) 12 ⁄20 d) 2 View Answer Answer: c

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Hence, by nal 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, \(x(0)=\lim_{x\rightarrow \infty} sX(s)=2\) advertisement

10. Find the inverse lapace of \(\frac{(s+1)}{[(s+1)^2+4][(s+1)^2+1]}\). 1

t

1

t

a) ⁄ 3 e [Cos(t) – Cos(2t)]. b) 1⁄3 e-t [Cos(t) + Cos(2t)]. c) ⁄3 e [Cos(t) + Cos(2t)]. d) 1⁄3 e-t [Cos(t) – Cos(2t)].

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

-t

1

t

1

1

a) ⁄ 2 e + ⁄2 Sin(-t) – ⁄2 Cos(-t) 1 t 1 1 b) ⁄2 e + ⁄2 Sin(t) – ⁄2 Cos(t) 1

1

c) ⁄2 e + ⁄2 Sin(t) + ⁄2 Cos(t) 1 t 1 1 d) ⁄2 e – ⁄2 Sin(t) – ⁄2 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)=\(\frac{1}{2}G(s)=>y(t)=\int_0^t g(t)dt=\frac{1}{2}e^t+\frac{1}{2}sin(t)-\frac{1}{2}cos(t)\)

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To practice all areas of Engineering Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers. Participate in the Sanfoundry Certication contest to get free Certicate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs! Telegram | Youtube | LinkedIn | Instagram | Facebook | Twitter | Pinterest « Prev - Engineering Mathematics Questions and Answers – Laplace Transform by Properties – 2 » Next - Ordinary Dierential Equations Questions and Answers – Laplace Transform of Periodic Function

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