Ordinary Differential Equations Questions and Answers – Laplace Transform of Periodic Function

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This set of Ordinary Differential Equations Multiple Choice Questions & Answers (MCQs) focuses on “Laplace Transform of Periodic Function”.

1. Find the laplace transform of f(t), where
f(t) = 1 for 0 < t < a
-1 for a < t < 2a
a) \(\frac{1}{s} coth⁡(\frac{as}{2})\)
b) \(\frac{1}{s} sinh⁡(\frac{as}{2})\)
c) \(\frac{1}{s} e^{-as}\)
d) \(\frac{1}{s} tanh⁡(\frac{as}{2})\)
View Answer

Answer: d
Explanation: In the given question
f(t) is a periodic function having a period 2a
The formula for Laplace Transform is given by:
\(L(f(t))=\frac{1}{1-e^{-2as}} \int_{0}^{2a}e^{-st} f(t)dt\)
\(L(f(t))=\frac{1}{1-e^{-2as}} \int_{0}^{a}e^{-st} (1)dt + \frac{1}{1-e^{-2as}} \int_{a}^{2a}e^{-st}(-1)dt\)
=\(\begin{bmatrix}\frac{1}{1-e^(-2as)}×\frac{e^{-as}}{-s} – \frac{1}{1-e^{-2as}} × \frac{-1}{s}\end{bmatrix} – \begin{bmatrix}\frac{1}{1-e^{-2as}} × \frac{e^{-2as}}{-s} – \frac{1}{1-e^{-2as}} × \frac{e^{-as}}{-s}\end{bmatrix}\)
= \(\frac{1}{1-e^{-2as}}×\frac{1}{s}×(1-e^{-as})^2\)
= \(\frac{1}{s}(\frac{1+e^{-as}}{1-e^{-as}})\)
Dividing both numerator and denominator by \(e^{\frac{-as}{2}}\)
= \(\frac{1}{s} tanh⁡(\frac{as}{2})\)
Thus, the correct answer is \(\frac{1}{s} tanh⁡(\frac{as}{2})\).
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2. Find the laplace transform of f(t), where f(t) = |sin(pt)| and t>0.
a) \(\frac{p}{s^2+p^2}×cosh⁡(\frac{s\pi}{2p})\)
b) \(\frac{p}{s^2+p^2}×sinh⁡(\frac{s\pi}{2p})\)
c) \(\frac{p}{s^2+p^2}×coth⁡⁡(\frac{s\pi}{2p})\)
d) \(\frac{p}{s^2+p^2}×tanh⁡⁡(\frac{s\pi}{2p})\)
View Answer

Answer: c
Explanation: From this question, we know –
Period of sin(t)=2π
Period of sin⁡(pt)=\(\frac{2\pi}{p}\)
Period of |sin⁡(pt)|=\(\frac{\pi}{p}\)
\(L(f(t))=\frac{1}{1-e^{\frac{-\pi}{ps}}} \int_{0}^{\frac{\pi}{p}}e^{-st} f(t)dt\)
Since |sin⁡(pt)| is positive in all quadrants
\(L(f(t))=\frac{1}{1-e^{\frac{-\pi}{ps}}} \int_{0}^{\frac{\pi}{p}}e^{-st} sin⁡(pt)dt\)
=\(\frac{1}{1-e^{\frac{-\pi}{ps}}}\begin{bmatrix}\frac{e^{\frac{-sπ}{p}}}{s^2+p^2}×p\end{bmatrix}-\begin{bmatrix}\frac{1}{s^2+p^2}×(-p)\end{bmatrix}\)
=\(\frac{1}{1-e^{\frac{-\pi}{ps}}}×\frac{p}{s^2+p^2}×(1+e^{\frac{-π}{ps}})\)
=\(\frac{p}{s^2+p^2}×coth⁡(\frac{s\pi}{2p})\), (Multiplying and dividing by \(e^{\frac{-sπ}{2p}}\))
Thus, the answer is \(\frac{p}{s^2+p^2}×coth⁡⁡(\frac{s\pi}{2p})\).

Sanfoundry Global Education & Learning Series – Ordinary Differential Equations.

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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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