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 \(\frac{p}{s^2+p^2}×coth⁡⁡(\frac{s\pi}{2p})\).

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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