# 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) $$\frac{1}{s} tanh⁡(\frac{as}{2})$$

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 $$\frac{1}{s} tanh⁡(\frac{as}{2})$$.

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) $$\frac{p}{s^2+p^2}×tanh⁡⁡(\frac{s\pi}{2p})$$

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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