250+ TOP MCQs on Laplace Transform of Periodic Function and Answers

Ordinary Differential Equations Multiple Choice Questions 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}))
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})).

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{spi}{2p}))
b) (frac{p}{s^2+p^2}×sinh⁡(frac{spi}{2p}))
c) (frac{p}{s^2+p^2}×coth⁡⁡(frac{spi}{2p}))
d) (frac{p}{s^2+p^2}×tanh⁡⁡(frac{spi}{2p}))
Answer: c
Explanation: From this question, we know –
Period of sin(t)=2π
Period of sin⁡(pt)=(frac{2pi}{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}×pend{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{spi}{2p})), (Multiplying and dividing by (e^{frac{-sπ}{2p}}))
Thus, the answer is (frac{p}{s^2+p^2}×coth⁡⁡(frac{spi}{2p})).

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