Home » Uncategorized » GATE 2016 EE - SET 2 Question 29

GATE 2016 EE - SET 2 Question 29

Using Laplace Transform, consider, 2\int_{-\infty }^{\infty }\frac{\sin 2\pi t}{\pi t}dt=I

Since \frac{\sin 2\pi t}{\pi t} is an even function, the integral now becomes,

2\int_{-\infty }^{\infty }\frac{\sin 2\pi t}{\pi t}dt\Rightarrow 2\left [ 2\int_{0}^{\infty }\frac{\sin 2\pi t}{\pi t} \right ]dt\Rightarrow 4\int_{0}^{\infty }\frac{\sin 2\pi t}{\pi t}dt\Rightarrow \frac{4}{\pi }\int_{0}^{\infty }\frac{\sin 2\pi t}{t}dt ________ (1)

Laplace transform of \frac{\sin 2\pi t}{t}=\int_{s}^{\infty }L[\sin 2\pi t]ds \because L[\frac{f(t)}{t}]=\int_{s}^{\infty }L[f(t)]ds

\int_{s}^{\infty }L[\sin 2\pi t]ds\Rightarrow \int_{s}^{\infty }\frac{2\pi }{s^{2}+(2\pi )^{2}}ds

\because L(\sin at)=\frac{a}{s^{2}+a^{2}} also,  \because \int \frac{a}{x^{2}+a^{2}}dx=\tan ^{-1}(\frac{x}{a})

\Rightarrow\tan ^{-1}\left [ \frac{s}{2\pi } \right ]_{s}^{\infty }\Rightarrow \tan ^{-1}\infty -\tan ^{-1}\left ( \frac{s}{2\pi } \right )

\Rightarrow \frac{\pi }{2}-\tan ^{-1}\left ( \frac{s}{2\pi } \right )\Rightarrow \cot ^{-1}\left ( \frac{s}{2\pi } \right )

We now have, L\left [ \frac{\sin 2\pi t}{t} \right ]=\cot ^{-1}\left ( \frac{s}{2\pi } \right )

Taking s = 0, \int_{0}^{\infty }\frac{\sin 2\pi t}{t}dt=\cot ^{-1}(0)=\frac{\pi }{2} _______(2)

From (1) and (2), I=\frac{4}{\pi }\times \frac{\pi }{2}\Rightarrow 2

Translate »