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GATE 2015 EE - SET 2 - Question 29

Solution :

Fourier  transform of  x(t) = \int_{-\infty }^{\infty}x(t)e^{-jwt}dt

\Rightarrow \int_{-1}^{1}e^{j10t}e^{-jwt}dt \Rightarrow\int_{-1}^{1}e^{(10-w)jt}dt

\left \| \frac{e^{10-w}jt}{(10-w)j} \right \|^{1}_{-1}

\Rightarrow \frac{e^{(10-w)j}}{(10-w)j} - \frac{e^{-(10-w)j}}{(10-w)j}

\Rightarrow \frac{1}{(10-w)}\left [ \frac{e^{(10-w)j- e^{-(10-w)j}}}{j} \right ]

Sinx = \frac{e^{xi-e^{-xi}}}{2i}

\Rightarrow \frac{2}{(10-w)}\left [ \frac{e^{(10-w)j- e^{-(10-w)j}}}{2j} \right ]

\Rightarrow\frac{-2}{(w-10)}\left [ \frac{-(e^{(w-10)j- e^{-(w-10)j})}}{j} \right ]

\Rightarrow \frac{2}{(w-10)}\left [ \frac{e^{(w-10)j- e^{-(w-10)j}}}{2j} \right ]

\frac{2}{w-10}sin(w-10)

 

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