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)