Question

Let X($\omega$) denote the Fourier transform of signal x(t) depicted in figure as shown below :

Then the value of ${\int }_{-\infty }^{\infty }X\left(\omega \right)\frac{2\sin \omega }{\omega }{e}^{j2\omega }d\omega$ is _____.

• A. 22

Answer 1

The correct option is A 22

Assume $Y\left(\omega \right)=\frac{2sin\omega }{\omega }{e}^{j2\omega }$

$rect\left(t/2\right)⟷\frac{2sin\omega }{\omega }$

$rect\left(\frac{t+2}{2}\right)⟷\frac{2\sin \omega e^{j2\omega }}{\omega }$
so, y(t) is

Now,
${\int }_{-\infty }^{\infty }X\left(\omega \right)\frac{2sin\omega }{\omega }{e}^{j2\omega }\delta \omega ={\int }_{-\infty }^{\infty }X\left(\omega \right)Y\left(\omega \right)d\omega$

$=2\pi {\left[x\right(t)\ast y(t\left)\right]}_{t=0}$

$=2\pi {\int }_{-\infty }^{\infty }x\left(\tau \right)y(t-\tau )\delta \tau {|}_{t=0}$

$=2\pi {\int }_{-\infty }^{\infty }x\left(\tau \right)y(-\tau )\delta \tau$

$x\left(\tau \right)\times y(-\tau )$ =


${\int }_{-\infty }^{\infty }x\left(\tau \right)y(-\tau )\delta \tau$ =Area under the curve


${\int }_{-\infty }^{\infty }x\left(\tau \right)y(-\tau )\delta \tau =1\times 1+\frac{1}{2}\times 1\times 1+1\times 2=3.5$

$2\pi {\int }_{-\infty }^{\infty }x\left(\tau \right)y(-\tau )\delta \tau =3.5\times 2\pi$
$=7\pi =22$