Question

$X\left(j\omega \right)$ denote the Fourier transform of the signal x(t) shown in below figure :

Then the value of $2{\int }_{-\infty }^{\infty }X\left(j\omega \right)cos\omega e^{j2\omega }d\omega$ is ____. (Answer upto two decimal places)

• A. 18.85

Answer 1

The correct option is A 18.85
We know that,

$x\left(t\right)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X\left(j\omega \right)e^{j\omega t}d\omega$

$2\pi x\left(t\right)={\int }_{-\infty }^{\infty }X\left(j\omega \right)e^{j\omega t}d\omega$ ....(i)

$\therefore 2{\int }_{-\infty }^{\infty }X\left(j\omega \right)cos\omega e^{j2\omega }d\omega ={\int }_{-\infty }^{\infty }X\left(j\omega \right)(e^{j\omega }+e^{-j\omega })e^{j2\omega }d\omega$

$={\int }_{-\infty }^{\infty }X\left(j\omega \right)e^{j3\omega }d\omega +{\int }_{-\infty }^{\infty }X\left(j\omega \right)e^{j\omega }d\omega$

$=2\pi x\left(3\right)+2\pi x\left(1\right)$ ...from equation(i),

$=2\pi \left(\frac{7}{4}\right)+2\pi \left(\frac{5}{4}\right)$

$=2\pi (\frac{7}{4}+\frac{5}{4})$

$=2\pi \left[\frac{12}{4}\right]=2\pi \left(3\right)=6\pi =18.85$