Question

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

Then the value of ${\int }_{-\infty }^{\infty }X\left(j\omega \right)cos\omega \cdot e^{j2\omega }d\omega$is ________.

• A. -18.845

Answer 1

The correct option is A -18.845
We know that,

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

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

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

$=\frac{1}{2}{\int }_{-\infty }^{\infty }\left[X\right(j\omega \left)e^{j3\omega }\right]d\omega +\frac{1}{2}{\int }_{-\infty }^{\infty }\left[X\right(j\omega \left)e^{j\omega }\right]d\omega$

$=\frac{1}{2}\left[2\pi x\right(3\left)\right]+\frac{1}{2}\left[2\pi x\right(1\left)\right]$

$=\frac{1}{2}\left[2\pi \right(-2\left)\right]+\frac{1}{2}\left[2\pi \right(-4\left)\right]$

$=-2\pi -4\pi =-6\pi$