Question

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


then ${\int }_{-\infty }^{\infty }|X\left(j\omega \right){|}^{2}d\omega$ is

• A. $\frac{38\pi }{3}$
• B. $\frac{64\pi }{5}$
• C. $\frac{38\pi }{5}$
• D. $\frac{64\pi }{3}$

Answer 1

The correct option is D $\frac{64\pi }{3}$

$E_{x\left(t\right)}={\int }_{-\infty }^{\infty }|x\left(t\right){|}^{2}dt$


$=\left[{\int }_0^1\right(2t{)}^{2}dt+{\int }_1^3(2{)}^{2}dt+{\int }_3^4(8-2t{)}^{2}dt]$

$=\frac{4}{3}+8+\frac{4}{3}=\frac{32}{3}$

So, ${\int }_{-\infty }^{\infty }|X\left(j\omega \right){|}^{2}d\omega =2\pi E_{x\left(t\right)}=\frac{2\pi \times 32}{3}=\frac{64\pi }{3}$