Question

$X\left(\omega \right)$ is the Fourier transform of $x\left(t\right)$ shown below. The value of $\underset{-\infty }{\overset{\infty }{\int }}|X\left(\omega \right){|}^{2}d\omega$ (rounded off to two decimal places) is

Answer 1

$\underset{-\infty }{\overset{\infty }{\int }}|X\left(\omega \right){|}^{2}d\omega =2\pi \underset{-\infty }{\overset{\infty }{\int }}|x\left(t\right){|}^{2}dt=2\pi \underset{-\infty }{\overset{\infty }{\int }}|y\left(t\right){|}^{2}dt$
$=2\times 2\pi \underset{-2}{\overset{0}{\int }}|y\left(t\right){|}^{2}dt$
$=2\times 2\pi \left[\underset{-2}{\overset{-1}{\int }}(t+2{)}^{2}d+\underset{-1}{\overset{0}{\int }}(2t+3{)}^{2}dt\right]$
$=4\pi [{\left\{\frac{(t+2{)}^3}{3}\right\}}_{-2}^{-1}+{\left\{\frac{(2t+3{)}^3}{3\times 2}\right\}}_{-1}^0]$
$=4\pi [\frac{1-0}{3}+\frac{3^3-1}{6}]=4\pi [\frac{1}{3}+\frac{26}{6}]$
$=4\pi \times [\frac{1}{3}+\frac{26}{6}]=4\pi \times [\frac{1}{3}+\frac{13}{3}]$
$=4\pi \times \frac{14}{3}=\frac{56\pi }{3}$