Question

An ideal second order low pass filter shown below with cutoff frequency of 1 rad/sec is supplied with input Energy present in the response of the system is _____________

$x\left(t\right)=e^{-t}u\left(t\right)$

• A. 0.25

Answer 1

The correct option is A 0.25



${\left|\gamma \right(\omega \left)\right|}^2={\left|H\right(\omega \left)\right|}^2{\left|X\right(\omega \left)\right|}^2$

${\left|\gamma \right(\omega \left)\right|}^{2}for-1rad/sec\le \omega \le 1rad/sec$

${\left|\gamma \right(\omega \left)\right|}^2=\frac{1}{(1+{\omega }^2)}$

Output energy, $E_y={\int }_{-\infty }^{\infty }{\left|y\right(t\left)\right|}^{2}dt=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{\left|\gamma \right(\omega \left)\right|}^{2}d\omega$

$E_y=\frac{1}{\pi }{\int }_0^1\frac{1}{1+{\omega }^2}d\omega =\frac{1}{\pi }\left(ta{n}^{-1}\omega {\mid }_0^1\right)$

$=\frac{1}{\pi }\left[ta{n}^{-1}\right(1)-ta{n}^{-1}(0\left)\right]$

$=\frac{1}{\pi }\times \frac{\pi }{4}$

$E_y=\frac{1}{4}=0.25J$