Question

$\text{A signal}x\left(t\right)=e^{-t}u\left(t\right)\text{is passed through an ideal low pass filter with cut-off frequency of}{\omega }_c\text{rad/sec, then the value of}{\omega }_c\text{for the output energy of low pass filter is 90\% of the}\text{input energy will be \_\_\_\_\_\_\_\_rad/sec}$

• A. 6.314

Answer 1

The correct option is A 6.314
$\text{Given input signal}x\left(t\right)=e^{-t}u\left(t\right)\text{Energy of the input signal}{E}_i={\int }_{-\infty }^{\infty }|x\left(t\right){|}^{2}dt={\int }_{-\infty }^{\infty }|e^{-t}u\left(t\right){|}^{2}dt={\int }_0^{\infty }{e}^{-2t}dt={\left[\frac{e^{-2t}}{-2}\right]}_0^{\infty }=\frac{1}{2}=0.5J\text{Now, ESD [Energy spectral density] of the output y(t) is given by}{E}_y\left(\omega \right)=|H\left(\omega \right){|}^2{E}_x\left(\omega \right)\therefore \text{Input ESD,}{E}_x\left(\omega \right)=|X\left(\omega \right){|}^{2}X\left(\omega \right)=FT\left[x\right(t\left)\right]=FT\left[e^{-t}u\right(t\left)\right]=\frac{1}{1+j\omega }{E}_x\left(\omega \right)=|X\left(\omega \right){|}^2={\frac{1}{1+\omega }}^2\text{For a LPF,the square of the transfer function is given as}|H\left(\omega \right){|}^2=\{\begin{array}{cc}1,& |\omega <{\omega }_c|\\ 0,& \text{otherwise}\end{array}{E}_y\left(\omega \right)=|H\left(\omega \right){|}^2{E}_x\left(\omega \right)=\{\begin{array}{cc}\frac{1}{1+{\omega }^2}& |\omega |<{\omega }_c\\ 0& \text{otherwise}\end{array}\therefore \text{The total energy of the output signal is}{E}_0{E}_0=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{E}_y\left(\omega \right)d\omega [\therefore E_y(\omega \left)\text{is an even function of}\right(\omega \left)\right]=\frac{1}{\pi }{\int }_0^{\infty }{E}_y\left(\omega \right)d\omega =\frac{1}{\pi }{\int }_0^{{\omega }_c}\frac{1}{1+{\omega }^2}d\omega 0.90\left[\frac{1}{2}\right]=\frac{1}{\pi }ta{n}^{-1}\left({\omega }_c\right)ta{n}^{-1}{\omega }_c=0.90\times \frac{\pi }{2}{\omega }_c=tan[0.90\times \frac{\pi }{2}]{\omega }_c=6.314rad/sec$