Question

A real and stationary process X(t) is passed through the system shown below.

$\text{If the power spectral density (PSD) of the process X(t) is}{X}_x\left(f\right)\text{, then the PSD of the output process Y(t) will be}$

• A. $4S_X\left(f\right)[1+\cos (4\pi f\left)\right]$​​​
• B. $2S_X\left(f\right)\cos \left(4\pi f\right)$
  
• C. $2S_X\left(f\right){\cos }^2\left(2\pi f\right)$
• D. $4S_X\left(f\right){\cos }^2\left(2\pi f\right)$

Answer 1

The correct option is D $4S_X\left(f\right){\cos }^2\left(2\pi f\right)$

Ans : (c)

$y\left(t\right)=x\left(t\right)+x(t-2)$
$H\left(j\omega \right)=\frac{Y\left(j\omega \right)}{X\left(j\omega \right)}=1+e^{-j2\omega }$
$|H\left(f\right){|}^2=|1+e^{-j4\pi f}{|}^2=(1+e^{-j4\pi f})(1+e^{j4\pi f})$
$=1+1+e^{-j4\pi f}+e^{j4\pi f}=2+2\cos \left(4\pi f\right)$
$=2[1+\cos (4\pi f\left)\right]=4{\cos }^2\left(2\pi f\right)$
$\text{So,}{S}_Y\left(f\right)=S_X\left(f\right)|H\left(f\right){|}^2=4S_X\left(f\right){\cos }^2\left(2\pi f\right)$