Question

$\text{A zero mean white Gaussian process X(t), with a two-sided power spectral density of}20\mu \text{W/Hz is passed through the system shown in the figure below.}$

$\text{The bandpass filter (BPF) has ideal characteristics with a passband bandwidth of 2B centered around}{f}_0\text{and the low-pass filter (LPF) has ideal characteristics with a passband bandwidth of B. If}{f}_0=1\text{MHz and B=5 kHz. then the average power present in the output process Y(t) will be \_\_\_\_\_W.}$

• A. 0.1

Answer 1

The correct option is A 0.1
The filtered noise signal n(t) can be treated as narrowband noise which can be represented as,

$n\left(t\right)=n_c\left(t\right)\cos \left(2\pi f_{0}t\right)-n_s\left(t\right)\sin \left(2\pi f_{0}t\right)$

$z\left(t\right)=n\left(t\right)\cos \left(2\pi f_{0}t\right)$

$=n_c\left(t\right){\cos }^2\left(2\pi f_{0}t\right)-n_s\left(t\right)\cos \left(2\pi f_{0}t\right)\sin \left(2\pi f_{0}t\right)$

$=\frac{1}{2}{n}_c\left(t\right)+\frac{1}{2}{n}_c\left(t\right)\cos \left(4\pi f_{0}t\right)-\frac{1}{2}{n}_s\left(t\right)\sin \left(4\pi f_{0}t\right)$

After passing through LPF, we get.
$Y\left(t\right)=\frac{1}{2}{n}_c\left(t\right)$

$\text{Average power of}{n}_c\left(t\right)=\text{average power of}n\left(t\right)$

$P_{nc}=2[2B\times \frac{N_0}{2}]=2[10\times {10}^3\times 20\times {10}^{-6}]=0.4W$

$\text{Average power of}Y\left(t\right)=\frac{1}{4}{P}_{n_c}=\frac{1}{4}(0.4=0.1W$