Topic: Electronics Engineering
A uniformly distributed random signal $x[n]$ with mean $m_{x}=2$ and variance $\sigma_{x}^{2}=3,$ is passed through a 3-point moving-average filter having an impulse response $\{h[n]\}=\{\frac{1}{3},\frac{1}{3},\frac{1}{3}\}.$ What will be the mean and variance of output?
- A. $m_{y}=2,$ $\sigma_{y}^{2}=1$
- B. $m_{y}=2,$ $\sigma_{y}^{2}=3$
- C. $m_{y}=1,$ $\sigma_{y}^{2}=1$
- D. $m_{y}=3,$ $\sigma_{y}^{2}=3$
Answer: A. $m_{y}=2,$ $\sigma_{y}^{2}=1$
Explanation: $m_{y}=m_{x}\sum_{n=-\infty}^{\infty}h[n]=2\sum_{n=-\infty}^{\infty}h[n] = 2[\frac{1}{3}+\frac{1}{3}+\frac{1}{3}] = 2$. Variance $\sigma_{y}^{2} = \sigma_{x}^{2}\sum_{n=-\infty}^{\infty}|h[n]|^{2} = 3[\frac{1}{9}+\frac{1}{9}+\frac{1}{9}] = 3[\frac{3}{9}] = 1$.