Variance Due to white noise input

Question

I have the problem below. It sounds simple but for some reason I have been stuck on it for a long time and don't know what am doing wrong.

Am trying to solve this using correlations.

So we all know that the variance $\sigma _y^2\:=\:R_y\left(0\right)$ which is basically the correlation of $y$ evaluated at zero. Now we know the convolution $y(n)\:=\:x(n)* h(n) = \sum _{m\:=-\infty }^{\infty }h\left(m\right)x\left(n-m\right)$

Using the idea of correlation, I get the following:

$R_y\left(k\right)\:=\:E\left(y\left(k\right)y\left(n+k\right)\right)\:=\:E\left[\sum _{m=-\infty }^{\infty }h\left(m\right)x\left(k-m\right)\sum _{m=-\infty }^{\infty }h\left(m\right)x\left(n+k-m\right)\right]$

Now am so stuck here and just don't know what to do. Please help me by directing me in the right direction and by providing clear and coherent steps and explanations. Thank you very much in advance.

Answer 1

You have a typo in your definition of $R_y(k)$, and an error in the time indices when developing the equation. The proper definition of autocorrelation for complex signals (in the case of wide-sense stationary processes) is $$R_y(k)=\mathbb{E}\left[y(n)\overline{y(n+k)}\right],$$

and setting $k=0$, we obtain $$\begin{align} R_y(0)&=\mathbb{E}\left[y(n)\overline{y(n+0)}\right]=\mathbb{E}\left[\left|y(n)\right|^2\right]\\ &=\mathbb{E}\left[\left|\sum_{m=-\infty}^{\infty}h(m)x(n-m)\right|^2\right]\\ &=\mathbb{E}\left[\left(\sum_{m=-\infty}^{\infty}h(m)x(n-m)\right)\overline{\left(\sum_{i=-\infty}^{\infty}h(i)x(n-i)\right)}\right]\\ &=\mathbb{E}\left[\sum_{m=-\infty}^{\infty}\sum_{i=-\infty}^{\infty}h(m)\overline{h(i)}x(n-m)\overline{x(n-i)}\right]\\ &=\sum_{m=-\infty}^{\infty}\sum_{i=-\infty}^{\infty}h(m)\overline{h(i)}\,\mathbb{E}\left[x(n-m)\overline{x(n-i)}\right]\\ &=\sum_{m=-\infty}^{\infty}\sum_{i=-\infty}^{\infty}h(m)\overline{h(i)}\,R_x(m-i)\\ \end{align}$$

where I first use the fact that the product of sums is the sum of products (and I use a different summation index to better show the cross-products), and second, that the impulse response of the system is deterministic and therefore constant for the purposes of taking expectation.

Now, we can group those terms where $m=i$ and those where $m\neq i$, obtaining $$\begin{align} R_y(0)&=\sum_{m=-\infty}^{\infty}\sum_{i=-\infty}^{\infty}h(m)\overline{h(i)}\,R_x(m-i)\\ &=\sum_{m=-\infty}^{\infty}h(m)\overline{h(m)}\,R_x(0)+\sum_{m=-\infty}^{\infty}\sum_{i=-\infty\\ i\neq m}^{\infty}h(m)\overline{h(i)}\,R_x(m-i)\\ &=R_x(0)\sum_{m=-\infty}^{\infty}|h(m)|^2+\sum_{m=-\infty}^{\infty}\sum_{i=-\infty\\ i\neq m}^{\infty}h(m)\overline{h(i)}\,R_x(m-i)\\ \end{align}$$

Finally, we know that the input noise is white, and therefore $R_x(k)=0$ for $k\neq0$. We also know that $R_x(0)=\sigma_x^2$. Thus, the second term vanishes, and we obtain $$\begin{align} R_y(0) &=R_x(0)\sum_{m=-\infty}^{\infty}|h(m)|^2=\sigma_x^2\sum_{m=-\infty}^{\infty}|h(m)|^2 \end{align}.$$