Calculating a NRR of an IIR I'd like to calculate the Noise Reduction Ratio of a system.

The system I have is $$ H(z)= \frac{z^{-5}}{1-0.1z^-1}.$$
For FIR Filters this can easily done by $$ \sum h[n].$$

For an IIR Filter this gets more complicated. I thought about making it a recursive formula for h(n) and then calculate some sort of sum which might converge (without any result yet).

Another way would be integrating $ |H(w)|$ by replacing z with $e^{jw}$. But there has to be an easier way. Some sort of trick. 

I can imagine two things:


The system can be splited into two or more known systems. The NRR can than somehow be added togehter. (How would one do that?)
$z^{-5}$ is just a delay which shouldn't contribute to the NRR. The remaining part has maybe a well known NRR. 


EDIT: UPDATE. I made a mistake

NRR for an FIR Filter is given by: $$NRR = \frac{\sigma^2_{Yv}}{\sigma^2_{Xv}} = \frac{1}{2\pi}\int_{-\pi}^{\pi}|H(\omega)|^2d\omega = \sum_n h(n)^2$$

EDIT 2: What's an NRR

Since I left my textbook in the office I can only show you some definitions from powerpoint slides:


  The noise reduction ratio NRR measures, how much the noise power at the output is less than the noise power at the input
  –
  The last equal sign follows from Parseval’stheorem
  –
  Note that small NRR means large noise reduction
  –
  Small NRR are good NRR


and


  $$\frac{SNR_{out}}{SNR_{in}} = \frac{E[y_s(n)^2]}{E[y_v(n)^2]} \cdot \frac{E[v(n)^2]}{E[s(n)^2]} = \frac{1}{NRR} \cdot \frac{E[y_s(n)^2]}{E[s(n)^2]}$$
  If $x_S(n)$ isn't changed by Filter $H(\omega)$ then:
  $$ \frac{SNR_{out}}{SNR_{in}} = \frac{1}{NRR} $$
Let's find the impulse response for this system:

$$H(z)= \frac{z^{-5}}{1-0.1z^-1} = z^{-5}\frac{1}{1-0.1z^-1}$$

You didn't specify the ROC, so I'll assume that we take it so that the filter is stable. Namely:

$$|z|&gt;\frac{1}{10}$$

I've split the expression of $H(z)$ to notice that the impulse response basically consists of an exponential with a delay. We know that

$$\frac{1}{1-0.1z^-1}\xrightarrow{\mathcal{Z}^{-1}}(0.1)^nu[n]$$

where $u[n]$ is the unitary step. The $z^{-5}$ factor translates to a delay in the time domain. So the impulse response is:

$$h[n]=(0.1)^{n-5} u[n-5]$$

If we want to find the sum of all the terms squared in the sequence:

$$\begin{align}
\sum_{n=-\infty}^{\infty}h^2[n]
&amp;=\sum_{n=-\infty}^{\infty}\left[(0.1)^{n-5} u[n-5]\right]^2\\
&amp;=\sum_{n=5}^{\infty}\left[(0.1)^{n-5}\right]^2\\
&amp;=\sum_{m=0}^{\infty}(0.1)^{2m}\\
&amp;=\sum_{m=0}^{\infty}(0.01)^{m}\\
&amp;=\frac{1}{1-0.01}\\
&amp;=\frac{100}{99}\\
\end{align}$$

Where I've used the formula for the infinite geometric sum.

EDIT:

Note that a smaller NRR implies a worse performance of the filter at reducing noise. In this case, NRR turned out to be almost $1$ (or $0 \ \mathrm{dB}$), so the variance of the noise at the output is practically the same as in the input. As Fat32 pointed out in the comments, this depends on the position of the pole of your filter. The noise is white (as far as I understood from your comments), so it has infinite bandwidth. If the filter's bandwidth is too large, then "a lot" of noise will pass through it, appearing at the output. On the contrary, if it is narrow, then it will "filter out" most of it, leading to a decrease in the variance at the output. In this case, the filter's bandwidth is rather wide due to the pole being near the origin, and then the NRR approaches $1$: not a really good filter to reduce noise.