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


$$\frac{SNR_{out}}{SNR_{in}} = \frac{E[y_s(n)^2]}{E[y_v(n)^2]} \cdot \frac{E[v(n)^2]}{E[s(n)^2]} = \frac{1}{NRR} \cdot \frac{E[y_s(n)^2]}{E[s(n)^2]}$$ If $x_S(n)$ isn't changed by Filter $H(\omega)$ then: $$ \frac{SNR_{out}}{SNR_{in}} = \frac{1}{NRR} $$ enter image description here

  • $\begingroup$ Welcome to DSP.SE! So basically what you want to calculate is $\sum_{k=-\infty}^{\infty} h[k]$ for the system $H(z)$ you wrote, right? $\endgroup$
    – Tendero
    Aug 28, 2018 at 13:37
  • $\begingroup$ Hi! Thank you. Yes, but thats not the only way. This Formula is actually for FIR Filters. If you know any faster way for IIR Filters this would be great :) $\endgroup$
    – Mr.Sh4nnon
    Aug 28, 2018 at 13:40
  • $\begingroup$ Hi new contributor! Do you mind if I add a comment? I would like to kindly ask, how did you arrive or derive the formula for NRR (noise reduction ratio?) as the sum of impulse response samples (or the DC frequency response) of an LTI system? $\endgroup$
    – Fat32
    Aug 28, 2018 at 14:15
  • $\begingroup$ @Fat32 Hi! Thank you for the nice welcome. Just realised I made a huge mistake. The formula is derived as follow: $$ NRR = \frac{\sigma^2_{Yv}}{\sigma^2_{Xv}} = \frac{1}{2\pi}\int_{-\pi}^{\pi}|H(\omega)|^2d\omega = \sum_n h(n)^2$$ $\endgroup$
    – Mr.Sh4nnon
    Aug 28, 2018 at 18:27
  • $\begingroup$ Oh your welcome dear mr. Shannon! So you assume a white noise at the input and hence simplify the power ratios. Please indicate this to @Tendero as well so that he could modify his answer. We are all looking forward to getting new questions from our new contributors! ;-) $\endgroup$
    – Fat32
    Aug 28, 2018 at 18:34

2 Answers 2


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:


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


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:

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

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


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.

  • $\begingroup$ Thank you for that fast reply. Actually I made a huge mistake.Check my latest edit? Since you gave an error despite this answer, I assume you don't know a rule or a trick to calculate an IIR's NRR, and you're just got with sum limits? $\endgroup$
    – Mr.Sh4nnon
    Aug 28, 2018 at 18:30
  • 1
    $\begingroup$ @Mr.Sh4nnon I don't know of any trick, but I've updated the answer to match your new equation. $\endgroup$
    – Tendero
    Aug 28, 2018 at 18:36
  • $\begingroup$ @Mr.Sh4nnon there's no fixed thesarus definition of NRR. One should define it by clarifying what he consideres as input noise, output noise and the mechanism of input output transfer. $\endgroup$
    – Fat32
    Aug 28, 2018 at 18:38
  • 1
    $\begingroup$ So this system does not achieve any considerable noise reduction. Which can also be understood by loking at its Frequency response which is a very very mild low pass filter indicated by its faded pole at a radius of $r=0.1$ and an angle of $0$ radian (DC). Instead had you chosen a radius of $r=0.9$ you would get a considerable noise reduction as it will be a narrow band filter. Yet these systems are emplyed in hearing aid applications where various other considerations are made for chosing the appropriate filters. $\endgroup$
    – Fat32
    Aug 28, 2018 at 18:51

So today I verified Tendero's solution. The integral

$$NRR = \frac{\sigma^2_{Yv}}{\sigma^2_{Xv}} = \frac{1}{2\pi}\int_{-\pi}^{\pi}|H(\omega)|^2d\omega = \sum_n h(n)^2$$

gives for the same problem the exact same result as the Infinite Geometric Sum.

In my textbook the author uses the same technique from time to time. Since this is very impractical he showed a shortcut for some given filters. E.g. for a IIR-Smoother first order $$ H(z) = \frac{b}{1-az^-1}$$ with $$b = 1-a$$ one can make use of this shortcut $$ NRR = \frac{1-a}{1+a}$$ which can be derived with the Infinite Geometric Sum.

One can now calculate some NRRs for the most used IIR-Filters and use them by manipulating coefficients like a and b.

E.g. for my original problem

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

one can alter the numerator to receive a form like $H(z)$ above . $z^{-5}$ is just a delay an does not contribute to the NRR. Since $a=0.1$, $b$ should be $1-0.1 = 0.9$. By excluding the term $\frac{1}{0.9}$ $H(z)$ becomes $$ \frac{1}{0.9} \cdot \frac{1}{1-0.1z^-1}$$ which leads to $$ NRR_2 = \frac{1-0.1}{1+0.1}$$ for the second part. $\frac{1}{0.9}$ is just a constant which gets multiplied. But it has to be squared first, since it's about power. This leads to

$$NRR = \frac{1}{0.9^2} \cdot \frac{0.9}{1.1}=\frac{100}{99}$$

As @Fat32 mentioned before, this filter doesn't reduce much noise. The opposite is the case. NRR 1 means by definition output noise power = input noise power. Therefore small numbers are better. The $\frac{100}{99}$ indicates that this filter even amplifies noise, since the gain is bigger than the 0.9 which would be needed for a good IIR-Smoother.


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