What are the formulas for signal and noise power gain of digital filters (FIR and IIR)?

For a FIR, I've seen in Harris' windowing paper that the DC gain is the sum of the filter weights.

$$ G = \sum_{i=0}^{N-1} w_i $$

For a FIR, I've seen that the noise gain is the square root of the sum of the square of the weights. $$ G = (\sum_{k=0}^{N-1} w_i^2 )^\frac{1}{2} $$

What about an IIR?

  • 2
    $\begingroup$ Welcome to DSP.SE. Participants are more likely to help you out if you show some effort and tell us what you've found out on your own so far. $\endgroup$
    – Phonon
    Commented Nov 1, 2013 at 1:09
  • $\begingroup$ @Phonon - I've clarified this question. $\endgroup$ Commented Nov 4, 2013 at 17:04

1 Answer 1


Is there a reason you don't think that the general formulae are:

$$ G = \sum_{k=0}^{\infty} h[k] $$


$$ G = (\sum_{k=0}^{\infty} h^2[k] )^\frac{1}{2} $$

These will work, provided the system is linear and time-invariant (and BIBO stable).

To work with a rational transfer function: $$ H(z) = \frac{\displaystyle\sum_{m=0}^{M} b_m z^{-m}}{\displaystyle\sum_{n=0}^{N} a_n z^{-n}} $$ then the DC gain (at $z = e^{j0}$) will be: $$ G_{DC} = \left. H(z) \right|_{z=1 = e^{j0}} = \frac{\displaystyle\sum_{m=0}^{M} b_m}{\displaystyle\sum_{n=0}^{N} a_n } $$

I'll have to think a bit more about the second gain in your question.

  • $\begingroup$ These are only for FIR, right? What about IIR? $\endgroup$ Commented Nov 6, 2013 at 18:11
  • $\begingroup$ @Seth: FIR means finite (duration) impulse response. IIR means infinite (duration) impulse response. The equations in my answer do not assume that the impulse response is finite in duration. Or do you mean that there is some other form of IIR (e.g. a rational function of $z$) that you are thinking should be used? $\endgroup$
    – Peter K.
    Commented Nov 6, 2013 at 20:34
  • $\begingroup$ Since an IIR has poles, these equations do not work. Summing the taps for the DC gain of a FIR is fine, but not for a IIR. (to my knowledge) $\endgroup$ Commented Nov 6, 2013 at 20:51
  • $\begingroup$ @Seth Of course they work, provided the system is linear and time invariant and the system is BIBO stable. If the IIR filter has poles outside (or on) the unit circle, it will not be stable. Is your system stable? If it is, these equations are correct. Again: Are you thinking of using a rational function of $z$ and those coefficients, rather than using the impulse response directly? You seem to have a different understanding of what IIR is from the usual definition. $\endgroup$
    – Peter K.
    Commented Nov 6, 2013 at 20:54
  • $\begingroup$ Yes I have the poles and zeros; not the impulse response. (I used the wrong notation) $\endgroup$ Commented Nov 6, 2013 at 21:05

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