I am studying for my exam in signal processing. For one of the old exam sets, a discrete impulse response of a filter is given as $h[n]$ \begin{bmatrix} -3&0&-3&0 \end{bmatrix}

With a frequency response $H[k]$. What is the easiest way of calculating the DC gain from this information? I have tried using a DFT but did not see any usable conclusion.

  • $\begingroup$ DFT works just fine. The DC value of the DFT is -6 as well. What didn't work for you? $\endgroup$
    – Hilmar
    Dec 27, 2020 at 13:08
  • $\begingroup$ I just don’t think I made the correct connection between the things in my mind. $\endgroup$
    – Mag
    Dec 27, 2020 at 14:16

2 Answers 2


The DC gain is simply the sum of filter taps or coefficients. This is the value of the frequency response at DC (i.e. $0\ \rm Hz$), or equivalently $$ H(0) = \sum_{n = 0}^{N - 1}h[n]\tag{1} $$ Because, for a digital FIR filter of length $N$ with impulse response given by Equation $(2)$ $$ \big\{h[n]\big\}, \quad\text{with}\quad 0\le n\le N -1\tag{2} $$ the frequency response is the $z$-transform at the unit circle (i.e. $z = e^{j2\pi f}$), or in this case, equivalently $$ H(e^{j\omega}) \equiv H(\omega) = \mathcal F \left\{h[n]\right\} \triangleq \sum_{n = 0}^{N - 1}h[n]e^{-j\omega n } = \sum_{n = 0}^{N - 1}h[n]e^{-j2\pi n f}\tag{3} $$ Then you can think at DC (i.e. at $0\ \rm Hz$ or $f = 0$ or $\omega = 0$) to see the why of Equation $(1)$.

From DTFT to DFT

The DFT is the one practically computed in place of the DTFT. For finite-length signals (as is the case here), it provides the frequency-domain samples of the DTFT. Then $H[k]$, which is the DFT of the $h[n]$, is computed for $N$-point as shown in Equation $(4)$ $$ H[k] = H(e^{j\omega}) \bigg\vert_{\omega = 2\pi k/N}\quad\text{with}\quad 0\leq k \leq N - 1\tag{4} $$ i.e. $H[k]$ consists of equally-spaced (by $2\pi/N$) samples of $H(e^{j\omega})$.

Noting that $$\omega \equiv 2\pi f \quad\text{with}\quad -\pi \leq \omega \leq \pi\quad\text{and}\quad -\frac 12 \leq f \leq \frac 12$$ With $\omega$ in [radians/sample] and $f$ in [cycles/sample].


This can be done using MATLAB's freqz function as follows:

>> [h, w] = freqz([-3 0 -3 0], 1, [0 , pi/4])

h =

 -6.000000000000000 + 0.000000000000000i -3.000000000000000 + 3.000000000000000i

w =

                   0   0.785398163397448


Giving you the flexibility to specify at whatever angular frequencies $\omega$ (in example above the evaluation is at two frequencies $0$ and $\pi/4$ rad/samples) you would want to evaluate the frequency response, (Here we're interested in the value at $\omega = 0$). The magnitude response with these two points is shown below

enter image description here

Note that the two points in magenta correspond exactly to

>> 20*log10(abs(h))

ans =

  15.563025007672874  12.552725051033063


The values at $\omega = 0$ and $\omega = \pi/4$ respectively.

  • 1
    $\begingroup$ Thank you, I understand! $\endgroup$
    – Mag
    Dec 27, 2020 at 10:32
  • $\begingroup$ @user54828, you're welcome :). You can also upvote the answer (up arrow) if you found it useful. $\endgroup$
    – Gilles
    Dec 27, 2020 at 10:37
  • $\begingroup$ my understanding though was that the frequency response would be defined as the DFT and thereby $H[k]=\sum_{n=0}^{N-1}h[n]e^{-j(2\pi /N)kn}$? Would you mind explaining why $e^{-j2\pi nF}$ works? $\endgroup$
    – Mag
    Dec 27, 2020 at 15:59
  • 1
    $\begingroup$ @Mag , please see my edits and the new section From DTFT to DFT. $\endgroup$
    – Gilles
    Dec 27, 2020 at 23:08
  • $\begingroup$ @Mag. Your H[k] equation is the most common (typical) version of the DFT equation. Gilles' DFT expression is non-standard in that his 'F' variable is a "relative frequency", where F = k/N, that is always less than one. $\endgroup$ Dec 28, 2020 at 11:24

Giles' answer applies to tapped delay-line digital filters. A more general answer is: The DC gain of a digital system is the sum of the system's impulse response.


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