Let's consider an IIR filter with transfer function: $H(z)$.

Given the sampling frequency $F_s$ how can I calculate gain at say $F$ ?

When I was dealing with analog systems when I wanted to calculate gain at say $F=0 Hz$ (DC gain) I would simply put $S=0$ and I would find out what DC gain is.

Now I'm not sure if the same technique applies here.

My inital thought was to put $z= e^{j2\pi}=1$ into the equation but it won't work.

The problem boils down to finding $|{H(z)}|$ which I don't know how to do.

How can I approach this problem?

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    $\begingroup$ You're on the right track... Set z = 1 In your case, the DC gain should be equal to 0.8 $\endgroup$ – Ben Jul 4 at 14:06
  • $\begingroup$ Yes, I see that's right for 0Hz,but how would it be for 4000Hz? $\endgroup$ – Healow Jul 4 at 15:11
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    $\begingroup$ set z = -1 as proposed in the answer below $\endgroup$ – Ben Jul 4 at 15:26
  • $\begingroup$ w = 2*pi*f/fs = 2*pi*8000/4000 = 4*pi; e^j*4*pi = 1? I'm confused between 1 and -1 $\endgroup$ – Healow Jul 4 at 15:28
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    $\begingroup$ Sorry I misundersootd, in that case z = 1 $\endgroup$ – Ben Jul 4 at 15:32

For continuous-time systems, you obtain the frequency response by evaluating the transfer function $H(s)$ on the imaginary axis $s=j\omega$ (assuming stability). In discrete-time, you get the $\mathcal{Z}$-transform instead of the Laplace transform, and the imaginary axis is replaced by the unit circle. So the frequency response is obtained by evaluating the transfer function $H(z)$ for $z=e^{j\omega}$. Note that for discrete-time systems, $\omega$ is a normalized frequency (in radians):

$$\omega=\frac{2\pi f}{f_s}\tag{1}$$

where $f_s$ is the sampling frequency.

Coming to you example, a frequency of $8000$ Hz with a sampling frequency of $4000$ Hz is the same as DC (Eq. $(1)$ should make that clear). Computing the frequency response at any frequency means to determine the desired $\omega$ from Eq. $(1)$, and then evaluate $H(z)$ with $z=e^{j\omega}$. In the case of DC ($\omega=0$) and Nyquist ($\omega=\pi$) this becomes especially easy: just evaluate $H(1)$ for DC, and $H(-1)$ for Nyquist.

For most other values of $\omega$, the frequency response is most likely complex-valued, so after computing $H(e^{j\omega})$ you just compute the magnitude of that complex number to obtain the gain at the given frequency. You don't need to find a general expression for $|H(z)|$.

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  • $\begingroup$ Ok, but when you say w=pi than it means e^j*pi which is -1? And you said it's like computing Z=1, but shouldn't it be -1? But when I compute with -1 i get gain of 0, is this correct? $\endgroup$ – Healow Jul 4 at 15:17
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    $\begingroup$ @Healow: No, I said Nyquist ($\omega=\pi$) corresponds to $z=-1$ (and DC corresponds to $z=1$). Plugging values for $z$ into $H(z)$ should be easy enough, so I'll leave that up to you. A gain of zero is of course possible. $\endgroup$ – Matt L. Jul 4 at 15:29
  • $\begingroup$ w = 2*pi*f/fs = 2*pi*8000/4000 = 4*pi; e^j*4*pi = 1? I'm confused between 1 and -1 $\endgroup$ – Healow Jul 4 at 15:30
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    $\begingroup$ @Healow: If you read my answer, you'll see that I actually mention the fact that 8000 Hz with a sampling frequency of 4000 Hz is the same as DC, so $z=1$ is correct. $\endgroup$ – Matt L. Jul 4 at 15:32
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    $\begingroup$ @Healow: No, you just evaluate $H(e^{j\omega})$ and then take the absolute value of the outcome. $\endgroup$ – Matt L. Jul 4 at 15:37

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