IIR - How the gain is calculated?

Question

I try to extend my understanding of digital filter design, so I play around with some YouTube videos and the MicroModeler DSP with the following IIR design (based on a video):

I use the following parameters:

Zeros: 0.57 ± 0.78i
Poles: 0.51 ± 0.7i

How does the calculator generate the additional gain with value 0.90134? As I understand it so far the gain value is generated by the equation of the zeros, but I can´t calculate the value for it.

You show us an IIR filter, not an FIR filter – Ben Apr 02 '20 at 12:08 — Ben, Apr 02 '20 at 12:08
Yes, you are right. I fixed it. – Kampi Apr 02 '20 at 12:44 — Kampi, Apr 02 '20 at 12:44

score 3 · Accepted Answer · answered Apr 02 '20 at 11:40

Apparently it's normalized so that the gain at Nyqusit is 0 dB. Potentially they just normalize to the maximum being 0 dB. Gain at Nyquist is given by

$$H(\omega = \pi) = \frac{b_0-b_1+b_2}{a_0-a_1+a_2}$$

whereas gain at DC is

$$H(\omega = 0) = \frac{b_0+b_1+b_2}{a_0+a_1+a_2}$$

Both values are real.

IIR - How the gain is calculated?

1 Answers1