I'm trying to design a digital Chebyshev filter of order 2. This gives the general transfer function

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If I transform this and simplify I get enter image description here

If I then expand the denominator and then normalize so that a0 = 1 I get:

$$a_1 = \frac{2p_0p_1-8f_s^2}{4f_s^2 - 2f_s(p_0+p_1)+p_0p_1}$$


$$a_2 = \frac{2f_s(p_0+p_1)+p_0p_1+4f_s^2}{4f_s^2 - 2f_s(p_0+p_1)+p_0p_1}$$

Where $a_n$ are the coefficients of the denominator for the transfer function as $[1, a_1, a_2]$.

The issue here is that as the sampling rate grows don't the a coefficients become independent of the poles of the analog filter? This is what I find atleast, when using a sampling rate of 10^5. No matter the poles I get a = [1 -2 1].

Is there a way around this?

  • $\begingroup$ Please specify what poles you are trying to get that results in the two poles at z=1. Also are you trying to implement this in fixed point? $\endgroup$ Sep 30, 2023 at 17:26


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