I'm trying to design digital Butterworth filter from scratch. First I use butterap to design an analog prototype to get zeros, poles and gain

[za, pa, ka] = buttap(n);

Here I set n = 5 for example and then convert it to transfer function to check its frequency response

[b, a] = zp2tf(za, pa, ka);
freqs(b, a);

enter image description here

From the figure above I know it's a lowpass filter with cutoff frequency $\Omega_c=1$.

Now I want to convert this analog filter to a half-band lowpass digital filter using bilinear transform, according to the documentation of MATLAB's bilinear.

% Strip any zeros at infinity
za = za(isfinite(za));
% Do bilinear transformation
pd = (2 * fs + pa) ./ (2 * fs - pa);
zd = (2 * fs + za) ./ (2 * fs - za);
kd = real(ka * prod(2 * fs - za) ./ prod(2 * fs - pa));
% Add extra zeros at -1 so the resulting system has equivalent numerator and denominator order.
zd = [zd; -ones(length(pd) - length(zd), 1)];

Then convert it to transfer function and look at the frequency response

[b, a] = zp2tf(zd, pd, kd);
figure; freqz(b, a);

My question is about the sampling frequency $f_s$. I want a half-band lowpass filter meaning that $\omega_c = 0.5\pi$. From the relationship between digital frequency and analog frequency $\omega = \Omega T$ I can derive that $T = 0.5\pi\ \text{s}$ and $f_s = 1/T = 2/\pi\ \text{Hz}$. The following figure shows the frequency response when I set fs = 2/pi

enter image description here

Apparently it doesn't have a -3 dB attenuation at $\omega = 0.5\pi$. Only when I set fs = 0.5 can I get the desired result:

enter image description here

Anyone can help me to find where am I wrong? Thank you in advance.


1 Answer 1


Bilinear transform has a nonlinear frequency mapping $$ \varOmega = \frac{2}{T}\tan{\frac{\omega}{2}} $$ Substituting $\varOmega = 1$ and $\omega = 0.5\pi$ into it and you can derive that $$ T = 2 $$ and thus $$ f_s = 0.5 $$

  • $\begingroup$ Thank you very much! $\endgroup$
    – DSP novice
    Jul 28 at 7:58

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