My professor mentioned that the order of a band-pass and a notch filter must always be even, when showing an example of designing a digital filter using the bilinear transformation.
Then he also mentions that the MATLAB code ellipord returns half the actual order, so I figured the two statements must be connected somehow. I can't ask him since it's currently 3 AM and I couldn't find anything alone as to why these two statements are true.
Any thoughts? Thank you!
I believe your thinking is correct. For bandpass filters, for each z-plane pole in the positive-frequency range there's a conjugate pole in the z-plane's negative-frequency range. So for bandpass filters there will all be an even number of total z-plane poles (two poles, four poles, six poles, etc.). When using MATLAB's ellipord command for bandpass filters that command returns the number of poles in the z-plane's positive-frequency range which is half the actual number of z-plane poles.
Practical bandpass filters will be even-ordered. Below Matt L points out that modifying an even-ordered filter’s transfer function by placing a z-plane zero at DC can make a bandpass filter odd-ordered, which is true. But placing a z-plane zero at DC so badly distorts the original even-ordered filter’s frequency response that the new odd-ordered filter becomes unusable.
In principle there is no reason why the filter order of a general bandpass or bandstop filter must be even. Such a restriction is a consequence of a specific design procedure. In classic IIR filter design (Butterworth, Chebyshev, Cauer) you start with an analog prototype lowpass filter. Bandpass or bandstop filters are then obtained by a frequency transformation. And it is this frequency transformation that doubles the order of the prototype lowpass filter, hence the even filter order for that specific design method. Note that the bilinear transform has nothing to do with that restriction on the filter order.
Of course, there are specific designs for which an odd filter order doesn't make much sense. E.g., for a notch filter (with a notch frequency greater than zero) you want exactly one zero at the notch frequency plus its mirror image at the negative notch frequency. So for each notch frequency you get two zeros, and, consequently, two poles.
Maybe that's a just a matter of semantics. You can certainly cascade an even order high pass with an odd order lowpass and you get something that's an odd order filter that sure looks like a bandpass.
%% odd order bandpass
fs = 44100;
fc = 1000;
[z,p,k] = butter(2,fc/sqrt(2)/fs*2,'high');
sos = zp2sos(z,p,k);
[z,p,k] = butter(3,fc*sqrt(2)/fs*2);
sos = [sos; zp2sos(z,p,k)];
nx = 8192;
f = logspace(log10(20),log10(20000),1000);
h = freqz(sos,f,fs);
semilogx(f,20*log10(abs(h)));
xlabel('Frequency in Hz'); ylabel('Level in dB'); title('odd order bandpass');
grid('on');
set(gca,'xlim',[f(1) f(end)]); set(gca,'ylim',[-60 3]);
