% error: [Zz, Zp, Zg] = bilinear(Sz, Sp, Sg, T) % [Zb, Za] = bilinear(Sb, Sa, T) % % Transform a s-plane filter specification into a z-plane % specification. Filters can be specified in either zero-pole-gain or % transfer function form. The input form does not have to match the % output form. 1/T is the sampling frequency represented in the z plane. % % Note: this differs from the bilinear function in the signal processing % toolbox, which uses 1/T rather than T. % % Theory: Given a piecewise flat filter design, you can transform it % from the s-plane to the z-plane while maintaining the band edges by % means of the bilinear transform. This maps the left hand side of the % s-plane into the interior of the unit circle. The mapping is highly % non-linear, so you must design your filter with band edges in the % s-plane positioned at 2/T tan(w*T/2) so that they will be positioned % at w after the bilinear transform is complete. % % The following table summarizes the transformation: % % +---------------+-----------------------+----------------------+ % | Transform | Zero at x | Pole at x | % | H(S) | H(S) = S-x | H(S)=1/(S-x) | % +---------------+-----------------------+----------------------+ % | 2 z-1 | zero: (2+xT)/(2-xT) | zero: -1 | % | S -> - --- | pole: -1 | pole: (2+xT)/(2-xT) | % | T z+1 | gain: (2-xT)/T | gain: (2-xT)/T | % +---------------+-----------------------+----------------------+ % % With tedious algebra, you can derive the above formulae yourself by % substituting the transform for S into H(S)=S-x for a zero at x or % H(S)=1/(S-x) for a pole at x, and converting the result into the % form: % % H(Z)=g prod(Z-Xi)/prod(Z-Xj) % % Please note that a pole and a zero at the same place exactly cancel. % This is significant since the bilinear transform creates numerous % extra poles and zeros, most of which cancel. Those which do not % cancel have a 'fill-in' effect, extending the shorter of the sets to % have the same number of as the longer of the sets of poles and zeros % (or at least split the difference in the case of the band pass % filter). There may be other opportunistic cancellations but I will % not check for them. % % Also note that any pole on the unit circle or beyond will result in % an unstable filter. Because of cancellation, this will only happen % if the number of poles is smaller than the number of zeros. The % analytic design methods all yield more poles than zeros, so this will % not be a problem. % % References: % % Proakis & Manolakis (1992). Digital Signal Processing. New York: % Macmillan Publishing Company.

- butter % Generate a butterworth filter.
- cheby1 % Generate an Chebyshev type I filter with Rp dB of pass band ripple.
- cheby2 % Generate an Chebyshev type II filter with Rs dB of stop band attenuation.
- ellip % N-ellip 0.2.1

Generated on Sat 16-May-2009 00:04:49 by