Function File: [Zb, Za] = bilinear (Sb, Sa, T) ¶

Function File: [Zb, Za] = bilinear (Sz, Sp, Sg, T) ¶

Function File: [Zz, Zp, Zg] = bilinear (…) ¶

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.

Package: signal