[Sz, Sp, Sg] = sftrans (Sz, Sp, Sg, W, stop) ¶Transform band edges of a generic lowpass filter (cutoff at W=1) represented in splane zero-pole-gain form. W is the edge of the target filter (or edges if band pass or band stop). Stop is true for high pass and band stop filters or false for low pass and band pass filters. Filter edges are specified in radians, from 0 to pi (the nyquist frequency).
Theory: Given a low pass filter represented by poles and zeros in the splane, you can convert it to a low pass, high pass, band pass or band stop by transforming each of the poles and zeros individually. The following table summarizes the transformation:
Transform Zero at x Pole at x
---------------- ------------------------- ------------------------
Low Pass zero: Fc x/C pole: Fc x/C
S -> C S/Fc gain: C/Fc gain: Fc/C
---------------- ------------------------- ------------------------
High Pass zero: Fc C/x pole: Fc C/x
S -> C Fc/S pole: 0 zero: 0
gain: -x gain: -1/x
---------------- ------------------------- ------------------------
Band Pass zero: b +- sqrt(b^2-FhFl) pole: b +- sqrt(b^2-FhFl)
S^2+FhFl pole: 0 zero: 0
S -> C -------- gain: C/(Fh-Fl) gain: (Fh-Fl)/C
S(Fh-Fl) b=x/C (Fh-Fl)/2 b=x/C (Fh-Fl)/2
---------------- ------------------------- ------------------------
Band Stop zero: b +- sqrt(b^2-FhFl) pole: b +- sqrt(b^2-FhFl)
S(Fh-Fl) pole: +-sqrt(-FhFl) zero: +-sqrt(-FhFl)
S -> C -------- gain: -x gain: -1/x
S^2+FhFl b=C/x (Fh-Fl)/2 b=C/x (Fh-Fl)/2
---------------- ------------------------- ------------------------
Bilinear zero: (2+xT)/(2-xT) pole: (2+xT)/(2-xT)
2 z-1 pole: -1 zero: -1
S -> - --- gain: (2-xT)/T gain: (2-xT)/T
T z+1
---------------- ------------------------- ------------------------
where C is the cutoff frequency of the initial lowpass filter, Fc is the edge of the target low/high pass filter and [Fl,Fh] are the edges of the target band pass/stop filter. With abundant 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(S)=g prod(S-Xi)/prod(S-Xj)
The transforms are from the references. The actual pole-zero-gain changes I derived myself.
Please note that a pole and a zero at the same place exactly cancel. This is significant for High Pass, Band Pass and Band Stop filters which create 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 and the filter is high pass or band pass. 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