Examples
[Sz, Sp, Sg] = sftrans (5, 10, 15, 20, 30) Sz = 4 Sp = 2 Sg = 7.5000 |  |  |
Transform band edges of a generic lowpass filter (cutoff at W=1) represented in splane zero-pole-gain form.
[Sz, Sp, Sg] = sftrans (Sz, Sp, Sg, W, stop) [Sz, Sp] = sftrans (Sz, Sp, Sg, W, stop) [Sz] = sftrans (Sz, Sp, Sg, W, stop)
Zeros.
Poles.
Gain.
Edge of target filter.
True for high pass and band stop filters or false for low pass and band pass filters.
This is an Octave function. 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)
[Sz, Sp, Sg] = sftrans (5, 10, 15, 20, 30) Sz = 4 Sp = 2 Sg = 7.5000 | | |