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Bilinear transformation method of IIR design

Short answer

Bilinear transformation substitutes s=(2/T)(1-z^-1)/(1+z^-1) into the analog transfer function. It maps the jΩ axis to the unit circle with Ω=(2/T)tan(ω/2), eliminating aliasing and preserving stability; pre-warping the critical frequencies compensates its nonlinear frequency warping.

Explain briefly about bilinear transformation method of IIR filter design.

The bilinear transformation designs a digital IIR filter from an analog prototype H_a(s) by the substitution

s = (2/T) · (1 − z⁻¹)/(1 + z⁻¹), so H(z) = H_a(s) at s = (2/T)(1 − z⁻¹)/(1 + z⁻¹)

It comes from approximating an integral by the trapezoidal rule.

Mapping:

  • With s = jΩ and z = e^jω, the relation becomes Ω = (2/T) tan(ω/2), which maps Ω = −∞…∞ onto ω = −π…π once only.
  • The left half of the s-plane maps inside the unit circle, so a stable analog filter gives a stable digital filter.
  • The jΩ-axis maps onto the unit circle.

Advantage: no aliasing, and it works for low-pass, high-pass, band-pass and band-stop filters.

Disadvantage: the relation is nonlinear (frequency warping), so the phase response is distorted and the edges are shifted. This is compensated by pre-warping: the digital edges are converted to analog edges by Ω = (2/T) tan(ω/2) before designing H_a(s).

Steps: (1) pre-warp ωp, ωs; (2) design H_a(s) (Butterworth: find N and Ωc); (3) substitute the bilinear relation to get H(z).

✓ Answer checked by Claude Opus 5.5 (high effort)

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