Bilinear Butterworth, 0.25π passband, −15 dB at 0.45π

Step 1: specifications

Pass-band edge 0.25π with a maximum drop of 1 dB, so 20 log|H(0.25π)| ≥ −1 dB (|H| ≥ 0.89125). Stop-band: gain at most −15 dB (|H| ≤ 0.17783) at 0.45π. Sampling frequency 1 Hz gives T = 1 s.

Step 2: pre-warp the edges (bilinear: Ω = (2/T) tan(ω/2), T = 1 s)

• Ωp = (2/1) tan(0.25π/2) = 0.8284 rad/s
• Ωs = (2/1) tan(0.45π/2) = 1.7082 rad/s

Step 3: order of the Butterworth filter

N ≥ log[(1/0.17783² − 1)/(1/0.89125² − 1)] / (2 log(Ωs/Ωp)) = log[30.6228/0.2589] / (2 log 2.0619) = 3.2979

So N = 4.

Step 4: cut-off frequency

Using the pass-band condition (1 + (Ωp/Ωc)^(2N)) = 1/0.89125²: Ωc = Ωp / (0.2589)^(1/8) = 0.9809 rad/s (This meets the pass-band exactly and exceeds the stop-band requirement.)

Step 5: analog transfer function

Poles of the normalised filter lie on the left half of the circle of radius Ωc. Pairing complex poles:

H(s) = 0.9256 / [(s² + 0.7507 s + 0.9621) (s² + 1.8124 s + 0.9621)]

Step 6: bilinear transformation

Replace s by (2/T)(1 − z⁻¹)/(1 + z⁻¹) = 2(1 − z⁻¹)/(1 + z⁻¹) in each factor:

• H1(z) = 0.14885(1 + 2z⁻¹ + z⁻²) / (1 − 0.9400 z⁻¹ + 0.5354 z⁻²)
• H2(z) = 0.11204(1 + 2z⁻¹ + z⁻²) / (1 − 0.7076 z⁻¹ + 0.1557 z⁻²)

H(z) = H1(z) · H2(z)

Multiplied out:

        0.01668 0.06671 0.10006 0.06671 0.01668   (numerator coefficients, powers z⁰ … z⁻4)
H(z) = ---------------------------------------------
        1.00000 -1.64759 1.35628 -0.52524 0.08338   (denominator coefficients)

Step 7: check

|H(e^jω)| at ωp = 0.8913 (required ≥ 0.89125); at ωs = 0.1081 (required ≤ 0.17783). Both conditions are satisfied.

Answer: N = 4, Ωc = 0.9809 rad/s, H(z) = H₁(z)·H₂(z). The cascade of two second-order sections is the usual realisation.