Bilinear Butterworth, 0.2π passband, −15 dB at 0.4π

Step 1: specifications

Pass-band edge 0.2π with a maximum drop of 1 dB (|H| ≥ 0.89125); stop-band gain at most −15 dB (|H| ≤ 0.17783) at 0.4π. 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(0.2π/2) = 0.6498 rad/s
• Ωs = (2/1) tan(0.4π/2) = 1.4531 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 2.2361) = 2.9656

So N = 3.

Step 4: cut-off frequency

Using the pass-band condition (1 + (Ωp/Ωc)^(2N)) = 1/0.89125²: Ωc = Ωp / (0.2589)^(1/6) = 0.8140 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) = 0.5393 / [(s + 0.8140) (s² + 0.8140 s + 0.6626)]

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:

• H₀(z) = 0.2893(1 + z⁻¹) / (1 − 0.4215 z⁻¹)
• H1(z) = 0.10533(1 + 2z⁻¹ + z⁻²) / (1 − 1.0611 z⁻¹ + 0.4824 z⁻²)

H(z) = H0(z) · H1(z)

Multiplied out:

        0.03047 0.09140 0.09140 0.03047   (numerator coefficients, powers z⁰ … z⁻3)
H(z) = ---------------------------------------------
        1.00000 -1.48258 0.92964 -0.20333   (denominator coefficients)

Step 7: check

|H(e^jω)| at ωp = 0.8913 (required ≥ 0.89125); at ωs = 0.1731 (required ≤ 0.17783). Both conditions are satisfied.

Answer: N = 3, Ωc = 0.8140 rad/s, H(z) = H₀(z)·H₁(z).