Bilinear Butterworth, 0.27π and 0.58π

Step 1: specifications

Low-pass Butterworth, ωp = 0.27π, ωs = 0.58π. δp = 0.11 and δs = 0.21 are read as |H(ωp)| ≥ 0.89 and |H(ωs)| ≤ 0.21. Sampling frequency 0.5 Hz gives T = 2 s.

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

• Ωp = (2/2) tan(0.27π/2) = 0.4515 rad/s
• Ωs = (2/2) tan(0.58π/2) = 1.2892 rad/s

Step 3: order of the Butterworth filter

N ≥ log[(1/0.21² − 1)/(1/0.89² − 1)] / (2 log(Ωs/Ωp)) = log[21.6757/0.2625] / (2 log 2.8552) = 2.1035

So N = 3.

Step 4: cut-off frequency

Using the pass-band condition (1 + (Ωp/Ωc)^(2N)) = 1/0.89²: Ωc = Ωp / (0.2625)^(1/6) = 0.5643 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) = 0.1797 / [(s + 0.5643) (s² + 0.5643 s + 0.3184)]

Step 6: bilinear transformation

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

• H₀(z) = 0.3607(1 + z⁻¹) / (1 − 0.2785 z⁻¹)
• H1(z) = 0.16913(1 + 2z⁻¹ + z⁻²) / (1 − 0.7241 z⁻¹ + 0.4006 z⁻²)

H(z) = H0(z) · H1(z)

Multiplied out:

        0.06101 0.18303 0.18303 0.06101   (numerator coefficients, powers z⁰ … z⁻3)
H(z) = ---------------------------------------------
        1.00000 -1.00260 0.60224 -0.11157   (denominator coefficients)

Step 7: check

|H(e^jω)| at ωp = 0.8900 (required ≥ 0.89); at ωs = 0.0836 (required ≤ 0.21). Both conditions are satisfied.

Answer: N = 3, Ωc = 0.5643 rad/s, H(z) = H₀(z)·H₁(z) as above.