Bilinear Butterworth with realization, T = 1 s

Step 1: specifications

|H| ≥ 0.707 for 0 ≤ ω ≤ π/2 and |H| ≤ 0.2 for 3π/4 ≤ ω ≤ π; T = 1 s.

Step 2: pre-warping (Ω = (2/T) tan(ω/2))

• Ωp = 2 tan(π/4) = 2 rad/s
• Ωs = 2 tan(3π/8) = 4.828 rad/s

Step 3: order

N ≥ log[(1/0.2² − 1)/(1/0.707² − 1)] / (2 log(Ωs/Ωp)) = log(24/1.0006) / (2 log 2.414) = 1.380/0.7658 = 1.80

N = 2.

Step 4: cut-off

Since 0.707 = 1/√2, the pass-band edge is the 3 dB point, so Ωc = Ωp = 2 rad/s. (Check at the stop-band edge: |H| = 1/√(1 + (4.828/2)⁴) = 0.169 ≤ 0.2.)

Step 5: analog filter

H(s) = Ωc² / (s² + √2 Ωc s + Ωc²) = 4 / (s² + 2.828 s + 4)

Step 6: bilinear transformation

s = 2(1 − z⁻¹)/(1 + z⁻¹). Let u = (1 − z⁻¹)/(1 + z⁻¹); then H = 4/(4u² + 4√2 u + 4) = 1/(u² + √2 u + 1).

Multiply the numerator and denominator by (1 + z⁻¹)²:

H(z) = (1 + z⁻¹)² / [(1 − z⁻¹)² + √2(1 − z⁻²) + (1 + z⁻¹)²] = (1 + 2z⁻¹ + z⁻²) / [(2 + √2) + (2 − √2) z⁻²]

H(z) = 0.2929 (1 + 2z⁻¹ + z⁻²) / (1 + 0.1716 z⁻²)

Check: |H| at ω = π/2 is 0.7071; at ω = 3π/4 it is 0.169 (≤ 0.2); at ω = 0 it is 1.

Realization

A second-order system is most conveniently realised in Direct Form II (canonic: 2 delays). With the intermediate signal w[n]:

• w[n] = x[n] − 0.1716 w[n−2]
• y[n] = 0.2929 (w[n] + 2w[n−1] + w[n−2])

 x[n] ──►(+)──────┬──────►[×0.2929]──►(+)────► y[n]
          ▲       ▼ z⁻¹                  ▲  ▲
          │       ├──►[×2 ×0.2929 = 0.5858]┘  │
          │       ▼ z⁻¹                       │
          └[−0.1716]┴──►[×0.2929]─────────────┘

(w[n−2] feeds back through −0.1716 and forward through 0.2929; w[n−1] feeds forward through 0.5858; w[n] goes forward through 0.2929.)

Answer: N = 2, Ωc = 2 rad/s, H(z) = 0.2929(1 + 2z⁻¹ + z⁻²)/(1 + 0.1716z⁻²), realised in Direct Form II.