Impulse invariance Butterworth, 200 Hz and 500 Hz

Step 1: specifications

Pass-band edge 200 Hz with −5 dB (|H| ≥ 0.5623), stop-band edge 500 Hz with −12 dB (|H| ≤ 0.2512), Fs = 5 kHz, T = 0.2 ms.

Step 2: analog edge frequencies (impulse invariance: Ω = ω/T)

Ωp = 1256.6371 rad/s, Ωs = 3141.5927 rad/s (T = 0.0002 s)

Step 3: order

N ≥ log[(1/0.2512² − 1)/(1/0.5623² − 1)] / (2 log(Ωs/Ωp)) = log[14.8489/2.1623] / (2 log 2.5000) = 1.0514

N = 2

Step 4: cut-off

Ωc = Ωp / (2.1623)^(1/4) = 1036.2916 rad/s

Step 5: analog poles and H(s)

Poles: s_k = Ωc exp[jπ(2k + N + 1)/(2N)], k = 0 … 1 (left half plane):

• s0 = -732.7689 + j732.7689
• s1 = -732.7689 − j732.7689

H(s) = Ωc^2 / Π(s − s_k) = Σ A_k/(s − s_k), with residues

• A0 = 0.0000 − j732.7689
• A1 = -0.0000 + j732.7689

Step 6: impulse invariant transformation

Each term A/(s − s_k) becomes T·A/(1 − e^(s_k T) z⁻¹), so h[n] = T·h_a(nT). Combining each complex-conjugate pair into a real second-order section:

• (0.00000 + 0.03697z⁻¹) / (1 − 1.70884z⁻¹ + 0.74594z⁻²)

H(z) is the sum of these terms (parallel form). Here the factor T = 0.0002 is included in the numerator so that the DC gain of the digital filter equals that of the analog filter (if T is left out, as some textbooks do, the numerator becomes 1/T = 5000 times larger, i.e. 184.8 z⁻¹).

Step 7: check

|H(e^jωp)| = 0.5633 (required ≥ 0.5623); |H(e^jωs)| = 0.1114 (required ≤ 0.2512); |H(1)| = 0.9964.

Answer: N = 2, Ωc = 1036.3 rad/s, H(s) = Ωc²/(s² + 1465.5 s + Ωc²), H(z) = 0.03697 z⁻¹ / (1 − 1.7088z⁻¹ + 0.7459z⁻²).