Impulse invariance Butterworth, 0.7 dB and 14 dB

Step 1: specifications

Pass band within 0.7 dB of 0 dB up to 0.15π (|H| ≥ 10^(−0.7/20) = 0.9226); stop-band attenuation at least 14 dB (|H| ≤ 0.1995) from 0.6π to π. T is not given; take T = 1 s.

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

Ωp = 0.4712 rad/s, Ωs = 1.8850 rad/s (T = 1 s)

Step 3: order

N ≥ log[(1/0.1995² − 1)/(1/0.9226² − 1)] / (2 log(Ωs/Ωp)) = log[24.1189/0.1749] / (2 log 4.0000) = 1.7769

The analog formula gives N = 2. However, impulse invariance causes aliasing: the usual N = 2 design (Ωc = 0.7287 rad/s, H(z) = 0.3033z⁻¹/(1 − 1.0396z⁻¹ + 0.3568z⁻²)) gives only 0.899 at 0.15π, below the required 0.9226, and a numerical search shows that no choice of Ωc lets a second-order design meet both edges. So take N = 3.

Step 4: cut-off

Ωc = Ωp / (0.1749)^(1/6) = 0.6301 rad/s

Step 5: analog poles and H(s)

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

• s0 = −0.3151 + j0.5457
• s1 = −0.6301
• s2 = −0.3151 − j0.5457

H(s) = Ωc³ / [(s + Ωc)(s² + Ωc s + Ωc²)] = 0.2502 / [(s + 0.6301)(s² + 0.6301s + 0.3971)] = Σ A_k/(s − s_k), with residues

• A0 = −0.3151 − j0.1819
• A1 = 0.6301
• A2 = −0.3151 + j0.1819

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 the complex-conjugate pair into a real second-order section:

• 0.63015 / (1 − 0.53251z⁻¹)
• (−0.63015 + 0.53085z⁻¹) / (1 − 1.24748z⁻¹ + 0.53251z⁻²)

H(z) is the sum of these two terms (parallel form).

Step 7: check

|H(e^jωp)| = 0.9231 (required ≥ 0.9226); |H(e^jωs)| = 0.0360 (required ≤ 0.1995); |H(1)| = 0.9996. Both conditions are satisfied.

Answer: N = 3, Ωc = 0.6301 rad/s, H(s) = 0.2502/[(s + 0.6301)(s² + 0.6301s + 0.3971)], H(z) = 0.63015/(1 − 0.53251z⁻¹) + (−0.63015 + 0.53085z⁻¹)/(1 − 1.24748z⁻¹ + 0.53251z⁻²).