--- title: "Bilinear Butterworth with realization, T = 1 s" subject: "Digital Signal Analysis and Processing" chapter: 6 key: "semester_7/digital_signal_analysis_and_processing:bilinear-butter-0-707-realize" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/bilinear-butter-0-707-realize site: BCT Engineering --- # Bilinear Butterworth with realization, T = 1 s Digital Signal Analysis and Processing · Chapter 6: IIR Filter Design - Asked in: 2073 Chaitra, 11+4 marks - Video: [Digital Butterworth Filter using Bilinear Transformation - Problem Solved - DTSP - DSP](https://www.youtube.com/watch?v=4yNIBnYwLuQ) — Padmasri Naban - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/bilinear-butter-0-707-realize - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-6 ## Question Design a digital Butterworth low pass filter satisfying the constraints 0.707 ≤ |H(e^jw)| ≤ 1 for 0 ≤ w ≤ π/2; |H(e^jw)| ≤ 0.2 for 3π/4 ≤ w ≤ π. With T = 1 sec using bilinear transformation method. Realize the filter using the most convenient realization form. ## Short answer Filter order N=2, analog cut-off Ωc=2 rad/s. The digital transfer function is H(z)=0.2929(1+2z⁻¹+z⁻²)/(1+0.1716z⁻²), realized in Direct Form II. ## Answer #### 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]) ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is Omega_c equal to Omega_p here? A. Because N = 2 forces it B. Because T = 1 C. 0.707 = 1/sqrt(2) is the 3 dB point, so the pass-band edge is the cut-off D. Because the stop-band is 0.2 Answer: C. At Omega_c a Butterworth filter has gain 1/sqrt(2). (Key point: Cut-off from 0.707) 2. Which values are Omega_p and Omega_s? A. pi/2 and 3 pi/4 B. 1 and 2.414 C. 0.5 and 1 D. 2 and 4.828 rad/s Answer: D. 2 tan(pi/4) = 2 and 2 tan(3pi/8) = 4.828. (Key point: Pre-warped edges) 3. What is the order? A. N = 2 (1.80 rounded up) B. N = 1 C. N = 3 D. N = 4 Answer: A. The formula gives 1.80, rounded up to 2. (Key point: Order) 4. Why Direct Form II? A. It needs no multipliers B. It is canonic, using only 2 delays for a second-order system C. It has no feedback D. It needs 4 delays Answer: B. Direct Form II shares delays between the poles and zeros. 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