--- title: "Magnitude response of second order system" subject: "Digital Signal Analysis and Processing" chapter: 3 key: "semester_7/digital_signal_analysis_and_processing:magnitude-response-0-5-0-3" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/magnitude-response-0-5-0-3 site: BCT Engineering --- # Magnitude response of second order system Digital Signal Analysis and Processing · Chapter 3: Analysis of LTI System in Frequency Domain - Asked in: 2072 Kartik, 6 marks - Video: [Problem No. 2 on Frequency Response in Discrete Time Signal Processing](https://www.youtube.com/watch?v=tH3lYfKqOnE) — Ekeeda - 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/magnitude-response-0-5-0-3 - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-3 ## Question Plot magnitude response (not to the scale) of the system described by difference equation y[n] − 0.5y[n−1] + 0.3y[n−2] = x[n] + 0.7x[n−1]. ## Short answer H(z) = (1+0.7z⁻¹)/(1−0.5z⁻¹+0.3z⁻²). Real zero at −0.7; pole pair 0.25±0.487j (radius 0.548). Low‑pass response: maximum 2.44 near ω≈0.28π, |H(0)|=2.125, falls to 0.167 at ω=π. ## Answer #### System function Taking the z-transform of y[n] − 0.5y[n−1] + 0.3y[n−2] = x[n] + 0.7x[n−1] with zero initial conditions (x[n−k] ↔ z⁻ᵏ X(z)): H(z) = Y(z)/X(z) = (1 + 0.7z⁻¹) / (1 − 0.5z⁻¹ + 0.3z⁻²) Multiplying numerator and denominator by z^2: H(z) = N(z)/D(z) with - **Poles** (roots of D(z)): 0.250 + 0.487j and its conjugate (radius 0.548) - **Zeros** (roots of N(z)): -0.700 (real); z = 0 (order 1) All poles are inside the unit circle (largest radius 0.548), so the causal system is stable. #### Pole-zero plot ```text | | | ....... ...... | ...... ... | ... ... | ... .. | x .. .. | .. . | . --------.---o----------o--------------.-------- . | . .. | .. .. | x .. ... | ... ... | ... ...... | ...... ....... | | | ``` (x = pole, o = zero, dotted circle = unit circle; Re axis horizontal, Im axis vertical; range ±1.5) #### Magnitude response Put z = e^(jω): |H(e^(jω))| = (product of distances from e^(jω) to the zeros) / (product of distances to the poles). The response is large where e^(jω) passes close to a pole and small near a zero. | w | 0 | pi/6 | pi/3 | pi/2 | 2pi/3 | 5pi/6 | pi | |---|---|---|---|---|---|---|---| | \|H\| | 2.125 | 2.293 | 2.370 | 1.419 | 0.684 | 0.317 | 0.167 | ```text 2.44 | ****** 2.17 |*************** 1.90 |***************** 1.63 |****************** 1.36 |******************** 1.09 |********************** 0.81 |************************* 0.54 |***************************** 0.27 |************************************* +------------------------------------- 0 pi/2 pi (w) ``` **Shape:** a low-pass response. |H(0)| = 2.125; there is a gentle peak of 2.44 near ω ≈ 0.28π, caused by the pole pair at radius 0.548 and angle 62.8° (about 0.35π); then the response falls steadily to |H(π)| = 0.167 (the zero at −0.7 pulls the high-frequency end down). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is H(z) for y[n] - 0.5y[n-1] + 0.3y[n-2] = x[n] + 0.7x[n-1]? A. (1 - 0.7z^-1)/(1 + 0.5z^-1 - 0.3z^-2) B. (1 + 0.7z^-1)/(1 - 0.5z^-1 + 0.3z^-2) C. (1 - 0.5z^-1 + 0.3z^-2)/(1 + 0.7z^-1) D. (1 + 0.7z)/(1 - 0.5z + 0.3z^2) Answer: B. Output coefficients form the denominator, input coefficients the numerator. (Key point: System function) 2. Where are the poles and is the system stable? A. 0.7 and 0, so stable B. 0.5 +/- 0.3j, radius 0.58, unstable C. 0.25 +/- 0.487j, radius 0.548, so stable D. 0.25 +/- 0.487j, radius 1.1, unstable Answer: C. Roots of z^2 - 0.5z + 0.3 have radius sqrt(0.3) = 0.548 < 1. (Key point: Poles) 3. How is |H(e^(jw))| found graphically? A. Sum of distances to poles divided by sum to zeros B. Product of distances to poles divided by product to zeros C. Distance to the nearest pole only D. Product of distances from e^(jw) to the zeros divided by product of distances to the poles Answer: D. |H| = (zero distances)/(pole distances) evaluated on the unit circle. (Key point: Computing magnitude) 4. What is the shape of the response? A. Low-pass with a gentle peak near 0.28 pi, falling to 0.167 at w = pi B. High-pass rising to a maximum at pi C. Notch at pi/3 D. Flat at 2.125 Answer: A. The pole pair gives a mild peak and the zero at -0.7 lowers the high-frequency end. 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