--- title: "Magnitude response, y−0.3y+0.225y" subject: "Digital Signal Analysis and Processing" chapter: 3 key: "semester_7/digital_signal_analysis_and_processing:magnitude-response-0-3-0-225" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/magnitude-response-0-3-0-225 site: BCT Engineering --- # Magnitude response, y−0.3y+0.225y Digital Signal Analysis and Processing · Chapter 3: Analysis of LTI System in Frequency Domain - Asked in: 2070 Chaitra, 6 marks - Video: [Frequency Response Magnitude and Poles and Zeros](https://www.youtube.com/watch?v=8jNjVkoZQCU) — Barry Van Veen - 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-0-3-0-225 - 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] − 0.3y[n−1] + 0.225y[n−2] = x[n] + 0.5x[n−1]. ## Short answer H(z) = (1+0.5z⁻¹)/(1−0.3z⁻¹+0.225z⁻²). Poles 0.15±0.45j (r=0.474); zero at −0.5. Low‑pass with peak 1.80 around ω=0.30π, |H(0)|=1.62, ending at |H(π)|=0.33. ## Answer #### System function Taking the z-transform of y[n] − 0.3y[n−1] + 0.225y[n−2] = x[n] + 0.5x[n−1] with zero initial conditions (x[n−k] ↔ z⁻ᵏ X(z)): H(z) = Y(z)/X(z) = (1 + 0.5z⁻¹) / (1 − 0.3z⁻¹ + 0.225z⁻²) Multiplying numerator and denominator by z^2: H(z) = N(z)/D(z) with - **Poles** (roots of D(z)): 0.150 + 0.450j and its conjugate (radius 0.474) - **Zeros** (roots of N(z)): -0.500 (real); z = 0 (order 1) All poles are inside the unit circle (largest radius 0.474), 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\| | 1.622 | 1.704 | 1.787 | 1.345 | 0.765 | 0.438 | 0.328 | ```text 1.80 | ********* 1.60 |***************** 1.40 |******************* 1.20 |********************* 1.00 |*********************** 0.80 |************************* 0.60 |***************************** 0.40 |************************************* 0.20 |************************************* +------------------------------------- 0 pi/2 pi (w) ``` **Shape:** low-pass with a mild peak. |H(0)| = 1.62, a maximum of 1.80 near ω ≈ 0.30π (the poles 0.15 ± 0.45j have angle 71.6°, radius 0.474), then a steady fall to |H(π)| = 0.328 (the zero at −0.5 attenuates high frequencies). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is H(z)? A. (1 - 0.5z^-1)/(1 + 0.3z^-1 + 0.225z^-2) B. (1 + 0.3z^-1)/(1 - 0.5z^-1 + 0.225z^-2) C. (1 - 0.3z^-1 + 0.225z^-2)/(1 + 0.5z^-1) D. (1 + 0.5z^-1)/(1 - 0.3z^-1 + 0.225z^-2) Answer: D. Numerator from the x terms, denominator from the y terms. (Key point: System function) 2. Where are the poles? A. 0.15 +/- 0.45j, radius 0.474, angle about 71.6 degrees B. 0.3 +/- 0.225j, radius 0.375 C. -0.5 and 0 D. 0.15 +/- 0.45j, radius 1.5 Answer: A. Roots of z^2 - 0.3z + 0.225 give 0.15 +/- 0.45j. (Key point: Pole location) 3. Is the causal system stable and why? A. No, because there is a zero at -0.5 B. Yes, both poles lie inside the unit circle C. No, because the poles are complex D. Yes, because the numerator has fewer terms Answer: B. Stability depends only on poles being inside the unit circle. (Key point: Stability) 4. Describe the magnitude response. A. High-pass with maximum at pi B. Sharp resonance of more than 10 C. Low-pass with a mild peak of about 1.8 near 0.30 pi, falling to 0.328 at pi D. Constant magnitude Answer: C. The pole pair gives a mild peak; the zero at -0.5 attenuates high frequencies. 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