--- title: "Pole-zero and magnitude response, y−0.4y+0.1y" subject: "Digital Signal Analysis and Processing" chapter: 3 key: "semester_7/digital_signal_analysis_and_processing:magnitude-response-0-4-0-1" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/magnitude-response-0-4-0-1 site: BCT Engineering --- # Pole-zero and magnitude response, y−0.4y+0.1y Digital Signal Analysis and Processing · Chapter 3: Analysis of LTI System in Frequency Domain - Asked in: 2072 Chaitra, 2+4 marks - Video: [How to find Pole Zero Values from Difference Equation/ DSP](https://www.youtube.com/watch?v=Z2-8ZNMru88) — Dr.Jayaudhaya ,Simple and Easy Way - 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-0-4-0-1 - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-3 ## Question Plot the pole-zero in z-plane and draw magnitude response (not to the scale) of the system described by difference equation y[n] − 0.4y[n−1] + 0.1y[n−2] = x[n] + 0.6x[n−1]. ## Short answer Poles at 0.2±0.245j (radius0.316); zero at −0.6. Magnitude is low‑pass, |H(0)|=2.286, smooth fall to |H(π)|=0.267. No resonance due to small pole radius. ## Answer #### System function Taking the z-transform of y[n] − 0.4y[n−1] + 0.1y[n−2] = x[n] + 0.6x[n−1] with zero initial conditions (x[n−k] ↔ z⁻ᵏ X(z)): H(z) = Y(z)/X(z) = (1 + 0.6z⁻¹) / (1 − 0.4z⁻¹ + 0.1z⁻²) Multiplying numerator and denominator by z^2: H(z) = N(z)/D(z) with - **Poles** (roots of D(z)): 0.200 + 0.245j and its conjugate (radius 0.316) - **Zeros** (roots of N(z)): -0.600 (real); z = 0 (order 1) All poles are inside the unit circle (largest radius 0.316), 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\| | 2.286 | 2.173 | 1.764 | 1.184 | 0.709 | 0.397 | 0.267 | ```text 2.29 |******* 2.03 |*********** 1.78 |************** 1.52 |**************** 1.27 |******************* 1.02 |********************** 0.76 |************************** 0.51 |******************************* 0.25 |************************************* +------------------------------------- 0 pi/2 pi (w) ``` **Shape:** low-pass. The maximum is |H(0)| = 2.286 and the response falls smoothly (no resonance, because the poles at radius 0.316 are far from the unit circle) to |H(π)| = 0.267. The zero at −0.6 lies near ω = π and keeps the high-frequency gain small. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the denominator of H(z)? A. 1 + 0.4z^-1 + 0.1z^-2 B. 1 - 0.4z^-1 + 0.1z^-2 C. 1 - 0.4z^-1 - 0.1z^-2 D. 1 + 0.6z^-1 Answer: B. The y coefficients give the denominator with their signs as written on the left side. (Key point: System function) 2. What are the pole radius and stability? A. Radius 0.6, stable B. Radius 1.0, marginally stable C. Radius 0.316, inside the unit circle, so stable D. Radius 3.16, unstable Answer: C. |p|^2 = 0.1 so |p| = 0.316. (Key point: Poles) 3. Why is there no resonance peak? A. The zero at -0.6 cancels the poles B. The poles lie on the real axis C. The numerator order is too low D. The poles at radius 0.316 are far from the unit circle Answer: D. A peak needs poles close to the unit circle. (Key point: No resonance) 4. What does the zero at -0.6 do? A. It lies near w = pi and keeps the high-frequency gain small B. It creates a peak at w = 0 C. It makes the system unstable D. It has no effect on the response Answer: A. A zero near the unit circle at angle pi reduces |H| at high frequency. (Key point: Zero effect) ## Close variants set in other exams - [Pole-zero and magnitude response, third order](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Amagnitude-response-0-4-0-2-0-5) — Digital Signal Analysis and Processing, chapter 3, asked 3× - [Difference equation need; pole-zero, magnitude response](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Awhy-difference-eq-magnitude-response) — Digital Signal Analysis and Processing, chapter 3, asked 3× ## Related in Digital Signal Analysis and Processing - [Pole-zero, magnitude response of y=x+x(n−2)](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Asystem-y-x-plus-x-delay-2) — Digital Signal Analysis and Processing, chapter 3 - [Magnitude response of second order system](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Amagnitude-response-0-5-0-3) — Digital Signal Analysis and Processing, chapter 3, asked 2× - [Magnitude response, y−0.3y+0.225y](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Amagnitude-response-0-3-0-225) — Digital Signal Analysis and Processing, chapter 3, asked 2× - [Difference equation example; poles/zeros response](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Apole-zero-define-difference-eq) — Digital Signal Analysis and Processing, chapter 3 ## Related in other subjects - [Nyquist criterion from given pole-zero and Nyquist plot](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Anyquist-criteria-from-plot) — Control Systems, chapter 6