--- title: "Pole-zero, magnitude response of y=x+x(n−2)" subject: "Digital Signal Analysis and Processing" chapter: 3 key: "semester_7/digital_signal_analysis_and_processing:system-y-x-plus-x-delay-2" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/system-y-x-plus-x-delay-2 site: BCT Engineering --- # Pole-zero, magnitude response of y=x+x(n−2) Digital Signal Analysis and Processing · Chapter 3: Analysis of LTI System in Frequency Domain - Asked in: 2070 Asar, 2+6+2 marks - Video: [Problem No. 1 on Frequency Response in Discrete Time Signal Processing](https://www.youtube.com/watch?v=9x_LMbWI7LM) — 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/system-y-x-plus-x-delay-2 - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-3 ## Question Given H(z) for a system with the following difference equation: y(n) = x(n) + x(n−2). Plot its poles and zeros in Z plane. Determine its magnitude response. Also, determine whether system is causal and stable. ## Short answer H(z)=1+z^{-2} has zeros at ±j and a double pole at z=0. Magnitude response is 2|cos ω|, a notch filter with null at ω=π/2 and maximum gain 2. The system is causal (h[n]={1,0,1}) and stable (FIR, poles inside unit circle). ## Answer #### System function and pole-zero plot y(n) = x(n) + x(n − 2). Taking the z-transform: Y(z) = X(z)(1 + z⁻²), so **H(z) = 1 + z⁻² = (z² + 1)/z²** - Zeros: z² + 1 = 0, so z = ±j (on the unit circle, at angles ±90°). - Poles: z² = 0, a double pole at z = 0. - Impulse response: h(n) = {1, 0, 1}, n = 0, 1, 2 (FIR). ```text Im 1 | o (0,+j) | ---+---(2)x---------- Re x (double) = pole at origin | o = zeros at +j and −j -1 | o (0,−j) ``` (the dashed unit circle passes through both zeros) #### Magnitude response H(e^(jω)) = 1 + e^(−j2ω) = e^(−jω)(e^(jω) + e^(−jω)) = 2cos(ω) e^(−jω) |H(e^(jω))| = 2|cos ω|, phase = −ω (linear, with jumps of π where cos ω changes sign). | ω | 0 | π/4 | π/2 | 3π/4 | π | |---|---|---|---|---|---| | \|H\| | 2 | 1.414 | 0 | 1.414 | 2 | ```text |H| 2 |* * | * * 1 | * * | * * 0 +-------------*------------> w 0 pi/4 pi/2 3pi/4 pi ``` It is a **notch (band-stop) filter** with a null at ω = π/2 (the zeros ±j), maximum gain 2 at ω = 0 and ω = π. #### Causality and stability - **Causal:** h(n) = 0 for n < 0, and the output depends only on x(n) and x(n−2) (past and present). ROC: |z| > 0, which is the exterior of a circle (including ∞). - **Stable:** Σ|h(n)| = 2 < ∞; equivalently the only poles are at z = 0, inside the unit circle, and the ROC includes |z| = 1. Every FIR system is stable. ## 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) = x(n) + x(n-2)? A. 1 + z^-2 = (z^2 + 1)/z^2 B. 1 - z^-2 C. 1 + z^2 D. 1/(1 + z^-2) Answer: A. A delay of 2 gives z^-2. (Key point: System function) 2. Where are the poles and zeros? A. Zeros at z = +/- 1, poles at z = +/- j B. Zeros at z = +/- j, double pole at z = 0 C. Zeros at z = 0, poles at +/- j D. Zeros at z = -1 (double), poles at 0 Answer: B. z^2 + 1 = 0 gives z = +/- j; z^2 = 0 gives the double pole at origin. (Key point: Poles and zeros) 3. What is |H(e^(jw))|? A. 2|sin w| B. 2 cos(2w) C. 2|cos w| D. 1 + cos w Answer: C. 1 + e^(-j2w) = e^(-jw)(2cos w). (Key point: Magnitude response) 4. Why is the system causal and stable? A. It has poles outside the unit circle but is still causal B. It is non-causal because of the x(n-2) term C. It is unstable since zeros lie on the unit circle D. h(n) = {1, 0, 1} is zero for n < 0 and finite, with poles only at z = 0 Answer: D. FIR systems with h(n) = 0 for n < 0 are causal, and sum |h| = 2 is finite. 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