--- title: "Map given poles and zeros, magnitude response" subject: "Digital Signal Analysis and Processing" chapter: 3 key: "semester_7/digital_signal_analysis_and_processing:pole-zero-map-given-poles" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/pole-zero-map-given-poles site: BCT Engineering --- # Map given poles and zeros, magnitude response Digital Signal Analysis and Processing · Chapter 3: Analysis of LTI System in Frequency Domain - Asked in: 2071 Chaitra, 2+8 marks; 2074 Chaitra, 2+8 marks (asked 2 times) - 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/pole-zero-map-given-poles - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-3 ## Question The poles of a system are located at 0.45+0.77i and −2±0.3i and zeroes at 1.2±3i. Map the poles and zero in the z-plane and plot the magnitude response of the system. ## Short answer Poles: 0.45±j0.77 (radius 0.89, inside unit circle) and -2±j0.3 (radius 2.02, outside); zeros: 1.2±j3 (radius 3.23). Magnitude response peaks near ω=π/3 (~7) from inner pole near unit circle, rises near ω=π from outer poles. Causal realization is unstable. ## Answer **Reading:** a real system has poles and zeros in complex-conjugate pairs. I read the poles as 0.45 ± 0.77i and −2 ± 0.3i, and the zeros as 1.2 ± 3i. No gain is given, so take K = 1: H(z) = K (z − z₁)(z − z₁*) / [(z − p₁)(z − p₁*)(z − p₂)(z − p₂*)] #### Locations (polar form) | Point | Value | Radius | Angle | |---|---|---|---| | Pole p₁, p₁* | 0.45 ± 0.77i | 0.892 | ±59.7° | | Pole p₂, p₂* | −2 ± 0.3i | 2.022 | ±171.5° | | Zero z₁, z₁* | 1.2 ± 3i | 3.231 | ±68.2° | #### Pole-zero map ```text ^ Im | o z1 = 1.2 + j3 | .------|------. x / | x \ ----------+-------+-------+-----> Re x \ | x / '------|------' | | o z1* = 1.2 − j3 ``` (x = pole, o = zero; the circle is the unit circle |z| = 1, crossing the Re axis at −1 and +1; the vertical scale is compressed.) The poles p₁, p₁* = 0.45 ± 0.77i are just inside the unit circle. The poles p₂, p₂* = −2 ± 0.3i (far left) and both zeros are outside it. #### Magnitude response |H(e^jω)| = |e^jω − z₁||e^jω − z₁*| / (product of the distances to the four poles), evaluated at points on the unit circle. Computed (K = 1): | ω | 0 | π/6 | π/3 | π/2 | 2π/3 | 5π/6 | π | |---|---|---|---|---|---|---|---| | \|H\| | 1.11 | 1.57 | 7.04 | 2.09 | 2.09 | 3.36 | 4.71 | ```text 7.04 | * 6.26 | * 5.48 | *** 4.69 | *** *** 3.91 | ***** ****** 3.13 | ******* ********* 2.35 | ************ ************** 1.56 | ********************************** 0.78 |************************************* +------------------------------------- 0 pi/2 pi (w) ``` - A sharp **peak (about 7.0) near ω ≈ π/3 (60°)**, because the pole 0.45 + 0.77i (radius 0.89) is very close to the unit circle at that angle. - A **smaller rise towards ω = π** (about 4.7), caused by the poles at −2 ± 0.3i (angle about 171°); they are outside the circle but are nearest to it at ω ≈ π. - The lowest value (≈ 1.1) is at ω = 0. The zeros (radius 3.2) are far from the circle, so they produce no notch. **Stability:** the poles with |p| = 2.02 > 1 lie outside the unit circle, so a causal realisation is unstable. The magnitude response above exists only for the (non-causal) ROC that contains the unit circle. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why are the poles read as 0.45 +/- 0.77i and -2 +/- 0.3i? A. A real system has poles and zeros in complex-conjugate pairs B. Poles must always be in pairs of equal radius C. Poles must lie on the unit circle D. The system must be unstable Answer: A. Real coefficients force complex roots to appear as conjugate pairs. (Key point: Conjugate pairs) 2. Which pole/zero locations are inside the unit circle? A. Only the poles at -2 +/- 0.3i B. Only the poles at 0.45 +/- 0.77i C. The zeros at 1.2 +/- 3i D. All poles and zeros Answer: B. Radius 0.892 is below 1; the other poles (2.02) and zeros (3.23) are outside. (Key point: Location relative to unit circle) 3. Why does the magnitude response peak near w = pi/3? A. The zeros lie on the unit circle at that angle B. The poles at -2 +/- 0.3i are at that angle C. The pole 0.45 + 0.77i (radius 0.89, angle about 60 degrees) is close to the unit circle there D. The gain K is largest there Answer: C. A pole near the unit circle makes the distance in the denominator small, producing a peak. (Key point: Cause of peak) 4. What does a pole of magnitude 2.02 imply for a causal realisation? A. It is stable, since the zeros are even farther out B. It is stable if K = 1 C. It is marginally stable D. It is unstable, since a causal system needs all poles inside the unit circle Answer: D. A causal system is stable only if all poles satisfy |p| < 1. 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