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- 2072 Chaitra · 2+4 marks
Pole-zero and magnitude response, y−0.4y+0.1y
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.
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].
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
|
|
|
.......
...... | ......
... | ...
... | ...
.. | ..
.. | x ..
. | .
--------.-----o--------o--------------.--------
. | .
.. | x ..
.. | ..
... | ...
... | ...
...... | ......
.......
|
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|
(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 |
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.
✓ Answer checked by Claude Opus 5.5 (high effort)
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