- 2073 Chaitra · 2+8 marks
Difference equation example; poles/zeros response
Short answer
A difference equation relates present output to past outputs and inputs, e.g., y[n]=0.5y[n-1]+x[n]. For the given poles and zeros, the magnitude response has a sharp resonance peak of about 2.09 at ω≈π/3 from the pole pair 0.45±0.77i near the unit circle; the system is unstable because poles 2±0.7i lie outside the unit circle.
Define the difference equation with example. The poles of a system are located at 0.45 + 0.77i and 2 ± 0.7i and zeros at 1.2 ± 0.43i. Plot the magnitude response of this system.
Difference equation
A difference equation relates the present output to present and past inputs and past outputs. A linear constant-coefficient difference equation of order N is
Σ(k=0 to N) aₖ y[n − k] = Σ(k=0 to M) bₖ x[n − k]
Example: y[n] = 0.5y[n − 1] + x[n]. For the input x[n] = δ[n], y[0] = 1, y[1] = 0.5, y[2] = 0.25, …, so h[n] = (0.5)ⁿ u[n] and H(z) = 1/(1 − 0.5z⁻¹). It is the standard way to describe and implement a digital filter.
Magnitude response from the given poles and zeros
Reading: real filters have conjugate pairs, so I take poles at 0.45 ± 0.77i and 2 ± 0.7i, and zeros at 1.2 ± 0.43i. No gain is given, so K = 1:
H(z) = (z − z₁)(z − z₁*) / [(z − p₁)(z − p₁*)(z − p₂)(z − p₂*)]
| Point | Radius | Angle |
|---|---|---|
| Poles 0.45 ± 0.77i | 0.892 | ±59.7° |
| Poles 2 ± 0.7i | 2.119 | ±19.3° |
| Zeros 1.2 ± 0.43i | 1.275 | ±19.7° |
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...........
.... | x.... x
.. | .. o
. | ..
-------------.---------|---------.-------------
. | ..
.. | .. o
.... | x.... x
...........
|
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|
|
|
|
(x = pole, o = zero, dotted circle = unit circle; Re axis horizontal, Im axis vertical; range ±2.4)
|H(e^(jω))| = (distances to the two zeros) / (distances to the four poles), computed with K = 1:
| w | 0 | pi/6 | pi/3 | pi/2 | 2pi/3 | 5pi/6 | pi |
|---|---|---|---|---|---|---|---|
| |H| | 0.169 | 0.268 | 2.095 | 0.506 | 0.281 | 0.214 | 0.196 |
2.09 | *
1.86 | *
1.63 | ***
1.40 | ***
1.16 | *****
0.93 | ******
0.70 | *********
0.47 | ***************
0.23 |*************************************
+-------------------------------------
0 pi/2 pi (w)
Observations:
- A sharp resonance peak (about 2.09) at ω ≈ π/3 (60°) because the pole pair 0.45 ± 0.77i (radius 0.89) lies close to the unit circle at that angle.
- A small dip (about 0.16) near ω ≈ 0.08π, and the response then settles to a flat, low value (about 0.2) at high frequencies.
- The zeros at angle 19.7° almost coincide with the poles at angle 19.3° (outside the circle), so their effects nearly cancel and no extra peak appears near ω ≈ 0.1π.
- The poles at radius 2.12 are outside the unit circle, so a causal realisation is unstable; the response above is for the ROC containing the unit circle (non-causal system). The pole pair at radius 0.89 alone would be stable.
✓ Answer checked by Claude Opus 5.5 (high effort)
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