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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*)]

PointRadiusAngle
Poles 0.45 ± 0.77i0.892±59.7°
Poles 2 ± 0.7i2.119±19.3°
Zeros 1.2 ± 0.43i1.275±19.7°
                       |
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                  ...........
               ....    |   x....          x
              ..       |       .. o
              .        |        ..
-------------.---------|---------.-------------
              .        |        ..
              ..       |       .. o
               ....    |   x....          x
                  ...........
                       |
                       |
                       |
                       |
                       |
                       |

(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:

w0pi/6pi/3pi/22pi/35pi/6pi
|H|0.1690.2682.0950.5060.2810.2140.196
 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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