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Pole-zero and magnitude response, third order

Short answer

H(z) = (1+0.5z⁻¹+0.6z⁻²+0.8z⁻³)/(1−0.4z⁻¹+0.2z⁻²). Poles at 0.2±0.4j (radius 0.447) and z=0; zeros at 0.184±0.942j (radius 0.959) and −0.869. Magnitude is low-pass with max 3.625 at ω=0, a deep dip to ≈0.14 near ω≈0.44π, a small bump, and |H(π)|=0.19.

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] + 0.2y[n−2] = x[n] + 0.5x[n−1] + 0.6x[n−2] + 0.8x[n−3].

System function

Taking the z-transform of y[n] − 0.4y[n−1] + 0.2y[n−2] = x[n] + 0.5x[n−1] + 0.6x[n−2] + 0.8x[n−3] with zero initial conditions (x[n−k] ↔ z⁻ᵏ X(z)):

H(z) = Y(z)/X(z) = (1 + 0.5z⁻¹ + 0.6z⁻² + 0.8z⁻³) / (1 − 0.4z⁻¹ + 0.2z⁻²)

Multiplying numerator and denominator by z^3: H(z) = N(z)/D(z) with

  • Poles (roots of D(z)): 0.200 + 0.400j and its conjugate (radius 0.447); z = 0 (order 1)
  • Zeros (roots of N(z)): 0.184 + 0.942j and its conjugate (radius 0.959); -0.869 (real)

All poles are inside the unit circle (largest radius 0.447), so the causal system is stable.

Pole-zero plot

                       |
                       |
                       |
                    .......
               ......  |  o......
            ...        |        ...
          ...          |          ...
         ..            |  x         ..
        ..             |             ..
        .              |              .
--------.-o------------x--------------.--------
        .              |              .
        ..             |             ..
         ..            |  x         ..
          ...          |          ...
            ...        |        ...
               ......  |  o......
                    .......
                       |
                       |
                       |

(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.

w0pi/6pi/3pi/22pi/35pi/6pi
|H|3.6253.1011.3370.5591.0300.6800.187
 3.62 |****
 3.22 |*******
 2.82 |*********
 2.42 |**********
 2.01 |***********
 1.61 |************
 1.21 |*************          ***
 0.81 |***************    ************
 0.40 |**************** *******************
      +-------------------------------------
       0            pi/2          pi  (w)

Shape: low-pass with maximum |H(0)| = 3.625 at ω = 0, falling quickly to a deep dip (about 0.14) near ω ≈ 0.44π, because the zero pair 0.184 ± 0.942j has radius 0.96, almost on the unit circle at angle 79°. There is a small bump (about 1.0 near 0.63π) and the response ends low, |H(π)| = 0.19. The poles (radius 0.447) are far from the unit circle, so there is no sharp peak.

✓ Answer checked by Claude Opus 5.5 (high effort)

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