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Magnitude response of second order system

Short answer

H(z) = (1+0.7z⁻¹)/(1−0.5z⁻¹+0.3z⁻²). Real zero at −0.7; pole pair 0.25±0.487j (radius 0.548). Low‑pass response: maximum 2.44 near ω≈0.28π, |H(0)|=2.125, falls to 0.167 at ω=π.

Plot magnitude response (not to the scale) of the system described by difference equation y[n] − 0.5y[n−1] + 0.3y[n−2] = x[n] + 0.7x[n−1].

System function

Taking the z-transform of y[n] − 0.5y[n−1] + 0.3y[n−2] = x[n] + 0.7x[n−1] with zero initial conditions (x[n−k] ↔ z⁻ᵏ X(z)):

H(z) = Y(z)/X(z) = (1 + 0.7z⁻¹) / (1 − 0.5z⁻¹ + 0.3z⁻²)

Multiplying numerator and denominator by z^2: H(z) = N(z)/D(z) with

  • Poles (roots of D(z)): 0.250 + 0.487j and its conjugate (radius 0.548)
  • Zeros (roots of N(z)): -0.700 (real); z = 0 (order 1)

All poles are inside the unit circle (largest radius 0.548), so the causal system is stable.

Pole-zero plot

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

(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|2.1252.2932.3701.4190.6840.3170.167
 2.44 |       ******
 2.17 |***************
 1.90 |*****************
 1.63 |******************
 1.36 |********************
 1.09 |**********************
 0.81 |*************************
 0.54 |*****************************
 0.27 |*************************************
      +-------------------------------------
       0            pi/2          pi  (w)

Shape: a low-pass response. |H(0)| = 2.125; there is a gentle peak of 2.44 near ω ≈ 0.28π, caused by the pole pair at radius 0.548 and angle 62.8° (about 0.35π); then the response falls steadily to |H(π)| = 0.167 (the zero at −0.7 pulls the high-frequency end down).

✓ Answer checked by Claude Opus 5.5 (high effort)

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