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- 2072 Kartik · 6 marks
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
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.......
...... | ......
... | ...
... | ...
.. | 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| | 2.125 | 2.293 | 2.370 | 1.419 | 0.684 | 0.317 | 0.167 |
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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