- Asked 3/11
- 2074 Asoj · 3+7 marks
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.
| w | 0 | pi/6 | pi/3 | pi/2 | 2pi/3 | 5pi/6 | pi |
|---|---|---|---|---|---|---|---|
| |H| | 3.625 | 3.101 | 1.337 | 0.559 | 1.030 | 0.680 | 0.187 |
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)
Related questions
- Pole-zero and magnitude response, y−0.4y+0.1y · Digital Signal Analysis and Processing, chapter 3 · asked 3×
- Difference equation need; pole-zero, magnitude response · Digital Signal Analysis and Processing, chapter 3 · asked 3×
- Magnitude response of second order system · Digital Signal Analysis and Processing, chapter 3 · asked 2×
- Pole-zero, magnitude response of y=x+x(n−2) · Digital Signal Analysis and Processing, chapter 3
- Magnitude response, y−0.3y+0.225y · Digital Signal Analysis and Processing, chapter 3 · asked 2×
- Lattice structure of second order pole-zero system · Digital Signal Analysis and Processing, chapter 4
- Difference equation example; poles/zeros response · Digital Signal Analysis and Processing, chapter 3
- Characteristics and response of zero-order system · Instrumentation and Measurement, chapter 4
- Third-order difference equation: limit x(k+2)/x(k) · Digital Control System, chapter 2 · asked 2×
- Third-order difference equation: limit x(k+2)/x(k+1) · Digital Control System, chapter 2
- Equation and length of axes of x²−4xy−2y²+10x+4y=0 · Mathematics I, chapter 3
All Digital Signal Analysis and Processing chapter 3 questions →