Pole-zero analysis for stability in Signal Processing - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding Stability from Pole Locations

In discrete-time signal processing, which condition on the poles of a system's transfer function ensures the system is stable?

A All poles lie on the real axis regardless of their magnitude.

B All poles lie on or outside the unit circle in the complex plane.

C All poles lie strictly outside the unit circle in the complex plane.

D All poles lie strictly inside the unit circle in the complex plane.

💡 Hint

Think about where poles must be located so that the system's response does not grow over time.

For discrete-time systems, stability requires all poles to be inside the unit circle (magnitude less than 1). Poles outside or on the unit circle cause instability or marginal stability.

❓ Predict Output

intermediate

Identify Stable System from Poles

Given the poles of a system as complex numbers, which set corresponds to a stable system?

Signal Processing

poles_sets = {
    'A': [0.5 + 0.5j, 0.3 - 0.4j],
    'B': [1.0 + 0j, 0.8 + 0.1j],
    'C': [1.1 + 0j, 0.9 - 0.2j],
    'D': [0.9 + 0.9j, 0.7 + 0j]
}

# Which set is stable?

A Set A

B Set D

C Set C

D Set B

💡 Hint

Check if all poles have magnitude less than 1.

Set A poles have magnitudes less than 1, so the system is stable. Set B has a pole exactly on the unit circle (magnitude 1), which is marginally stable. Set C has a pole outside the unit circle (magnitude > 1), unstable. Set D has a pole with magnitude sqrt(0.9^2 + 0.9^2) ≈ 1.27 > 1, unstable.

❓ data_output

advanced

Calculate Stability from Given Poles

Given the following poles of a discrete-time system, determine how many poles lie outside the unit circle.

Signal Processing

import numpy as np

poles = np.array([0.9 + 0.1j, 1.05 + 0j, 0.7 - 0.7j, 1.0 + 0j, 0.5 + 0j])
magnitudes = np.abs(poles)
outside_count = np.sum(magnitudes > 1)
outside_count

A 2

B 1

C 3

D 0

💡 Hint

Count poles with magnitude strictly greater than 1.

Poles magnitudes: 0.9+0.1j ≈ 0.905, 1.05+0j = 1.05, 0.7-0.7j ≈ 0.99, 1.0+0j = 1.0, 0.5+0j = 0.5. Only one pole (1.05) is outside the unit circle.

❓ visualization

advanced

Plot Poles and Determine Stability

Given the poles plotted on the complex plane, which statement about system stability is correct?

The unit circle is shown as a dashed line.

Signal Processing

import matplotlib.pyplot as plt
import numpy as np

poles = np.array([0.8 + 0.6j, 0.9 - 0.1j, 1.1 + 0j, 0.5 + 0j])

fig, ax = plt.subplots()
unit_circle = plt.Circle((0, 0), 1, color='gray', fill=False, linestyle='dashed')
ax.add_artist(unit_circle)
ax.plot(poles.real, poles.imag, 'ro', label='Poles')
ax.set_xlim(-1.5, 1.5)
ax.set_ylim(-1.5, 1.5)
ax.set_aspect('equal')
ax.axhline(0, color='black', linewidth=0.5)
ax.axvline(0, color='black', linewidth=0.5)
ax.legend()
plt.show()

A The system is stable because all poles are inside the unit circle.

B The system is marginally stable because all poles lie on the unit circle.

C The system is unstable because at least one pole lies outside the unit circle.

D The system's stability cannot be determined from the plot.

💡 Hint

Look for any pole outside the dashed circle.

One pole at 1.1 + 0j lies outside the unit circle, making the system unstable.

🚀 Application

expert

Determine System Stability from Transfer Function Coefficients

Given the transfer function denominator coefficients of a discrete-time system:

den = [1, -1.5, 0.7]

Which statement about the system's stability is correct?

A The system is unstable because at least one pole has magnitude greater than 1.

B The system is stable because all poles have magnitude less than 1.

C The system is marginally stable because a pole lies exactly on the unit circle.

D The system's stability cannot be determined without numerator coefficients.

💡 Hint

Find the roots of the denominator polynomial and check their magnitudes.

Roots of the polynomial 1 - 1.5z + 0.7z^2 are approximately 1.071 and 0.654. The root at approximately 1.071 is outside the unit circle, making the system unstable.