Stability analysis (pole-zero plot) in Signal Processing - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding system stability from pole locations

Consider a discrete-time system with poles at 0.5, 0.8 + 0.3j, and 1.1. Which statement correctly describes the system's stability?

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 one pole lies outside the unit circle.

D The system is stable because poles on the unit circle do not affect stability.

💡 Hint

Recall that for discrete-time systems, stability requires all poles to be strictly inside the unit circle.

Poles inside the unit circle mean the system's response decays over time. A pole at 1.1 lies outside the unit circle, causing instability.

❓ data_output

intermediate

Number of poles inside the unit circle

Given the poles of a system as [-0.9, 0.7 + 0.6j, 0.7 - 0.6j, 1.0], how many poles lie strictly inside the unit circle?

A 1

B 2

C 4

D 3

💡 Hint

Calculate the magnitude of each pole and count those with magnitude less than 1.

The pole at 1.0 lies exactly on the unit circle and is not inside. The other three poles have magnitudes less than 1.

❓ Predict Output

advanced

Output of pole-zero plot code snippet

What does the following Python code output?

import numpy as np
poles = np.array([0.5, 0.8 + 0.3j, 1.1])
inside_unit_circle = np.sum(np.abs(poles) < 1)
print(inside_unit_circle)

A 2

B 3

C 1

D 0

💡 Hint

Check the magnitude of each pole and count how many are less than 1.

Poles 0.5 and 0.8+0.3j have magnitudes less than 1, so count is 2.

❓ visualization

advanced

Interpreting a pole-zero plot for stability

You see a pole-zero plot with poles marked as 'x' and zeros as 'o'. Poles are at 0.9, -0.95, and 1.05 on the real axis. Which statement is true about the system's stability?

A The system is unstable because a pole lies outside the unit circle.

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

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

D The system is stable because zeros outside the unit circle do not affect stability.

💡 Hint

Focus on the poles' locations relative to the unit circle.

The pole at 1.05 lies outside the unit circle, causing instability regardless of zeros.

🚀 Application

expert

Determining system stability from transfer function poles

Given the transfer function H(z) = (z - 0.5) / (z^2 - 1.2z + 0.36), determine the system's stability by analyzing the poles.

Signal Processing

import numpy as np
coefficients = [1, -1.2, 0.36]
poles = np.roots(coefficients)
stable = all(np.abs(poles) < 1)
print(f"Poles: {poles}")
print(f"Is system stable? {stable}")

A Poles are [0.6, 0.6], system is unstable.

B Poles are [0.6, 0.6], system is stable.

C Poles are [0.9, 0.4], system is stable.

D Poles are [1.2, -0.3], system is unstable.

💡 Hint

Find roots of denominator polynomial and check if their magnitudes are less than 1.

The roots of z^2 - 1.2z + 0.36 are both 0.6, which are inside the unit circle, so the system is stable.

Practice

1. What does it mean for a system to be stable in terms of its poles on the pole-zero plot?

easy

A. All poles must lie in the left half of the complex plane (negative real parts).

B. All poles must lie on the imaginary axis.

C. All zeros must lie in the right half of the complex plane.

D. Poles and zeros must be equal in number.

Solution

1. Step 1: Understand pole location and stability

   Stability depends on poles, not zeros. Poles with negative real parts mean the system's response decays over time.

2. Step 2: Identify correct pole location for stability

   Poles in the left half-plane (negative real parts) ensure the system is stable.

3. Final Answer:

   All poles must lie in the left half of the complex plane (negative real parts). -> Option A

4. Quick Check:

   Stable system = poles with negative real parts [OK]

Hint: Stable system means poles left of imaginary axis [OK]

Common Mistakes:

• Confusing zeros with poles for stability
• Thinking poles on imaginary axis are stable
• Assuming zeros affect stability directly

2. Which of the following is the correct Python syntax to plot poles and zeros using the scipy.signal library?

easy

A. signal.pzmap(system, plot=True)

B. signal.pzplot(system)

C. signal.pzplot(system, plot=True)

D. signal.pzmap(system, plot=False)

Solution

1. Step 1: Recall scipy.signal function for pole-zero plot

   The function to plot poles and zeros is pzmap, not pzplot.

2. Step 2: Check correct usage with plotting enabled

   Using pzmap with plot=True plots the pole-zero map correctly.

3. Final Answer:

   signal.pzmap(system, plot=True) -> Option A

4. Quick Check:

   Use pzmap with plot=True for pole-zero plot [OK]

Hint: Use pzmap with plot=True to plot poles and zeros [OK]

Common Mistakes:

• Using non-existent function pzplot
• Forgetting plot=True argument
• Confusing function names

3. Given the poles of a system as poles = [-1+2j, -1-2j, 0.5], what can you say about the system's stability?

medium

A. The system is stable because poles are complex conjugates.

B. The system is stable because all poles have negative real parts.

C. The system is marginally stable because poles are on the imaginary axis.

D. The system is unstable because one pole has a positive real part.

Solution

1. Step 1: Analyze the real parts of each pole

   Poles are -1+2j (real part -1), -1-2j (real part -1), and 0.5 (real part 0.5).

2. Step 2: Determine stability from pole real parts

   Since one pole has positive real part (0.5), the system is unstable.

3. Final Answer:

   The system is unstable because one pole has a positive real part. -> Option D

4. Quick Check:

   Positive real pole means unstable system [OK]

Hint: Check if any pole has positive real part for instability [OK]

Common Mistakes:

• Ignoring positive real part pole
• Assuming complex conjugates guarantee stability
• Confusing marginal stability with instability

4. The following Python code attempts to plot poles and zeros but gives an error:

from scipy import signal
system = signal.TransferFunction([1, 3], [1, 2, 1])
signal.pzplot(system)

What is the error and how to fix it?

medium

A. TransferFunction requires discrete=True for pzplot to work.

B. TransferFunction arguments are reversed; swap numerator and denominator.

C. Function pzplot does not exist; use pzmap(system, plot=True) instead.

D. Import statement is wrong; use import scipy.signal as signal.

Solution

1. Step 1: Identify the function causing error

   The function pzplot does not exist in scipy.signal, causing the error.

2. Step 2: Use correct function to plot poles and zeros

   Use pzmap(system, plot=True) to plot poles and zeros correctly.

3. Final Answer:

   Function pzplot does not exist; use pzmap(system, plot=True) instead. -> Option C

4. Quick Check:

   Use pzmap, not pzplot, for pole-zero plots [OK]

Hint: Use pzmap with plot=True, not pzplot [OK]

Common Mistakes:

• Using non-existent pzplot function
• Swapping numerator and denominator unnecessarily
• Wrong import statement causing confusion

5. You have a system with poles at -0.5, -1+1j, -1-1j and zeros at 0, -2. Which statement best describes the system's stability and response?

hard

A. The system is marginally stable because of zeros at zero.

B. The system is stable with oscillatory decay due to complex poles.

C. The system is unstable because zeros are on the right half-plane.

D. The system is unstable due to poles with positive real parts.

Solution

1. Step 1: Check pole locations for stability

   All poles have negative real parts (-0.5, -1), so system is stable.

2. Step 2: Analyze pole types for response behavior

   Complex conjugate poles (-1±1j) cause oscillations that decay over time.

3. Step 3: Consider zeros' effect on stability

   Zeros do not affect stability directly; zeros at 0 and -2 are not in right half-plane.

4. Final Answer:

   The system is stable with oscillatory decay due to complex poles. -> Option B

5. Quick Check:

   Negative real poles + complex = stable oscillatory decay [OK]

Hint: Stable if poles negative; complex poles cause oscillations [OK]

Common Mistakes:

• Confusing zeros with poles for stability
• Thinking zeros at zero cause instability
• Ignoring oscillatory nature of complex poles