Transfer function H(z) in Signal Processing - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Transfer Function Evaluation

Given the transfer function H(z) = (1 - 0.5z^-1) / (1 - 0.8z^-1), what is the output of H(z) at z = 2?

Signal Processing

def H(z):
    return (1 - 0.5 / z) / (1 - 0.8 / z)

output = H(2)
print(f'{output:.3f}')

A 0.429

B 1.250

C 0.625

D 0.875

💡 Hint

Calculate numerator and denominator separately before dividing.

At z=2, numerator = 1 - 0.5/2 = 0.75, denominator = 1 - 0.8/2 = 0.6, so H(2) = 0.75/0.6 = 1.25. The code formats output to 3 decimal places: 1.250.

❓ data_output

intermediate

Frequency Response Magnitude at Specific Frequencies

Calculate the magnitude of the frequency response |H(e^jω)| for ω = π/4 and ω = π/2 for the transfer function H(z) = (1 - 0.3z^-1 + 0.4z^-2) / (1 - 0.5z^-1 + 0.25z^-2). What are the magnitudes?

Signal Processing

import numpy as np

def H(w):
    z = np.exp(1j * w)
    numerator = 1 - 0.3 / z + 0.4 / (z**2)
    denominator = 1 - 0.5 / z + 0.25 / (z**2)
    return numerator / denominator

mag_pi_4 = abs(H(np.pi/4))
mag_pi_2 = abs(H(np.pi/2))
print(f'{mag_pi_4:.3f}, {mag_pi_2:.3f}')

A 1.045, 1.000

B 0.987, 1.234

C 1.123, 0.875

D 1.000, 1.045

💡 Hint

Use Euler's formula to compute z and carefully calculate numerator and denominator.

Calculate z = e^(jω), then compute numerator and denominator complex values, finally take magnitude of their ratio.

❓ visualization

advanced

Plot the Pole-Zero Diagram of H(z)

Which plot correctly shows the poles and zeros of the transfer function H(z) = (z^2 - 0.6z + 0.08) / (z^2 - 1.2z + 0.36)?

Signal Processing

import matplotlib.pyplot as plt
import numpy as np
from numpy import roots

zeros = roots([1, -0.6, 0.08])
poles = roots([1, -1.2, 0.36])

plt.figure(figsize=(5,5))
plt.scatter(np.real(zeros), np.imag(zeros), marker='o', facecolors='none', edgecolors='b', label='Zeros')
plt.scatter(np.real(poles), np.imag(poles), marker='x', color='r', label='Poles')
unit_circle = plt.Circle((0,0), 1, color='gray', fill=False, linestyle='dashed')
plt.gca().add_artist(unit_circle)
plt.axhline(0, color='black', linewidth=0.5)
plt.axvline(0, color='black', linewidth=0.5)
plt.title('Pole-Zero Plot')
plt.xlabel('Real')
plt.ylabel('Imaginary')
plt.legend()
plt.grid(True)
plt.axis('equal')
plt.show()

A Poles at 0.6 and 0.6, zeros at 0.8 and 0.1

B Poles at 0.6 and 0.6 (complex conjugates), zeros at 0.4 and 0.2 (complex conjugates)

C Poles at 0.6 and 0.6, zeros at 0.4 and 0.2 (complex conjugates)

D Poles at 0.6 and 0.6, zeros at 0.4 and 0.2

💡 Hint

Find roots of numerator and denominator polynomials to locate zeros and poles.

The roots of denominator polynomial z^2 - 1.2z + 0.36 = 0 are 0.6 (repeated). Numerator roots z^2 - 0.6z + 0.08 = 0 are 0.4 and 0.2. All real, matching A.

🧠 Conceptual

advanced

Effect of Pole Location on Stability

Which statement correctly describes the effect of pole locations of H(z) on the stability of a discrete-time system?

A The system is stable if all poles lie inside or on the unit circle in the z-plane.

B The system is stable if all poles lie outside the unit circle in the z-plane.

C The system is stable if all poles lie strictly inside the unit circle in the z-plane.

D The system is stable if all poles lie on the real axis regardless of their magnitude.

💡 Hint

Recall the condition for bounded output in discrete-time systems.

For discrete-time systems, stability requires all poles strictly inside the unit circle (magnitude <1). Poles on the unit circle lead to marginal stability, outside to instability.

🔧 Debug

expert

Identify the Error in Transfer Function Computation

The following code attempts to compute the transfer function H(z) = (1 - 0.7z^-1 + 0.2z^-2) / (1 - 1.1z^-1 + 0.3z^-2) at z=0.9 but raises an error. What is the cause?

Signal Processing

def H(z):
    numerator = 1 - 0.7*z**-1 + 0.2*z**-2
    denominator = 1 - 1.1*z**-1 + 0.3*z**-2
    return numerator / denominator

output = H(0.9)
print(output)

A No error; the code runs correctly and outputs a float

B SyntaxError due to incorrect use of negative exponents

C TypeError because z is a float and cannot be raised to negative powers

D ZeroDivisionError because denominator evaluates to zero at z=0.9

💡 Hint

Check if negative powers are valid for float inputs in Python.

Python supports float ** negative integer powers (e.g., 0.9 ** -1 = 1/0.9). Denominator at z=0.9: 1 - 1.1/0.9 + 0.3/0.81 ≈ 1 - 1.222 + 0.370 ≈ 0.148 ≠ 0. No error.