Why Z-transform is used in DSP in Signal Processing - Challenge Your Understanding

Challenge - 5 Problems

🧠 Conceptual

intermediate

Why is the Z-transform preferred over the Fourier transform for discrete signals?

In digital signal processing, the Z-transform is often used instead of the Fourier transform. Why is this the case?

A Because the Z-transform can analyze signals that are not absolutely summable, unlike the Fourier transform.

B Because the Z-transform only works for continuous signals, making it more general.

C Because the Fourier transform cannot represent frequency components of discrete signals.

D Because the Z-transform ignores the time domain and focuses only on frequency.

💡 Hint

Think about the types of signals each transform can handle and their convergence.

The Z-transform can handle a wider class of discrete signals, including those that are not absolutely summable, by considering complex values of z. The Fourier transform is a special case of the Z-transform evaluated on the unit circle and requires absolute summability for convergence.

🧠 Conceptual

intermediate

What advantage does the Z-transform provide in analyzing system stability?

How does the Z-transform help in determining the stability of a discrete-time system?

A It allows checking if all poles of the system's transfer function lie inside the unit circle in the z-plane.

B It shows if the system's impulse response is periodic.

C It converts the system into a continuous-time equivalent for stability analysis.

D It ignores poles and focuses only on zeros for stability.

💡 Hint

Consider how poles in the z-plane relate to system behavior over time.

The Z-transform represents the system's transfer function in the z-domain. Stability requires all poles to be inside the unit circle, ensuring the system's output does not grow unbounded.

❓ data_output

advanced

Output of Z-transform for a given discrete signal

Given the discrete signal x[n] = (0.5)^n for n ≥ 0 and 0 otherwise, what is the Z-transform X(z)?

Signal Processing

import sympy as sp
n, z = sp.symbols('n z')
x = 0.5**n
Xz = sp.summation(x * z**(-n), (n, 0, sp.oo))
Xz_simplified = sp.simplify(Xz)
print(Xz_simplified)

A z / (z + 0.5)

B 1 / (1 - 0.5 / z)

C z / (z - 0.5)

D 1 / (1 - 0.5 * z)

💡 Hint

Recall the formula for the sum of a geometric series and the definition of the Z-transform.

The Z-transform of x[n] = a^n u[n] is X(z) = z / (z - a) for |z| > |a|. Here, a = 0.5.

❓ visualization

advanced

Visualizing the region of convergence (ROC) for a Z-transform

Which plot correctly shows the region of convergence (ROC) for the Z-transform of the signal x[n] = (2)^n u[-n-1]?

Signal Processing

import matplotlib.pyplot as plt
import numpy as np
from matplotlib.patches import Circle

fig, ax = plt.subplots()
circle = Circle((0, 0), 2, color='blue', fill=False, linewidth=2)
ax.add_artist(circle)
filled_circle = Circle((0, 0), 2, color='lightblue', alpha=0.5)
ax.add_artist(filled_circle)
ax.set_xlim(-3, 3)
ax.set_ylim(-3, 3)
ax.set_aspect('equal')
plt.title('ROC: |z| < 2')
plt.xlabel('Real')
plt.ylabel('Imaginary')
plt.grid(True)
plt.show()

A ROC is the entire z-plane except the circle of radius 2.

B ROC is the region inside the circle of radius 2 (|z| < 2).

C ROC is the region outside the circle of radius 2 (|z| > 2).

D ROC is only the circle boundary where |z| = 2.

💡 Hint

Consider the signal's definition and whether it is right-sided or left-sided.

For left-sided signals like x[n] = (2)^n u[-n-1], the ROC is inside the circle with radius equal to the magnitude of the base (2), so |z| < 2.

🔧 Debug

expert

Identify the error in Z-transform calculation code

Consider the following Python code to compute the Z-transform of x[n] = (0.3)^n u[n]. What error will this code produce?

Signal Processing

import sympy as sp
n, z = sp.symbols('n z')
x = 0.3**n
Xz = sp.summation(x * z**n, (n, 0, sp.oo))
print(Xz)

A The code raises a SyntaxError due to incorrect summation syntax.

B The code runs correctly and outputs z / (z - 0.3).

C The code raises a TypeError because z is not numeric.

D The code produces an incorrect formula because the power of z should be negative.

💡 Hint

Check the definition of the Z-transform and the powers of z in the summation.

The Z-transform is defined as sum of x[n] * z^{-n}. Using z^{n} instead causes the formula to be incorrect.