Inverse Z-transform in Signal Processing - Practice Problems & Coding Challenges

Challenge - 5 Problems

🧠 Conceptual

intermediate

Understanding the Region of Convergence (ROC) in Inverse Z-transform

Which statement correctly describes the role of the Region of Convergence (ROC) in the inverse Z-transform?

A The ROC defines the frequency components present in the original signal but does not affect the inverse transform.

B The ROC is irrelevant for the inverse Z-transform and only matters in the forward Z-transform.

C The ROC determines the stability and causality of the time-domain signal obtained from the inverse Z-transform.

D The ROC is a graphical tool used only to plot the Z-transform and has no mathematical significance.

💡 Hint

Think about how the ROC affects the nature of the time-domain signal, such as whether it is causal or stable.

The ROC is crucial because it determines whether the inverse Z-transform corresponds to a causal, anti-causal, or two-sided sequence and whether the sequence is stable. Without the ROC, the inverse transform is not uniquely defined.

❓ Predict Output

intermediate

Output of Inverse Z-transform for a Simple Rational Function

What is the output sequence x[n] for the inverse Z-transform of the function X(z) = 1 / (1 - 0.5z-1) assuming the ROC is |z| > 0.5?

Signal Processing

from sympy import symbols, inverse_z_transform
z, n = symbols('z n', integer=True)
X = 1 / (1 - 0.5*z**-1)
x_n = inverse_z_transform(X, z, n)
print(x_n)

A 0.5**n if n >= 0 else 0

B 0.5**n

C (-0.5)**n if n >= 0 else 0

D n * 0.5**n

💡 Hint

Recall the inverse Z-transform of a simple first-order pole with ROC outside the pole radius corresponds to a causal geometric sequence.

The function corresponds to a causal sequence x[n] = (0.5)^n for n >= 0. The ROC |z| > 0.5 ensures causality and convergence.

❓ data_output

advanced

Inverse Z-transform of a Two-Pole System

Given the Z-transform X(z) = (1 - 0.3z-1) / ((1 - 0.5z-1)(1 - 0.8z-1)) with ROC |z| > 0.8, what is the first five values of the time-domain sequence x[n]?

Signal Processing

from sympy import symbols, inverse_z_transform
z, n = symbols('z n', integer=True)
X = (1 - 0.3*z**-1) / ((1 - 0.5*z**-1)*(1 - 0.8*z**-1))
x_n = inverse_z_transform(X, z, n)
sequence = [x_n.subs(n, i).evalf() for i in range(5)]
print(sequence)

A [1.0, 0.5, 0.25, 0.125, 0.0625]

B [1.0, 1.0, 0.65, 0.365, 0.2045]

C [1.0, 0.8, 0.64, 0.512, 0.4096]

D [1.0, 1.3, 1.69, 2.197, 2.856]

💡 Hint

Use partial fraction expansion and inverse Z-transform properties for each term.

The sequence is a sum of two weighted geometric sequences derived from the poles at 0.5 and 0.8 with coefficients from the numerator. The first five values are as in option B.

❓ visualization

advanced

Plotting the Time-Domain Sequence from Inverse Z-transform

Which plot correctly shows the sequence x[n] obtained from the inverse Z-transform of X(z) = 1 / (1 - 0.7z-1) with ROC |z| > 0.7 for n = 0 to 9?

Signal Processing

import matplotlib.pyplot as plt
import numpy as np
n = np.arange(10)
x_n = 0.7**n
plt.stem(n, x_n, use_line_collection=True)
plt.xlabel('n')
plt.ylabel('x[n]')
plt.title('Sequence from Inverse Z-transform')
plt.grid(True)
plt.show()

A A stem plot showing values decreasing exponentially from 1 at n=0 to about 0.04 at n=9.

B A stem plot showing values increasing exponentially from 1 at n=0 to about 10 at n=9.

C A line plot showing constant values of 1 for all n from 0 to 9.

D A scatter plot showing random values between 0 and 1 for n from 0 to 9.

💡 Hint

Recall that the sequence is geometric with ratio 0.7, so values decrease as n increases.

The sequence x[n] = 0.7^n starts at 1 and decreases exponentially. The stem plot in option A correctly represents this behavior.

🔧 Debug

expert

Identifying the Error in Inverse Z-transform Calculation

Consider the code below attempting to compute the inverse Z-transform of X(z) = z / (z - 0.6). What error will this code produce?

from sympy import symbols, inverse_z_transform
z, n = symbols('z n', integer=True)
X = z / (z - 0.6)
x_n = inverse_z_transform(X, z, n)
print(x_n)

A TypeError: 'z' is a symbol and cannot be used directly in inverse_z_transform without substitution.

B SyntaxError due to missing parentheses in the expression.

C No error; the output is (0.6)**n for n >= 0.

D ValueError: The expression is not a proper rational function in z^-1.

💡 Hint

Check the form of the expression passed to inverse_z_transform; it expects a function in terms of z^-1.

The inverse_z_transform function expects the input as a function of z^-1. Here, the expression is in terms of z, not z^-1, causing a ValueError.