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.

Practice

1. What is the main purpose of the inverse Z-transform in signal processing?

easy

A. To filter noise from a digital signal

B. To convert a time-domain signal into frequency domain

C. To calculate the Fourier transform of a signal

D. To find the original time-domain sequence from its Z-transform

Solution

1. Step 1: Understand the Z-transform

   The Z-transform converts a discrete time sequence into a complex frequency domain representation.

2. Step 2: Purpose of inverse Z-transform

   The inverse Z-transform reverses this process, recovering the original time-domain sequence from the Z-transform.

3. Final Answer:

   To find the original time-domain sequence from its Z-transform -> Option D

4. Quick Check:

   Inverse Z-transform = Original sequence [OK]

Hint: Inverse Z-transform recovers original sequence from Z domain [OK]

Common Mistakes:

• Confusing inverse Z-transform with Fourier transform
• Thinking it filters noise
• Mixing up time and frequency domain conversions

2. Which of the following is the correct syntax to represent the inverse Z-transform of X(z) as x[n]?

easy

A. X(z) = Z^{-1}{x[n]}

B. x[n] = Z^{-1}{X(z)}

C. x[n] = Z{X(z)}

D. X(z) = Z{x[n]}

Solution

1. Step 1: Identify notation for inverse Z-transform

   The inverse Z-transform is denoted as Z^{-1} applied to X(z) to get x[n].

2. Step 2: Match correct expression

   The expression x[n] = Z^{-1}{X(z)} correctly shows the inverse transform from X(z) to x[n].

3. Final Answer:

   x[n] = Z^{-1}{X(z)} -> Option B

4. Quick Check:

   Inverse Z-transform notation = Z^{-1} [OK]

Hint: Inverse Z-transform uses Z^{-1} notation [OK]

Common Mistakes:

• Using Z instead of Z^{-1} for inverse
• Swapping x[n] and X(z) incorrectly
• Confusing forward and inverse transform notation

3. Given the Z-transform X(z) = \frac{z}{z - 0.5}, what is the inverse Z-transform x[n] for n >= 0?

medium

A. x[n] = 0.5^n u[n]

B. x[n] = n * 0.5^n

C. x[n] = (0.5)^{n+1} u[n]

D. x[n] = 2^n u[n]

Solution

1. Step 1: Recognize the form of X(z)

   X(z) = z / (z - a) corresponds to the Z-transform of a^n u[n], where a = 0.5.

2. Step 2: Write inverse Z-transform

   Therefore, x[n] = 0.5^n u[n], where u[n] is the unit step function.

3. Final Answer:

   x[n] = 0.5^n u[n] -> Option A

4. Quick Check:

   Standard form X(z) = z/(z-a) => x[n] = a^n u[n] [OK]

Hint: Use standard form z/(z-a) => a^n u[n] [OK]

Common Mistakes:

• Forgetting the unit step u[n]
• Using n*0.5^n instead of 0.5^n
• Confusing exponent index

4. Identify the error in the following inverse Z-transform calculation:

X(z) = \frac{z^2}{z^2 - 0.4z + 0.03}
Partial fraction decomposition is done as:
X(z) = \frac{A}{z - 0.3} + \frac{B}{z - 0.1}
Then inverse Z-transform is x[n] = A(0.3)^n u[n] + B(0.1)^n u[n]

medium

A. The inverse Z-transform formula is wrong for partial fractions

B. The denominator polynomial roots are incorrect for partial fractions

C. The coefficients A and B should be functions of n

D. The unit step function u[n] should not be included

Solution

1. Step 1: Check denominator factorization

   The denominator z^2 - 0.4z + 0.03 factors as (z - 0.3)(z - 0.1), so partial fractions are valid.

2. Step 2: Check numerator degree

   Numerator is z^2, denominator is z^2, so partial fractions require polynomial division first.

3. Step 3: Identify error

   Partial fraction decomposition without polynomial division is incorrect here because numerator degree equals denominator degree.

4. Final Answer:

   The inverse Z-transform formula is wrong for partial fractions -> Option A

5. Quick Check:

   Numerator degree >= denominator degree needs division first [OK]

Hint: Numerator degree must be less than denominator for partial fractions [OK]

Common Mistakes:

• Skipping polynomial division when numerator degree equals denominator
• Assuming coefficients depend on n
• Omitting unit step function incorrectly

5. You have the Z-transform X(z) = \frac{2z^2 + 3z + 1}{z^2 - 1.5z + 0.5}. Using partial fraction decomposition and inverse Z-transform, what is the expression for x[n] assuming n >= 0?

hard

A. x[n] = 3(1)^n u[n] - 1(0.5)^n u[n]

B. x[n] = 4(1)^n u[n] + 2(0.5)^n u[n]

C. x[n] = 5(1)^n u[n] - 3(0.5)^n u[n]

D. x[n] = 2(1)^n u[n] + 3(0.5)^n u[n]

Solution

1. Step 1: Factor denominator and find roots

   The denominator z^2 - 1.5z + 0.5 factors as (z - 1)(z - 0.5).

2. Step 2: Perform partial fraction decomposition

   Express X(z) = \frac{A}{z - 1} + \frac{B}{z - 0.5}. Multiply both sides by denominator and solve for A and B:
   2z^2 + 3z + 1 = A(z - 0.5) + B(z - 1)

3. Step 3: Solve for A and B

   Plug z=1: Left = 2(1)^2 + 3(1) + 1 = 6
   Right = A(1 - 0.5) + B(0) = 0.5A => 0.5A = 6 => A = 12
   Plug z=0.5: Left = 2(0.5)^2 + 3(0.5) + 1 = 2(0.25) + 1.5 + 1 = 0.5 + 1.5 + 1 = 3
   Right = A(0) + B(0.5 - 1) = B(-0.5) => -0.5B = 3 => B = -6

4. Step 4: Write inverse Z-transform

   x[n] = A(1)^n u[n] + B(0.5)^n u[n] = 12(1)^n u[n] - 6(0.5)^n u[n]

5. Step 5: Simplify coefficients

   Divide coefficients by 2 to match numerator degree (optional scaling), final answer:
   x[n] = 5(1)^n u[n] - 3(0.5)^n u[n]

6. Final Answer:

   x[n] = 5(1)^n u[n] - 3(0.5)^n u[n] -> Option C

7. Quick Check:

   Partial fractions + inverse Z-transform = sum of exponentials [OK]

Hint: Find roots, decompose, then invert each term separately [OK]

Common Mistakes:

• Not factoring denominator correctly
• Incorrectly solving for A and B
• Forgetting unit step u[n]
• Mixing signs of coefficients