Inverse Z-transform in Signal Processing - Step-by-Step Execution

Concept Flow - Inverse Z-transform

Start with Z-transform X(z)

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Choose method: Partial Fraction / Power Series / Contour Integral

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Apply chosen method to X(z)

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Calculate coefficients x[n

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Obtain time-domain sequence x[n

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End

The inverse Z-transform converts a function from the Z-domain back to the time domain by applying methods like partial fractions or power series to find sequence coefficients.

Execution Sample

Signal Processing

X_z = "(z^2 + 0.5*z) / (z^2 - 0.8*z + 0.15)"
# Find x[n] using partial fraction expansion
# x[0], x[1], x[2], ...

This code represents a Z-transform function and outlines the process to find its inverse sequence x[n].

Execution Table

Step	Action	Expression	Result/Value
1	Start with given X(z)	X(z) = (z^2 + 0.5z) / (z^2 - 0.8z + 0.15)	Given function
2	Factor denominator	z^2 - 0.8z + 0.15	(z - 0.5)(z - 0.3)
3	Set partial fractions	X(z)/z = A/(z - 0.5) + B/(z - 0.3)	Unknown A, B
4	Multiply both sides by denominator	(z + 0.5) = A(z - 0.3) + B(z - 0.5)	Equation for A, B
5	Plug z=0.5 to solve for A	z=0.5: 0.5 + 0.5 = A(0.5 - 0.3) + B(0.5 - 0.5)	1.0 = 0.2 A => A = 5
6	Plug z=0.3 to solve for B	z=0.3: 0.3 + 0.5 = A(0.3 - 0.3) + B(0.3 - 0.5)	0.8 = B*(-0.2) => B = -4
7	Calculate A and B	A = 5, B = -4	Coefficients found
8	Write inverse Z-transform terms	x[n] = 5*(0.5)^n u[n] - 4*(0.3)^n u[n]	Time domain sequence
9	Calculate first values	x[0] = 5 - 4 = 1.0	x[0] = 1.0
10	Calculate x[1]	5*0.5 - 4*0.3 = 2.5 - 1.2	x[1] = 1.3
11	Calculate x[2]	5*0.5^2 - 4*0.3^2 = 1.25 - 0.36	x[2] = 0.89
12	End	Sequence x[n] found	x[n] = 5*(0.5)^n - 4*(0.3)^n for n>=0

💡 Partial fraction coefficients found and sequence x[n] computed for n>=0

Variable Tracker

Variable	Start	After Step 2	After Step 7	Final
X(z)	(z^2 + 0.5z) / (z^2 - 0.8z + 0.15)	(z^2 + 0.5z) / ((z - 0.5)(z - 0.3))	5/(z - 0.5) - 4/(z - 0.3)	5 z/(z - 0.5) - 4 z/(z - 0.3)
A	unknown	unknown	5	5
B	unknown	unknown	-4	-4
x[n]	unknown	unknown	unknown	5*(0.5)^n - 4*(0.3)^n for n>=0

Key Moments - 3 Insights

Why do we factor the denominator before partial fraction expansion?

Factoring the denominator (Step 2) reveals the poles which are needed to set up the partial fractions correctly (Step 3). Without factoring, we cannot write the terms for A and B.

How do we find coefficients A and B in partial fractions?

We multiply both sides by the denominator (Step 4) and then substitute values of z equal to the poles (Steps 5 and 6) to solve for A and B. This isolates each coefficient.

Why does the inverse Z-transform sequence involve terms like (0.5)^n and (0.3)^n?

These terms come from the poles found in the denominator factors (Step 2). Each pole corresponds to a geometric sequence term in time domain (Step 8).

Visual Quiz - 3 Questions

Look at the execution table, what is the value of coefficient A after Step 7?

A -2.5

B 0.5

C 5

D Unknown

💡 Hint

Check the 'Calculate A and B' row in the execution_table.

At which step do we substitute z=0.3 to solve for coefficient B?

A Step 4

B Step 6

C Step 5

D Step 7

💡 Hint

Look for the step mentioning plugging z=0.3 in the execution_table.

If the denominator had a repeated root, how would the partial fraction setup change?

A Add terms with powers of (z - root) in denominator

B Use only one term per root

C No partial fractions needed

D Change numerator only

💡 Hint

Recall partial fraction rules for repeated roots in the concept_flow.

Concept Snapshot

Inverse Z-transform converts X(z) back to x[n].
Common method: partial fraction expansion.
Steps: factor denominator, set partial fractions, solve coefficients.
Result: sum of terms A*(pole)^n u[n].
Each pole corresponds to geometric sequence in time domain.

Full Transcript

The inverse Z-transform process starts with a given Z-transform function X(z). We factor the denominator to find poles, then express X(z) as partial fractions with unknown coefficients. By multiplying both sides by the denominator and substituting the poles, we solve for these coefficients. Finally, the inverse transform is a sum of geometric sequences based on these poles and coefficients, giving the time-domain sequence x[n]. This step-by-step method helps convert from frequency domain back to time domain signals.

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