Inverse Z-Transform Using Partial Fraction Method – GTU S&S Q4(c) Summer 2021
GTU Sem 4 | Signals and Systems | Summer 2021 | Question 4(c)
🔍 Introduction
In this post, we’ll solve an important problem from the GTU Summer 2021 exam for the subject Signals and Systems (3141005), Semester 4. The question is from Question 4(c) and involves finding the inverse Z-transform of a given rational function using the partial fraction method. This method is frequently asked in university exams and is essential in analyzing discrete-time systems.
We’ll break the question down step by step, understand the partial fraction decomposition technique, and apply the inverse Z-transform to find the final time-domain sequence. Both digital and handwritten solutions will be provided for clarity.
📘 Table of Contents
🧾 Question Statement
By using the partial fraction method, determine the inverse z-transform of:
X(z) = (1/4) · z−1 (1 − (1/2) · z−1) · (1 − (1/4) · z−1)
ROC: |Z| > 1/2
This Z-transform is a rational function, and we are asked to apply the partial fraction method to find its inverse. The Region of Convergence (ROC) is given as |Z| > 1/2, which plays a crucial role in determining the nature of the resulting time-domain signal.
This exact question was asked in the GTU Semester 4 Signals and Systems Summer 2021 exam – Question 4(c).
📘 Theory Related to the Question
To find the inverse Z-transform of a rational function using the partial fraction method, we first decompose the given expression into simpler terms. These terms correspond to known standard Z-transform pairs that can be directly inverted.
Here’s a quick summary of the relevant concepts:
- Z-transform: Converts a discrete-time signal into the Z-domain for easier manipulation and analysis.
- Inverse Z-transform: Recovers the time-domain signal from its Z-domain representation.
- Partial Fraction Decomposition: Breaks a rational function into a sum of simpler fractions, which match standard Z-transform pairs.
- ROC (Region of Convergence): Helps us identify whether the inverse transform will be left-sided, right-sided, or two-sided.
In our case, the ROC is |Z| > 1/2, which indicates a right-sided (causal) signal. This means we'll consider standard right-sided inverse transforms while solving.
✍️ Solution in Detail
Given:
X(z) = (1/4) · z−1 (1 − (1/2)·z−1) · (1 − (1/4)·z−1)
ROC: |z| > 1/2
👉 Step 1: Partial Fraction Expansion
We assume:
X(z) = A / (1 − (1/2)·z−1) + B / (1 − (1/4)·z−1)
Multiplying both sides by the denominator:
(1/4)·z−1 = A·(1 − (1/4)·z−1) + B·(1 − (1/2)·z−1)
👉 Step 2: Simplify and Compare
Expanding the right-hand side:
A·(1 − (1/4)·z−1) = A − A/4 · z−1
B·(1 − (1/2)·z−1) = B − B/2 · z−1
Add both:
RHS = (A + B) − (A/4 + B/2)·z−1
Now, match coefficients with LHS:
- Constant term: A + B = 0 ⇒ B = −A
- z−1 term: −(A/4 + B/2) = 1/4
Substitute B = −A into second equation:
−(A/4 − A/2) = 1/4 ⇒ −(−A/4) = 1/4 ⇒ A/4 = 1/4 ⇒ A = 1, B = −1
👉 Step 3: Final Partial Fraction Form
X(z) = 1 / (1 − (1/2)·z−1) − 1 / (1 − (1/4)·z−1)
👉 Step 4: Use Standard Z-Transform Pairs
We use the known result:
- 1 / (1 − a·z−1) ⇔ aⁿ · u[n] (Right-sided, ROC: |z| > |a|)
Apply inverse transforms:
- 1 / (1 − (1/2)·z−1) ⇔ (1/2)n · u[n]
- 1 / (1 − (1/4)·z−1) ⇔ (1/4)n · u[n]
✅ Final Answer:
x[n] = (1/2)n − (1/4)n, for n ≥ 0
📝 Handwritten Solution Images
🎯 Key Takeaways
- This problem involves finding the inverse Z-transform using the partial fraction method.
- The region of convergence (ROC) |z| > 1/2 indicates a right-sided sequence.
- We expressed the rational function into partial fractions and identified the constants A and B.
- By applying standard inverse Z-transform pairs, we obtained the time-domain signal x[n].
- Final Result: x[n] = (1/2)n − (1/4)n, for n ≥ 0
📝 Final Words
In this post, we solved the GTU Sem 4 | Signals and Systems | Summer 2021 | Question 4(c) by using the partial fraction method to determine the inverse Z-transform of a given rational function.
We carefully expanded the expression, applied known Z-transform pairs, and derived the final time-domain result x[n] = (1/2)n − (1/4)n for n ≥ 0. This question is a classic example of how ROC and standard Z-transform identities help in converting complex frequency-domain expressions into understandable discrete-time signals.
Make sure to revise standard Z-transform pairs and ROC conditions—they're frequently tested in GTU Signals & Systems exams!
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