📘 Inverse Z-Transform – Partial Fraction Method
Inverse Z-transform converts frequency domain expression back to time domain sequence. Most problems are solved using partial fraction expansion.
🔹 1. General Approach
If:X(z) = Rational Function
Steps: 1. Express in z⁻¹ form 2. Do partial fraction expansion 3. Compare with standard pairs 4. Apply ROC to decide signal type🔹 2. Example 1 (Simple Case)
Given:X(z) = 1 / (1 − a z⁻¹)
We know directly:x[n] = aⁿ u[n]
If ROC: |z| > |a|🔹 3. Example 2 (Two Poles)
Given:X(z) = 1 / [(1 − 0.5z⁻¹)(1 − 0.2z⁻¹)]
Step 1: Partial Fraction
Assume:X(z) = A/(1 − 0.5z⁻¹) + B/(1 − 0.2z⁻¹)
Solve for A and B. Result:A = 2 B = -1
Step 2: Apply Inverse
x[n] = 2(0.5)ⁿ u[n] − (0.2)ⁿ u[n]
🔹 4. Example 3 (Repeated Pole)
Given:X(z) = 1 / (1 − az⁻¹)²
Standard result:x[n] = n aⁿ u[n]
Repeated poles introduce multiplication by n.🔹 5. Important ROC Cases
If ROC is: |z| > |a| → Right sided (causal) |z| < |a| → Left sided Between poles → Two sided signal🔹 6. Shortcut Table
1/(1 − az⁻¹) ↔ aⁿ u[n]
z⁻¹/(1 − az⁻¹) ↔ aⁿ⁻¹ u[n−1]
1/(1 − az⁻¹)² ↔ n aⁿ u[n]
🎯 GATE Important Tips
- Always check ROC before writing answer
- Repeated poles → n term appears
- Stability depends on pole location
- Unit circle inside ROC → Stable system
Inverse Z = Algebra + ROC Logic
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