Sunday, 22 February 2026

 

📘 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

No comments:

Post a Comment

  Operational Amplifiers – Complete Theory Page 15 – Active Low Pass Filter An Active Low Pass Filter allows low-frequency sig...