Inverse Z-Transform

The inverse Z-transform recovers a discrete-time sequence x[n] from its Z-domain representation X(z). While the Z-transform is straightforward to compute using summation formulas and standard pairs, the inverse requires more structured techniques. Mastering partial fraction expansion, power series, and the residue method is essential for GATE DSP problems involving system analysis and difference equations.

Core Concept Explanation

The inverse Z-transform recovers x[n] from X(z) given the ROC. The formal definition is the contour integral x[n] = (1/2πj) ∮ X(z) z^(n-1) dz, where the contour is a closed counterclockwise path within the ROC. In practice, three computational methods avoid direct evaluation of this complex integral.

The most widely used technique for rational X(z) is partial fraction expansion. The standard approach is to first compute X(z)/z (not X(z) directly), perform partial fraction decomposition, and then multiply the result by z. This technique produces terms of the form Az/(z-a), which directly match standard Z-transform pairs and can be inverted by inspection using the pair z/(z-a) ↔ aⁿ u[n] for ROC |z| > |a|.

The power series method (long division) expands X(z) as a series in powers of z⁻¹: X(z) = x[0] + x[1]z⁻¹ + x[2]z⁻² + .... The coefficients of z⁻ⁿ directly give the sequence values x[n]. This method is most convenient when only a few values of x[n] are needed, or when X(z) represents a finite impulse response (FIR) filter that terminates after a finite number of terms.

The residue method is the most rigorous approach, applying the Cauchy residue theorem to evaluate the contour integral. For a simple pole at z = pₖ, the residue of X(z)z^(n-1) is computed as (z - pₖ)·X(z)·z^(n-1) evaluated at z = pₖ. Then x[n] equals the sum of all residues at poles enclosed by the contour C. The ROC determines which poles are enclosed, connecting back to the fundamental role of the ROC in uniquely identifying x[n].

Mathematical Expression

For partial fractions with distinct poles p₁ and p₂, the decomposition of X(z)/z gives A/(z-p₁) + B/(z-p₂). The residues are found as A = (z-p₁)·(X(z)/z) evaluated at z = p₁, and similarly for B. After multiplying by z, the result is A·z/(z-p₁) + B·z/(z-p₂). Using the causal inverse pair z/(z-a) ↔ aⁿ u[n] (valid for ROC |z| > |a|), the final answer is x[n] = A·p₁ⁿ u[n] + B·p₂ⁿ u[n] for a causal ROC.

For repeated poles, if X(z)/z has a second-order pole at z = p, the partial fraction includes A/(z-p) + B/(z-p)². The residue for the repeated pole requires differentiation: B = d/dz [(z-p)² · X(z)/z] evaluated at z = p. Each such term introduces a ramp-like factor n·pⁿ in the time domain.

Practical Understanding

In digital filter implementation, the inverse Z-transform connects the transfer function H(z) to the impulse response h[n]. If H(z) is known from design specifications, applying the inverse Z-transform with the causal ROC gives the causal impulse response that can be directly programmed as a digital filter. This is the fundamental link between the Z-transform and the actual computations performed by a DSP processor.

Given:
X(z) = z² / [(z - 0.5)(z - 0.25)]
ROC: |z| > 0.5 (causal sequence)

Why this formula applies:
Two distinct simple poles at z = 0.5 and z = 0.25.
Use partial fraction method on X(z)/z.

Formula:
X(z)/z = z / [(z - 0.5)(z - 0.25)]
       = A/(z - 0.5) + B/(z - 0.25)

Substitution:
A = (z - 0.5) · X(z)/z at z = 0.5
  = 0.5 / (0.5 - 0.25) = 0.5 / 0.25 = 2

B = (z - 0.25) · X(z)/z at z = 0.25
  = 0.25 / (0.25 - 0.5) = 0.25 / (-0.25) = -1

Calculation:
X(z)/z = 2/(z - 0.5) - 1/(z - 0.25)
X(z)   = 2z/(z - 0.5) - z/(z - 0.25)

Using z/(z-a) ↔ aⁿ u[n] for ROC |z| > |a|:
x[n] = 2·(0.5)ⁿ·u[n] - (0.25)ⁿ·u[n]

Final Answer:
x[n] = [2·(0.5)ⁿ - (0.25)ⁿ] · u[n]

Exam Tip: Always perform partial fractions on X(z)/z, not on X(z) directly. This avoids messy remainder terms and ensures every term in the result has the standard form z/(z-a), which maps directly to known Z-transform pairs.

• Always decompose X(z)/z (not X(z)) to produce terms of the form constant/(z-pole).
• After finding residues, multiply entire result by z to get standard form Aₖz/(z-pₖ).
• Causal ROC gives aⁿu[n]; anti-causal ROC gives -aⁿu[-n-1] for the same pole.
• Power series (long division) gives x[n] values directly but is impractical for infinite sequences.
• Residue method is the most general but requires computing (z-pₖ)X(z)z^(n-1) at each pole.

Quick Revision

• Inverse Z-transform: x[n] = (1/2πj) ∮ X(z) z^(n-1) dz evaluated using residues.
• Partial fractions: always expand X(z)/z, then multiply by z to get standard pairs.
• Key pair: z/(z-a) ↔ aⁿu[n] for causal (ROC |z| > |a|).
• Power series: divide N(z) by D(z) in powers of z⁻¹; coefficients give x[n].
• Residue at simple pole p: (z-p)·X(z)·z^(n-1) evaluated at z = p.
• Trap: Do not apply causal pair to anti-causal ROC — gives the wrong x[n].
• Trap: For repeated poles, residue requires differentiation — do not use cover-up alone.