Z-transform and inverse z-transform

Why it matters

The z-transform does for sequences and difference equations what the Laplace transform does for continuous signals. Digital filters in a DAQ card, digital controllers in a PLC and recursive smoothing of sensor data are all designed and checked for stability with H(z), its poles and zeros and its region of convergence (ROC).

Key ideas

Definition. The z-transform weights a sequence by z^(−n), where z = r·e^(jω) is a complex variable. On the unit circle (r = 1) it becomes the DTFT. The extra factor r^(−n) lets the transform exist for sequences, such as 2ⁿ u[n], whose DTFT does not.

Region of convergence. The ROC depends only on |z|, so it is a ring (annulus) centred on the origin, possibly extending to 0 or to ∞. It never contains a pole. The ROC must be stated — the same X(z) can come from different sequences.

• Right-sided (e.g. causal) sequence: ROC is outside the outermost pole, |z| > rmax (a causal sequence's ROC also includes z = ∞).
• Left-sided sequence: ROC is inside the innermost pole, |z| < rmin.
• Two-sided sequence: ROC is a ring between two poles.
• Finite-length sequence: the whole z-plane, except possibly z = 0 and/or z = ∞.

Causality and stability of an LTI system with H(z)

• Causal ⇔ ROC is the exterior of a circle and includes z = ∞.
• BIBO stable ⇔ ROC includes the unit circle |z| = 1.
• Causal and stable ⇔ all poles strictly inside the unit circle.

Pole position and time behaviour (causal): a real pole at z = p gives pⁿ; |p| < 1 decays, |p| > 1 grows, p negative alternates in sign. A complex pair r·e^(±jθ) gives rⁿ cos(θn + φ). Poles near the unit circle give long, lightly damped responses.

Inverse z-transform methods

• Partial fractions (most common): expand X(z)/z in powers of z, or X(z) in powers of z⁻¹, then use standard pairs, choosing right- or left-sided terms from the ROC.
• Power-series (long) division: gives the first few samples directly — useful for checking.
• Residues/contour integration (rarely needed in exams).

Key properties. Linearity; time shift x[n − k] ↔ z^(−k) X(z); scaling aⁿ x[n] ↔ X(z/a); convolution x * h ↔ X(z)H(z); differentiation n·x[n] ↔ −z dX/dz; time reversal x[−n] ↔ X(1/z).

Initial and final values (causal sequences): x[0] = lim(z→∞) X(z); x[∞] = lim(z→1) (z − 1) X(z), valid only when the poles of (z − 1)X(z) are inside the unit circle.

Formulas

• X(z) = Σ x[n] z^(−n) (n from −∞ to ∞) — bilateral z-transform.
• aⁿ u[n] ↔ 1/(1 − a z⁻¹) = z/(z − a), ROC |z| > |a|.
• −aⁿ u[−n − 1] ↔ 1/(1 − a z⁻¹), ROC |z| < |a|.
• δ[n] ↔ 1 (all z); δ[n − k] ↔ z^(−k); u[n] ↔ z/(z − 1), |z| > 1.
• n aⁿ u[n] ↔ a z⁻¹/(1 − a z⁻¹)², |z| > |a|.
• rⁿ cos(ω0n) u[n] ↔ (1 − r cos ω0 z⁻¹)/(1 − 2r cos ω0 z⁻¹ + r² z⁻²), |z| > r.
• x[n − k] ↔ z^(−k) X(z); x * h ↔ X(z) H(z).
• x[0] = lim(z→∞) X(z); x[∞] = lim(z→1) (z − 1) X(z).

Symbols: z complex variable (dimensionless); r = |z|; ω digital frequency (rad/sample); a, p pole values; k integer delay in samples; n sample index.

Worked examples

Example 1 (standard). Find the causal x[n] with X(z) = z/((z − 0.5)(z − 0.25)), and check the first samples by long division.

1. Expand X(z)/z = 1/((z − 0.5)(z − 0.25)) = A/(z − 0.5) + B/(z − 0.25).
2. A = 1/(0.5 − 0.25) = 4; B = 1/(0.25 − 0.5) = −4.
3. X(z) = 4z/(z − 0.5) − 4z/(z − 0.25), causal ⇒ ROC |z| > 0.5.
4. x[n] = 4[(0.5)ⁿ − (0.25)ⁿ] u[n]: x[0] = 0, x[1] = 4(0.5 − 0.25) = 1, x[2] = 4(0.25 − 0.0625) = 0.75.
5. Long division: X(z) = z/(z² − 0.75z + 0.125) = z⁻¹ + 0.75 z⁻² + … ✓. Initial-value check: lim(z→∞) X(z) = 0 = x[0] ✓.
6. x[n] = 4[(0.5)ⁿ − (0.25)ⁿ] u[n].

Example 2 (GATE level). H(z) = 1/((1 − 0.5z⁻¹)(1 − 2z⁻¹)). Which ROC gives a stable system? Find h[n] for it and say whether it is causal.

1. Poles at z = 0.5 and z = 2. Possible ROCs: |z| > 2, |z| < 0.5, 0.5 < |z| < 2.
2. Only 0.5 < |z| < 2 contains |z| = 1 ⇒ stable; it is a ring, so the system is non-causal (two-sided).
3. Partial fractions in z⁻¹: H = A/(1 − 0.5z⁻¹) + B/(1 − 2z⁻¹); A = 1/(1 − 2/0.5) = −1/3; B = 1/(1 − 0.5/2) = 4/3.
4. Pole 0.5 is inside the ring ⇒ right-sided: −(1/3)(0.5)ⁿ u[n].
5. Pole 2 is outside the ring ⇒ left-sided: (4/3) × [−2ⁿ u[−n − 1]].
6. h[n] = −(1/3)(0.5)ⁿ u[n] − (4/3)·2ⁿ u[−n − 1] — both parts decay away from n = 0, so Σ|h| is finite.

Common mistakes

• Writing X(z) without the ROC, or assuming every ROC is |z| > something.
• Doing partial fractions on X(z) in powers of z without first dividing by z — the result then lacks the z/(z − a) form.
• Forgetting the minus sign in the left-sided pair.
• Using "poles inside the unit circle" as the stability test for a non-causal system; the general test is "ROC contains the unit circle".
• Missing a pole-zero cancellation — (z − 1)/((z − 1)(z − 2)) has only one pole, at z = 2.
• Applying the final-value theorem when a pole is on or outside the unit circle (other than a single pole at z = 1).

For GATE IN

Expect: choosing the ROC of a given sequence (including two-sided sums), deciding causality and stability from poles plus ROC, inverse transforms by partial fractions (often asking for one sample value such as h[2]), initial- and final-value theorem questions, and transfer functions from difference equations. Practise the three-ROC case and long division to check a sample.

Quick check

1. ROC of x[n] = (0.8)ⁿ u[n] + (−1.2)ⁿ u[n]?
2. Z-transform of δ[n − 3]?
3. A causal system has poles at 0.9 and −1.1. Stable?
4. Causal X(z) = 1/((1 − 0.5z⁻¹)(1 − 0.25z⁻¹)) — value of x[2]?
5. What does z^(−1) represent in a block diagram? Answers: 1. |z| > 1.2. 2. z⁻³, ROC z ≠ 0. 3. No — the pole at −1.1 lies outside the unit circle. 4. x[n] = 2(0.5)ⁿ − (0.25)ⁿ, so x[2] = 0.5 − 0.0625 = 0.4375. 5. A one-sample delay.