Z-Transform

The Z-transform is the Laplace transform for discrete-time systems. Learn the definition and ROC, common pairs, the delay property z⁻¹, difference equations and transfer functions H(z), poles and zeros in the z-plane, stability (|p| < 1), frequency response on the unit circle, and mapping s to z (bilinear transform).

Digital filters and controllers are difference equations running on processors. The Z-transform turns them into polynomial ratios, revealing stability, frequency response and how to design them from analog prototypes.

An analogy

z⁻¹ is a one-sample memory cell: every z⁻¹ in a formula means 'use the value from one tick ago'.

From difference equations

Replace each delay by z⁻¹ and solve for Y(z)/X(z). The transfer function is a ratio of polynomials — FIR filters have only zeros; IIR filters have poles (feedback).

Stability

A causal system is stable if all poles lie inside the unit circle |z| < 1. Poles near the circle give sharp resonances and slow decay; poles outside grow without bound.

Frequency response

Evaluate H(z) on the unit circle z = e^(jω). Zeros near the circle create notches; poles near it create peaks. ω = π corresponds to f_s/2 (Nyquist).

Design mappings

Bilinear transform maps the entire analog frequency axis onto the unit circle (with pre-warping); impulse invariance matches impulse responses. Fixed-point implementations must manage coefficient quantisation and limit cycles.

Formulas

• Definition: X(z) = Σ x[n] z^(−n)
• Delay: x[n − k] ↔ z^(−k) X(z)
• First-order IIR: y[n] = a y[n−1] + b x[n] ↔ H(z) = b/(1 − a z⁻¹)
• Frequency response: H(e^(jω)), ω = 2πf/f_s
• Bilinear transform: s = (2/T)(1 − z⁻¹)/(1 + z⁻¹)

Worked example: Transfer function

Given: y[n] = 0.9y[n − 1] + 0.1x[n]

Find: H(z), pole, DC gain.

• H(z) = 0.1/(1 − 0.9z⁻¹)
• Pole at z = 0.9 (stable)
• DC (z = 1): 0.1/0.1 = 1

Answer: Stable low-pass with unity DC gain

Worked example: Cut-off frequency

Given: Same filter, f_s = 1 kHz

Find: −3 dB frequency.

• |H|² = 0.01/(1.81 − 1.8 cos ω) = 0.5 → cos ω = 0.9944
• ω = 0.1055 → f = 16.8 Hz

Answer: ≈ 16.8 Hz

Common mistakes

• Wrong: Using s-plane stability rules in z. — Right: In z, stable means inside the unit circle.
• Wrong: Ignoring coefficient quantisation. — Right: Poles near |z| = 1 are sensitive in fixed point.

Applications

• Digital filters: IIR biquads in audio and sensors.
• Digital control: Discrete PID and compensators.
• Sensor smoothing: Exponential moving averages.

Key takeaways

• Delays → z⁻¹; H(z) = B(z)/A(z).
• Stable if |poles| < 1.
• Frequency response on unit circle.
• Bilinear transform from analog designs.

Going further

• Pre-warping: Match critical frequencies in bilinear designs.
• Limit cycles: Fixed-point oscillations in IIR.
• Lattice structures: Robust filter implementations.