Digital control: sampling and z-transform

Why it matters

Almost every modern mechatronic controller — a motor drive, a 3-D printer axis, a drone flight controller — is code running on a microcontroller. It sees the plant only at sampling instants and drives it through a DAC or PWM that holds each value until the next update. Sampling adds delay and can destabilise a loop that is perfectly stable in continuous time, so you need the z-transform to analyse, tune and implement digital loops properly.

Key ideas

The sampled-data loop. Sensor → anti-aliasing filter → ADC (sampler, period T) → algorithm (difference equation) → DAC with zero-order hold (ZOH) → plant. The controller works on sequences x[k] = x(kT); the plant still runs in continuous time.

Sampling theorem and aliasing. A signal band-limited to f_max can be reconstructed from samples taken at f_s > 2·f_max (the Nyquist rate). Components above f_s/2 (the Nyquist frequency) fold back and appear as false low frequencies — aliasing — which no later processing can remove; hence the analogue anti-aliasing filter before the ADC. For control, the theorem is only a minimum: a common rule is to sample 10–30 times faster than the closed-loop bandwidth, because the ZOH adds roughly T/2 of delay (phase lag ωT/2) that erodes phase margin.

Zero-order hold. Holds each sample constant for one period: transfer function G_h(s) = (1 − e^(−sT))/s. The combination ZOH + plant has the exact pulse transfer function G(z) = (1 − z⁻¹)·Z{G_p(s)/s}.

z-transform. For a sequence x[k], k ≥ 0: X(z) = Σ x[k]·z^(−k). It is the discrete counterpart of the Laplace transform, related by z = e^(sT). Key pairs: δ[k] → 1; unit step → z/(z − 1); aᵏ → z/(z − a) for |z| > |a|; sampled e^(−at) → z/(z − e^(−aT)). Key property: x[k − 1] → z⁻¹X(z) (a one-sample delay), which turns difference equations into algebra.

Mapping s to z (z = e^(sT)).

• The left half s-plane maps inside the unit circle; the jω-axis maps onto the unit circle; the right half-plane maps outside.
• s = 0 maps to z = 1; poles move toward z = 0 as they get faster (e^(−aT) → 0).
• Stability: all closed-loop poles strictly inside the unit circle |z| < 1. A pole on the circle is marginal; outside is unstable.

Analysis tools. Final value theorem: lim x[k] = lim(z→1) (z − 1)·X(z) (for stable sequences). Stability tests: Jury's test, or the bilinear (w-) transformation followed by Routh. For first-order loops, simply require |pole| < 1.

Discretising a controller. Options: backward Euler s ≈ (1 − z⁻¹)/T; forward Euler s ≈ (z − 1)/T (can turn stable poles unstable); Tustin (bilinear) s ≈ (2/T)·(z − 1)/(z + 1), which maps the whole left half-plane inside the unit circle so stability is preserved, but warps frequencies (pre-warp at a critical frequency if needed). The digital PID in the PID topic is a backward-Euler discretisation.

Practical issues. Computation delay (output applied one sample late), quantisation of ADC and DAC (resolution limits and limit cycles), integer overflow and wind-up all appear in real implementations.

Formulas

X(z) = Σₖ₌₀^∞ x[k]·z^(−k) — z-transform.

z = e^(sT) — s-to-z mapping; T: sampling period (s).

f_s > 2·f_max — sampling theorem; f in Hz, f_s = 1/T.

G_h(s) = (1 − e^(−sT))/s — zero-order hold.

G(z) = (1 − z⁻¹)·Z{G_p(s)/s} — ZOH + plant.

For G_p(s) = K/(s + a): G(z) = (K/a)·(1 − e^(−aT))/(z − e^(−aT)).

lim(k→∞) x[k] = lim(z→1) (z − 1)·X(z) — final value theorem.

s ≈ (2/T)·(z − 1)/(z + 1) — Tustin (bilinear) approximation.

Worked examples

Example 1 (standard — discretising a plant). The plant G_p(s) = 1/(s + 2) is driven through a ZOH with T = 0.1 s. Find G(z), the difference equation, the first three samples of the unit-step response and the final value.

1. With K = 1, a = 2: e^(−aT) = e^(−0.2) = 0.8187.
2. G(z) = (1/2)(1 − 0.8187)/(z − 0.8187) = 0.0906/(z − 0.8187).
3. Difference equation: y[k + 1] = 0.8187·y[k] + 0.0906·u[k].
4. With y[0] = 0 and u[k] = 1: y[1] = 0.0906, y[2] = 0.8187 × 0.0906 + 0.0906 = 0.1648, y[3] = 0.2256.
5. Final value: G(1) × 1 = 0.0906/(1 − 0.8187) = 0.500 — equal to the continuous DC gain 1/2, as it must be.
6. The pole z = 0.8187 is inside the unit circle: stable, and it equals e^(−2 × 0.1), the image of s = −2.

Example 2 (GATE level — sampling limits the gain). A digital proportional controller of gain K, a ZOH with T = 0.1 s and the plant 1/(s + 1) form a unity-feedback loop. Find the range of K for stability. Compare with the continuous loop.

1. ZOH + plant: e^(−0.1) = 0.9048, so G(z) = 0.0952/(z − 0.9048).
2. Closed-loop characteristic equation 1 + K·G(z) = 0 → z − 0.9048 + 0.0952K = 0.
3. Closed-loop pole: z = 0.9048 − 0.0952K.
4. Stability: −1 < 0.9048 − 0.0952K < 1.
5. Upper limit (pole reaches −1): 0.0952K < 1.9048 → K < 20.0. Lower limit gives K > −1.0.
6. For K > 0: stable for 0 < K < 20.0. (Exactly K < (1 + e^(−T))/(1 − e^(−T)) = 20.02.)
7. The continuous loop K/(s + 1 + K) is stable for every K > 0. Sampling and the hold introduced delay, which created a finite gain limit — and it shrinks as T grows.

Common mistakes

• Using the left-half-plane test on z-plane poles. In z, stability means |z| < 1, not Re(z) < 0.
• Forgetting the ZOH and taking Z{G_p(s)} directly for a plant driven by a DAC.
• Sampling at just above twice the bandwidth and expecting good control; aim for 10–30 times.
• Applying the final value theorem with z → 0 instead of z → 1.
• Assuming a design that is stable in continuous time stays stable after discretisation by forward Euler.
• Confusing f in Hz with ω in rad/s in the sampling theorem.

For GATE ME

Expect: minimum sampling frequency or maximum sampling period for a given signal; z-transforms of simple sequences and evaluating X(z) at a point; pulse transfer function of ZOH + first-order plant; pole location and stability in the z-plane; range of gain for a first-order sampled loop; and final value of a sequence. Practise e^(−aT) evaluations and the s-to-z mapping rules.

Quick check

1. A signal's highest component is 500 Hz. What is the minimum sampling rate to avoid aliasing?
2. What is the z-transform of δ[k]?
3. For x[k] = {1, 2, 3, 4} (k = 0 to 3), what is X(2)?
4. Where must the poles of a stable discrete system lie?
5. A sampled loop's closed-loop pole is z = −1.2. Is it stable?

Answers: 1. More than 1000 Hz. 2. 1. 3. 1 + 1 + 0.75 + 0.5 = 3.25. 4. Strictly inside the unit circle. 5. No — |z| = 1.2 > 1.