Sampling and digital control

Software reads measurements and changes actions at discrete instants; the plant continues evolving between them. We use the exact update from Dynamic models to separate sampling, hold, computation delay and numerical integration.

One sample, one held action

At $t_k=k{T}_s$, measure $y_k$, calculate $u_k$, and hold it until the next update: a zero-order hold. Initially assume no additional computation or transport delay. $T_s$ is the physical control period, not screen refresh time.

For $\dot{x}=-x+u$, the exact held-input model is

$$x_{k+1}=a{x}_k+b{u}_k,a=e^{-T_s},b=1-a.$$

The current action applies during $[k{T}_s,(k+1)T_s)$ to produce the next state. Do not use a future state to calculate the current action. This deviation model allows positive and negative actions without saturation.

Continuous stability need not survive sampling

Continuous feedback $u=-Kx$ gives $\dot{x}=-(1+K)x$, stable for $K>0$. Sampled feedback gives

$$x_{k+1}=p_d{x}_k,p_d=a-(1-a)K.$$

The state evolves as $p_d^k$, so asymptotic stability requires $|p_d|<1$. For $K=2$, $p_d=3e^{-T_s}-2$, yielding

$$0<T_s<\ln 3\approx 1.099\mathrm{s}.$$

At 0.2 s, the multiplier is approximately 0.456: monotone decay. At 1 s it is −0.896: alternating decay. At 1.2 s it is −1.096: alternating growth. This is a genuine hold-and-sample effect, present even with exact plant integration.

Start at $x_0=1$. Change the period and compare sample points with the continuous intersample response over the same 12 s window. Curves use the exact constant-input solution, not straight interpolation between samples.

A one-sample delay adds memory

If this period executes the previous action, $u_k=-K{x}_{k-1}$, then

$$x_{k+1}=a{x}_k-bK{x}_{k-1},z^2-az+bK=0.$$

An old state or pending action must be stored. For $K=2,T_s=0.5$, the delay-free pole is −0.180; delayed conjugate roots have magnitude $\sqrt{2(1-e^{-0.5})}\approx 0.887$. At $T_s=0.8$, the delay-free pole is −0.652 but delayed roots have magnitude 1.049 and are unstable. Frequent measurements do not imply immediate actuation.

For discrete LTI systems, all closed-loop eigenvalues must lie strictly inside the unit circle. The map $z=e^{s{T}_s}$ describes exact sampling of a continuous free mode. A sampled-feedback loop generally cannot be obtained by directly mapping continuous closed-loop poles because the action is held between samples.

Integration step is different

Euler integration of continuous $\dot{x}=-3x$ gives $x_{j+1}=(1-3h)x_j$. With numerical step $h=1$, the multiplier −2 diverges even though the exact solution $x=e^{-3t}{x}_0$ decays. This is a numerical artifact.

Keep control period and delays fixed while reducing integration step to check convergence. Exact zero-order-hold discretization removes this integration error for the stated linear plant; it does not remove measurement aliasing or computation delay.

Implement discrete PID consistently

For example, forward integration is $z_{k+1}=z_k+K_i{T}_s{e}_k$. A held-measurement filter can use $y_{f,k+1}=\alpha y_{f,k}+(1-\alpha )y_{m,k}$, $\alpha =e^{-T_s/{\tau }_f}$. Specify whether current action uses old or updated states.

An anti-alias filter acts before sampling; smoothing afterward cannot recover already aliased information. Filtering also adds dynamics and phase cost. Choose periods using required bandwidth, computation/communication budgets and relevant plant modes, then test worst-case delays and jitter. No single sampling multiplier is universal.

Check your understanding

At $K=2,T_s=\ln 3$, is multiplier −1 asymptotically stable?

Should reducing browser frame rate change model poles or sample period?

Further reading

python-control sample_system offers several discretization methods. Match the actual hold mechanism before designing the discrete loop.