Digital PI design, plant e^(−5s)/(s+0.4), step response

A digital PI controller $G_D(z)=K_p+\dfrac{K_i}{1-z^{-1}}=K\dfrac{z-a}{z-1}$ is designed by the root-locus (angle and magnitude) method in the z-plane. The plant is $G_p(s)=\dfrac{e^{-5s}}{s+0.4}$ (time constant 2.5 s, dead time 5 s).

Choice of sampling period

• The dead time should be a whole number of sampling periods, so $T = 5/N$.
• Ten samples per cycle fixes the angle of the desired pole at $\omega_d T = 2\pi/10 = 36^\circ$, whatever T is. Each sample of delay ($z^{-1}$) then adds $36^\circ$ of phase lag at that pole.
• With $T = 1$ s ($N = 5$) the delay alone gives $5\times36^\circ = 180^\circ$. The poles at $z=1$ and $z=e^{-0.4}$ add more than $180^\circ$ again, and one PI zero cannot make up the difference, so no PI controller exists.
• Choose $T = 2.5$ s: this equals the plant time constant, the dead time is $2T$, and the angle condition can be met.

Pulse transfer function of the plant (with ZOH)

With $e^{-0.4T}=e^{-1}=0.3679$:

$$\begin{aligned} G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+0.4)}\right] = z^{-2}\,\frac{(1-e^{-1})/0.4}{z-e^{-1}}\\ &= \frac{1.5803}{z^{2}(z-0.3679)} \end{aligned}$$

Desired closed-loop poles

$$\begin{aligned} |z| &= \exp\left(-\frac{2\pi\zeta}{10\sqrt{1-\zeta^2}}\right) = \exp\left(-\frac{0.6283\times0.5}{0.8660}\right) = 0.6958\\ z_1 &= 0.6958\angle 36^\circ = 0.5629 + j0.4090 \end{aligned}$$

(This gives $\omega_d = 0.2513$ rad/s and $\omega_n = 0.2902$ rad/s.)

Angle condition

Open loop: $G_D(z)G(z) = \dfrac{1.5803K(z-a)}{z^2(z-0.3679)(z-1)}$. Pole angles at $z_1$:

The zero must give $273.4^\circ - 180^\circ = 93.4^\circ$:

$$a = 0.5629 - \frac{0.4090}{\tan 93.4^\circ} = 0.5873$$

Magnitude condition

$$K = \left|\frac{z_1^2(z_1-0.3679)(z_1-1)}{1.5803(z_1-0.5873)}\right| = 0.2028$$ $$G_D(z) = \frac{0.2028(z-0.5873)}{z-1} = 0.1191 + \frac{0.0837}{1-z^{-1}}$$

So $K_p = Ka = 0.1191$ and $K_i = K(1-a) = 0.0837$.

Unit-step response

$$\frac{C(z)}{R(z)} = \frac{0.3204(z-0.5873)}{z^4-1.3679z^3+0.3679z^2+0.3204z-0.1882}$$

Closed-loop poles: $0.5629\pm j0.4090$ (designed), $0.7562$ and $-0.5141$. Solving the difference equation with $r(k)=1$:

• Output is zero for 3 samples (2 for the dead time, 1 for ZOH plus computation).
• The extra real pole at 0.7562 slows the final approach, so the response rises to about 0.96, dips slightly and creeps up to 1 with almost no overshoot.
• Because of the integral action, steady-state error to a step is zero: $c(\infty)=1$. $K_v = K(1-a)(1.5803)/[T(1-0.3679)] = 0.0837\ \text{s}^{-1}$.

Answer: $T = 2.5$ s, $G_D(z) = 0.2028(z-0.5873)/(z-1)$ ($K_p = 0.1191$, $K_i = 0.0837$); the step response settles at 1 with zero steady-state error.