Digital PI design, plant e^(−2s)/(s+2)

A digital PI controller $G_D(z) = K_p + \dfrac{K_i}{1-z^{-1}} = K\dfrac{z-a}{z-1}$ is designed with the root-locus angle and magnitude conditions. The sampling period is not given; $T = 1$ s is assumed. It makes the 2 s dead time exactly two samples and allows a feasible PI design.

Plant pulse transfer function

With $e^{-2T} = e^{-2} = 0.1353$:

$$\begin{aligned} G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+2)}\right] = z^{-2}\,\frac{(1-e^{-2})/2}{z-e^{-2}}\\ &= \frac{0.4323}{z^2(z-0.1353)} \end{aligned}$$

Desired dominant poles

10 samples per cycle gives $\omega_dT = 2\pi/10 = 36^\circ$:

$$\begin{aligned} |z| &= \exp\left(-\frac{\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{2\pi}{10}\right) = \exp\left(-\frac{0.55\times0.6283}{0.8352}\right) = 0.6611\\ z_1 &= 0.6611\angle36^\circ = 0.5349 + j0.3886 \end{aligned}$$

Angle condition

The PI zero must contribute $256.33^\circ - 180^\circ = 76.33^\circ$:

$$a = 0.5349 - \frac{0.3886}{\tan 76.33^\circ} = 0.4403$$

Magnitude condition

$$K = \left|\frac{z_1^2(z_1-0.1353)(z_1-1)}{0.4323(z_1-0.4403)}\right| = 0.8540$$ $$G_D(z) = 0.8540\,\frac{z-0.4403}{z-1} = 0.3760 + \frac{0.4780}{1-z^{-1}}$$

So $K_p = Ka = 0.3760$ and $K_i = K(1-a) = 0.4780$.

Check

Characteristic equation: $z^4 - 1.1353z^3 + 0.1353z^2 + 0.3692z - 0.1626 = 0$, with roots $0.5349 \pm j0.3886$ (designed), $0.6435$ and $-0.5780$. Step response: $c(k) = 0, 0, 0, 0.369, 0.626, 0.867, 0.970, 1.020, 1.015, 1.003, \ldots \to 1$. The peak is about 2% and there is zero steady-state error.

Velocity error constant

$$\begin{aligned} K_v &= \frac{1}{T}\lim_{z\to1}(1-z^{-1})G_D(z)G(z) = \frac{K(1-a)\times0.4323}{T(1-0.1353)}\\ &= \frac{0.4780\times0.4323}{0.8647} = 0.239\ \text{s}^{-1} \end{aligned}$$

The steady-state error to a unit ramp is $1/K_v = 4.18$.

Answer: $G_D(z) = 0.8540\,\dfrac{z-0.4403}{z-1}$ ($K_p = 0.376$, $K_i = 0.478$, with T = 1 s); $K_v = 0.239\ \text{s}^{-1}$.