Digital PD controller design for plant 1/s²

Step 1: Plant with ZOH ($T = 0.1$ s)

$$G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^3}\right] = (1-z^{-1})\frac{T^2z(z+1)}{2(z-1)^3} = \frac{T^2(z+1)}{2(z-1)^2} = \frac{0.005(z+1)}{(z-1)^2}$$

Step 2: Desired closed-loop poles

With $\zeta = 0.5$ and $\omega_n = 4$:

$$\begin{aligned} \zeta\omega_n &= 2, \qquad \omega_d = 4\sqrt{0.75} = 3.464\ \text{rad/s} \\ |z| &= e^{-\zeta\omega_nT} = e^{-0.2} = 0.8187 \\ \angle z &= \omega_dT = 0.3464\ \text{rad} = 19.85^\circ \\ z_1 &= 0.7701 + j0.2780 \end{aligned}$$

Step 3: Digital PD controller

$$G_D(z) = K_P + K_D(1-z^{-1}) = (K_P+K_D)\frac{z - \alpha}{z} = K\frac{z-\alpha}{z}, \qquad \alpha = \frac{K_D}{K_P+K_D}$$

Step 4: Angle condition at $z_1$

For a total of $-180^\circ$, the controller zero must give $+90.11^\circ$:

$$\alpha = 0.7701 - \frac{0.2780}{\tan 90.11^\circ} = 0.7706$$

The zero lies almost directly below the desired pole.

Step 5: Magnitude condition

$$K = \frac{|z_1|\,|z_1-1|^2}{0.005\,|z_1+1|\,|z_1-0.7706|} = \frac{0.8187 \times 0.3607^2}{0.005 \times 1.7918 \times 0.2780} = 42.78$$ $$G_D(z) = 42.78\,\frac{z - 0.7706}{z} = 42.78\,(1 - 0.7706z^{-1})$$

Step 6: PD gains

$$K_P + K_D = 42.78, \qquad K_D = K\alpha = 32.97, \qquad K_P = K(1-\alpha) = 9.81$$

Step 7: Check

$$1 + G_DG = 0 \;\Rightarrow\; z^3 - 1.7861z^2 + 1.0491z - 0.1648 = 0$$

The roots are $z = 0.7701 \pm j0.2780$ (the desired pair) and $z = 0.2459$. The third pole is much closer to the origin, so it decays fast, and the complex pair is dominant. ✔

Answer: $G_D(z) = 42.78\dfrac{z-0.7706}{z}$, i.e. $K_P = 9.81$ and $K_D = 32.97$. The difference equation is $u(k) = 42.78e(k) - 32.97e(k-1)$.