Controller design with 8 samples per cycle, step response and Kv

Step 1: Plant pulse transfer function ($T = 1$ s)

$$G(z) = (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+1)}\right] = \frac{0.3679z + 0.2642}{(z-1)(z-0.3679)} = \frac{0.3679(z + 0.7183)}{(z-1)(z-0.3679)}$$

Step 2: Desired poles

With 8 samples per cycle, $\omega_dT = 2\pi/8 = 45^\circ$, and

$$\begin{aligned} |z| &= \exp\left(-\frac{2\pi\zeta}{\sqrt{1-\zeta^2}}\cdot\frac{1}{8}\right) = e^{-0.4534} = 0.6354 \\ z_{1,2} &= 0.6354\angle \pm 45^\circ = 0.4493 \pm j0.4493 \end{aligned}$$

Step 3: Angle condition

The controller must add $+19.47^\circ$, so a lead compensator is needed.

Step 4: Lead controller $G_D(z) = K\dfrac{z-0.3679}{z-\beta}$

The zero cancels the plant pole at 0.3679 and adds $+79.73^\circ$. The controller pole must contribute $79.73^\circ - 19.47^\circ = 60.26^\circ$:

$$\beta = 0.4493 - \frac{0.4493}{\tan 60.26^\circ} = 0.1926$$

Step 5: Gain

$$K = \frac{|z_1-1|\,|z_1-0.1926|}{0.3679\,|z_1+0.7183|} = \frac{0.7107 \times 0.5175}{0.3679 \times 1.2511} = 0.7991$$ $$G_D(z) = 0.7991\,\frac{z - 0.3679}{z - 0.1926}$$

Step 6: Closed-loop pulse transfer function

$$\begin{aligned} G_DG &= \frac{0.2940(z + 0.7183)}{(z-1)(z-0.1926)} \\ \frac{C(z)}{R(z)} &= \frac{0.2940z + 0.2112}{z^2 - 0.8986z + 0.4038} \end{aligned}$$

The poles are $0.4493 \pm j0.4493$, as desired. ✔

Step 7: Unit step response

With $R(z) = z/(z-1)$, the difference equation is $c(k) = 0.8986c(k-1) - 0.4038c(k-2) + 0.2940r(k-1) + 0.2112r(k-2)$, where $r(k) = 1$:

The maximum overshoot is about 16.3% at $k = 4$, which matches $\zeta = 0.5$. One damped cycle takes about 8 samples, and $c(\infty) = 1$ (type-1 system).

Step 8: Static velocity error constant

$$K_v = \lim_{z\to1}\frac{(1-z^{-1})G_DG}{T} = \frac{0.2940 \times 1.7183}{1 \times (1 - 0.1926)} = 0.626\ \text{s}^{-1}$$

Answer: $G_D(z) = 0.7991\dfrac{z-0.3679}{z-0.1926}$. The step response overshoots by 16.3% and settles to 1, and $K_v = 0.626\ \text{s}^{-1}$.