z-plane design with Kv, plant 1/(s(s+2))

Step 1: Desired closed-loop poles

With $\zeta = 0.5$ and $t_s = 4/(\zeta\omega_n) = 2$ s:

$$\zeta\omega_n = 2, \quad \omega_n = 4, \quad \omega_d = 3.464\ \text{rad/s}$$

With $T = 0.2$ s:

$$\begin{aligned} |z| &= e^{-0.4} = 0.6703, \qquad \angle z = \omega_dT = 39.69^\circ \\ z_{1,2} &= 0.5158 \pm j0.4281 \end{aligned}$$

Step 2: ZOH + plant $1/[s(s+2)]$

$$G(z) = \frac{(2T-1+e^{-2T})z + (1-e^{-2T}-2Te^{-2T})}{4(z-1)(z-e^{-2T})} = \frac{0.01758(z + 0.8753)}{(z-1)(z-0.6703)}$$

Step 3: Angle deficiency

At $z_1$: the zero gives $+17.11^\circ$, the pole at 1 gives $-138.52^\circ$, and the pole at 0.6703 gives $-109.85^\circ$. The total is $-231.26^\circ$. The controller must supply $+51.26^\circ$ (lead).

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

Choose $\alpha = 0.6703$ to cancel the plant pole. This contributes $+109.85^\circ$. The controller pole must then contribute $109.85^\circ - 51.26^\circ = 58.59^\circ$:

$$\beta = 0.5158 - \frac{0.4281}{\tan 58.59^\circ} = 0.2543$$

Step 5: Gain

$$K = \frac{|z_1-1|\,|z_1-0.2543|}{0.01758\,|z_1+0.8753|} = \frac{0.6464 \times 0.5017}{0.01758 \times 1.4555} = 12.67$$ $$G_D(z) = 12.67\,\frac{z - 0.6703}{z - 0.2543}$$

Step 6: Check

$$G_DG = \frac{0.2228(z + 0.8753)}{(z-1)(z-0.2543)}$$

The characteristic equation $z^2 - 1.0316z + 0.4493 = 0$ gives $z = 0.5158 \pm j0.4281$. ✔

Step 7: Static velocity error constant

$$\begin{aligned} K_v &= \lim_{z\to1}\frac{(1-z^{-1})G_D(z)G(z)}{T} = \frac{1}{0.2}\cdot\frac{0.2228(1 + 0.8753)}{1 - 0.2543} \\ &= 5 \times \frac{0.4178}{0.7457} = 2.80\ \text{s}^{-1} \end{aligned}$$

Answer: $G_D(z) = 12.67\dfrac{z-0.6703}{z-0.2543}$ and $K_v = 2.80\ \text{s}^{-1}$. The steady-state error for a unit ramp is $1/K_v = 0.357$.