Controller with integral action, plant 10/((s+1)(s+5))

Step 1: Plant with ZOH ($T = 0.2$ s)

$$\begin{aligned} \frac{10}{s(s+1)(s+5)} &= \frac{2}{s} - \frac{2.5}{s+1} + \frac{0.5}{s+5} \\ G(z) &= (1-z^{-1})\left[\frac{2z}{z-1} - \frac{2.5z}{z-e^{-0.2}} + \frac{0.5z}{z-e^{-1}}\right] \\ &= \frac{0.1371z + 0.0921}{(z-0.8187)(z-0.3679)} = \frac{0.1371(z + 0.6714)}{(z-0.8187)(z-0.3679)} \end{aligned}$$

Step 2: Desired poles

With $\zeta = 0.5$ and 8 samples per cycle:

$$|z| = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac18\right) = 0.6354, \quad \angle z = 45^\circ, \quad z_1 = 0.4493 + j0.4493$$

Step 3: Try PI alone

Integral action requires a controller pole at $z = 1$. With a PI controller $K(z-0.8187)/(z-1)$ cancelling the slow plant pole, the angle at $z_1$ is:

$$\underbrace{+21.85^\circ}_{\text{zero } -0.6714} \underbrace{-140.79^\circ}_{\text{pole } 1} \underbrace{-79.73^\circ}_{\text{pole } 0.3679} = -198.67^\circ$$

This is not $-180^\circ$. There is $18.67^\circ$ too much lag, so PI alone cannot place the poles. A lead part is added, which makes the controller PID-like.

Step 4: Controller with integral + lead action

$$G_D(z) = K\frac{(z - 0.8187)(z - 0.3679)}{(z-1)(z-\beta)}$$

Both plant poles are cancelled. The angle condition at $z_1$ is:

$$21.85^\circ - 140.79^\circ - \angle(z_1 - \beta) = -180^\circ \;\Rightarrow\; \angle(z_1-\beta) = 61.06^\circ$$ $$\beta = 0.4493 - \frac{0.4493}{\tan 61.06^\circ} = 0.2009$$

Step 5: Gain

$$K = \frac{|z_1-1|\,|z_1-0.2009|}{0.1371\,|z_1+0.6714|} = \frac{0.7107 \times 0.5134}{0.1371 \times 1.2074} = 2.204$$ $$G_D(z) = 2.204\,\frac{(z-0.8187)(z-0.3679)}{(z-1)(z-0.2009)}$$

Step 6: Check

$$G_DG = \frac{0.3022(z + 0.6714)}{(z-1)(z-0.2009)}$$

The characteristic equation is $z^2 - 0.8986z + 0.4038 = 0$, with roots $z = 0.4493 \pm j0.4493$. ✔ That is $\zeta = 0.5$ with exactly 8 samples per cycle.

$$K_v = \frac{1}{T}\cdot\frac{0.3022 \times 1.6714}{1 - 0.2009} = 3.16\ \text{s}^{-1}$$

The pole at $z = 1$ gives zero steady-state error for a step input.

Answer: $G_D(z) = 2.204\dfrac{(z-0.8187)(z-0.3679)}{(z-1)(z-0.2009)}$, with $K_v = 3.16\ \text{s}^{-1}$.