Digital PI design, plant e^(−5s)/(s+0.4), Kv

Step 1: Choice of sampling period

The plant has time constant $1/0.4 = 2.5$ s and dead time 5 s. Choose $T = 2.5$ s:

• The dead time is then exactly $2T$, giving a clean $z^{-2}$.
• $e^{-0.4T} = e^{-1}$, which gives simple numbers.
• With 10 samples per cycle, the oscillation period is 25 s. This is reasonable for a process with 5 s dead time.

A smaller $T$ such as 1 s is not suitable. The dead time becomes $z^{-5}$, the phase lag at the desired point exceeds what a single PI zero can make up, and the angle condition cannot be met.

$$\begin{aligned} G(z) &= z^{-2}(1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s(s+0.4)}\right] = z^{-2}\cdot\frac{(1-e^{-1})}{0.4}\cdot\frac{z^{-1}}{1-e^{-1}z^{-1}} \\ &= \frac{1.5803}{z^2(z - 0.3679)} \end{aligned}$$

Step 2: Desired poles (ζ = 0.5, 10 samples per cycle)

$$|z| = \exp\left(-\frac{2\pi(0.5)}{0.866}\cdot\frac{1}{10}\right) = 0.6958, \qquad \angle z = 36^\circ, \qquad z_1 = 0.5629 + j0.4090$$

Step 3: PI controller $G_D(z) = K\dfrac{z-\alpha}{z-1}$

Angles at $z_1$:

• double pole at origin: $-72^\circ$
• pole at 0.3679: $-64.51^\circ$
• pole at 1: $-136.91^\circ$

The sum is $-273.41^\circ$, so the PI zero must add $+93.41^\circ$:

$$\alpha = 0.5629 - \frac{0.4090}{\tan 93.41^\circ} = 0.5873$$

Step 4: Gain

$$K = \frac{|z_1|^2\,|z_1-0.3679|\,|z_1-1|}{1.5803\,|z_1-0.5873|} = \frac{0.4841 \times 0.4531 \times 0.5986}{1.5803 \times 0.4097} = 0.2028$$ $$G_D(z) = 0.2028\,\frac{z - 0.5873}{z - 1}$$

so $K_P = K\alpha = 0.1191$ and $K_I = K(1-\alpha) = 0.0837$.

Step 5: Static velocity error constant

$$\begin{aligned} K_v &= \lim_{z\to1}\frac{(1-z^{-1})G_DG}{T} = \frac{1}{2.5}\cdot K(1-\alpha)\cdot\frac{1.5803}{1-0.3679} \\ &= \frac{0.0837 \times 2.5}{2.5} = 0.0837\ \text{s}^{-1} \end{aligned}$$

The steady-state error for a unit ramp is $1/K_v = 11.9$.

Check: the closed-loop roots are $0.5629 \pm j0.4090$ (as designed), $0.7562$ and $-0.5141$. The extra real pole at 0.7562 is not cancelled by the PI zero (0.5873), so it adds a slow tail to the response.

Answer (with $T = 2.5$ s): $G_D(z) = 0.2028\dfrac{z-0.5873}{z-1}$ and $K_v = 0.0837\ \text{s}^{-1}$.