Bode design: PM 55°, GM 10 dB, Kv = 5

The design is done in the w-plane (bilinear transformation), where the Bode diagram can be used as in the s-plane. A lead compensator $G_D(w) = K\dfrac{1+\tau w}{1+\alpha\tau w}$ is used.

Step 1: Pulse transfer function of the plant

With $T = 0.1$ s and $e^{-0.2} = 0.8187$:

$$\begin{aligned} G(z) &= (1-z^{-1})\,\mathcal{Z}\left[\frac{1}{s^2(s+2)}\right]\\ &= \frac{0.004683z + 0.004381}{(z-1)(z-0.8187)} = \frac{0.004683(z+0.9355)}{(z-1)(z-0.8187)} \end{aligned}$$

Step 2: Transform to the w-plane

Substitute $z = \dfrac{1+0.05w}{1-0.05w}$:

$$G(w) = \frac{0.5\left(1-\frac{w}{20}\right)\left(1+\frac{w}{600.4}\right)}{w\,(1+0.5017w)}$$

Step 3: Gain for $K_v = 5$

$$K_v = \lim_{w\to0} w\,G_D(w)G(w) = K\times0.5 = 5 \;\Rightarrow\; K = 10$$

Step 4: Uncompensated Bode data ($10\,G(j\nu)$)

Gain crossover is at $\nu = 2.88$ rad/s and the phase margin is only 26.8°. About 28° or more of lead is needed. The w-plane zero at $w=20$ adds extra lag at high frequency, so a larger lead is needed than in a continuous design.

Step 5: Lead compensator

Choose the compensator zero to cancel the plant pole at $w = -1.993$, so $\tau = 0.5017$. Then reduce $\alpha$ until PM = 55°:

$$G_D(w) = 10\,\frac{1 + 0.5017w}{1 + 0.0853w}$$

Compensated loop: $G_D(w)G(w) = \dfrac{5(1-w/20)(1+w/600.4)}{w(1+0.0853w)}$. Phase margin = 55°, phase crossover at $\nu = 15.7$ rad/s, gain margin = 12.3 dB (≥ 10 dB).

Step 6: Back to the z-plane

Substitute $w = \dfrac{2}{T}\dfrac{z-1}{z+1} = 20\dfrac{z-1}{z+1}$:

$$\begin{aligned} G_D(z) &= 10\,\frac{1+10.033\frac{z-1}{z+1}}{1+1.706\frac{z-1}{z+1}} = 40.78\,\frac{z-0.8187}{z-0.2608} \end{aligned}$$

Check of $K_v$

$$K_v = \frac{1}{T}\lim_{z\to1}(z-1)G_D(z)G(z) = \frac{1}{0.1}\times40.78\times\frac{0.1813}{0.7392}\times\frac{0.009063}{0.1813} = 5.0\ \text{s}^{-1}$$

Answer: $G_D(z) = 40.78\,\dfrac{z-0.8187}{z-0.2608}$, giving PM ≈ 55°, GM ≈ 12.3 dB and $K_v = 5\ \text{s}^{-1}$.