Lead compensator for k/[s²(s+5)], Ka = 5, PM 30°

The plant is type 2 with an extra pole, so it is unstable for every $k$. A phase-lead compensator $G_c(s) = \dfrac{1+Ts}{1+\alpha Ts}$ (unity DC gain) is designed on the Bode plot.

Step 1: Gain for $K_a$

$$K_a = \lim_{s\to0} s^2\,\frac{k}{s^2(s+5)} = \frac{k}{5} = 5 \;\Rightarrow\; k = 25$$ $$G(s) = \frac{25}{s^2(s+5)} = \frac{5}{s^2(0.2s+1)}$$

Step 2: Uncompensated phase margin

$$\begin{aligned} \frac{25}{\omega^2\sqrt{\omega^2+25}} &= 1 \;\Rightarrow\; \omega_{gc} = 2.14\ \text{rad/s} \\ \angle G &= -180^\circ - \tan^{-1}(2.14/5) = -203.2^\circ,\qquad PM = -23.2^\circ \end{aligned}$$

Step 3: Phase lead required

$$\phi_m = 30^\circ - (-23.2^\circ) + 5^\circ = 58.2^\circ$$

A single stage ($\alpha = 0.081$) moves the crossover to 3.75 rad/s, where the plant phase is already $-216.9^\circ$, so the PM is only $21.3^\circ$. Hence two identical lead stages of $35^\circ$ each are used:

$$\alpha = \frac{1-\sin35^\circ}{1+\sin35^\circ} = 0.271 \text{ per stage}$$

Step 4: New gain crossover

Two stages add $2\times10\log(1/\alpha) = 11.34$ dB at $\omega_m$, so $\omega_m$ is where $|G| = \alpha = 0.271$:

$$\frac{25}{\omega^2\sqrt{\omega^2+25}} = 0.271 \;\Rightarrow\; \omega_m = 3.83\ \text{rad/s}$$

Step 5: Corner frequencies

$$T = \frac{1}{3.83\sqrt{0.271}} = 0.502\ \text{s},\quad \frac{1}{T} = 1.99\ \text{rad/s},\quad \frac{1}{\alpha T} = 7.35\ \text{rad/s},\quad \alpha T = 0.136\ \text{s}$$ $$G_c(s) = \left(\frac{1+0.502s}{1+0.136s}\right)^2 = 13.6\left(\frac{s+1.99}{s+7.35}\right)^2$$ $$G_c(s)G(s) = \frac{25(1+0.502s)^2}{s^2(s+5)(1+0.136s)^2}$$

PM check at 3.83 rad/s: $180^\circ - 180^\circ - \tan^{-1}(3.83/5) + 2\times35^\circ = -37.45^\circ + 70^\circ = 32.55^\circ \approx 32.6^\circ$.

Step 6: Bode plots after compensation

Asymptotic magnitude slopes: $-40$ dB/dec up to 1.99 rad/s (14 dB at $\omega = 1$), $0$ dB/dec from 1.99 to 5 rad/s (double zero; the flat level is about 2 dB), $-20$ dB/dec from 5 to 7.35 rad/s (plant pole), and $-60$ dB/dec beyond 7.35 rad/s (double pole).

 dB
 54 |\
    | \ -40
 2  |  \______ 0
 0  |---------\--------- 0 dB
    |          \ -20
-20 |           \
    |            \ -60
    +---+-----+--+------> w (log)
       1.99   5 7.35

 deg
-140|      .--.
-160|     /    \
-180|----'------\------- -180
-220|            \__
-250|               \___
    +---+-----+--+------> w (log)
       1.99 3.83 7.89

Phase rises from $-180^\circ$ to a peak of about $-142^\circ$ near 2 rad/s, is $-147.4^\circ$ at the crossover 3.83 rad/s, and crosses $-180^\circ$ at 7.89 rad/s, where the magnitude is $-9.5$ dB.

Answer: $k = 25$, $G_c(s) = \left(\dfrac{1+0.502s}{1+0.136s}\right)^2$; $K_a = 5\ \text{s}^{-2}$, $PM = 32.6^\circ\ (\ge 30^\circ)$, $GM = 9.5$ dB.