Lag compensator for K/[s(s+4)(s+80)], Kv = 30

A lag compensator $G_c(s) = \dfrac{1+Ts}{1+\beta Ts}$ ($\beta > 1$) is designed with the Bode plot method: the gain is fixed by $K_v$, and the lag network lowers the high-frequency gain so the crossover moves to a frequency with enough phase margin.

Step 1: Gain for $K_v$

$$K_v = \lim_{s\to0} s\,\frac{K}{s(s+4)(s+80)} = \frac{K}{320} = 30 \;\Rightarrow\; K = 9600$$ $$G(s) = \frac{9600}{s(s+4)(s+80)} = \frac{30}{s(0.25s+1)(0.0125s+1)}$$

Step 2: Uncompensated margins

• Gain crossover: $|G(j\omega)| = 1$ at $\omega_{gc} = 10.55$ rad/s; $\angle G = -90^\circ - \tan^{-1}\frac{10.55}{4} - \tan^{-1}\frac{10.55}{80} = -166.7^\circ$, so $PM = 13.3^\circ$.
• Phase crossover: $\omega_{pc} = \sqrt{4\times80} = 17.9$ rad/s, $GM = 8.9$ dB.

The PM is far below $33^\circ$.

Step 3: New gain crossover frequency

Allow $5^\circ$ for the lag of the compensator at the new crossover. Required plant phase:

$$\angle G(j\omega_{gc}') = -180^\circ + 33^\circ + 5^\circ = -142^\circ$$ $$-90^\circ - \tan^{-1}\frac{\omega}{4} - \tan^{-1}\frac{\omega}{80} = -142^\circ \;\Rightarrow\; \omega_{gc}' = 4.56\ \text{rad/s}$$

Step 4: Value of $\beta$

At 4.56 rad/s:

$$|G(j4.56)| = \frac{9600}{4.56\times\sqrt{4.56^2+16}\times\sqrt{4.56^2+6400}} = 4.33 \;(12.7\ \text{dB})$$

The lag network must reduce the gain by this amount: $\beta = 4.33$.

Step 5: Corner frequencies

Place the zero one decade below the new crossover:

$$\begin{aligned} \frac{1}{T} &= \frac{4.56}{10} = 0.456\ \text{rad/s},\quad T = 2.19\ \text{s} \\ \frac{1}{\beta T} &= \frac{0.456}{4.33} = 0.105\ \text{rad/s},\quad \beta T = 9.50\ \text{s} \end{aligned}$$

Step 6: Compensator and check

$$G_c(s) = \frac{1 + 2.19s}{1 + 9.50s} = 0.231\,\frac{s+0.456}{s+0.105}$$ $$G_c(s)G(s) = \frac{9600(1+2.19s)}{s(s+4)(s+80)(1+9.50s)}$$

The lag network adds only $-4.4^\circ$ at the new crossover, within the $5^\circ$ allowed.

Answer: $K = 9600$, $G_c(s) = \dfrac{1+2.19s}{1+9.50s}$; $K_v = 30\ \text{s}^{-1}$ and $PM = 33.6^\circ$.