Lag compensator for k/[s(s+2)(s+20)], Kv = 20

A lag network $G_c(s) = \dfrac{1+Ts}{1+\beta Ts}$ ($\beta > 1$) is designed by the Bode plot method.

Step 1: Gain for $K_v$

$$K_v = \lim_{s\to0} s\,\frac{k}{s(s+2)(s+20)} = \frac{k}{40} = 20 \;\Rightarrow\; k = 800$$ $$G(s) = \frac{800}{s(s+2)(s+20)} = \frac{20}{s(0.5s+1)(0.05s+1)}$$

Step 2: Uncompensated margins

• $|G(j\omega)| = 1$ at $\omega_{gc} = 6.03$ rad/s; $\angle G = -90^\circ - \tan^{-1}\frac{6.03}{2} - \tan^{-1}\frac{6.03}{20} = -178.4^\circ$, so $PM = 1.6^\circ$.
• $\omega_{pc} = \sqrt{40} = 6.32$ rad/s, $GM = 0.8$ dB.

The system is on the verge of instability.

Step 3: New gain crossover frequency

Allow $10^\circ$ for the lag of the network:

$$\angle G(j\omega_{gc}') = -180^\circ + 35^\circ + 10^\circ = -135^\circ$$ $$-90^\circ - \tan^{-1}\frac{\omega}{2} - \tan^{-1}\frac{\omega}{20} = -135^\circ \;\Rightarrow\; \omega_{gc}' = 1.69\ \text{rad/s}$$

Step 4: Value of $\beta$

$$|G(j1.69)| = \frac{800}{1.69\times\sqrt{1.69^2+4}\times\sqrt{1.69^2+400}} = 9.02\;(19.1\ \text{dB}) \;\Rightarrow\; \beta = 9.02$$

Step 5: Corner frequencies

$$\begin{aligned} \frac{1}{T} &= \frac{1.69}{10} = 0.169\ \text{rad/s},\quad T = 5.92\ \text{s} \\ \frac{1}{\beta T} &= \frac{0.169}{9.02} = 0.0187\ \text{rad/s},\quad \beta T = 53.4\ \text{s} \end{aligned}$$

Step 6: Compensator and check

$$G_c(s) = \frac{1+5.92s}{1+53.4s} = 0.111\,\frac{s+0.169}{s+0.0187}$$ $$G_c(s)G(s) = \frac{800(1+5.92s)}{s(s+2)(s+20)(1+53.4s)}$$

(With only $5^\circ$ allowance the PM comes out at $35.0^\circ$, just on the limit; the $10^\circ$ allowance gives a safe margin.)

Answer: $k = 800$, $G_c(s) = \dfrac{1+5.92s}{1+53.4s}$ (zero at $-0.169$, pole at $-0.0187$); $K_v = 20\ \text{s}^{-1}$, $PM = 39.9^\circ \ge 35^\circ$.