Cascade lag compensator for 50K/[s(s+5)(s+10)]

A cascade lag network $G_c(s) = \dfrac{1+Ts}{1+\beta Ts}$ ($\beta > 1$, unity DC gain) is designed with the Bode plot. A lag network is suitable because the plant has enough phase at a lower frequency; we only need to lower the gain there.

Step 1: Gain for $K_v$

$$K_v = \lim_{s\to0} s\,\frac{50K}{s(s+5)(s+10)} = \frac{50K}{50} = K = 30$$ $$G(s) = \frac{1500}{s(s+5)(s+10)} = \frac{30}{s(0.2s+1)(0.1s+1)}$$

Step 2: Uncompensated margins

• $|G(j\omega)| = 1$ at $\omega_{gc} = 9.77$ rad/s; $\angle G = -90^\circ - \tan^{-1}\frac{9.77}{5} - \tan^{-1}\frac{9.77}{10} = -197.2^\circ$, so $PM = -17.2^\circ$.
• $\omega_{pc} = \sqrt{50} = 7.07$ rad/s, $|G(j7.07)| = 2$, $GM = -6.0$ dB.

The uncompensated system is unstable.

Step 3: New gain crossover frequency

Allow $10^\circ$ for the lag of the network (needed for this large shift):

$$\angle G(j\omega_{gc}') = -180^\circ + 45^\circ + 10^\circ = -125^\circ$$ $$-90^\circ - \tan^{-1}\frac{\omega}{5} - \tan^{-1}\frac{\omega}{10} = -125^\circ \;\Rightarrow\; \omega_{gc}' = 2.12\ \text{rad/s}$$

Step 4: Value of $\beta$

$$|G(j2.12)| = \frac{1500}{2.12\times\sqrt{2.12^2+25}\times\sqrt{2.12^2+100}} = 12.72 \;(22.1\ \text{dB}) \;\Rightarrow\; \beta = 12.72$$

Step 5: Corner frequencies

$$\begin{aligned} \frac{1}{T} &= \frac{\omega_{gc}'}{10} = 0.212\ \text{rad/s},\quad T = 4.71\ \text{s} \\ \frac{1}{\beta T} &= \frac{0.212}{12.72} = 0.0167\ \text{rad/s},\quad \beta T = 59.9\ \text{s} \end{aligned}$$

Step 6: Compensator and check

$$G_c(s) = \frac{1 + 4.71s}{1 + 59.9s} = 0.0786\,\frac{s+0.212}{s+0.0167}$$ $$G_c(s)G(s) = \frac{1500(1+4.71s)}{s(s+5)(s+10)(1+59.9s)}$$

A practical RC lag network ($R_1$ in series, $R_2$–$C$ in shunt) gives $T = R_2C$ and $\beta = (R_1+R_2)/R_2$; e.g. $C = 10\ \mu$F gives $R_2 = 471$ k$\Omega$ and $R_1 = 5.52$ M$\Omega$.

Answer: $K = 30$, $G_c(s) = \dfrac{1+4.71s}{1+59.9s}$; $K_v = 30\ \text{s}^{-1}$, $PM = 49.6^\circ \ (\ge 45^\circ)$.