Lag compensator for 1/[s(s+1)(0.5s+1)], Kv = 5

A lag compensator $G_c(s)=K_c\beta\dfrac{Ts+1}{\beta Ts+1}$ ($\beta>1$) is used: it supplies the low-frequency gain for $K_v$ and attenuates by $20\log\beta$ at higher frequencies, moving the gain crossover down to where the plant phase gives the required margins.

Step 1: Gain for $K_v$

$$K_v=\lim_{s\to0}s\,G_c(s)G(s)=K_c\beta=5$$

Let $K=K_c\beta=5$:

$$G_1(s)=\frac{5}{s(s+1)(0.5s+1)}$$

Step 2: Bode plot of $G_1$

Corners at 1 and 2 rad/s; slopes $-20$, $-40$, $-60$ dB/dec; 14 dB at $\omega=1$ on the first asymptote.

$$\phi=-90^\circ-\tan^{-1}\omega-\tan^{-1}0.5\omega$$

$\omega_{gc}=1.80$ rad/s, PM $=-13^\circ$; $\omega_{pc}=1.414$ rad/s, GM $=-4.4$ dB: unstable.

Step 3: New gain crossover

Required phase $=-180^\circ+40^\circ+12^\circ=-128^\circ$ (12° allowed for the lag network). Solving gives $\omega\approx0.47$ rad/s; it is rounded to $\omega_{gc}'=0.5$ rad/s ($\phi=-130.6^\circ$ there).

Step 4: Zero and β

• Corner of the zero well below $\omega_{gc}'$: $\frac1T=0.1$ rad/s (one-fifth of 0.5) $\Rightarrow T=10$ s.
• Attenuation needed at 0.5 rad/s is about 19–20 dB (18.8 dB from the table), so take $20\log\beta=20$ dB: $\beta=10$.
• Pole at $\frac{1}{\beta T}=0.01$ rad/s.

$$K_c=\frac{K}{\beta}=\frac{5}{10}=0.5$$ $$G_c(s)=0.5\times10\,\frac{10s+1}{100s+1}=5\,\frac{10s+1}{100s+1}=0.5\,\frac{s+0.1}{s+0.01}$$

Step 5: Compensated system

$$G_c(s)G(s)=\frac{5(10s+1)}{s(100s+1)(s+1)(0.5s+1)}$$ $$PM=180^\circ-138.4^\circ=41.6^\circ\ (\ge40^\circ),\qquad GM=14.3\ \text{dB}\ (\ge10\ \text{dB})$$

 dB 54 |\___ uncompensated
       |    \___
     0 |--x-----\----------- w
       | 0.45    \ 1.8
       | compensated: 20 dB lower
       | above 0.1 rad/s

Closed-loop poles: $-0.286\pm j0.520$, $-0.123$, $-2.32$ (all in LHP).

Answer: $G_c(s)=5\,\dfrac{10s+1}{100s+1}$ (β = 10, T = 10 s): Kv = 5 s⁻¹, PM ≈ 42°, GM ≈ 14 dB.