Module 5 : Design of Sampled Data Control Systems

Lecture 8 : Lag-lead Compensator

1.1 Lag-lead compensator design

Example 1 Consider the following system with transfer function

 

$\displaystyle G(s) = \frac{1}{s(1+0.1s)(1+0.2s)} $


Design a lag-lead compensator C(s) such that the phase margin of the compensated system is at least 45° at gain crossover frequency around 10 rad/sec and the velocity error constant Kv is 30.

The lag-lead compensator is given by

$\displaystyle C(s) = K \; \frac{1+\tau_1s}{1+\alpha_1 \tau_1 s} \;  \frac{1+\tau_2s}{1+\alpha_2 \tau_2 s}$ where, $\displaystyle \;  \alpha_1>1, \;\; \alpha_2<1 $


When $ s\to 0$

$\displaystyle K_v=\lim_{s\to 0} sG(s)C(s)=C(0)=30 $


Thus K = 30 . Bode plot of the modified system KG(s) is shown in Figure 2. The gain crossover frequency and phase margin of KG(s) are found out to be 9.77 rad/sec and -17.2° respectively.

 

\includegraphics[width=5.0in]{m5l8fig2.eps}

Figure 2: Bode plot of the uncompensated system for Example 1

Since the PM of the uncompensated system with K is negative, we need a lead compensator to compensate for the negative PM and achieve the desired phase margin.

However, we know that introduction of a lead compensator will eventually increase the gain crossover frequency to maintain the low frequency gain.

Thus the gain crossover frequency of the system cascaded with a lead compensator is likely to be much above the specified one, since the gain crossover frequency of the uncompensated system with K is already 9.77 rad/sec.

Thus a lag-lead compensator is required to compensate for both.

We design the lead part first.

From Figure 2, it is seen that at 10 rad/sec the phase angle of the system is -198°.