Module 5 : Design of Sampled Data Control Systems

Lecture 6 : Compensator Design Using Bode Plot

Phase angle at ωg = 3.1 is -90 - tan -1 3.1 = - 162° . Thus the PM of the uncompensated system with K is 18°.

If it was possible to add a phase without altering the magnitude, the additional phase lead required to maintain PM= 45° is 45° - 18° = 27° at ωg = 3.1 rad/sec.

However, maintaining same low frequency gain and adding a compensator would increase the crossover frequency. As a result of this, the actual phase margin will deviate from the designed one. Thus it is safe to add a safety margin of ε to the required phase lead so that if it devaites also, still the phase requirement is met. In general ε is chosen between 5° to 15°.

So the additional phase requirement is 27° + 10° = 37° , The lead part of the compensator will provide this additional phase at ωmax .

Thus



The only parameter left to be designed is τ. To find τ, one should locate the frequency at which the uncompensated system has a logarithmic magnitude of $ -20 \log_{10} \dfrac{1}{\sqrt \alpha}$.

Select this frequency as the new gain crossover frequency since the compensator provides a gain of $ 20 \log_{10} \dfrac{1}{\sqrt \alpha}$ at ωmax. Thus

$\displaystyle \omega_{\max} = \omega_{g_{new}}= \frac{1}{\tau\sqrt{\alpha}} $


In this case ωmax = ωg new = 4.41 . Thus

$\displaystyle \tau = \frac{1}{4.41\sqrt \alpha}= 0.4535 $


The lead compensator is thus

$\displaystyle C(s) = 10 \; \frac{0.4535 s+1}{0.1134 s+1} $


With this compensator actual phase margin of the system becomes 49.6° which meets the design criteria.

The corresponding Bode plot is shown in Figure 2

Figure 2: Bode plot of the compensated system for Example 1
\includegraphics[width=4.in]{m5l6fig1c.eps}