Lag-Lead Compensation Design

PROBLEM: Given a unity feedback system where G(s) = K/[s(s+1)(s+4)] , design a passive lag-lead compensator using Bode diagrams to yield a 13.25% overshoot, a peak time of 2 seconds, and $K_{\nu } =40$ .

Question

Lag-Lead Compensation Design

PROBLEM: Given a unity feedback system where G(s) = K/[s(s+1)(s+4)] , design a passive lag-lead compensator using Bode diagrams to yield a 13.25% overshoot, a peak time of 2 seconds, and [latex]K_{ u } =40[/latex].

Accepted Answer

We will follow the steps previously mentioned in this section for lag-lead
design.

1. The bandwidth required for a 2-seconds peak time is 2.29 rad/s.

2. In order to meet the steady-state error requirement, [latex]K_{ u } =12[/latex], the value of K is 48.

3. The Bode plots for the uncompensated system with K = 48 are shown in Figure 11.12. We can see that the system is unstable.

4. The required phase margin to yield a 13.25% overshoot is 55°.

5. Let us select ω = 1.8 rad/s as the new phase-margin frequency.

6. At this frequency, the uncompensated phase is -176° and would require, if we add a -5° contribution from the lag compensator, a 56° contribution from the lead portion of the compensator.

7. The design of the lag compensator is next. The lag compensator allows us to keep the gain of 48 required for [latex]K_{ u } =12[/latex] and not have to lower the gain to stabilize the system. As long as the lag compensator stabilizes the system, the design parameters are not critical, since the phase margin will be designed with the lead compensator. Thus, choose the lag compensator so that its phase response will have minimal effect at the new phase-margin frequency. Let us choose the lag compensator’s higher break frequency to be 1 decade below the new phase-margin frequency, at 0.18 rad/s. Since
we need to add 56° of phase shift with the lead compensator atω = 1.8 rad/s,we estimate fromFigure 11.8 that, if γ= 10.6 (since γ = 1/β, β =0.094), we can obtain about 56° of phase shift from the lead compensator. Thus with γ = 10.6 and a new phase-margin frequency of ω = 1.8 rad/s, the transfer function of the lag compensator is

[latex]G_{lag }(s)=\frac{1}{\gamma } \frac{\left(s+\frac{1}{T_{2} } ight) }{\left(s+\frac{1}{\gamma T_{2} } ight) } =\frac{1}{10.6} \frac{(s+0.183)}{(s+0.0172)}[/latex]

where the gain term, 1=γ, keeps the dc gain of the lag compensator at 0 dB. The lag-compensated system’s open-loop transfer function is

[latex]G_{lag-comp}(s)=\frac{4.53(s+0.183)}{s(s+1)(s+4)(s+0.0172)}[/latex]

8. Now we design the lead compensator. At ω = 1.8, the lag- compensated system has a phase angle of 180°. Using the values of [latex]\omega _{\max } =1.8[/latex] and β = 0.094, Eq. (11.9)

[latex]\omega _{\max } =\frac{1}{T\sqrt{\beta } }[/latex]

yields the lower break, [latex]1/T_{1} =0.56[/latex] rad/s. The higher break is then [latex]1/\beta T_{1} = 5.96[/latex] rad/s. The lead compensator is

[latex]G_{lead}(s)=\gamma \frac{\left(s+\frac{1}{T_{1}} ight) }{\left(s+\frac{\gamma }{T_{1}} ight) } = 10.6\frac{(s+0.56)}{(s+5.96)}[/latex]

The lag-lead-compensated system’s open-loop transfer function is

[latex]G_{lag-lead-comp}(s)=\frac{48(s+0.183)(s+0.56)}{s(s+1)(s+4)(s+0.0172)(s+5.96)}[/latex]

9. Now check the bandwidth. The closed-loop bandwidth is equal to that frequency where the open-loop magnitude response is approximately -7 dB. From Figure 11.12, the magnitude is -7 dB at approximately 3 rad/s. This bandwidth exceeds that required to meet the peak time requirement.

The design is now checked with a simulation to obtain actual performance values. Table 11.4 summarizes the system’s characteristics. The peak time requirement is also met. Again, if the requirements were not met, a redesign would be necessary.

Parameter	Proposed specification	Actual value
[latex]K_{ u }[/latex]	12	12
Phase margin	55°	59.3°
Phase-margin frequency	_	1.63 rad/s
Closed-loop bandwidth	2.29 rad/s	3 rad/s
Percent overshoot	13.25	10.2
Peak time	2.0 seconds	1.61 seconds

Students who are using MATLAB should now run ch11p4 in Appendix B. You will learn how to use MATLAB to design a lag-lead compensator.
You will enter the desired percent overshoot, peak time, and [latex] K_{ u }[/latex]. MATLAB then designs a lag-lead compensator using Bode plots, evaluates [latex]K_{ u }[/latex], and generates a closed-loop step response. This exercise solves Example 11.4 using MATLAB.

Question 11.4: Lag-Lead Compensation Design PROBLEM: Given a unity feedback...

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