Lag Compensation Design

PROBLEM: Given the system of Figure 11.2, use Bode diagrams to design a lag compensator to yield a tenfold improvement in steady-state error over the gain-compensated system while keeping the percent overshoot at 9.5%.

Question

Lag Compensation Design

PROBLEM: Given the system of Figure 11.2, use Bode diagrams to design a lag compensator to yield a tenfold improvement in steady-state error over the gain-compensated system while keeping the percent overshoot at 9.5%.

Accepted Answer

We will follow the previously described lag compensation design procedure.

1. From Example 11.1 a gain, K, of 583.9 yields a 9.5% overshoot. Thus, for this system, [latex]K_{ u } =16.22[/latex]. For a tenfold improvement in steady-state error,[latex]K_{ u }[/latex] must increase by a factor of 10, or [latex]K_{ u } =162.2[/latex] Therefore, the value of K in Figure 11.2 equals 5839, and the open-loop transfer function is

[latex]G(s)=\frac{583.900}{s(s+36)(s+100)}[/latex]

The Bode plots for K = 5839 are shown in Figure 11.6.

2. The phase margin required for a 9.5% overshoot (ζ = 0.6) is found from Eq. (10.73)

[latex]\Phi _{M} =90- an ^{-1} \frac{\sqrt{-2\zeta ^{2}+\sqrt {1+4\zeta ^{4} } } }{2\zeta }[/latex]

[latex]= an ^{-1}\frac{2\zeta }{\sqrt{-2\zeta ^{2}+\sqrt{1+4\zeta ^{4} } } }[/latex]

to be 59.2°. We increase this value of phase margin by 10° to 69.2° in order to compensate for the phase angle contribution of the lag compensator. Now find the frequency where the phase margin is 69.2°. This frequency occurs at a phase angle of -180°+69.2° = -110.8° and is 9.8 rad/s. At this frequency, the magnitude plot must go through 0 dB. The magnitude at 9.8 rad/s is now +24 dB (exact, that is, nonasymptotic). Thus, the lag compensator must provide -24 dB attenuation at 9.8 rad/s.

3.&4. We now design the compensator. First draw the high-frequency asymptote at -24 dB. Arbitrarily select the higher break frequency to be about one decade below the phase-margin frequency, or 0.98 rad/s. Starting at the intersection of this frequency with the lag compensator’s high-frequency asymptote, draw a -20 dB/decade line until 0 dB is reached. The compensator must have a dc gain of unity to retain the value of Kv that we have already designed by setting K = 5839. The lower break frequency is found to be 0.062 rad/s. Hence, the lag compensator’s transfer function is

[latex]G_{c}(s)=\frac{0.063(s+0.98)}{(s+0.062)} [/latex]

where the gain of the compensator is 0.063 to yield a dc gain of unity. The compensated system’s forward transfer function is thus

[latex]G(s)G_{c}(s)=\frac{36,786(s+0.98)}{s(s+36)(s+100)(s+0.062)}[/latex]

The characteristics of the compensated system, found from a simulation and exact frequency response plots, are summarized in Table 11.2.

Parameter	Proposed specification	Actual value
[latex]K_{ u }[/latex]	162.2	161.5
Phase margin	59.2°	62°
Phase-margin frequency	_	11 rad/s
Percent overshoot	9.5	10
Peak time	_	0.25 second

Students who are using MATLAB should now run ch11p2 in Appendix B. You will learn how to use MATLAB to design a lag compensator. You
will enter the value of gain to meet the steady-state error requirement as well as the desired percent overshoot. MATLAB then designs a lag compensator using Bode plots, evaluates [latex]K_{ u }[/latex], and generates a closed-loop step response. This exercise solves Example 11.2 using MATLAB.

Question 11.2: Lag Compensation Design PROBLEM: Given the system of Figure ...

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