Lead Compensator Design

PROBLEM: Design three lead compensators for the system of Figure 9.17 that will reduce the settling time by a factor of 2 while maintaining 30% overshoot. Compare the system characteristics between the three designs.

Question

Lead Compensator Design

PROBLEM: Design three lead compensators for the system of Figure 9.17 that will reduce the settling time by a factor of 2 while maintaining 30% overshoot. Compare the system characteristics between the three designs.

Accepted Answer

First determine the characteristics of the uncompensated system operating at 30% overshoot to see what the uncompensated settling time is. Since 30% overshoot is equivalent to a damping ratio of 0.358, we search along the ζ=0.358 line for the uncompensated dominant poles on the root locus, as shown in Figure 9.26. From the pole’s real part, we calculate the uncompensated settling time as[latex]T_{s} =4/1.007=3.972 [/latex] seconds. The remaining characteristics of the uncompensated system are summarized in Table 9.4.

Next we find the design point. A twofold reduction in settling time yields [latex]T_{s} =3.972/2=1.986[/latex] seconds, from which the real part of the desired pole location is [latex]-\zeta \omega _{n} =-4/T_{s} =-2.014 [/latex]. The imaginary part is [latex]\omega _{d} -2.014 an (110.98°)= 5.252[/latex].

We continue by designing the lead compensator. Arbitrarily assume a compensator zero at -5 as a possible solution. Using the root locus program, sum the angles from both this zero and the uncompensated system’s poles and zeros, using the design point as a test point. The
resulting angle is -179.69°. The difference between this angle and 180° is the angular contribution required from the compensator pole in order to place the design point on the root locus. Hence, an angular contribution of -7.31° is required from the compensator pole.

	Uncompensated	Compensation a	Compensation b	Compensation c
Plant and compensator	[latex]\frac{K}{s(s+4)(s+6)}[/latex]	[latex]\frac{K(s+5)}{s(s+4)(s+6)(s+42.96)}[/latex]	[latex]\frac{K(s+4)}{s(s+4)(s+6)(s+20.09)}[/latex]	[latex]\frac{K(s+2)}{s(s+4)(s+6)(s+8.971)}[/latex]
Dominant poles	-1.007±j2.627	-2.014±j5.252	-2.014±j5.252	-2.014±j5.252
K	63.21	1423	698.1	345.6
ζ	0.358	0.358	0.358	0.358
[latex]\omega _{n}[/latex]	2.813	5.625	5.625	5.625
[latex]\%OS^{\ast }[/latex]	30 (28)	30 (30.7)	30 (28.2)	30 (14.5)
[latex]T_{s}[/latex]*	3.972 (4)	1.986 (2)	1.986 (2)	1.986 (1.7)
[latex]T_{p}[/latex]*	1.196 (1.3)	0.598 (0.6)	0.598 (0.6)	0.598 (0.7)
[latex]K_{ u }[/latex]	2.634	6.9	5.791	3.21
[latex]e(\infty )[/latex]	0.380	0.145	0.173	0.312
Other poles	-7.986	-43.8, -5.134	-22.06	-13.3, -1.642
Zero	None	-5	None	-2
Comments	Second-order approx. OK	Second-order approx. OK	Second-order approx. OK	No pole-zero cancellation

The geometry shown in Figure 9.27 is used to calculate the location of the compensator pole. From the figure,

[latex]\frac{5.252}{p_{c}-2.014 } = an 7.31°[/latex]

from which the compensator pole is found to be

[latex]p_{c}= 42.96[/latex]

The compensated system root locus is sketched in Figure 9.28. In order to justify our estimates of percent overshoot and settling time, we must show that the second-order approximation is valid. To perform this validity check, we search for the third and fourth closed-loop poles found beyond -42.96 and between -5 and -6 in Figure 9.28. Searching these regions for the gain equal to that of the compensated dominant
pole, 1423, we find that the third and fourth poles are at -43.8 and -5.134, respectively.

Since -43.8 is more than 20 times the real part of the dominant pole, the effect of the third closed-loop pole is negligible. Since the closed-loop pole at -5.134 is close to the zero at -5, we have pole-zero cancellation, and the second-order approximation is valid.

All results for this design and two other designs, which place the compensator zero arbitrarily at -2 and -4 and follow similar design techniques, are summarized in Table 9.4. Each design should be verified by a simulation, which could consist of using MATLAB
(discussed at the end of this example) or the state-space model and the step-response program discussed in Appendix H.1 at www.wiley.com/college/nise.We have performed a simulation for this design problem, and the results are shown by parenthetical entries next to the estimated values in the table. The only design that disagrees with the simulation is the case where the compensator zero is at -2. For this case the closed-loop pole and zero do not cancel.

A sketch of the root locus, which you should generate, shows why the effect of the zero is pronounced, causing the response to be different from that predicted. Placing the zero to the right of the pole at -4 creates a portion of the root locus that is between the origin and the zero. In other words, there is a closed-loop pole closer to the origin than the dominant poles, with little chance of pole-zero cancellation except at high gain. Thus, a quick sketch of the root locus gives us information from which we can make better design decisions. For this example, we want to place the zero on, or to the left of, the pole at -4, which gives a better chance for pole-zero cancellation and for a higher-order pole that is to the left of the dominant poles and subsequently faster. This is verified by the fact that our results show good second-order approximations for the cases where the zero was placed at -4 and -5. Again, decisions about where to place the zero are based on simple rules of thumb and must be verified by simulations at the end of the design.

Let us now summarize the results shown in Table 9.4. First we notice differences in the following:

1. The position of the arbitrarily selected zero
2. The amount of improvement in the steady-state error
3. The amount of required gain, K
4. The position of the third and fourth poles and their relative effect upon the second-order approximation. This effect is measured by their distance from the dominant poles or the degree of cancellation with the closed-loop zero.

Once a simulation verifies desired performance, the choice of compensation can be based upon the amount of gain required or the improvement in steady-state error that can be obtained without a lag compensator.

The results of Table 9.4 are supported by simulations of the step response, shown in Figure 9.29 for the uncompensated system and the three lead compensation solutions.

Students who are using MATLAB should now run ch9p2 in Appendix B.
MATLAB will be used to design a lead compensator. You will input the desired percent overshoot from the keyboard. MATLAB will plot the root locus of the uncompensated system and the percent overshoot line. You will interactively select the gain, after which MATLAB will display the performance characteristics of the uncompensated system and plot its step response. Using these characteristics, you will input the desired settling time and a zero value for the lead compensator. You will then interactively select a value for the compensator pole. MATLAB will respond with a root locus. You can then continue selecting pole values until the root locus goes through the desired point. MATLAB will display the lead compensator ,enumerate its performance characteristics, and plot a step response. This exercise solves Example 9.4 using MATLAB.

Question 9.4: Lead Compensator Design PROBLEM: Design three lead compensat...

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