Question 8.8: Consider the system shown in Figure 8.21. Design the value o...

The root locus is shown in Figure 8.22. Notice that this is a third-order system with one zero. Breakaway points on the real axis can occur between 0 and −1 and between −1.5 and −10, where the gain reaches a peak. Using the root locus program and searching in these regions for the peaks in gain, breakaway points are found at −0.62 with a gain of 2.511 and at −4.4 with a gain of 28.89. A break-in point on the real axis can occur between −1.5 and −10, where the gain reaches a local minimum. Using the root locus program and searching in these regions for the local minimum gain, a break-in point is found at −2.8 with a gain of 27.91.

Next assume that the system can be approximated by a second-order, under-damped system without any zeros. A 1.52% overshoot corresponds to a damping ratio of 0.8. Sketch this damping ratio line on the root locus, as shown in Figure 8.22.
Use the root locus program to search along the 0.8 damping ratio line for the point where the angles from the open-loop poles and zeros add up to an odd multiple of 180°. This is the point where the root locus crosses the 0.8 damping ratio or 1.52% overshoot line. Three points satisfy this criterion: −0.87 ±j0.66, − 1.19 ±j0.90, and −4.6 ±j 3.45 with respective gains of 7.36, 12.79, and 39.64. For each point the settling time and peak time are evaluated using

$T_{s} = \frac{4}{ζ ω_{n} }$                    (8.54)

where $ζ ω_{n}$ is the real part of the closed-loop pole, and also using

$T_{p} = \frac{π}{ω_{n} \sqrt{1  –  ζ²} }$                  (8.55)

where $ω_{n} \sqrt{1  –  ζ²}$ is the imaginary part of the closed-loop pole.

To test our assumption of a second-order system, we must calculate the location of the third pole. Using the root locus program, search along the negative extension of the real axis between the zero at −1.5 and the pole at −10 for points that match the value of gain found at the second-order dominant poles. For each of the three crossings of the 0.8 damping ratio line, the third closed-loop pole is at −9.25, −8.6, and −1.8, respectively. The results are summarized in Table 8.4.

TABLE 8.4 Characteristics of the system of Example 8.8

Finally, let us examine the steady-state error produced in each case. Note that we have little control over the steady-state error at this point. When the gain is set to meet the transient response, we have also designed the steady-state error. For the example, the steady-state error specification is given by $K_{v}$ and is calculated as

$K_{v} = \underset{s \rightarrow 0}{lim} sG (s) = \frac{K (1.5)}{(1)(10)}$                      (8.56)

The results for each case are shown in Table 8.4.
How valid are the second-order assumptions? From Table 8.4, Cases 1 and 2 yield third closed-loop poles that are relatively far from the closed-loop zero. For these two cases there is no pole-zero cancellation, and a second-order system approximation is not valid. In Case 3, the third closed-loop pole and the closed-loop zero are relatively close to each other, and a second-order system approximation can be considered valid. In order to show this, let us make a partial-fraction expansion of the closed-loop step response of Case 3 and see that the amplitude of the exponential decay is much less than the amplitude of the underdamped sinusoid. The closed-loop step response, $C_{3}$ (s), formed from the closed-loop poles and zeros of Case 3 is

$C_{3} = \frac{39.64 (s  +  1.5)}{s (s  +  1.8)(s  +  4.6  +  j3.45) (s  +  4.6  −  j3.45)}$                      (8.54)

$= \frac{39.64 (s  +  1.5)}{s (s  +  1.8) (s²  +  9.2s  +  33.06)}$

$=\frac{1}{s}  +  \frac{0.3}{s (s  +  18)}  –  \frac{1.3 (s  +  4.6)  +  1.6 (3.45)}{(s  +  4.6)²  +  3.45²}$

Thus, the amplitude of the exponential decay from the third pole is 0.3, and the amplitude of the underdamped response from the dominant poles is $\sqrt{ 1.3²  + 1.6²}$ = 2.06. Hence, the dominant pole response is 6.9 times as large as the nondominant exponential response, and we assume that a second-order approximation is valid.

Using a simulation program, we obtain Figure 8.23, which shows comparisons of step responses for the problem we have just solved. Cases 2 and 3 are plotted for both the third-order response and a second-order response, assuming just the dominant pair of poles calculated in the design problem. Again, the second-order approximation was justified for Case 3, where there is a small difference in percent overshoot. The second-order approximation is not valid for Case 2. Other than the excess overshoot, Case 3 responses are similar.

Students who are using MATLAB should now run ch8apB2 in Appendix B. You will learn how to use MATLAB to enter a value of percent overshoot from the keyboard. MATLAB will then draw the root locus and overlay the percent overshoot line requested. You will then interact with MATLAB and select the point of intersection of the root locus with the requested percent overshoot line. MATLAB will respond with the value of gain, all closed-loop poles at that gain, and a closed-loop step response plot corresponding to the selected point. This exercise solves Example 8.8 using MATLAB.