Open-loop poles

In Figure 5.9, note that the loei eommenees at the open-loop poles (.v = 0, —4) when A" = 0. At A" = 4 they braneh into the eomplex spaee. This is ealled a breakaway point and eorresponds to eritieal damping. 

Equation (5.55) may be interpreted as For a point s to he on the loeus, the sum of all angles for veetors between open-loop poles (positive angles) and zeros (negative angles) to point. vi must equal 180°. ... 

Figure 5.12 shows veetors from open-loop poles and zeros to a trial point. vi. From Figure 5.12 and equation (5.57), for. vi to lie on a loeus, then... 

Starting points K = 0) The root loei start at the open-loop poles. 

S open-loop poles — S open-loop zeros =------------------7------1----------------... 

Root locus locations on real axis. A point on the real axis is part of the loei if the sum of the number of open-loop poles and zeros to the right of the point eoneerned is odd. 

Angles of departure and arrival Computed using the angle eriterion, by positioning a trial point at a eomplex open-loop pole (departure) or zero (arrival). 

Note that equations (5.72) and (5.75) are identieal, and therefore give the same roots. The first root, —3.79 lies at a point where there are an even number of open-loop poles to the right, and therefore is not valid. The seeond root, —0.884 has odd open-loop poles to the right, and is valid. In general, method (a) requires less eomputation than method (b). 

Breakaway points None, due to complex open-loop poles. ... 

Radius = open-loop poles Centre = (Open-loop zero, 0)... 

Find the values of K and h if a is seleeted to eaneel the non-unity open-loop pole. 

The roots of equation (8.95) are the open-loop poles or eigenvalues. For the closed-loop system described by equation (8.94), the characteristic equation is... 

The idea is that we may cancel the (undesirable open-loop) poles of our process and replace them with a desirable closed-loop pole. Recall in Eq. (6-20) that Gc is sort of the reciprocal of Gp. The zeros of Gc are by choice the poles of Gp. The product of GcGp cancels everything out—hence the term pole-zero cancellation. To be redundant, we can rewrite the general design equation as... 

Under normal circumstances, we would pick a x which we deem appropriate. Now if we pick x to be identical to xp, the zero of the controller function cancels the pole of the process function. We are left with only one open-loop pole at the origin. Eq. (6-29), when x = xp, is reduced to... 

Do not be confused by the function of integral control its pole at the origin is an open-loop pole. This point should clear up when we get to the root locus section. 

Example 7.2 If we have only a proportional controller (i.e., one design parameter) and real negative open-loop poles, the Routh-Hurwitz criterion can be applied to a fairly high order system with ease. For example, for the following closed-loop system characteristic equation ... 

Note that with this very specific case by choosing x = 1, the open-loop zero introduced by the PI controller cancels one of the open-loop poles of the process function at -1. If we do a root locus plot later, we d see how the root loci change to that of a purely second order system. With respect to this example, the value is not important as long as x > 1/2. 

In other words, on a root locus plot, we expect the "trace" of the root loci to begin at the open-loop poles and terminate at the open-loop zeros (if there is one). For real systems, m > n, and n>... 

In these cases, the (m - n) root loci will originate from an open-loop pole and extend toward infinity somehow, depending on the specific problem. 

With only open-loop poles, examples (a) to (c) can only represent systems with a proportional controller. In case (a), the system contains a first orders process, and in (b) and (c) are overdamped and critically damped second order processes. 

The number of loci equals the number of open-loop poles (or the order of the system). 

A locus (closed-loop root path) starts at an open-loop pole and either terminates at an open-loop zero or extends to infinity. 

Nyquist stability criterion Given the closed-loop equation 1 + Gol (joi) = 0, if the function G0l(J ) has P open-loop poles and if the polar plot of GOL(](o) encircles the (-1,0) point... 

The shape of the root locus plot resembles that of a PI controller, except of course we do not have an open-loop pole at the origin anymore. The root loci approach one another from -xp and -p0, then break away from the real axis to form a circle which breaks in to the left of the open-loop zero at -z0. One locus approaches negative infinity and the other toward -z0. One may design the controller with an approach similar to that in Example 7.7 (p. 7-16). 

To find the new state feedback gain is a matter of applying Eq. (9-29) and the Ackermann s formula. The hard part is to make an intelligent decision on the choice of closed-loop poles. Following the lead of Example 4.7B, we use root locus plots to help us. With the understanding that we have two open-loop poles at -4 and -5, a reasonable choice of the integral time constant is 1/3 min. With the open-loop zero at -3, the reactor system is always stable, and the dominant closed-loop pole is real and the reactor system will not suffer from excessive oscillation. 

Try an open-loop zero at -1 % to cancel the open-loop pole at -1... 

Optional reading In the initial learning stage, it can be a bad habit to rely on MATLAB too much. Hence the following tutorial goes the slow way in making root locus plots, which hopefully may make us more aware of how the loci relate to pole and zero positions. The first thing, of course, is to identify the open-loop poles. 

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