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4. (Rule 4) Since n - m = 3, we have three asymptotes and the center of gravity is y = (-3 - 2 - 1)/3 = -2. Angles which the asymptotes make with the real axis are 7r/3, 31rl3, and 51~13. These asymptotes are shown in Fig. 15.5a. With these few steps completed, a rough sketch of the root-locus diagram can be made as follows: Since the real axis to the left of -3 is an asymptote and one branch emerges from the pole at -3, it should be clear that one entire branch is the real axis to the left of the -3. Furthermore, from the fact that two loci must emerge from the poles - 1 and -2 and that the real axis between these poles is part of the locus, we see that two loci move toward each other along the real axis between - 1 and -2 and eventually meet at some common point. Since the location of the asymptotes is known, it is therefore necessary that the two loci that meet on the real axis must break away and eventually follow the asymptotes. From these observations, we could sketch a root-locus diagram that closely resembles that of Fig. 15.5~. If the breakaway point and the crossings of the imaginary axis were known, the sketch could be made with considerable accuracy. We now continue the example by applying Rule 5 to find the breakaway point and the Routh test to find the crossings of the imaginary axis. 5. Breakaway point. (Rule 5) The roots emerging from - 1 and - 2 move toward each other until they meet, at which point the loci leave the real axis at angles of + 7r/2. The breakaway point is found from Eq. (15.16) as follows 1 1 1 o = -+ -+s -p3 s -Pl s-p:!
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Solving this by trial and error gives s = -1.42 6. To find the points at which the loci cross the imaginary axis, the Routh test (theorem 3) of Chap. 14 may be used. Writing the characteristic equation D + KN = 0 in polynomial form gives
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D + KN = (s + l)(s + 2)(s -t 3)+ K = 0
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s2 +6s2 + 11s + K +6 = 0
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from which we can write the Routh array:
--I-bl
2 3 6
K-k6
ROOT LOCUS
The theorem states that, if one pair of roots are on the imaginary axis and all others in the left half plane, all the elements of the nth row must be zero. From this we obtain for the element bl
b, =
(6)Ul)
Solving for K, K = 60
A root on the imaginary axis is expressed as simply ju. Substituting s = j a and K = 60 into the polynomial gives -ju3 - 6a2 + llaj + 66 = 0 (66 - 6a*) + (11~ - u3)j = 0 Equating the real part or the imaginary part to zero gives a = r fi = 23.32 Therefore the ioci intersect the imaginary axis at + j fi and - j fi. 7. Having found these general features of the root-locus plot, we can sketch the root locus. If it is desirable to have a more accurate plot of the loci, the construction is continued by the trial-and-error method described earlier in this chapter. + To illustrate the method of finding roots, suppose the trial point, si = -0.75 + 1.5 j of Fig. 15.5b, is selected. This point is checked by the angle criterion [Eq. (15.14), which for this example may be written
&(s + 1) + i$(s + 2) + &(s + 3) = (2i + 1)Tr or
I31 + I32 + 03 = (2i + l) r From Fig. 15.5b, these angles are found to be 19~ = 81 and we have 81 + 51 + 34 = 166 + (2i + 1)7r e* = 51 03 = 34O
+Several computer software packages are now available for plotting the root-locus diagram. For example, the program CC is especially useful for root-locus plotting. Details on CC and other software packages are given in Appendix 34A (of Chap. 34). Evans (1954, 1948), who developed the root-locus method, produced an instrument for plotting root-locus diagrams called the Spirule. The Sprirule was essentially a drawing instrument that was used to add angles by rotating an arm with respect to a disk. The Sprirule, which is no longer available, is now obsolete as a result of the availability of computer programs for plotting root-locus diagrams.
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