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Shifting the trial point horizontally to the left will increase the sum of the angles. As a second trial point, sy = - 0.95 + 1.5 j gives for the sum of the angles 88 + 56 + 37 = 181 z TT This result is sufficiently close to V, which is (2; + 1)~ with i = 0, and we accept the point as one on the locus. In this manner, more points on the locus can be found and a curve drawn through them. 8. Gain. To determine the gain at various points along the loci, the magnitude criterion [Eq. (15.13)] is used. For example, if the gain at s = -0.95 + jl.5 (labeled si in Fig. 15.56), is wanted, we measure the distances directly with a ruler; thus Is -p1 1 = 1.50
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1 s -p2 1 =
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Is-p31
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(It is important to measure the vector lengths in units that are consistent with those used on the axes of the graph.) Substituting these values into Eq. (15.13) gives
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(1.50)(1.82)(2.52) = or K = (1.50)(1.82)(2.52) = 6.8. To find the point corresponding to K = 6.8 on the branch along the real axis to the left of p3 requires a trial-and-error solution if the graphical approach is used. For example, if s = -4.5 is tried, we obtain
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Is-plI=3.5 Is-p21 = 2.5
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1 s -p3 I=
from which we get
K = (1.5)(2.5)(3.5) = 13.1
We see that s = -4.5 does not correspond to a gain of 6.8. It is therefore necessary to try other values of s greater than -4.5 until the desired value of K = 6.8 is obtained. Although this procedure may seem very tedious, the actual calculations go quite quickly as the reader will discover while working out this example. We also may find the root on the real axis more directly by applying the following theorem from algebra: The sum of the roots (rl + r2 + *. *+ rn) of the nth-order polynomial equation
aox + alx -l + ..* + a, = 0
is given by
(rl +
12 + ..+r,)
= -z
RooTLmus
In this case, we have just found the complex roots for
r2, r3
K = 6.8 to be
= -0.95 -c j 1.5
The polynomial equation is (s + l)(s + Z)(s + 3) + K which can be expanded into s3 + 6s2 + 11s + (K + 6) = 0 According to the theorem r1 + (-0.95 + j1.5) + (-0.95 - j1/5) = -4 or 6 = -[rl - 2(0.95)] or rl = -4.10 All the detailed steps needed to plot the root locus for this problem have been discussed. The complete locus is shown in Fig. 15.5~. This same plot is also shown in mom detail in Fig. 15.2. Example 15.2. Consider the block diagram for the control system shown in Fig. 15.6. This system may represent a two-tank, liquid-level system having a PID controller and a first-order measuring lag. The open-loop transfer function is 1 + 2~13 + 113s G = K (20s + l)(lOs + 1)(0.5s + 1) Rearranging this into the standatd form,
KNID, gives
G= where K = Kc/150
Zl = - 0 . 5 .Q = - 1 p1 = -0.05 p2 = -0.1 p3 = -2
K(s - ZI)(S -
es - PlXS - PZ)(S - P3)
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FIGURE 15-7 Root-locus diagram for Example 15.2.
ln this case, them am four poles at 0, -0.05, -0.1, and -2 and two zems at -0.5 and -1. These am plotted in Fig. 15.7. Note that the three-action controller
contributes the pole at the origin and the zeros, -0.5 and - 1. The steps for plotting the root-locus diagram are as follows: 1. Since them am four poles, there are four branches emerging from them.
2. Three portions of the root locus are on the real axis between 0 and -0.05,
between -0.10 and -0.5, and between -1 and -2.
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