# Root locus 5. tw4 = 450. Root Locus S5-1 S O L U T I O N S

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1 Root Locus S5-1 S O L U T I O N S Root locus 5 Note: All references to Figures and Equations whose numbers are not preceded by an "S" refer to the textbook. (a) Rule 2 is all that is required to find the branches, and Rule 1 tells us that the branches terminate on the two zeros. Thus the root locus is as sketched in Figure S5. 1a. Solution 5.1 (P4.5) tw4 Figure S5.1 Root loci for Problem 5.1 (P4.5). (a) Root locus for Figure 4.27a. s plane -4 -~ ON (a) (b) Rule 4 tells us that the average distance of the poles from the origin remains constant at s -%. By Rule 3 5, the two complex poles approach asymptotes of ± 60*. By Rule 6, the angle of the branch in the vicinity of the upper complex pole is,= 180* + E < z -- E < p = (S5.1) = 450 At the lower complex pole the angle will be -45* The point at which the complex pair crosses the imaginary axis can be estimated by using the asymptotes that intersect the real axis at s = -- %. They intersect the imaginary axis at s = 1% tan 60 = ±j2.31. The complex pole pair will cross the imaginary axis near this point.

2 S5-2 Electronic Feedback Systems The exact intersection of the root loci with the imaginary axis can be solved for using the Routh criterion. By inspection of the pole-zero diagram the characteristic equation (after clearing fractions) is P(s) = (s + 2)(s j)(s j) + a 0 f 0 The Routh array is = (s + 2)(s 2 + 2s + 2) + a 0 fo (S5.2) = S3 + 4S2+ 6s aofo aofo (S5.3) 5 a o aofo 0 The value a 0 fo = 20 will make the third row all zero. With this value of a 0 fo, we use the second row to write the auxiliary equation This has solutions at 4S = 0 s = ij\/6 = ±j2.45 (S5.4) (S5.5) This exact value is close to our earlier estimate of s = ±j2.31, as expected. In many cases, such an estimate will be sufficiently accurate, and the Routh computation may be avoided. Using all of the above information we sketch the root locus in Figure S5. lb. (c) Following the development in the textbook of Rule 7, for moderate values of a 0 f, we may ignore the pole at s = , and sketch the locus of the three other poles. Using Rule 3, the breakaway point between the poles at s = - 1 and s = -2 will occur at the solution of Multiplying gives d((s + 1)(s + 2)(s + 3)) 0 (S5.6) ds d(s 3 + 6s 2 + lls + 6) 0 (S5.7) ds

3 Root Locus S5-3 Figure S5.1 Figure 4.27b. (b) Root locus for ji>j j2.31 j2 j -2 U -jil -j2 -j2.45 -j2.31 (b) -j3

4 S5-4 ElectronicFeedbackSystems Figure S5.1 Figure 4.27c. (c) Root locus for joi j3.4 j3 j2 jl s plane po. -jl -j2 -j3.46 -j3 (c)

5 Root Locus S5-5 which gives This is solved by and 3s2+ 12s + 11 = 0 (S5.8) s = s = (S5.9) Only the solution at s = is meaningful here (see p. 127 in the textbook), so the breakaway point is at s = By Rule 5, the loci approach asymptotes of + 600, and these asymptotes intersect the real axis at s = -2. The asymptotes cross the imaginary axis at s = ±j2 tan 60 = ±j3.46. Using this information, we sketch the root locus as shown in Figure S5.1c. (d) By Rule 7, for low values of af,, the root locus will be the same as in part c. However, by Rule 2, we know that there is a branch of the locus to the left of the zero at s = Thus, while the two complex poles initially enter the right halfof the s plane, they must at some point turn around, reenter the left half of the s plane, and rejoin the real axis at some point to the left of the zero at s = After rejoining the real axis, one pole will move to the right to the zero at s = -2000, and the other pole will move off to the left to infinity. Thus, the root locus appears as sketched in Figure S5. ld. This system has the interesting property of being stable for low and high values of af, but unstable for intermediate values of af,.

6 S5-6 ElectronicFeedbackSystems Figure S5.1 Figure 4.27d. (d) Root locus for 0* (d) (e) By Rule 2, branches exist on the real axis between s = 0 and s = -2. By Rule 5, because there are four poles and no zeros, all four poles eventually (for large aofo) approach infinity along asymptotes of ± 450 and ± The asymptotes intersect at s = -2. By Rule 6, the loci leave the pole at s = -1 + j at an angle of - 90*, and leave the pole at s = -1 - j at an angle of Combining this information we sketch the root locus as in Figure S5.le.

7 Root Locus S5-7 Figure S5.1 Figure 4.27e. (e) Root locus for j " jl s plane -2 0~ -jl (e)

8 S5-8 Electronic Feedback Systems Solution 5.2 (P4.7) The noninverting configuration is as shown in Figure S5.2. The loop transmission for this topology is -a(s)f4, where f4 R 2. As long as the a 0 fo product is large, that is, as long as 4 R, + R2 >> 10-6, the closed-loop gain is well approximated by - = R, + R 2 R2~ Figure S5.2 amplifier. Noninverting V, + R, At this point, a sketch of the root locus is helpful for understanding the system behavior as f4 is varied. The closed-loop poles start at the poles of a(s) for small values off4, and move as shown in Figure S5.3 as f is increased. (Note that the axes are not drawn to scale.) The root locus has been drawn by recognizing that two of the poles will approach asymptotes of ± 60, and eventually enter the right half of the s plane, as f4 grows. The third pole follows the real axis off to the left. Other details, such as breakaway points, are not important for this analysis, and we do not solve for them. The dashed lines indicate points where the damping ratio for a complex pair is equal to 0.5. From the root-locus sketch it is apparent that there is some value of f4 for which the poles will lie on this line. Following the development on p. 128 of the textbook, when the system damping ratio is 0.5, its characteristic equation is given by Equation 4.62 as P'(s) = s 3 + (y + 2#)s #(y + 2#)s + 4Y# 2 (S5.10) The closed-loop characteristic equation for the amplifier is 1minus the loop transmission, which is (when cleared of fractions)

9 Root Locus S5-9 Figure S5.3 Root locus for Problem 5.2 (P4.7). IW a 600 P(s) = (O.ls + 1)(10-6s + 1) f 0 (S5.11) ~ 10-"s X 10~ 7 s ls f 0 where three insignificant terms have been dropped. Now, we multiply P(s)through by 10" and equate with P'(s) s3 + 2 X 106s S + 101'f = S 3 (S5.12) + (y + 20)s (y + 2#)s + 4,Y#2 Equating the coefficients of s 2 yields -y+ 20 = 2 X 106 (S5.13)

10 S5-10 ElectronicFeedbackSystems Equating the coefficients of s, and substituting in from Equation S5.13 gives Thus, and 2#(2 X 106) = 1012 (S5.14) = 2.5 X 10' y = 1.5 X 106 Then the remaining term is used to find that (S5.15) fo = 3.75 X 10-2 (S5.16) With this value off 0, the low-frequency loop transmission is large, and the low-frequency closed-loop gain is given by - which is fo equal to Resistor values to realize this gain can be chosen by setting or R, + R2 = 26.7 (S5.17) R2 R, = 25.7R 2

11 MIT OpenCourseWare RES Electronic Feedback Systems Spring 2013 For information about citing these materials or our Terms of Use, visit:

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