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ME410_HWSoln8

# ME410_HWSoln8 - SOLUTIONS T0 PROBLEMS a No Not symmetric On...

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Unformatted text preview: SOLUTIONS T0 PROBLEMS a. No: Not symmetric; On real axis to left of an even number of poles and zeros b. No: On real axis to left of an even number of poles and zeros c. No: On real axis to left of an even number of poles and zeros d. Yes e. No: Not symmetric; Not on real axis to left of odd number of poles and/or zeros 1‘. Yes g. No: Not symmetric; real axis segment is not to the left of an odd number of poles h. Yes jco j“) s—plane s—plane (a) (b) jco lo) s-plane S-plane (c) (d) jc) jm s-plane s-plane (E) (f) lu: 'jl a. C 5. K - 1 2 a. The characteristic equation is given by 1— 2(5—4—) = 0 01' S + 25 + 2 (1— Ky;2 + (2 — 2K)S + (2 — K) = 0 . The Routh array is s 2 1— K 2 — K S 2 — 2K 1 2 — K For K > 0 . the ﬁrst column of the Routh array will have no Sign changes when either K < 101' when K > 2 . The system is closed loop unstable in the range 1 < K < 2 . I}. There are no asymptotes in this root locus. To calculate the break-in and breakaway points, let 02+20+2 K=Z—.'I'hen 0' +20'+1 dK (0'2+20+112CF+2)—(O'2+20+212t7+2) —2(0'+1) S th ”I : _ o eo y do- (02 +2o'+1)2 (0'2 +2o‘+1)2 break-in point occurs when 0' = —1. It is helpful to calculate directly the root positions from the characteristic equation. The closed loop —2+2K:J4a KY 40 K12 K) ILK: K—l poles are located at 512 = : Zﬂ—K) 1—K It can be seen that when K < 1 , both poles are complex conjugate with a real part =-1_: when K D 2 the two poles are real. The root locus is: 1441 c. When K = lthe poles are at 00 . When K = 2 . the solution of the quadratic equation above gives S = 01—2 Break-in: a" = 4.5608 for K: 61.986: Breakaway: U = 6.43? for K: 0.01613. 11. Rum Locus Ed'rtur [C] I I I? Imag .ﬂ-ocia Paul Ilvic' (2X2) 2 4 — (23in at the m‘igin). (3X3) 9 L Closed-loop poles ‘Will be in the right-half-plane for K > Therefore, stable for K < 4E9; unstable for K > 4?? . 14. a. . - . . , . 2.50 —2 . 5-: —lU.UU —5.UU .00 5.00 10.00 Heal Root locus crosses the imaginaiy axis at the origin for K = 6. Thus the system is stable for K P 6. b. 1.50 .75 .00 1'- -.T-‘5 E 5 E 5 i 5 2 —L50 -3.00 -I.50 .00 1.50 3.00 Heal Root locus crosses the imaginaiy axis at j0.65 for K = 0.79. Thus. the system is stable for K 4: 0.?9. 20. - . 0 - 3 - 6 - -u Assume that root locus is epsulon away 00111 the asymptotes. Thus. 0'3 = j = -1'. g2k+1 gr: It 311: , Angle = 2 = q = ? . Hence 0'. = 7’. Checking assumption at —1 :I: leO yields -1 80“ With K = 9997.02. 26. .00 j- -2.5D -3.?5 E i i i i 5 E -5_DU -5.DD -2.5D .DD 2.50 5.90 Heal b. Root locus crosses 20% overshoot line at 1.8995r 4 117.1260 = - 0.866 ::_j 1.69 for K = 9.398. 4 TE c. T5 = 0.866 = 4.62 seconds. Tp =1.69 = 1.859 seconds 11. Other poles with same gain as dominant poles: G = 4.27. e. Root locus crosses inlaginaiy axis at :3'332 for K = 60. Therefore stability for K < 60. 37. Root Locus with 15 Percent Ovarshool Line 3. Searching the 15% overshoot line (a; = 0.51.7; 0 = 121.131“) for 180°. we ﬁnd the point 2.404 4121.1310 = 4.243 +j2.058. b. K= 11.09. 1:. Another pole is located Ieﬁ of -3. Searching for a gain of 11.09 in that region. we ﬁnd the third pole at 4.514. d. The third pole is not 5 times farther than the dominant pair from the 3'0) axis. the second-order approximation is estimated to be invalid. ...
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