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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) splane Splane (c) (d) jc) jm splane splane (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 breakin 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
breakin 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 Breakin: 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 .ﬂocia Paul Ilvic' (2X2) 2 4 — (23in at the m‘igin). (3X3) 9 L Closedloop poles ‘Will be in the righthalfplane 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 secondorder approximation is estimated to be invalid. ...
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 Spring '08
 Wang

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