eee 350 test solution
DESCRIPTION
eee350 testTRANSCRIPT
EEE 350 – Test II Solution
1.
1 ;
114
92
sH
sss
sKsG
a) P1 = 0; P2 = ‐2 + 2.65j; P3 = ‐2 – 2.65j; n = 3;
b) z1 = ‐9; m = 1;
c) n – m = 2
d) Intersection with real axis (asymptote), = 2.5; angle of asymptote = 90 (k=0) & ‐90 (k=1) e) B/away and B/in points
S1 = ‐13.03 Real roots, however, not on locus, Not a breakaway point
S2 = ‐1.24+1.5j Complex roots, lack of breakaway
S3 = ‐1.24‐1.5j Complex roots, lack of breakaway
The system has no breakaway point.
f) Point where locus cross imaginary axis
Use Routh table for 09114 23 KsKss
K = 44/5 @ 8.8; s = j4.45 g) Angle of departure
Upper half pole on imaginary axis:
17
2090127180
180
zerospoles
The other pole on the lower half will have angle of departure = 17
h) Damping ratio, = 0.5 will have an angle of [cos‐1(0.5)] = 60 (from the horizontal axis)
i) Coordinate of intersection between locus and damping ratio line (1.5,2.6)
18420419
5.12
2.5tan0
6.2
5.1tan90
5.19
6.2tan 111
3211
PPPz
poleszeros
j) The dominant closed‐loop poles, s = ‐1.5 2.6j
k)
js
s
sssK
6.25.1
2
9
114
; K = 1
d
2.
16
13
1
181
16
613
31
631
sss
K
sss
K
sss
KsG
a) Gain, K
Assume K = 1,
Magnitude for gain: 20 log (1/18) = ‐25 dB
Phase for gain: 0
b) (s+ 1); Pole at = 1
Magnitude: 21log20 ; ‐20 dB/dec;
Phase: 1tan ; ‐45/dec (0.1 10)
c) (s+3); Pole at = 3
Magnitude: 211.01log20 ; ‐20 dB/dec;
Phase: 33.0tan 1 ; ‐45/dec (0.3 30)
d) (s+6); Pole at = 6
Magnitude: 2028.01log20 ; ‐20 dB/dec;
Phase: 17.0tan 1 ; ‐45/dec (0.6 60)
1 3 6
Pole at ‐1 ‐20 ‐20 ‐20
Pole at ‐3 0 ‐20 ‐20
Pole at ‐6 0 0 ‐20
Total slope (dB/dec) ‐20 ‐40 ‐60
0.1 0.3 0.6 10 30 60
Pole at ‐1 ‐45 ‐45 ‐45 0
Pole at ‐3 0 ‐45 ‐45 ‐45 0
Pole at ‐6 0 0 ‐45 ‐45 ‐45 0
Total slope (deg/dec) ‐45 ‐90 ‐135 ‐90 ‐45 0
e) Based on equation, open loop system is stable
Based on Bode plot, positive gain margin obtained. The phase margin will always be positive
Closed‐loop system stable
At phase of ‐180, the frequency is 5.4 rad/sec. Gain margin = 45 dB
The range of K that yield stability
20 log K = 45 dB, K = 178. Range of K: 0 < K < 178
Additional:
If student assume K = 18, the magnitude of Bode plot will start at 0 dB.
It will give a gain margin of approximately 20 dB. The range of K that yield stability
20 log K = 20 dB, K = 10. Since initial we set K = 18, thus range of K: 0 < K < 180
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