Análise de Estabilidade do Sistema

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Problem 5.32PP
For the system in Fig.,
(a) Find the locus of closed-loop roots with respect to K.
(b) Find the maximum value of K for which the system is stable. Assume /C = 2 for the remaining 
parts of this problem.
(c) What is the steady-state error {e = r - y) for a step change in r?
(d) What is the steady-state error in y fo r a constant disturbance w1?
(d) What is the steady-state error in y fo r a constant disturbance w1?
(e) What is the steady-state error in y for a constant disturbance w2?
(f) If you wished to have more damping, what changes would you make to the system? 
Figure Control system
S tep -by-s tep s o lu tio n
(a) D O H {s) =
Step 1 of 8
iooii:(s+i)
s ’ (s+ 6 + 2 j) ( s + 6 0 -2 j) 
K = 0 points: 0, 0 , -6 ± J 2 
K = a points: -1 , ot, a a 
Asyn^totcs: 60®, 180®, 300®
Centroid = --------- —̂ - = -3 .66
3
Break away point: s= -5 .8 
Characteristic equation: 
s '(s '+ 12a+ 40)+ 100J!:(s+ l) = 0 
s‘ +12s’+40s^+ 100rs+ 100r = 0
Step 2 of 8
7 1 40 100 K
7 12 100 K
7 480-lOOJi:
12
100 K
7 I00 ji:(4 8 0 -i0 0 ji:)- i4 4 0 0 i:
480-100^:
7 100 K
Step 3 of 8
From the above array,
K >0
iooi!:(480-iooj!:) > 14400̂
48000^-10000 i:^ -14400£ ’ > 0 
-1 0 ,0 0 0 ^ “+33600^: > 0 
10,000A: > 33600
K > 3.36
M 80-100^>1 a ^
— - — j s ^ + m x = 0
At K = 3.36,
M 8 0 - 336^ a
I 12 r
+336 = 0
12s^+336 = 0
s = ±J5.29
The root loci intersect the imaginary axis at 5 s ±^5.29.
Step 4 of 8
Angles of departure:
= 180®-
Where = 2 -Z 'fe 
2