1 (142)

Prévia do material em texto

P ro b le m
iltiple poles at the origin. Sketch the root locus with respect to K for the equation 1 + KL(s) 
d the listed choices for L(s). Be sure to give the asymptotes and the arrival and departure 
gles at any complex zero or pole. After completing each hand sketch, verify your results u 
itlab. Turn in your hand sketches and the Matlab results on the same scales.
t ( j ) = '5+S)
r (1+1)^
- ? c r n «
Step-by-step solution
Step 1 of 23
U s ) :
^ (s + 8 )
Step 2 of 23
K = 0 points: 0, 0, -8 
K = apoints: a, a, a 
Asymptotes: 60®, 180®, 300®
Centroid = — = -2 .6 6 
3
Intersection o f root loci on the Imaginary axis:
Characteristic equation:
s ^ + S s ^ + K = 0
Step 3 of 23
1 0
s’ 8 K
s’ -K
8
0
s“ E
Step 4 of 23
As the necessary condition o f Routh’s criterion itself is not satisfied, we can say 
that the system is unstable.
As the roots are not complex conjugate, angles o f departure or arrival need not be 
calculated.
Step 5 of 23
I £ (S ) :
f f^ (s + 8 )
Step 6 of 23
K = 0 points: 0, 0, 0, -8 
K = apoints: a, ot, a, a 
Asymptotes: 45®, 135®, 225®, 315®
Centroid = — = - 2 
4
Characteristic equation:
s * + S ^ + K = 0
As all the coefficients o f the characteristic equation are not present, we can say 
that the system is not stable.
Step 7 of 23
s* (s+ 8 )
Step 8 of 23
K = 0 points: 0, 0, 0 ,0 , -8 
K = apoints: a, ot, a, a 
Asymptotes: 36®, 108®, 180®, 252®, 324®
Centroid = — = -1 .6 . 
5
Step 9 of 23
U s ) :
s + 2
^ (s + 8 )
Step 10 of 23
K = 0 points: 0, 0, -8 
K = apoints: ot» ot* -3 
Asymptotes: 90®, 270®
^ - 8 - ( - 3 )Centroid = ------ i— i =
2
Characteristic equation:
-2.5
s ^ ( s + 8 ) + i: ( s + 3 ) = 0 
s ^ + S s ^ + K s + 3 K = 0
Step 11 of 23 A.
s^ 1 0
s’ 8 3 K
s’ 8E ’- 3 ^
8
0
s“ 3 K
Step 12 of 23
For system stability, 
K >0
— > 0 i.e ., > 0
8
Therefore, in order to determine, the point at which, the root loci cross the 
imaginary axis.
We have to substitute K = 0 into the equation,
8 s“+3JT = 0 
8s“ = 0 
s = 0
Therefore, we can say that, the root loci start &om the origin and they do not 
intersect the imaginary axis anywhere.
Step 13 of 23 A
L ( S ) :
s+3
^ { s + A )
Step 14 of 23
K = 0 points: 0, 0, 0, -4 
K = apoints: a, a, a, -3
Asymptotes: 60®, 180®, 300®
Centroid = = z l = -0 .3 3
3 3
Characteristic equation: 
s*+ A s^+ K {s+ 3) = 0 
s * + A ^ + K s + 3 K = 0
As the coefficient o f s^ term is absent, we can say that the system is unstable.
Step 15 of 23
L ( S ) :
s ^ (s + 4 )
Step 16 of 23
K = 0 points: 0, 0, 0, -4 
K = apoints: -1 , -1 , a, a
Asymptotes: 90®, 270®
- ( - 1 - 1 ) .
Centroid = >
Characteristic equation: 
s ^ (s + 4 )+ A r (s + l)^ = 0 
f f * + 4 s ^ + J ^ (s ^ + 2 s + l) = 0
z 2 = _ ,
2
Step 17 of 23 ^
s* 1 K K
s’ 4 2 K
s’ AK-2K
4
K
s’ E ’ - 4 E 
XI2
0
s“ K
Step 18 of 23
> 0 
> AK
K > 4 (2)
The condition which satisfies both the equations (1) and (2) is K > A
Now, in order to get point at which the root loci intersect the imaginary axis, let 
us substitute K = 4 into the equatioa
— s’ + i r = 0
2 s ’ + 4 = 0 
s’ = - 2 
s = ±J^ .
Step 19 of 23
s’ (s-l-lO)
Step 20 of 23
Centroid = ' = - 6
K = 0 points: 0, 0, 0, -1 0 , -10 
K = apoints: -1 , -1 , a, a, a
Asymptotes: 60®, 180®, 300®
( - lO - lO ) - ( - l - l )
3
Characteristic equation: 
s’ ( s + io ) V j i : ( s + i ) “ = 0 
s’ (s’ -M 0 0 + 2 0 s )+ i:(s ’ +l-l-2s) = 0 
s’ + io o s ’ + 2 0 s*- i-J i:s ’ - i- 2 i! :s + j i : = o 
s’ + 2 0 s * - i- io o s ’ -i-Ji:s’ - i - 2 i : s + j i : = o
step 21 of 23
s’ 1 100 2 K
s* 20 K K
2000K-K
20
AOK-K
20
s’ 200K-K^-ia0K K
2000-K
s’ XY-WZ
r
0
s" K
step 22 of 23
■Where fT = 
X :
2000-K
20
40K-K
20
r :
Z
2000K-X̂ -7S0K
2000-K
Step 23 of 23
From the above array, 
K >0
39K {̂\220-K)-K{2000-K) > 0 
39 a : ’ (1 2 2 0 - j! :) > A : ( 2 0 o o - i: ) ^ 
47580 ^ -3 9 1 !:’ > 4xl0*+J!:’ -4000 
4 0 i : ’ -5 1 5 8 0 ii:+ 4 x l0 ‘ > 0 .