modern control system ekt 308 aspects of state equation & root locus method
DESCRIPTION
Transfer Function from State Equation (contd…)TRANSCRIPT
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Modern Control SystemEKT 308
Aspects of State Equation &
Root Locus Method
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Transfer Function from State Equation
.....(2) (1) ..... Given,
DuxCyBuxAx
....(4) )()()( (3) .... )()()(
condition) zero (initial Transform, Laplace Taking
sDUsXCsYsBUsXAsXs
1
1
)( ere, wh
)()()()(
)( )()()( (3), From
(sI-A)s
sBUssBU (sI-A)sX
sBU) (sI-A)X(ssBUsAXssX
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Transfer Function from State Equation (contd…)
DBsCsUsYsG
sUDBsCsYsDUsBUsC
sDUsXCsY
)()()()(
function, transfer theTherefore,)()( )(
)()()( )()()(
(4),Equation From
)()()( function, transfer theFind
0] [5y 10
4310
equations,matrix thehas system SISOA :
sUsYsG
x
uxx
Example
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Answer:
22
221
43433
431
434
431
)(
431
4310
1001
sss
ss
sssss
ss
s
ss
sAsI
345
10
43433
431
434
05
)()(
2
22
22
ss
sss
ss
sssss
DBsCsG
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Concept of Root Locus
)()(
)()()(
sqsp
sRsYsT
Characteristic equation (above fig), KsKG 0,0)(1
01)(/)(1)(
jsKGsKGsKG
,.......2,1,0
360.180)(/,1)(
k
ksKGsKG oo
Fig 1
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Concept of Root Locus (contd…)
The root locus is the path of the roots of the characteristic equation traced out in the s-plane as the system parameter varies from zero to infinity.
Root Locus ProcedureStep 1:
KsPKsF
0 where0)(1 as, arrange-reThen 0)(1equation, sticcharacteri theWrite
(1) ---- 0)(
)(1
follows, as zeros and poles of form in the writeRe
1
1
n
jj
M
ii
ps
zsK
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Root Locus Procedure (contd…)
Locate poles and zeros in the s-plane (‘x’ for poles, ‘o’ for zeros
as, written becan (1) Eq
(2) ... 0)()(11
M
ii
n
jj zsKps
Let us vary K from 0 to infinity. When K = 0,
n
n
jj ppppsps ,......,,, i.e. ,0)( 321
1
So, when K=0, values of s coincide with poles of P(s).
0)()(
,When
11
M
ii
n
jjK
zsKpsLt
K
Note:
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M
M
ii
M
ii
n
jjK
zzzzszs
zspsK
Ltei
,......,, giving ,0)(
)()(1..
3211
11
The locus of the roots of the characteristic equation 1+KP(s)=0 begins at the poles of P(s) and ends at the zeros of P(s) as K increases from 0 to infinity. If n > M, (n-M) branches of root locus approach the (n-M) zeros at infinity.
Step 2: Locate the segments of the real axis that are root loci. The root locus on the real axis lies in a segment of the real axis to the left of an odd number of poles and zeros.
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Example fo step 1 an 2.
0
41
121
.1)(1
equation sticcharacteri following thehas which system afor locusroot Find
2
ss
sKsGH
2- : eros Z4- 0, :Poles
)4()2(2
4)2(2)(
0)(14
)2(21 eq, sticCharacteri :1 Step
:
2
2
sss
ssssP
sKPss
sK
Ans
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