Modern Control System EKT 308 Root Locus Method (contd…)
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Transcript of Modern Control System EKT 308 Root Locus Method (contd…)
Modern Control SystemEKT 308
Root Locus Method (contd…)
Root Locus Procedure(contd…)
Root Locus ProcedureStep 1 (review):
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
Locate poles and zeros in the s-plane (‘x’ for poles, ‘o’ for zeros)
Step 2 (review): 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.
Magnitude and Angle Criterion
0)(1 equation, sticcharacteri theSuppose,
sKP
(1) ---- 0)(
)(1
1
1
n
jj
M
ii
ps
zsK
1)(...)()()(....)()(
Magnitude,21
21
n
M
pspspszszszs
K
integer.an is where,
360.180)(-....)(-)(-
)(...)( )( Angle,
21
21
l
lpspsps
zszszsoo
n
M
Magnitude and Angle Criterion (contd…)
432143
21
4433
2211
r.rr.r
:refresh Background
rrrr
1. figurein shown are )1(s toingcorrespond Angles
11 2
4.1.K-42-,
areeqn sitccharacteri theof roots theSay,
)02 ,( 0)2(
1 equation, sticCharacteri
Example,
1
21
2
K
Kss
KssorssK
Figure 1: Angle for s = s1
Note: Because complex roots appear as complex conjugate pairs, root loci must be symmetrical with respect to horizontal real axis.
Step 3:The loci proceed to the zeros at infinity along asymptotes centered at
Mn
zp
MnsPofzerossPofPoles
n
j
M
iij
A
AA
1 1
)()(
)()( centroid, Asymptote
where, angle with and
)1,.....(2,1,0 ,180.12
,asymptotes of angle And
MnkMn
k oA
Where n, the order of numerator polynomial and M is the order of denominator Polynomial
Example for step 3.
2 figin shown are axis real on the lociroot and zeros and Poles
0)4)(2(
)1(11
equation, sticcharacteri heConsider t
2
ssssKGH(s)
Figure 2: Root loci on real axis
ly.respective 2 and 1, 0,for 300 ,180 ,60
60).12(180.12
are, asymptotes of angles The
.339
14)1()4(2)2(
),asymptotes ofon intersecti called (also centroid Asymptote
ooo
k
kMn
k ooA
A
Asymptotes are shown in Figure 3
Figure 3: Asymptotes
Step 4:Determine where the locus crosses the imaginary axis (if it does so), using Routh-Hurwitz criterion. Hint: When root locus crosses the imaginary axis from left to right, the system moves from stability to instability.
Example: Complete first four steps of sketching root locus of the characteristic equation
0)44)(44)(4(
1
jsjsss
K
Step 1: Poles and zeros are shown in figure 4.
Figure 4: Poles and zeros
Step 2: There is a segment of root locus on the real axis between s=0 to s=-4 as shown in figure 4 above.
34
444
centroid, Asymptotes.315 ,225 ,135 ,45
3. 2, 1, 0,k ,180.4
12
are, asymptotes of Angles :3 Step
A
oo0
oA
oA
k
The asymptotes are drawn in figure 5.
Figure 5: Asymptotes
Step 4.
Kc
Kb
K
ss
RouthKssss
11
12812641
sss
array, 01286412 equation, sticCharacteri
0
1
2
3
4
234
568.8912/)128(33.53)128(33.5312
012)128(33.53 ,33.53
12)128(33.531 ,33.531
KK
KSo
Kcb
above. 5 figurein shown as 266.3at axisimaginary thecrosses locusroot the568.89, when So,
)266.3)(266.3(33.53))266.3((33.53
)67.10(33.5389.56853.33s
by,given areequation aux theof Roots
222
22
jsK
jsjsjs
s