cise302 081 root locus - kfupm1 cise302_ch 7 al-amer2008 ١ cise302: linear control systems dr....
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CISE302_ch 7 Al-Amer2008 ١
CISE302: Linear Control Systems
Dr. Samir Al-Amer
Chapter 7: Root locus
Term 081
CISE302_ch 7 Al-Amer2008 ٢
Learning Objectives
Understand the concept of root locus and its role in control system designBe able to sketch root locus and use MATLAB to plot the root locusRecognize the role of root locus in parameter design and sensitivity analysisBe able to design controllers using root locus
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CISE302_ch 7 Al-Amer2008 ٣
Outlines
The concept of root locus
CISE302_ch 7 Al-Amer2008 ٤
Root locus
The location of the roots of the characteristic equation determines the systems response.Modifying one or more of the system’s parameter cause the roots of the characteristic equation to change.Root locus is a graphical method that determines the location of the roots as one parameter change.
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Root locusRoot locus was developed by Evans in 1948.Used to analyze and design systemsWe use it to ensure that the roots of the characteristic equation are in a desirable regionCan be used to study sensitivity of the pole location w.r.t changes in the parameter.
Desirableregion
CISE302_ch 7 Al-Amer2008 ٦
Root Locus Using MATLAB
rlocus(1,[1 2 0])
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Example 1
KrKr
Kr
arerootsKs
−−−=−+−=
−±−=
=++
11,112
442
:02s equation sticcharacteri the
21
2,1
2 )(sR E(s)Y(s)
_ K )2(1+ss
XX
Root Locus
CISE302_ch 7 Al-Amer2008 ٨
Root locus can be used to study sensitivity of system w.r.t variations in the parameter
K=0:0.1:10; G=rlocus(1,[1 2 0],K); plot(G,’.’)
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CISE302_ch 7 Al-Amer2008 ٩
Mathematics of root locus c
)(sR E(s)Y(s)
_ K )(sG
integer:360180)(;1
01)(0)(1
kksKGKG
jsKGKGsKG
±=∠=⇒
+−=∠
=+
CISE302_ch 7 Al-Amer2008 ١٠
Question c
)(sR E(s)Y(s)
_ K )(sG
? and 0between variesK as change0)(1 equation
sticcharacteri theof roots thedoes How
∞=+ sKG
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CISE302_ch 7 Al-Amer2008 ١١
Example
200)2(0
12)s(s
K)(
−==→=+→=
=+
=
sorsssKWhen
sKG
)(sR E(s)Y(s)
_ K )2(1+ss
The locus of the roots of 1+KG(s)=0 starts at the poles of G(s)
CISE302_ch 7 Al-Amer2008 ١٢
Root Locus Procedure
Initial Step:Express the characteristic equation in the form
1+K P(s) = 0 Factor P(s) and rewrite the equation in the form
0)(
)(1
1
1 =−
−+
∏
∏
=
=n
ii
m
ii
ps
zsK
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CISE302_ch 7 Al-Amer2008 ١٣
Example 1
Apply the initial step of the root locus procedure for the system
sssK
23 23 ++_
CISE302_ch 7 Al-Amer2008 ١٤
Example
Step 1: Express the characteristics as
023
11
0)(1
23 =++
+
=+
sssK
sPK
sssK
23 23 ++_
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CISE302_ch 7 Al-Amer2008 ١٥
Example
Step 2: Factor P(s)
0)2)(1(
11
0)(1
=++
+
=+
sssK
sPK
sssK
23 23 ++_
n = 3
m = 0
CISE302_ch 7 Al-Amer2008 ١٦
Root Locus Procedure Locate the open loop poles and zeros of P(s)
Locate the poles and zeros of P(s) on the s-plane. Use X for poles and O for zeros
0)2(
)1(1 =+−
+sssK
XX-2 1
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The root locus starts at the poles of P(s) and terminates at the zeros of P(s) or at infinity
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Root Locus Procedure # of separate root loci
The number of separate root loci = The number of poles of P(s) Note that ( # poles ≥ # zeros)
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Example 2
How many separate root loci does the root locus of the system has?
sssK
23 23 ++_
0)2)(1(
11
0)(1
=++
+
=+
sssK
sPK
# of poles =3Three separate root loci
CISE302_ch 7 Al-Amer2008 ٢٠
Root Locus ProcedureSegments of real line that is part of root locus
2: Segments of real line:A segment of real line is a part of root locus if sum of # of poles plus # of zeros to right of that segment is odd.
XXX
3 2 1 0
Yes No Yes No
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CISE302_ch 7 Al-Amer2008 ٢١
ExampleA segment of real line is a part of root locus if sum of # of poles
plus # of zeros to right of that segment is odd.
XXX
6 5 3 2 1 0
No Yes Yes No Yes No
X
Double poles
CISE302_ch 7 Al-Amer2008 ٢٢
ExampleA segment of real line is a part of root locus if sum of # of poles
plus # of zeros to right of that segment is odd.
XXX
5 4 3 2 1 0 Yes No Yes No Yes No
X
X
Complex poles or zeros do not
affect the count
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CISE302_ch 7 Al-Amer2008 ٢٣
Example
Step 2: Segments of real axis that are parts of the root locus
0)2)(1(
11 =++
+sss
K
XXX
CISE302_ch 7 Al-Amer2008 ٢٤
Asymptotes
If n > m # of poles of P(s) ># of zeros
The root locus approaches infinity at n-mdirections.
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Asymptotes
Step 3: # of asymptotes = n – m =3
300,180,60
2,1,018012
103
}0{}210{
=Φ
=−+
=Φ
−=−
−−−=
−−
= ∑ ∑
A
A qmn
qmn
zerospolescentroid
XXX60
-2 -1
CISE302_ch 7 Al-Amer2008 ٢٦
Example
How many asymptotes?
0)1(
11 =+
+ss
K
XX
n=2, m=0, n-m=22 asymptotes
Centroid =(0+(-1))/2
=-0.5
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CISE302_ch 7 Al-Amer2008 ٢٧
Intersection with Imaginary axis
To find intersection with imaginary axis use Routh Hurwitz method
Ks
KKs
Kss
Kssssss
K
0
1
2
3
23
23
6onIntersecti3
63
21023
023
11equation sticCharacteri
=−
=+++
=++
+
XXX60
CISE302_ch 7 Al-Amer2008 ٢٨
Intersection with Imaginary axis
263
s Equation006321
2
1
2
3
jss
Auxiliarys
Kss
±=⇒+
=
XXX60
Intersection with imaginary axis
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CISE302_ch 7 Al-Amer2008 ٢٩
CISE302_ch 7 Al-Amer2008 ٣٠
Breakaway points
Breakaway points are points at which the root loci breakaway from real axis or the root loci return to real axis.At breakaway points
Solve the above equation to determine the breakaway point. Select solution that are in segments of real axis that is part of root locus
0)(=
dssdP
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CISE302_ch 7 Al-Amer2008 ٣١
Breakaway points
To find breakaway points
( )( )
422604226057741 pointsaway break
023
26310
023
1
223
2
23
.Select. and.
sssss
sssdsd
−−−
=++
++−
=⎟⎠⎞
⎜⎝⎛
++ XXX-1-2
CISE302_ch 7 Al-Amer2008 ٣٢
Symmetry of root locus
The root locus is symmetric with respect to the real axis.
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CISE302_ch 7 Al-Amer2008 ٣٣
Angle of departure
Sum of angle contributions of poles and zeros (measured with standard reference) = 180+360kIn the example all poles/zeros are realNo need to do this
CISE302_ch 7 Al-Amer2008 ٣٤
Angle criteria
At all points of the root loci : Sum of angle contributions of poles and zeros (measured with standard reference) = 180+360k
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CISE302_ch 7 Al-Amer2008 ٣٥
Example 8Draw root locus for the following system
2)1(
++
ssK
_
1,21)(
0211
:
==++
=
=++
+
mnsssP
ssK
EquationsticCharacteri
X
CISE302_ch 7 Al-Amer2008 ٣٦
02
)1(1 =++
+ssKofLocusRoot
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CISE302_ch 7 Al-Amer2008 ٣٧
Example 9Draw root locus for the following system
)3)(2)(1()5(+++
+sss
sK_
1,3,)3)(2)(1(
5)(
0)3)(2)(1(
51
:
==+++
+=
=+++
++
mnsss
ssP
ssssK
EquationsticCharacteri
CISE302_ch 7 Al-Amer2008 ٣٨
Example 9
21
13)5(321
270,90213#
5:,3,2,1:
1,3,)3)(2)(1(
5)(
−=
−−−−−−
=
=−=−−−−
==+++
+=
Centroid
Anglesymptotesof
ZerosPoles
mnsss
ssP
XXX
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CISE302_ch 7 Al-Amer2008 ٣٩
0)3)(2)(1(
51 =+++
++
ssssK
oflocusroot
CISE302_ch 7 Al-Amer2008 ٤٠
Example10
How do we draw root locus for this system?
Ksss +++ 231
23_
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Example10
How do we draw root locus for this system?
Ksss +++ 231
23_
Initial Step: Express the characteristics as
?)(0)(1
sPisWhatsPK =+
CISE302_ch 7 Al-Amer2008 ٤٢
Example10
Initial Step: Express the characteristics as
( )
1231)(
0123
11
1230123
023
11:Equation sticCharacteri
23
23
2323
23
+++=⇒
=+++
+
++++==++++
=+++
+
ssssP
sssK
KsssKsssKsss
Continue the Root Locus Procedure
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CISE302_ch 7 Al-Amer2008 ٤٣
CISE302_ch 7 Al-Amer2008 ٤٤
Example11
How do we draw root locus for this system?
)2(2
2 Kssss
+++
_
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CISE302_ch 7 Al-Amer2008 ٤٥
Example11
Initial Step: Express the characteristics as
( )
22)(
022
1
222202)2(
0)2(
21:Equation sticCharacteri
23
23
23
232
2
+++=⇒
=+++
+
++++
++++==++++
=++
++
sssssP
ssssK
KsssssKssssKsss
Kssss
Continue the Root Locus Procedure
CISE302_ch 7 Al-Amer2008 ٤٦
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CISE302_ch 7 Al-Amer2008 ٤٧
Controller Design Based on Root Locus
The idea here is to select the parameter K so that all the poles in the desired location.
Desiredregion
CISE302_ch 7 Al-Amer2008 ٤٨
The desired region is obtained to satisfy the given specificationsDraw the desired region on the root locusSelect the gain K such that all poles of the closed loop system are in the desired region
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CISE302_ch 7 Al-Amer2008 ٤٩
Desiredregion
CISE302_ch 7 Al-Amer2008 ٥٠
What can you tell about behavior of this system?
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CISE302_ch 7 Al-Amer2008 ٥١
CISE302_ch 7 Al-Amer2008 ٥٢
What can you tell about this system?
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Keywords
Root locusBreakaway pointsAsymptotesCentroidAngle of departureAngle of arrival
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