Upload
caraf
View
1.708
Download
124
Tags:
Embed Size (px)
DESCRIPTION
Chapter 5: Root Locus. Nov. 13-15, 2012. Two Conditions for Plotting Root Locus. Given open-loop transfer function G k (s). Characteristic equation. Magnitude Condition and Argument Condition. Rules for Plotting Root Locus. Break-in point. Breakaway point. - PowerPoint PPT Presentation
Citation preview
Chapter 5: Root LocusChapter 5: Root Locus
Nov. 13-15, 2012
Two Conditions for Plotting Root Locus
1
1
( )1
( )
m
g ii
k n
jj
K s zG s
s p
Given open-loop transfer function Gk(s)
Characteristic equation
m
ii
n
jj
g
zs
ps
K
1
1
|)(|
|)(|
..2,1,0,)12()()(11
kkpszsn
jj
m
ii
Magnitude Condition and Argument Condition
Content Rules1 Continuity and Symmetry Symmetry Rule
2
Starting and end points
Number of segments
n segments start from n open-loop poles, and end at m open-loop zeros and (n-m) zeros at infinity.
3 Segments on real axis On the left of an odd number of poles or zeros
4 Asymptote n-m segments:
5 Asymptote
3
Rules for Plotting Root Locus
mn
zpn
j
m
iij
1 1
)()(
,2,1,0,)12(
kmn
k =
4
7Angle of
emergence and entry
Angle of emergence
Angle of entry
8 Cross on the imaginary axis
Substitute s = j to characteristic equation and solve
Routh’s formula
n
j
m
zii
ijz k1 1
)12(
m
i
n
pjj
jip k1 1
)12(
6Breakaway
and break-in points
0
][
ds
sFd 0 sZKsPsF g
0 sZsPsZsP
m
i
n
j ii pz1 1
11
23/4/21 5
jBreak-in point
jBreakaway point
Rule 6: Breakaway and Break-in Points on the Real Axis
Use the following necessary condition
0 sZsPsZsP
0
d
dor 0
1
d
dor 0
d
d
s
K
sHsGss
sHsG g
ji pszs
11
23/4/21 6
3.3.Symmetry 5.Asymptote6.Breakaway and break-in points4. Segments on real axis1.Draw the open-loop poles and zeros
-1-2-3
2. Two segments
polepolezero
18012
)12(180
k
0
12
3)21(1 1
mn
zpn
j
m
iij
breaka
wayBreak-
in23a
aaa
2
1
1
1
3
1
Example 5.3.1: Given the open-loop transfer function, please draw the root locus.
)2)(1(
)3()()(
ss
sksHsG
3
21
1
1
a
aa
zs
ps
k m
ii
n
jj
=0.172k =5.818k
2
* )1()(
s
sKsGk
Example 5.3.2:
j
0
s1
s2
Conclusion: For the open-loop transfer function with one zero and two poles, the root locus of characteristic equation is probably a circle in the complex plane.
please prove that the root locus in the complex plane is a circle.
Given the open-loop transfer function
23/4/21 8
Example 5.3.3:
)2)(1()()(
sss
KsHsG
1. Open-loop poles and zeros
-1-2
2. Segments on real axis3. Asymptotes
103
210
180,6003
)12(180
k
4. Breakaway and break-in points
58.142.0 0]263[ 0
21
2
sssolutionssyieldssZsPsZsP
-0.42
5. Points across the imaginary axis
Ks
Ks
Ks
s
Ksss
0
1
2
3
23
3
63
21
023
K=6063 2 s
j1.414K=6
K=6-j1.414
23/4/21 9
3. Symmetry4. No segments on real axis5. Asymptote6. Breakaway and break-in points7.The point where the locus across the imaginary axis
Example 5.3.4:
1. Find poles and zeros
-1
2. 4 segments
14
11111 1
mn
zpn
j
m
iij
135454
)12(180或
k
311010123
0)]181)(181()1[(
3,21
23
2
jssSolutionssssyields
ds
jsjssd
Breakaway
ks
ks
ks
s
ks
kssss
19
014
44841914
0404
19241
01940244
0
1
2
3
4
234
)181)(181()1()()(
2 jsjss
ksHsG
484-4 =0 get =121k k
014014 2 s
16.3js =121-j3.16k
j3.16 =121k
Example 5.3.5
s
K
15.0
5.0
ss
sR sC
2
Please sketch the root locus with respect to K=[0,+∞).
23/4/21 11
Extension of Root Locus
Parameter Root Locus
Zero-degree Root Locus
Canonical form1
)(
)()(
1
1
n
jj
m
ii
gk
ps
zsKsG
Root locus gain
1. How to sketch the root locus for other parameters?
2. How to sketch if Gk(s)=1
ConventionalRoot Locus
23/4/21 12
Example 5.3.6: Ks is a ramp feedback
gain, please sketch the root locus with respect to Ks=[0,+∞).
010)102(2 sKs s
01102
102
ss
sK s
0)(')('1 sHsG
102
10)(')('
2
ss
sKsHsG s
1. Break-in and breakaway point2. Angle of emergence 198)10890(180
3.412.3
65.365.3
10
12.310
0
21
1
1
zs
psps
zs
ps
K
s
ds
sdF
m
ii
n
jj
s
90o
108o
198o
0101022 sKss s
5.3.2 Parameter Root Locus
23/4/21 13
Example: Sketch the root loci of the system with the open loop transfer function:
0 : )1(
)1()()( 1
1 Kss
sKsHsG
5.3.2 Zero-degree Root Locus
)1(
)1(
)1(
)1()()( 11
ss
sK
ss
sKsHsGAnalysis:
For this kind of systems, the characteristic equations are like as:
1)()( 0)()(1 111111 sHsGKsHsGK
0 : 1
)( gg
k KK
sG
,2 ,1 ,0 ,2)( kksGk
Magnitude equation
Argument equation
23/4/21 14
1 1
Root locus by using the sketching rules with the following modification:Real-axis: Left of even number of zeros or polesAsymptoteAngles of emergence and entry
1
1
K
js
)16.0(
3.0
1
K
s
)17.6(
3.3
1
K
s
For Kg varying from ∞- → 0 together with Kg =[0,+∞) simultaneously, the root loci are named the “complete root loci”.
m
i
n
pjj
jip k1 1
2
23/4/21 15
5-4 Application of Root Locus
Insert a zero )4)(1(1
)()(
)4)(1(1)(
sss
asKsG
sss
KsG g
kg
k
23/4/21 16
Add a pole to the open-loop transfer function
))(2(
)3()(
)2(
)3()( 1
21
1 asss
sKsGH
ss
sKsGH
Im
Re
Im
5
Re
Im
23 23
23
Re
Re
23
Im
1
23/4/21 17
Generally, adding an open zero in the left s-plane will lead the root loci to be bended to the left. The more closer to the imaginary axis the open zero is, the more prominent the effect on the system’s performance is.
Generally, adding an open pole in the left s-plane will lead the root loci to be bended to the right. The more closer to the imaginary axis the open pole is, the more prominent the effect on the system’s performance is.
5.4.1 The effects of Zeros and Poles
Attracting effect
Repelling effect
23/4/21 18
Example 5.4.1:
How to enlarge the stability region?)12)(15)(110(
6.3)(
sss
KsG c
k
)5.0)(2.0)(1.0(
036.0)(
sss
KsG c
k
)5.0)(2.0)(1.0(
sss
K
60o
j0.2ω1=0.17
Kc=0.778
-j0.2ω2=-0.17Kc=0.778
ω3=0Kc=0.278
)5.0)(2.0)(1.0(
)4.0()(
sss
sKsGk
ω1=0Kc=0.278
-0.06-0.027
278.0for 0axisimaginary with Intercept
06.0 :point Breakaway90 1.0:Asymptote
1
12
cK
a
278.0for 0778.0for 17.0
:axisimaginary with Intercept 0.027a:pointBreakaway
180,60 2.0:Asymptote
3
12
312
c
cK
K
23/4/21 19
5.4.2 Performance Analysis Based on Root Locus
Example 5.4.2: Given the open-loop transfer function
)15.0()(
ss
KsGk
please analyze the effect of open-loop gain K on the
system performance.
Calculate the dynamic performance criteria for K =5.
23/4/21 20
j
k¡ú¡
Þ
0
k £½0
k £½2
k £½0
k £½3
k £½10
k £½3
k £½2
3 j
k¡ú¡
Þ
[s ]
k £½1£ 2 £ 1
23/4/21 21
It is observed from the root locus that the system is
stable for any K.
For 0 < K < 0.5(0 < k < 1), there are two different
negative real roots.
For K=0.5(k=1), there are two same negative real roots.
For K > 0.5(k > 1), there are a pair of conjugate
complex poles.
For K=5( k=10), the closed-loop poles are 311 2
12 jjs nn
316.016.3
1,16.310 n
23/4/21 22
The criteria for transient performance can be given by
%35%100%100 05.11/ 2
eep
Peak time stn
p 05.11 2
Settling time %)5(33
stn
s
Overshoot
23/4/21 23
1. We can get the information of the system’s stability by checking whether or not the root loci are on the left half s-plane with the system’s parameter varying.
2. We can get some information of the system’s steady-state error in terms of the number of the open-loop poles at the origin of the s-plane.
3. We can get some information of the system’s transient performance in terms of the tendencies of the root loci with the system’s parameter varying.
4. If the root loci in the left half s-plane move to somewhere far away from the imaginary axis with the system’s parameter varying, the system’s response decays more rapidly and the system is more stable, vice versa.
Complement Use Matlab to sketch root locus
23/4/21 24
2
12
1
( ) +7 +12+3 +4= =
+1 +2 +3 +2( )
m
g iggi
k n
jj
K s z K s sK s sG s
s s s ss p
In Matlab:
num=[1 7 12]
den=[1 3 2]
rlocus(num,den)
sys=tf[num,den]
23/4/21 25