ieng3004 lecture 6-11-12 s1 root locus
TRANSCRIPT
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IENG3004IENG3004Control Systems TechnologyControl Systems Technology
Lecture 6Lecture 6
Root LocusRoot Locus
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Root Locus IntroductionRoot Locus Introduction
We know that the transient response of a system isgoverned by the roots of the characteristic equation
It is frequently necessary to analyse the variation intransient response due to changes in the values ofsystem parameters
2IENG3004 Control Systems Technology Lecture 6 Root Locus
cumbersome
Requires a graphical method to describe variation in transientresponse (Root Locus)
Root Locus traces the movement of the closed looproots as an OLTF parameter (usually gain) is changed
Roots can be real or complex, so the s-plane has real andcomplex dimensions
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Poles & ZerosPoles & Zeros
Zero:
A value of s which reduces the Transfer Function tozero
(a root of the numerator)
Pole
3IENG3004 Control Systems Technology Lecture 6 Root Locus
va
ue o s w c sen s e rans er unc on oinfinity
(a root of the denominator)
The poles of a Transfer Function are the same as the
roots of a the characteristic equation
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Poles & ZerosPoles & Zeros
Example: Determine the poles and zeros of theT.F.:
)22)(1(
)2(2 +++
+
ssss
s
ZEROS: s + 2 = 0 s = -2
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POLES: s = 0 s = 0s + 1 = 0 s = -1
s2
+ 2s + 2 = 0
s = -1 j
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Example: First Order SystemExample: First Order System
)3(
1
+s
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)3()3(1
)3(
R(s)
C(s)
Ks
K
sK
sK
++
=
++
+=
From the block diagram, the CLTF reduces to:
Location of the closed-loop pole is: s = -(3+K)
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Example: First Order SystemExample: First Order System
Root locus has one pole on the real axis:
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Example: First Order SystemExample: First Order System
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Simple example shows that: For K = 0, closed loop pole is located at -3
As K increases, the pole moves to the left for
increasing values of K As K increases, the transient response decays more
rapidly and thus steady-state is achieved morerapidly
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Root Locus Construction StepsRoot Locus Construction Steps
by way of an example
Kc(t)r(t)
+-)1(
1
+
ss
1
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+s
Step 1: Obtain Open Loop Poles & Zeros
(a) Root loci start at open-loop poles
(b) Root loci end at open-loop zeros or infinity
(c) The number of separate loci is equal to the number ofopen-loop poles
(d) Loci are symmetrical about the real axis
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Root Locus Construction StepsRoot Locus Construction Steps
)2)(1(OLTF ++== sss
K
KGH
Open loop poles at: 0, -1, -2
Step 1: Obtain Open Loop Poles & Zeros (cont)
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Real & Imaginary Axes Poles as x Zeros as o Locus is plotted
Step 2: Sketch Root Loci on Real Axis (if existing)
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Root Locus Construction StepsRoot Locus Construction Steps
Step 3: Derive asymptotes of Root Loci
The branches of a root locus approach a set of straight-
line asymptotes at large distances from the origin.(a) The asymptotes emanate from a point on the real
axis called the centre of asymptotes, given by:
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mn
zpm
i
i
n
i
i
c
=
== 11
where n = number of open-loop polesm = number of open-loop zerospi = location of i
th polezi = locaiton of i
th zero
103
)0()210(=
=c
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Root Locus Construction StepsRoot Locus Construction Steps
Step 3: Derive asymptotes of Root Loci (cont)
(b) The angle made by the asymptotes and the real
axis is given by:
k+=
)21( k = 0, 1, 2, 3
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mn
35,,
3
03
)21(
=
+=k
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Root Locus Construction StepsRoot Locus Construction Steps
Step 4: Determine breakaway points, b
A breakaway point (b) occurs on the real axis where two
or more branches depart from, or arrive at, the real axis.They occur where:
1= from characteristic=
dK
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0=dsdK
)(sGH
equation)
0)2)(1(
1 =++
+sss
K023 23 =+++ Ksss
0263 2 =+++dsdKss
if
1.58or42.06
24366=
=s
ds
then:
x
No RootLocus inthis area
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Root Locus Construction StepsRoot Locus Construction Steps
Step 5: Find K for marginal stability and c
Finding c will give us the intersection of the root locus
with the Imaginary Axis Use Routh-Hurwitz stability criterion to find this
(characteristic equation)023 23 =+++ Ksss
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Ks
Ks
Ks
s
0
1
2
3
03
3
213
21
Ks
Ks
Ks
s
0
1
2
3
03)6(
3
21
Therefore stability for 0 < K < 6
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Root Locus Construction StepsRoot Locus Construction Steps
Step 5: Find K for marginal stability and c (cont)
Form the Auxiliary Equation to find c
For the auxiliary equation (s2 row):
3s2 + K = 0
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3s2 + 6 = 0
s = j2
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Root Locus Construction StepsRoot Locus Construction Steps
Step 6: If there are no complex poles, plot Root Locus
2
jSymmetrical about the Real axis
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j
Re
XXX-2-3
-1
60o
-60o
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Root Locus Second Order ApproximationRoot Locus Second Order Approximation
Dominant Poles
The transient response of a system may be
equated to a sum of exponential terms Contribution of each term is determined by one of the
roots of the characteristic equation
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Some roots dominate the response
Example
)203)(4)(8()( 2 ++++=
ssss
K
sG
Gives a transient response in the form:
)2.4sin(.)(5.148
+++=
tCeBeAetyttt
Constantsdetermined byinitial conditions
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Root Locus Second Order ApproximationRoot Locus Second Order Approximation
)2.4sin(.)( 5.148 +++= tCeBeAety ttt
Pole Locations
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Transient response
Total response is
DOMINATED bydecaying sinusoidal
roots for this termmuch closer toImaginary axis
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Root Locus Second Order ApproximationRoot Locus Second Order Approximation
Since dominating roots are second order
the system may be approximated by an
equivalent second order system This gives us a system with well documented
time res onses.
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Performance given in normalised form in terms ofdamping ratio () and natural frequency (n)
The higher order system has been reduced to a
second order approximation, making it possible to useavailable performance data to effect a successfuldesign
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DeterminingDetermining andand nn of an Equivalent Secondof an Equivalent Second--OrderOrder
SystemSystem
Consider normalised form of the second order transferfunction:
22
2
2 nn
n
i
o
ssXX
++=
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The poles of this transfer function are located at:
)1( 2 = nn js
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DeterminingDetermining andand nn of an Equivalent Secondof an Equivalent Second--OrderOrder
SystemSystem
X
j
Pn
One pole from the complex conjugate pair is located at P,shown in this diagram:
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Re
n
0
d = n
cos
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DeterminingDetermining andand nn of an Equivalent Secondof an Equivalent Second--OrderOrder
SystemSystem
From the geometry of the diagram:
X
j
n
Pn
d = nnn
nn
OP
OP
+=
+=
)1(
)1()(22222
2222
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Re
0
cos
n=
==
n
ncos
In the figure, cos defines the damping line and thedistance of point P from the origin is equal to the naturalfrequency of the system
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DeterminingDetermining Gain KGain K
From the characteristic equation [ 1 + KGH (s) = 0 ]:
)(
1
sGHK
=)(
1
sGHK
=
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L
L
))()((
))()((
321
321
pspsps
zszszsKG(s)
+++
+++=
Then, from above, assuming unity feedback:
L
L
321
321
zszszs
pspspsK
+++
+++=
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DeterminingDetermining Gain KGain K
If the location of the closed-loop pole is known, then Kmay be determined graphically:
Q is the closedloop pole
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If the real and imaginary axes of the root locus diagramhave been drawn to the same scale then the lengths ofvectors P1Q, P2Q, P3Q, and ZQ can be measured
.
... 321
ZQ
QPQPQP
K=
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ExampleExample
)2)(1(
)(
++
=
sss
KsG
Determine the value of K whichwould give an equivalent
Sketch the root locus of the system with an open looptransfer function:
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o60)5.0cos( =
.
8.076.188.052.0
)2(.)1(.)(
1
==
++==
K
ssssGH
K
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Root Locus with Complex PolesRoot Locus with Complex Poles
In some circumstances the open-loop system hascomplex poles:
)2(1
22
nnsss ++
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In order to sketch the root locusfor the closed loop system, weneed to determine the angle of
emergenceof the locus from thecomplex poles
X
j
Re
is angle of
emergenceThe angle of emergencecan be determined usingthe angle condition
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Root Locus with Complex PolesRoot Locus with Complex Poles
Angle Condition
Key concept: if a point lies on the root locus, the argument
G(s) will be a multiple of 180o
.o
321 180ofmultipleofsum=++
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n)( 21.180321 +=
Since poles are on denominator,angles will be negative
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Root Locus with Complex PolesRoot Locus with Complex Poles
2s
+++
+
Angle Condition (cont)
If there is a zeroin the Transfer Function, then the zero will
be associated with a POSITIVE angle.e.g.:
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So the angle condition becomes:
32121.180
PPPZn)( =+
Or:)(arg180 i
o
D pHG +=
)(arg180 io
A zHG=
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Angle ConditionAngle Condition Example 1Example 1
Take an OLTF with four poles, two of which are complex,
and no zeros
Select a point just to the right
From Dunn Root Locus Tutorial (http://www.freestudy.co.uk)
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enough that angle 3 = 0
Measure or calculateother angles
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Angle ConditionAngle Condition Example 1Example 1
P1 = -10 P2 = -1P3 = -4 + 4j P4 = -4 4j
1 = tan-1 (4/6) = 33.7o
2 = 180 tan-1 (4/3) = 126.9o
= 90o
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180 = -33.7-126.9-90- 33 = -430.6
= -70.6
Apply:
432121.180 PPPPZn)( =+
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Angle ConditionAngle Condition Example 2Example 2
Similar OLTF as in Example 1, but introduce 1 zero.
P1 = 0 P2 = -10P
3= -4 + 4j P
4= -4 4j
Z1 = -1
1 = 180 tan-1 (4/4) = 135o
-1 o
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2 .
4 = 90o = 180 tan-1 (4/3) = 126.9o
180 = 126.9-135-22.7-90- 3
3 = -300.8 60
Apply:
32121.180 PPPZn)( =+
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Angle ConditionAngle Condition Example 3Example 3
P1 = 0P
2= -4 + 4j P
3= -4 4j
1 = 180 tan-1 (4/4) = 135o
o
Consider a 3-pole example:
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3
180 = 135-90- 3
3 = -120.8
Apply:
321
21.180PPPZ
n)( =+
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Angle ConditionAngle Condition Example 3Example 3
Consider the following system:
52
12
++ sss
1
X jP1
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P1 = -1 + 2j P2 = -1 2jP3 = 0
Re
Q
X
P3
P3
X
P1
P2P2
Set Q to be just to the right of P1
2 = 90o
3 = 180 tan-1 (2/1) = 117o
180 = -387- 1
1 = -27
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Angle ConditionAngle Condition Example 3Example 3
Resulting Root Locus plot:
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32
03
0)11(00=
=
= ==mn
ZP
m
ii
n
ii
c
35,,
3
03
)21(
=
+=k
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Effect of Gain on Second Order ApproximationEffect of Gain on Second Order Approximation
Kc(t)
+
-
r(t)
s
1
Controller Hydraulic
Actuator
Mass-spring-damper
(load)
22
2
1
nnss ++
Consider the following generalised third order system:
34IENG3004 Control Systems Technology Lecture 6 Root Locus
Looking at gain where < 1 (i.e.underdamped)
Varying K (0 K ), the closedsystem root locus looks like:
Is the second order approximation valid
for K giving pole set A? What about B?
Re
Xj
XP3
P1
P2 X
Dynamics
ActuatorPole set A
Pole set B
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Effect of adding Poles on Root LocusEffect of adding Poles on Root Locus
)()()(ass
KsHsG
+= 0>a
Take the function:
36IENG3004 Control Systems Technology Lecture 6 Root Locus
This gives two poles, one atthe origin and the other at a.
The following slides show theeffect of adding additionalpoles and zeros to the shapeof the Root Locus
From Glonaraghi & Kuo, Automatic Control Systems, 9th Edition, 2010, pp385-388
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Effect of adding Poles on Root LocusEffect of adding Poles on Root Locus
Add a pole at -b Add a pole at -c
y gy
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Effect of adding Poles on Root LocusEffect of adding Poles on Root Locus
Add a complexconjugate pair
y gy
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Effect of adding Zeros on Root LocusEffect of adding Zeros on Root Locus
Add a zero at -b
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Effect of adding Zeros on Root LocusEffect of adding Zeros on Root Locus
Add a complexzero with b realparts
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With three polesand one zero
42
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Open-Loop Pole-ZeroConfigurations & theCorresponding Root Loci
From Ogata, Modern ControlEngineering, 5th Edition, p289
T i l Q iT i l Q i43IENG3004 Control Systems Technology Lecture 6 Root Locus
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Tutorial QuestionTutorial Question
Plot the Root Locus by hand for both of these:
1.
2.
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