Post on 14-Jan-2016
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Root-locus Technique for
Control Design
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INTRODUCTION
Stability and the characteristics of transient response of closed-loop systems
Locations of the closed-loop poles
Problems to solve characteristic equation : 1. Difficult for a system of third or higher order. 2. Tedious for varying parameters.
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VARYING THE LOOP GAIN K
In many systems, simple gain adjustment may move the closed-loop poles to desired locations.
Then the design problem may become the selection of an appropriate gain value.
It is important to know how the closed-loop poles move in the s plane as the loop gain K is varied.
The open loop gain K is an important parameter that can affect the performance of a system
R(s)K
Y(s)
H(s)
G (s)-
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Root Locus:
the locus of roots of the characteristic equation of the closed-loop system as a specific parameter (usually, gain K) is varied form 0 to ∞.
The advantages of RL approach:
1. Avoiding tedious and complex roots-solving calculation
2. Clearly showing the contributions of each loop poles or zeros to the location of the closed-loop poles.
3. Indicating the manner in which the loop poles and zeros should be modified so that the response meets system performance specifications.
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START FROM AN EXAMPLE
Consider a second-order system shown as follows:
)1( sskR(
s) -
Y(s)
The roots of CE change as the value of k changes.
Closed-loop TF:
kss
ks
2)(
Characteristic equation (CE): 02 kss
Roots of CE: kk
s 412
1
2
1
2
4112,1
When k changes from 0 to ∞, how will the locus of the roots of CE move?
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1,2
1 11 4 , : 0
2 2s k k
1/ 4k= 2/121 ss
01 s 1s2 k= 0
0 1/ 4k As the value of k increases, the two negative real roots move closer to each other.
A pair of complex-conjugate roots leave the negative real-axis and move upwards and downwards following the line s=-1/2.
1/ 4 k
- 1/2- 1 0
On the s plane, using arrows to denote the direction of characteristic roots move when k increases, by numerical value to denote the gain at the poles.
0k 0k
1
4k
k
k
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By Root loci, we can analyze the system behaviors
(2)Steady-state performance: there ’s an open-loop pole at s=0, so the system is a type I system. The steady-state error is 0 under step input signal R/Kv under ramp signal v0t
∞ under parabolic signal.
(1)Stability: when Root loci are on the left half plane, then the system is definitely stable for all k>0.
(3)Transient performance: there’s a close relationship between root loci and system
behavior
on the real-axis: k<0.25 underdamped;
k=0.25 critically damped
k>0.25 underdamped.
However, it’s difficult to draw root loci directly by closed-
loop characteristic roots-solving method.
The idea of root loci : by loop transfer function, draw closed-loop root loci directly.
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RELATIONSHIP BETWEEN ZEROS AND POLES OF G(S)H(S) AND
CLOSED-LOOP ONES
G(s)
H(s)
-
R(S) Y(s)
Forward path TF: Closed-loop TF:( )G s
)()(1)(
)(sHsG
sGs
Feedback path TF: ( )H s
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Suppose that G(s)H(s) has m zeros (Zi) and n poles ( Pi), the above equation can be re-written as
1
1
( )( ) ( ) 1
( )
m
iin
ii
K s ZG s H s
s P
K* varies
from 0 to ∞
To draw Closed-loop root locus is to solve the CE
1 ( ) ( ) 0G s H s That is
( ) ( ) 1G s H s RL equation
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Magnitude equation (ME)
1
1
1
m
iin
ii
K s Z
s P
Angle equation (AE)Angle equation (AE)
1 1
( ) ( )
(2 1) , 0, 1, 2,
m n
i ii i
s Z s P
l l
Since G(s)H(s) is function of a complex variable s, the root locus equation can be described by the following two equations:
1
1
( )( ) ( ) 1
( )
m
iin
ii
K s ZG s H s
s P
Magnitude equation is related not only to zeros and poles of G(s)H(s), but also to RL gain ;
Angel equation is only related to zero and poles of G(s)H(s).
Use AE to draw root loci and use ME to determine the value of K on root loci.
RL equation:
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2 21 2 2
'
1
( 1)( ) ( )
( 1)( 2 1)
( 1/ )
( 1/ )( )( )n d n d
K sG s H s
s T s T s T s
K s
s s T s j s j
' 21 2/K K TT
21/n T 2
2d T1
/11 z
1 4
1 11 1
1 1 2 3 4
( ) ( ) ( ) ( )
( )
i ii i
G s H s s z s p
Poles of G(s)H(s) ( × )
Zeros of G(s)H(s) (〇)
Angel is in the direction of anti-clockwise
For a point s1 on the root loci, use AE
Use ME
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4131211'
zs
pspspssK
Example 1
2 11/p T dn4,3 jp 1 0p
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Unity-feedback transfer function:s
K)s(G
Test a point s1 on the negative real-axis
11 1
1 1
( ) ( ) |
180
m n
i i si i
s z s p
s p
Test a point outside the negative real-axis s2=-1-j
21 1
2 1
( ) ( ) |
135
m n
i i si i
s z s p
s p
All the points on the negative real-axis are on RL.
All the points outside the negative real-axis are not on RL.
Example 2
One pole of G(s)H(s): 1 0p
No zero.
1. Find all the points that satisfy the Angle Equation on the s-plane, and then link all these points into a smooth curve, thus we have the system root locus when k* changes from 0 to ∞;2. As for the given k, find the points that satisfy the Magnitude Equation on the root locus, then these points are required closed-loop poles.
However, it’s unrealistic to apply such “probe by each point” method. W.R. Evans (1948) proposed a set of root loci drawing rules which simplify the our drawing work.
Probe by each point:
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4-3 RULES TO DRAW REGULAR ROOT LOCI
(suppose the varying parameter is open-loop gain K )
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2 Number of Branches on the RL
3 Symmetry of the RL
4 Root Loci on the real-axis
5 Asymptotes of the RL
6 Breakaway points on the RL
7 Departure angle and arrival angle of RL
8 Intersection of the RL with the imaginary axis
9 The sum of the roots and the product of the roots of the closed-loop characteristic equation
PROPERTIES OF ROOT LOCI1 and points of Root Loci0K K
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Root loci originate on the poles of G(s)H(s) (for K=0) and terminates on the zeros of G(s)H(s) (as K=∞).
MagnitudeEquation:
1
1
n
iim
ii
s PK
s Z
=
0K Root loci start from poles of G(s)H(s)
K Root loci end at zeros of G(s)H(s).
1
1
1
m
iin
ii
K s Z
s P
1 and points0K K
is P
is Z
1
1
( )( ) ( ) 1
( )
m
iin
ii
K s ZG s H s
s P
RLEquation:
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2 Number of branches on the RLnth-order system, RL have n starting points and RL have n branches
For a real physical system, the number of poles of G(s)H(s) are more than zeros , i.e. n > m.
m root loci end at open-loop zeros ( finite zeros) ;( n - m ) root loci end at (n - m) infinite zeros.
The order of the characteristic equation is n as K varies from 0 to ∞ ,n roots changen root loci.
1
1
( )( ) ( ) 1
( )
m
iin
ii
K s ZG s H s
s P
RLEquation:
n root loci end at open-loop zeros ( finite zeros) ;
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3 Symmetry of the RL
The RL are symmetrical with respect to the real axis of the s-plane.
The roots of characteristic equation are real or complex-conjugate.
Therefore, we only need to draw the RL on the up half s-plane and on the real-axis, the rest can be obtained by plotting its mirror image.
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On a given section of the real axis, RL for k>0 are found in the section only if the total number of poles and zeros of G(s)H(s) to the right of the section is odd.
4 RL on the Real Axis
zero : z1 poles : p1 、 p2 、 p3 、 p4 、 p5
Pick a test point s1 on [p2 , p3]
The sum of angles provided by every pair of complex conjugate poles are 360°;
The angle provided by all the poles and zeros on the left of s1 is 0°.
The angle provided by all the poles and zeros on the right of s1 is 180°;
1 5
1 1 1 11 1
( ) ( ) ( ) ( )i ii i
G s H s s z s p
1 1( ) ( ) (2 1)180G s H s l = ?
jw
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Consider the following loop transfer function
2 2
( 1)( 4)( 6)( ) ( )
( 2)( 3)
K s s sG s H s
s s s
On the right of [-2 , -1] the number of real zeros and poles=3.
On the right of [-6 , -4], the number of real zeros and poles=7.
Example
Determine its root loci on the real axis.
Repeated poles:S-plane
0-1-2-3-4-5-6
Poles: Zeros:
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5 Asymptotes of RL
(2 1)180a
i
n m
mn
zpm
1ii
n
1ii
a
=
When n ≠ m, there will be 2|n-m| asymptotes that describe the behavior of the RL at |s|=∞.
The angles between the asymptotes and the real axis are( i= 0 , 1 , 2 ,… ,n-m-1 ) :
The asymptotes intersect the real axis at :
1
1
( )( ) ( ) 1
( )
m
iin
ii
K s ZG s H s
s P
RLEquation:
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*(0.25 1) ( 4)( ) ( )
( 1)(0.2 1) ( 1)( 5)
K s K sG s H s
s s s s s s
The angles between the asymptotes and the real axis are
3 poles : 0 、 -1 、 -5
1 zero : -4n-m = 3 -1 = 2
The asymptotes intersect the real axis at
113
)4()5()1()0(
mn
zpm
1ii
n
1ii
a
=
(2 1)180, 0,1a
ii
n m
90 ,270a
Example Consider the following loop transfer function
Determine the asymptotes of its root loci.
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2
( 1)( ) ( )
( 4)( 2 2)
K sG s H s
s s s s
4 poles : 0 、 -1+j 、 -1-j 、 -4
1 zero : -1
n-m=4-1=3
The asymptotes intersect the real axis at
1 1
(0) ( 1 ) ( 1 ) ( 4) ( 1) 5
4 1 3
n m
i ii i
a
p z
n mj j
=
The angles between the asymptotes and the real axis are
(2 1)180, 0,1,2a
ii
n m
60 ,180 ,300a
Example Consider the following loop transfer function
Determine the asymptotes of its root loci.
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6 Breakaway points on the RLBreakaway points on the RL correspond to multi-order roots of the RL equation.
j
4p
3p
1p2p
A
B
0
[ s]
The breakaway points on the RL are determined by finding the roots of dK/ds=0 or dG(s)H(s)=0.
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The poles and zeros of G(s)H(s) are shown in the following figure, determine its root loci.
Rule 1 、 2 、 3RL have three branches , starting from poles 0 、- 2 、- 3 , ending at on finite zero - 1 and two infinite zeros. The RL are symmetrical with respect to the real axis.
Ruel 4The intersections [-1,0] and[-3,-2] on the real axis are RL.
Rule 5The RL have two asymptotes(n - m = 2)
(2 1)18090 ,270
0,1
a
i
n mi
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)1()3()2(0
mn
zpm
1ii
n
1ii
a
=
Rule 6The RL have breakaway points on the real axis (within[-3,-2])
3
1
2
1
0
1
1
1
++
++=
+ bbbb 47.2b
0-1-2-3
jω
σ
Example
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2( ) ( )
( 4)( 4 20)
KG s H s
s s s s
Rule 1 、 2 、 3 、 4 n=4,m=0the RL are symmetrical with respect to the real axis;the RL have four branches which start from poles 0,-4 and -2±j4 ;the RL end at infinite zeros;the intersection [-4,0] on the real-axis is RL
0-2
-j4
-4
jω
σ
j4
(2 1)180( 0,1,2,3)
45 135 , 225 315
a
ll
n m
, ,
Rule 5 The RL have four asymptotes.
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)2()2()4(0
mn
zpm
1ii
n
1ii
a
+
=
Example
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21 ( ) ( ) 1 0
( 4)( 4 20)
KG s H s
s s s s
2
4 3 2
( 4)( 4 20)
( 8 36 80 )
K s s s s
s s s s
3 2(4 24 72 80) 0dK
s s sds
Rule 6the breakaway point of the RL
21 b
45.223,2 jb
2( ) ( )
( 4)( 4 20)
KG s H s
s s s s
0-2
-j4
-4
jω
σ
j4
Example
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7 Angles of departure and angles of arrival of the RL
The angle of departure or arrival of a root locus at a pole or zero, respectively, of G(s)H(s) denotes the angle of the tangent to the locus near the point.
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Angle of Departure:
1 1
(2 1) ( ) ( ), 0, 1,m n
pj j i j ii i
i j
l p z p p l
= +
Pick up a point s1 that is close to p1
Applying Angle Equation (AE)
1 1 1 1 1 2 1 3( ) ( ) ( ) ( )
(2 1)
s z s p s p s p
l
s1p1 )( 11 ps angle of departure θp1
1 1 1 1 2 1 3(2 1) ( ) ( ) ( )p l p z p p p p = +
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1 1
(2 1) ( ) ( ), 0, 1,n m
zj j i j ii i
i j
l z p z z l
= +
Angle of Arrival:
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2( ) ( )1 1
KsG s H s
s s
3 poles P1,2=-1(repeated poles) P3=1 ; 1 zero Z1=0 , n-m=2 。3 branches , 2 asymptotes
1 1 1 1 1 00.5
3 1
n m
i ii i
a
P Z
n m
(2 1) 3,
2 2a
l
n m
3(2 1) ,
2 2 2pl l
Angle of departure:
Example Consider the following loop transfer function
Determine its RL when K varies from 0 to ∞.
-1
j
-0.5 1
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8 Intersection of the RL with the Imaginary Axis
Intersection of the RL with Im-axis?
Method 1 Use Routh’s criterion to obtain the value of K when the system is marginally stable, the get ω from K.
The characteristic equation have roots on the Im-axis and the system is marginally stable.
Method 2
js 1 ( ) ( ) 0G j H j
0)()(1Im
0)()(1Re
jHjG
jHjG
1 ( ) ( ) 0G s H s
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( ) ( )( 1)( 2)
KG s H s
s s s
Closed-loop CE:3 2( 1)( 2) 3 2 0s s s K s s s K
3
2
1
0
1 2
3
60
3
s
s K
Ks
s K
Marginally stable: K =6
Auxiliary equation:
063 2 s
2js Intersection point:
Example Consider the following loop transfer function
Determine the intersection of the RL with Im-axis.
Routh’s Tabulation:
Method 1
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3 2 2 3( ) 3( ) 2( ) ( 3 ) (2 ) 0j j j K K j
2 3 0K Real part: =
3Imaginary part: 2 0 =2 6K
( ) ( )( 1)( 2)
KG s H s
s s s
js 1 ( ) ( ) 0G j H j → Closed-loop CE
Example Consider the following loop transfer function
Determine the intersection of the RL with Im-axis.
Method 2
Closed-loop CE:3 2( 1)( 2) 3 2 0s s s K s s s K