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CHAPTER 4: ROOT LOCUS METHOD
4.1 INTRODUCTION
What? Find the positions of the closed-loop poles when a
certain system parameter varies.
Why? Positions of closed-loop poles determine important
properties such as stability, damping, etc.
How? By using root locus method.
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A root locus of a system is a plot of the roots of the system characteristic
equation (poles of the closed-loop transfer function) as some parameters
of the system are varied.
Each root locus starts at an open-loop pole with K= 0 and terminates at
infinity as K.
Each root locus gives one characteristic root (closed-loop pole) for aspecific value ofK.
Note that information about damping can be obtained from the root
locus.
4.3 BASIC PROPERTIES OF ROOT LOCI
4
KG(s)
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Closed-loop transfer function: )()(1)(
)(
)(
sHsKG
sKG
sR
sY
+=
1 + KG(s)H(s) = 0
G(s)H(s) = -1/K
Any point on the root locus (with K> 0) must satisfy
Magnitude criterion: |G(s)H(s)| = 1/K
Angle criterion: G(s)H(s) = 180o
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KG(S)
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Let )())((
)())(()()(
21
21
n
m
pspsps
zszszsKsHsKG
+++
+++=
Magnitude criterion:( )
( ) Kps
zs
sHsGn
j
j
m
i
i1
)()(
1
1 =
+
+
=
=
=
Angle criterion:o
j
n
j
i
m
i
pszssHsG 180)()()()(11
=++= ==
Example: Find ifs = -1 and s = -3 are points on the root locus. If so, find
the value(s) ofKat these points.
6
-2
-1+j
-1-j
Re(s)
Im(
s)
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Fors = -1 Fors = -3
7
-2
-1+j
-1-j
a
b
c = 0o
-2
-1+j
-1-j
d
ef
-3
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Angle criterion:o
j
n
j
i
m
i
pszssHsG 180)()()()(11
=++= ==
Total phase = c a b = 0o Total phase = f d e = -180o
Point s = -1 is not on root locus Point s = -3 is on root locus
Find Kat s = -3.
8
-2
-1+j
-1-j
p
q
r-3
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Magnitude criterion:
( )
( ) Kps
zs
sHsGn
j
j
m
i
i 1)()(
1
1 =
+
+
=
=
=
Kqp
r 1=
K= 5
4.4 PROPERTIES AND CONSTRUCTION OF ROOT LOCI
RULES FOR CONSTRUCTION OF ROOT LOCI
Rule 1: Number of Root Loci (Branches)
The root locus plot consists of n root loci (branches) as K varies from 0to . The loci are symmetric with respect to the real axis.
Characteristic equation: 0)()()(11
=+++= ==
m
i
i
n
j
j zsKpss
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Degree n n roots
Complex roots occur in complex conjugate pairs
Root loci symmetrical about the real axis
Rule 2: Starting and Ending Points of Root Loci
As K increases from 0 to , each root locus starts from an open-loop
pole with K = 0 and ends on an open-loop zero with K = . The number
of root loci ending atequals the number of open-loop zeros at.
Example: For a system with )1)(1(2
)(++
+=
ss
ssG and 1)( =sH , find the starting and
ending points of the root loci.
Characteristic equation:0)2()1)(1( =++++ sKss
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Starting points: K= 0, s = -1, -1 (open-loop poles)
Ending points: K= , s = -2, (open-loop zeros)
Rule 3: Asymptotes to Root Loci (Behaviour at Infinity)
The (n-m) root loci which tend to do so along straight line asymptotes
radiating out from a single point s= - a on the real axis (called thecentroid) where
mna
=
)zerosloop-openofpartreal()polesloop-openofpartreal(
These (n-m) asymptotes have angles
)1(,,1,0;180)12( 0
=
+= mnq
mn
qa
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( )
( )0
)(11
1
1 =+
++
++
=
=
mn
a
n
j
j
m
ii
s
K
ps
zsK
12
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( )
( )
mn
am
i
i
n
j
j
s
zs
ps
=
= ++
+
)(
1
1
...)(...11
11
++=+
+
==
mn
a
mnmnm
i
i
n
j
j
mn smnsszps
mn
zpn
j
m
i
ij
a
=
= =1 1
)()(
From angle criterion:,1,0;180)12()(
0=+= qqmn
)1(,,1,0;180)12(
0
=
+= mnqmn
qa
Example: Find the centroid and asymptotes of the root locus for
)4)(3)(41)(41(
)2()()(
+++++
+=
ssjsjs
sKsHsKG .
Rule 1: 4 branches.
Rule 2: Starting points are open-loop poles s = -1 4j, -3, -4.
Ending points are open-loop zeros s = -2, , , .Rule 3: 3 asymptotes.
Centroid, 37
14
)2(4311=
= a
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Angles of asymptotes
)1(,,1,0;180)12( 0
=
+= mnq
mn
qa
2,1,0;14
180)12(0
=
+= q
qa a= 600, 1800, and 3000
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Rule 4: On-locus Segments on the Real Axis
A point on the real axis lies on the locus if the number of open-loop polesplus zeros on the real axis to the right of this point is odd.
Take a point s0 on the real axis.
(i)Poles and zeros on the real axis to the right of this point contribute an
angle of 1800 each.(ii) Poles and zeros to the left of this point contribute angle of 00 each.
(iii) The net angle contribution of a complex conjugate pole or zero
pair is always zero.
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G(s)H(s) = (mr+ nr)180
0
mr= number of open-loop zeros on the real axis to the right ofs0
nr= number of open-loop poles on the real axis to the right ofs0
From angle criterion: G(s)H(s) = 1800 = (2q + 1)1800, q = 0, 1, 2,
Angle criterion is satisfied if (nr+ mr) is odd.
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Rule 5: On-Locus Points of the Imaginary Axis
The intersections (if any) of root loci with the imaginary axis can be
determined by use of the Routh criterion.
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Stability limits Imaginary axis crossings
Example: Find the imaginary axis crossings for the root locus plot of
)4)(3)(41)(41()2()()(
++++++=
ssjsjssKsHsKG .
Characteristic equation02204)143(439
234=++++++ KsKsss
Routh Table:
s4 1 43 204+2K
s3 9 143+K
s2 (244 K)/9 204+2K
s
1
(18368 61KK
2
)/(244 K)s0 204+2K
244 K> 0 K< 244
18368 61KK2 > 0 -169.4 < K< 108.4
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204 + 2K> 0 K> -102
Hence, 0 < K< 108.4
ForK= 108.4, all the coefficients in s1
row are zero.Auxiliary equation formed from the coefficients ofs2 row.
0)2204(9
244 2=++
Ks
K s = j5.28
Root loci intersect the imaginary axis at s = j5.28, corresponding value ofK
is 108.4.
Rule 6: Angle of Departure from Complex PolesThe angle of departure, p, of a locus from a complex open-loop pole is
given by p = 1800+ , where is the net angle contribution at thispole of all other open-loop poles and zeros.
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Net angle contribution of all other open-loop poles and zeros at s0 is)( 4312 ++=
Total phase G(s)H(s) at s0 = - p.
From angle criterion, - p = 1800
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Angle of departure from the complex open-loop pole, p = 1800 + .
For this example,
1 = 90076)1/4(tan
1
2 ==
63)2/4(tan1
3 ==
53)3/4(tan1
4 ==
0
4312130)( =++=
p = 1800
+ = 500
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Rule 7: Angle of Arrival at Complex Zeros
The angle of arrival, z, of a locus at a complex zero is given by
z = 1800 - , where is the net angle contribution at this zero of allother open-loop poles and zeros.
Example:Let us consider the characteristic equation 0)2()1(
12
=+
++
ss
sK.
Rule 1: 2 branches.
Rule 2: Starting points are open-loop poles s = 0, -2.
Ending points are open-loop zeros s = j1.
Rule 3: Not applicable.
Rule 4: On-locus segments on the real axis between 0 and -2.
Rule 5: Root locus touches imaginary axis at s =
j1.Rule 6: Not applicable.
Rule 7:
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Points at which multiple roots of the characteristic equation occur
(breakaway/break-in points of root loci) are the solutions of 0=dsdK
where( )
( )
=
=
+
+
= m
i
i
n
jj
zs
ps
K
1
1
).
Example: Draw the root locus plot for the characteristic equation0
)1(
)3)(2(1 =
+
+++
ss
ssK
.
Rule 1: 2 branches.
Rule 2: Starting points are open-loop poles s = 0, -1.
Ending points are open-loop zeros s = -2, -3.
Rule 3: Not applicable.
Rule 4: On-locus segments on the real axis between 0 and -1, and
between -2 and -3.Rule 5: Not applicable.
Rule 6: Not applicable.
Rule 7: Not applicable.26
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Rule 8:
65)3)(2(
)1(2
2
++
+=
++
+=
ss
ss
ss
ssK
0)65(
)52)(()12)(65(22
22
=
++
+++++=
ss
ssssss
ds
dK
03622
=++ ss
s = -0.634 (breakaway point), s = -2.366 (break-in point).
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Condition for breakaway/break-in point (as derived above) is necessary
but not sufficient.
4.5 COMPLETE EXAMPLES
Question 1: Consider a feedback system with the characteristic equation0;0
)2)(1(1 =
+++ K
sss
K
. Plot the root locus for this system. Find the dominant
closed-loop poles for = 0.5.
Solution:
Rule 1: 3 branches.
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Rule 2: Starting points are open-loop poles s = 0, -1, -2.
Ending points are open-loop zeros s = , , .
Rule 3: 3 asymptotes.
Centroid, 103 12 == a Angles, )1(,,1,0;
180)12( 0=
+= mnq
mn
qa
2,1,0;03
180)12( 0=
+= q
qa a= 600, 1800, and 3000
Rule 4: On-locus segments on the real axis between 0 and -1, and
between -2 and -.
Rule 5: Characteristic equation: 023 23 =+++ Ksss
Routh Table:
s3 1 2
s2 3 K
s1 (6-K)/329
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s0 K
For stability, K> 0, (6 - K)/3 > 0
Crosses imaginary axis when K= 6.Auxiliary equation: 0633 22 =+=+ sKs 2js =
Rule 6: Not applicable.
Rule 7: Not applicable.
Rule 8: K= - (s3 + 3s2 + 2s).0)263(
2=++= ss
ds
dK
s = -0.4226 (breakaway point), -1.5774 (not valid)
If two loci break away from a breakaway point, their tangents will be
1800 apart.
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In general, if r loci break away from a breakaway point, then their
tangents will be 3600/rapart, i.e., the tangents will equally divide 3600.
= 0.5 line makes an angle cos-1 = 600 with the negative real axis.
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Points of intersection: s = -0.33 j0.58.
From magnitude criterion, { } 04.121 58.033.0 =++= += jssssK .
Question 2 (past year exam question 2004/2005): Sketch the root locus
of a unity feedback system with forward path transfer function G(s) givenas follows:
)54()(
2++
=sss
KsG
Solution:
Rule 1: 3 branches.Rule 2: Starting points are open-loop poles s = 0, -2+j, -2-j.
Ending points are open-loop zeros s = , , .
Rule 3: 3 asymptotes.
Centroid, 34
03
22=
=
a
Angles, )1(,,1,0;180)12( 0
=
+= mnq
mn
qa
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2,1,0;
03
180)12( 0=
+= q
qa a= 600, 1800, and 3000
Rule 4: On-locus segments on the real axis between 0 and -.
Rule 5: Characteristic equation: 05423
=+++ Ksss Routh Table:
s3 1 5
s2 4 K
s1 (20-K)/4
s0 K
For stability, K> 0, (20-K)/4 > 0
Crosses imaginary axis when K= 20.
Auxiliary equation: 02044 22 =+=+ sKs 5js =
Rule 6: For pole -2+j,
= -153.43 - 90 p = 1800 + = -63.430
For pole -2-j,
Due to symmetry properties: p = 63.430
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Rule 7: Not applicable.
Rule 8: Characteristic equation: K= - (s3 + 4s2 + 5s).0)583(
2=++= ss
ds
dK
s = -1 (breakaway point), -1.667 (break-in point)
From the characteristic equation,21 ==sK , 852.1667.1 ==sK
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4.6 EFFECTS OF ADDING POLES AND ZEROS TO G(s)H(s)
Controller design may be treated as an investigation of the effects to root
loci when poles and zeros are added to the loop transfer functionKG(s)H(s).
Addition of Poles:
Has the effect of pushing the root loci toward the right-half s-plane.
Example: Consider the loop transfer function given by )2()()( += ssK
sHsKG .
Rule 1: 2 branches.
Rule 2: Starting points are open-loop poles s = 0, -2.
Ending points are open-loop zeros s = , .
Rule 3: 2 asymptotes.
Centroid, 10220
=
= a
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Angles, )1(,,1,0;
180)12( 0=
+= mnq
mn
qa
1,0;02
180)12( 0=
+= q
qa a= 900, 2700
Rule 4: On-locus segments on the real axis between 0 and -2.Rule 5: Not applicable.
Rule 6: Not applicable.
Rule 7: Not applicable.
Rule 8: 0)22( =+= sdsdK
s = -1 (breakaway point)
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Stable for all K.
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Let us introduce a pole at s = -b (b > 2).
)3)(2())(2()()(
++=
++=
sss
K
bsss
KsHsKG , with b = 3
The centroid of asymptotes changes from -1 to -(2+b)/3.
The angles of asymptotes change from 90 to 60and 180. Root loci bend towards right-half s-plane unstable at large K.
Addition of Zeros
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Has the effect of pushing the root loci toward the left-half s-plane.
Let us introduce a zero at s = -3. )2()3(
)()(+
+=
ss
sKsHsKG .
Relative stability is improved.
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