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Lab 08
• Introduction
• The definition of a root locus
• Construction of root loci
• Closed loop stability via root locus

Introduction
• Consider a unity feedback control system shown
below.
• The open loop transfer function G(s) of the system
is
• And the closed transfer function is
K
s
K
s
G
s
G
s
R
s
C





1
)
(
1
)
(
)
(
)
(
1

s
K
)
(s
R )
(s
C
1
)
(


s
K
s
G

Introduction
• The open loop stability does not depend upon gain
K.
• Whereas, the location of closed loop poles vary
with the variation in gain.
K
s
K
s
R
s
C



1
)
(
)
(
1
)
(


s
K
s
G

Introduction
• Location of closed loop Pole for different values of K
(remember K>0).
K
s
K
s
R
s
C



1
)
(
)
(
K Pole
0.5 -1.5
1 -2
2 -3
3 -4
5 -6
10 -11
15 -16
Pole-Zero Map
Imaginary
Axis
-16 -14 -12 -10 -8 -6 -4 -2
-0.5
0
0.5
1
0

K


K

What is Root Locus?
• The root locus is the path of the roots of the
characteristic equation traced out in the s-plane
as a system parameter varies from zero to infinity.

How to Sketch root locus?
• One way is to compute the roots of the
characteristic equation for all possible values
of K.
K
s
K
s
R
s
C



1
)
(
)
(
K Pole
0.5 -1.5
1 -2
2 -3
3 -4
5 -6
10 -11
15 -16

How to Sketch root locus?
• Computing the roots for all values of K might
be tedious for higher order systems.
K
s
s
s
s
K
s
R
s
C





)
20
)(
10
)(
1
(
)
(
)
(
K Pole
0.5 ?
1 ?
2 ?
3 ?
5 ?
10 ?
15 ?

Construction of Root Loci
• Finding the roots of the characteristic equation of degree
higher than 3 is laborious and will need computer
solution.
• A simple method for finding the roots of the
characteristic equation has been developed by W. R.
Evans and used extensively in control engineering.
• This method, called the root-locus method, is one in
which the roots of the characteristic equation are plotted
for all values of a system parameter.

• The roots corresponding to a particular value of this
parameter can then be located on the resulting
graph.
• Note that the parameter is usually the gain, but any
other variable of the open-loop transfer function
may be used.
• By using the root-locus method the designer can
predict the effects on the location of the closed-loop
poles of varying the gain value or adding open-loop
poles and/or open-loop zeros.

Angle & Magnitude Conditions
• In constructing the root loci angle and magnitude
conditions are important.
• Consider the system shown in following figure.
• The closed loop transfer function is
)
(
)
(
1
)
(
)
(
)
(
s
H
s
G
s
G
s
R
s
C



• The characteristic equation is obtained by setting the
denominator polynomial equal to zero.
• Or
• Where G(s)H(s) is a ratio of polynomial in s.
• Since G(s)H(s) is a complex quantity it can be split
into angle and magnitude part.
0
)
(
)
(
1 
 s
H
s
G
1
)
(
)
( 

s
H
s
G

• The angle of G(s)H(s)=-1 is
• Where k=1,2,3…
• The magnitude of G(s)H(s)=-1 is
)
1
2
(
180
)
(
)
(
1
)
(
)
(








k
s
H
s
G
s
H
s
G

1
)
(
)
(
1
)
(
)
(



s
H
s
G
s
H
s
G

• Angle Condition
• Magnitude Condition
• The values of s that fulfill both the angle and
magnitude conditions are the roots of the
characteristic equation, or the closed-loop poles.
• A locus of the points in the complex plane satisfying
the angle condition alone is the root locus.
...)
3
,
2
,
1
(
)
1
2
(
180
)
(
)
( 



 k
k
s
H
s
G 
1
)
(
)
( 
s
H
s
G

Construction of root loci
• Step-1: The first step in constructing a root-locus plot
is to locate the open-loop poles and zeros in s-plane.
)
2
)(
1
(
)
(
)
(



s
s
s
K
s
H
s
G
-5 -4 -3 -2 -1 0 1 2
-1
-0.5
0
0.5
1
Pole-Zero Map
Real Axis
Imaginary
Axis

-5 -4 -3 -2 -1 0 1 2
-1
-0.5
0
0.5
1
Pole-Zero Map
Imaginary
Axis Construction of root loci
• Step-2: Determine the root loci on the real axis.

• Step-3: Determine the asymptotes of the root loci.
Asymptote is the straight line approximation of a curve
Actual Curve
Asymptotic Approximation

• where
• n-----> number of poles
• m-----> number of zeros
• For this Transfer Function
m
n
k
asymptotes
of
Angle






)
1
2
(
180

)
2
)(
1
(
)
(
)
(



s
s
s
K
s
H
s
G
0
3
)
1
2
(
180





k


• Since the angle repeats itself as k is varied, the distinct angles
for the asymptotes are determined as 60°, –60°, -180°and
180°.
• Thus, there are three asymptotes having angles 60°, –60°,
180°.
3
when
420
2
when
300
1
when
180
0
n
whe
60
















k
k
k
k


• Before we can draw these asymptotes in the complex
plane, we must find the point where they intersect the
real axis.
• Point of intersection of asymptotes on real axis (or
centroid of asymptotes) can be find as out
m
n
zeros
poles







• For
0
3
0
)
2
1
0
(






)
2
)(
1
(
)
(
)
(



s
s
s
K
s
H
s
G
1
3
3






-5 -4 -3 -2 -1 0 1 2
-1
-0.5
0
0.5
1
Pole-Zero Map
Real Axis
Imaginary
Axis


60

 60

180




 180
,
60
,
60

1




Home Work
• Consider following unity feedback system.
• Determine
– Root loci on real axis
– Angle of asymptotes
– Centroid of asymptotes

• Step-4: Determine the breakaway point.
• The breakaway point
corresponds to a point
in the s plane where
multiple roots of the
characteristic equation
occur.
• It is the point from
which the root locus
branches leaves real
axis and enter in
complex plane. -5 -4 -3 -2 -1 0 1 2
-1
-0.5
0
0.5
1
Pole-Zero Map
Real Axis
Imaginary
Axis

• Step-4: Determine the break-in point.
• The break-in point
corresponds to a point
in the s plane where
multiple roots of the
characteristic equation
occur.
• It is the point where the
root locus branches
arrives at real axis.
-5 -4 -3 -2 -1 0 1 2
-1
-0.5
0
0.5
1
Pole-Zero Map
Real Axis
Imaginary
Axis

• Step-4: Determine the breakaway point or break-in point.
• The breakaway or break-in points can be determined from the
roots of
• It should be noted that not all the solutions of dK/ds=0
correspond to actual breakaway points.
• If a point at which dK/ds=0 is on a root locus, it is an actual
breakaway or break-in point.
• Stated differently, if at a point at which dK/ds=0 the value of K
takes a real positive value, then that point is an actual breakaway
or break-in point.
0

ds
dK

• The characteristic equation of the system is
• The breakaway point can now be determined as
)
2
)(
1
(
)
(
)
(



s
s
s
K
s
H
s
G
0
)
2
)(
1
(
1
)
(
)
(
1 





s
s
s
K
s
H
s
G
1
)
2
)(
1
(



 s
s
s
K
 
)
2
)(
1
( 


 s
s
s
K
 
)
2
)(
1
( 


 s
s
s
ds
d
ds
dK

• Set dK/ds=0 in order to determine breakaway point.
 
)
2
)(
1
( 


 s
s
s
ds
d
ds
dK
 
s
s
s
ds
d
ds
dK
2
3 2
3




2
6
3 2



 s
s
ds
dK
0
2
6
3 2



 s
s
0
2
6
3 2


 s
s
5774
.
1
4226
.
0




s

• Since the breakaway point must lie on a root locus between 0
and –1, it is clear that s=–0.4226 corresponds to the actual
breakaway point.
• Point s=–1.5774 is not on the root locus. Hence, this point is
not an actual breakaway or break-in point.
• In fact, evaluation of the values of K corresponding to s=–
0.4226 and s=–1.5774 yields
5774
.
1
4226
.
0




s

-5 -4 -3 -2 -1 0 1 2
-1
-0.5
0
0.5
1
Pole-Zero Map
Real Axis
Imaginary
Axis


60

 60

180
4226
.
0


s

4226
.
0


s
-5 -4 -3 -2 -1 0 1 2
-1
-0.5
0
0.5
1
Pole-Zero Map
Real Axis
Imaginary
Axis

• Step-5: Determine the points where root loci cross the
imaginary axis.
-5 -4 -3 -2 -1 0 1 2
-1
-0.5
0
0.5
1
Pole-Zero Map
Imaginary
Axis


60

 60

180

imaginary axis.
– These points can be found by use of Routh’s stability criterion.
– Since the characteristic equation for the present system is
– The Routh Array Becomes

imaginary axis.
• The value(s) of K that makes the system
marginally stable is 6.
• The crossing points on the imaginary
axis can then be found by solving the
auxiliary equation obtained from the
s2 row, that is,
• Which yields

imaginary axis.
• An alternative approach is to let s=jω in the characteristic
equation, equate both the real part and the imaginary part to
zero, and then solve for ω and K.
• For present system the characteristic equation is
0
2
3 2
3



 K
s
s
s
0
2
)
(
3
)
( 2
3



 K
j
j
j 


0
)
2
(
)
3
( 3
2



 

 j
K

imaginary axis.
• Equating both real and imaginary parts of this equation
to zero
• Which yields
0
)
2
(
)
3
( 3
2



 

 j
K
0
)
3
( 2

 
K
0
)
2
( 3




-7 -6 -5 -4 -3 -2 -1 0 1 2
-5
-4
-3
-2
-1
0
1
2
3
4
5
Root Locus
Real Axis
Imaginary
Axis

Example#1
• Determine the value of K such that the damping ratio of
a pair of dominant complex-conjugate closed-loop poles
is 0.5.
)
2
)(
1
(
)
(
)
(



s
s
s
K
s
H
s
G

Example#1
• The damping ratio of 0.5 corresponds to

 cos


 1
cos



 
60
)
5
.
0
(
cos 1


Example#1
• The value of K that yields such poles is found from the
magnitude condition
1
)
2
)(
1
( 5780
.
0
3337
.
0


 

 j
s
s
s
s
K

Example#1
• The third closed loop pole at K=1.0383 can be obtained
as
0
)
2
)(
1
(
1
)
(
)
(
1 





s
s
s
K
s
H
s
G
0
)
2
)(
1
(
0383
.
1
1 



s
s
s
0
0383
.
1
)
2
)(
1
( 


 s
s
s

Home Work
• Determine the value of K such that the natural
undamped frequency of dominant complex-conjugate
closed-loop poles is 1 rad/sec.
)
2
)(
1
(
)
(
)
(



s
s
s
K
s
H
s
G

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Root Locus
Imaginary
Axis
-0.2+j0.96

Example#2
• Sketch the root locus of following system and
determine the location of dominant closed loop
poles to yield maximum overshoot in the step
response less than 30%.

Example#2
• Step-1: Pole-Zero Map
-5 -4 -3 -2 -1 0 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Pole-Zero Map
Real Axis
Imaginary
Axis

Example#2
• Step-2: Root Loci on Real axis
-5 -4 -3 -2 -1 0 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Pole-Zero Map
Real Axis
Imaginary
Axis

Example#2
• Step-3: Asymptotes
-5 -4 -3 -2 -1 0 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Pole-Zero Map
Real Axis
Imaginary
Axis


 90

2




Example#2
• Step-4: breakaway point
-5 -4 -3 -2 -1 0 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Pole-Zero Map
Real Axis
Imaginary
Axis
-1.55

Example#2
-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5
-8
-6
-4
-2
0
2
4
6
8
Root Locus
Real Axis
Imaginary
Axis

Example#2
• Mp<30% corresponds to
100
2
1

 



e
M p
100
%
30
2
1

 



e
35
.
0



Example#2
-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5
-8
-6
-4
-2
0
2
4
6
8
0.35
0.35
6
6
Root Locus
Real Axis
Imaginary
Axis

Example#2
-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5
-8
-6
-4
-2
0
2
4
6
8
0.35
0.35
6
6
System: sys
Gain: 28.9
Pole: -1.96 + 5.19i
Damping: 0.354
Overshoot (%): 30.5
Frequency (rad/sec): 5.55
Root Locus
Real Axis
Imaginary
Axis

Consider the following unity feedback system
num=[1 0];
den=[1 1];
G=tf(num,den);
rlocus(G)
sgrid
1

S
KS
)
(S
R )
(S
C



1

S
KS
)
(S
R )
(S
C


1
)
2
(


S
S
K
)
(S
R )
(S
C



num=1;
den1=[1 0];
den2=[1 3];
den=conv(den1,den2);
G=tf(num,den);
rlocus(G)
sgrid
)
3
( 
S
S
K
)
(S
R )
(S
C



)
1
(
)
3
(


S
S
S
K
)
(S
R )
(S
C


)
(S
R )
(S
C


)
1
(
)
3
)(
2
(



S
S
S
S
K
(1)
(2)

)
(S
R )
(S
C


)
2
)(
1
( 
 S
S
S
K
Determine the closed loop gain that would
make the system marginally stable.
num=1;
den1=[1 0];
den2=[1 1];
den3=[1 2];
den12=conv(den1,den2);
den=conv(den12,den3);
G=tf(num,den);
rlocus(G)
sgrid

)
(S
R )
(S
C


)
2
)(
1
(
)
3
(



S
S
S
S
K
)
(S
R )
(S
C


)
2
)(
1
(
)
5
)(
3
(




S
S
S
S
S
K
(1)
(2)

k=3;
z=-5;
p=[0 -1 -2];
G=zpk(z,p,k);
rlocus(G)
sgrid
)
(S
R )
(S
C


)
2
)(
1
(
)
5
(
3



S
S
S
S

1
4
3
3
2
)
( 2




S
S
S
S
G
num=[2 3];
den=[3 4 1];
G=tf(num,den);
[kd,poles]=rlocfind(G)
sgrid

• Consider the Following Unity feedback system
k=4
p=[0 -2 ]
z=[];
usys=zpk(z,p,k)
rlocus(usys)
csys=feedback(usys,1)
[wn b]=damp(csys)
sgrid(b,wn)
axis([-3 0 -3 3 ])
)
( 2
4

S
S
)
(S
R )
(S
C



Design Requirements
• It is desired to increase ωn to 4 rad/sec without changing
damping ratio of closed loop poles.
k=4
p=[0 -2 ]
z=[];
usys=zpk(z,p,k)
rlocus(usys)
wn=[2 4]; b=[0.5 0.5];
sgrid(b,wn)
axis([-5 0 -5 5 ])
)
( 2
4

S
S
)
(S
R )
(S
C



)
4
.
5
(
)
9
.
2
(
68
.
4
)
(



S
S
S
Gc
kc=4.68
pc=-5.4
zc=-2.9
lc=zpk(zc,pc,kc)
ltiview(lc)

Root locus of Compensated & uncompensated Systems
)
4
.
5
)(
2
(
)
9
.
2
(
7
.
18
)
(
)
(




S
S
S
S
S
G
S
G c
b=0.5; wn=[2 4];
ku=4
pu=[0 -2 ]
zu=[ ];
usys=zpk(zu,pu,ku)
k=18.7
p=[0 -2 -5.4]
z=-2.9
csys=zpk(z,p,k)
rlocus(usys, ‘g’,csys,’b’)
Sgrid(b,wn)
)
2
(
4
)
(


S
S
S
G
Compensated Vs Uncompensated System

Root Locus
Real Axis
Imag
Axis
-5 -4 -3 -2 -1 0
-10
-8
-6
-4
-2
0
2
4
6
8
10
4
2
4
2
0.5
0.5
Root locus of Compensated & uncompensated Systems

Response of compensated & Uncompensated system
23
.
54
5
.
29
4
.
7
23
.
54
7
.
18
)
(
)
(
1
)
(
)
(
2
3





 S
S
S
S
S
G
S
G
S
G
S
G
c
c
numu=4;
denu=[1 2 4];
numc=[18.7 54.23];
denc=[1 7.4 29.5 54.23];
csys=tf(numc,denc);
usys=tf(numu,denu);
ltiview(usys, ‘g:’,csys,’b’)
4
2
4
)
(
1
)
(
2



 S
S
S
G
S
G

Step Response
Time (sec)
Amplitude
0 1 2 3 4 5 6
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Step response of Compensated & Uncompensated System

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