Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
Root-Locus Analysis
INTRODUCTION:
The basic characteristic of the transient response of a closed loop system is
closely related to the location of the closed loop poles. If the system has a variable loop
gain, then the location of the closed loop poles depends on the value of the loop gain
chosen. It is important, therefore, that the designer know how the closed loop poles
move in the s-plane as the loop gain is varied.
The closed loop poles are the roots of the characteristic equation. 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 which the roots of the characteristic equation are plotted for all values of
a system parameter. The roots corresponding to 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. Unless otherwise
stated, we shall assume that the gain of the open loop transfer function is the parameter
to be varied through all values, from zero to infinity.
ROOT-LOCUS PLOTS
Ex.(6-1): Consider the system shown in Fig.(6-1). Sketch the root-locus plot and then
determine the value of K such that the damping ratio  of a pair of dominant complex
conjugate closed loop poles is 0.5.
Fig.(6-1)
Sol:
1
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
For this system
G( s) 
K
,
s( s  1)( s  2)
H ( s)  1
 Angle Conditions
For the given system, the angle condition becomes
G( s) H ( s) 
K
s( s  1)( s  2)
  s  s  1  s  2  180  (2k  1),
(k  0, 1, 2, 3, )
 Magnitude condition
G (s) H (s) 
K
1
s ( s  1)( s  2)
A typical procedure for sketching the root-locus plot is as follows:
 There are no finite zeros of G(s)H(s) (on graph represented by  )
(The number of zeros is 0)
 The finite poles of G(s)H(s) are (s = 0, s = -1, s = -2) (on graph represented by x)
(The number of poles is 3)
1. Determine the root locus on the real axis.
 The test point are selected at real positive real axis
s  s  1  s  2  0
This shows that the angle condition can not be satisfied. There is no root locus
on the positive real axis.
 The test point on the negative real axis between 0 and -1. then
s 180  ,
Thus,
s  1  s  2  0
 s  s  1  s  2   180 
2
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
The angle condition is satisfied Therefore, the portion of the negative real axis
between 0 and -1 a portion of a root locus.
 The test point on the negative real axis between -1 and -2. then
s  s  1 180  ,
s  2  0
 s  s  1  s  2   360 
The angle condition is not satisfied.Therefore; the portion of the negative real axis
between -1 and -2 is not a part of a root locus.
 The test point on the negative real axis between -2 and   . then
s  s  1  s  2  180 
 s  s  1  s  2   540 
The angle condition is satisfied Therefore, the portion of the negative real axis
between -2 and   a portion of a root locus.
2. Determine the asymptotes of the root loci.
The number of distinct asymptotes is n – m
 180  (2k  1)
Angle of Asymptotes 
,
nm
(k  0, 1, 2, 3, )
Where n number of finite poles of G(s)H(s)
And m number of finite zeros of G(s)H(s)
Since the angle repeats itself as k is varied. So, there are three asymptotes, and the
angles of asymptotes are 60  ,  60  , and 180 
All the asymptotes intersect on the real axis at (substituting the values without
there signs)
  p1  p 2    p n   z1  z 2    z m 

nm


  
 0  1  2  0
  1
30
  

Now, collect the obtained information and sketch Fig.(6-2)
3
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
Fig.(6-2)
3. Determine the breakaway and the break-in .points
The characteristic equation of the system is obtained by putting the denominator of
the closed loop transfer function equal to zero, for a negative feedback control
system we have
1  G( s) H ( s)  0
The characteristic equation for the given system is
K
1 0
s( s  1)( s  2)
or
s 3  3s 2  2s  K  0
so
K  ( s 3  3s 2  2s)
And
dK
 (3s 2  6s  2)  0
ds
Find the roots of the obtained equation as
s = -0.4226, and s = -1.5774
Since the breakaway is must be located between 0 and -1, so s = -0.4226 and s = 1.5774 is not actual a breakaway point.
4
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
Substitute the s = -0.4226 in the characteristic equation we get K = 0.3849
4. Determine the points where the root loci cross the imaginary axis
This points is found by using Routh’s stability criterion, from the characteristic
equation
s3
1
2
s2
3
6K
3
K
K
s1
s0
The value of K that makes the s 1 terms in the first column equal zero is K = 6. The
crossing points on the imaginary axis can then be found by solving the auxiliary
equation obtained from the s 2 row, that is
3s 2  K  3s 2  6  0
Yield
sj 2
The frequencies at the crossing points on the imaginary axis are thus    2 . The
gain value corresponding to the crossing point is K = 6.
5. Chose a test pint in the broad neighborhood of the j axis and the origin.
Fig.(6-3)
As shown in Fig.(6-3), any point on the root loci must satisfy the angle condition
5
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
1   2   3  180 
Continue this process and locate a sufficient number of points satisfy this condition
6. Draw the root loci, based on the information obtained in the foregoing steps, as
shown in Fig.(6-4)
Fig.(6-4)
After we have drawn the root loci, now we will solve the question objective
Determine the complex conjugate closed loop poles such that the damping ratio
is 0.5
The closed loop poles with   0.5 lie on lines passing through the origin and
making angles  cos   cos 0.5  60  , with the negative real axis as shown in
Fig.(6-4). The poles are
r1  0.3337  j 0.5780 ,
r2  0.3337  j 0.5780
The values of K that yields such poles is found from the magnitude condition as
follow
6
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
K  s( s  1)( s  2) s  0.3337  j 0.5780
= 1.0383
Using this value of K, the third pole as found at s = -2.3326
(r1 )( r2 )( r3 )  K
(0.3337  j 0.5780 )(0.3337  j 0.5780 )r3  1.0383
found at r3  -2.3326
Ex.(6-7): Draw the complete root locus for the system shown in Fig.(6-5). Where
K 0
Fig.(6-5)
And find the value of K at which the complex conjugate closed loop poles have the
damping ratio of 0.7
Sol:
For this system
G( s) 
K ( s  2)
s 2  2s  3
,
H ( s)  1
It is seen that G(s) has a pair of complex conjugate poles at
s  1  j 2 ,
s  1  j 2
A typical procedure for sketching the root-locus plot is as follows:
 There are one zero for G(s)H(s) (on graph represented by  )
 There are two poles of G(s)H(s) (on graph represented by x)
7
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
1. Determine the root locus on the real axis.
 For any test point s on the real axis, the sum of the angular contributions of the
complex conjugate poles is 360  . As shown in Fig.(6-6). Thus the net effect of
the complex conjugate poles is zero on the real axis
 The location of the root locus on the real axis is determined from the open loop
zero on the negative real axis. A section that between - 2   is a part of root
locus.
Fig.(6-6)
2. Determine the asymptotes of the root loci.
The number of distinct asymptotes is n – m
 180  (2k  1)
Angle of Asymptotes 
,
nm
(k  0, 1, 2, 3, )
Since the angle repeats itself as k is varied. So, there is one asymptote, and the angle
coincides with the negative real axis.
3. Determine angle of departure from the complex conjugate open loop poles
Angle of departure from a complex pole = 180  - (sum of the angles of vectors
to a complex pole from other poles) + (sum of the angles of vectors to a complex
zero in question from poles)
Referring to Fig.(6-7)
1  180    2  1
8
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
The angle of departure is then
1  180   90   55   145 
Fig.(6-7)
Since the root locus is symmetric about the real axis, the angle of departure from
the pole at s   p 2 is 145 
4. Determine the break-in point
The system characteristic equation is
s 2  2s  3  K ( s  2)  0
K 
or
s 2  2s  3
s2
 (2s  2)( s  2)  ( s 2  2s  3) 
dK
 
0
2
ds
( s  2)


This gives
s 2  4s  1  0
The roots of this equation
s  3.7320,
s  0.2680
9
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
Notice that point s  3.7320 is on the root locus. Hence this point is an actual break
in point. And the corresponding K value is 5.4641. Since the point s  0.2680 is
not on the root locus, it can not be a break in point
5. Sketch a root locus plot, based on the information obtained in the forgoing
steps.
To determine the accurate root loci, several points must be found by trail and
error between the break in point and the complex open loop poles. Fig.(6-8)
shows a complete root locus plot for the given system.
Fig.(6-8)
The value of K at which the complex conjugate closed loop poles have the
damping ratio of 0.7 can be found by locating the roots as shown in Fig.(6-8) ,
and computing the value of K from the magnitude condition as follow
K
( s  1  j 2 )( s  1  j 2 )
s2
 1.34
s 1.67  j1.7
10
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
Typical Pole-Zero Configurations and Corresponding Root Loci. In summarize,
we show several open loop pole zero configurations and their corresponding root loci in
Table (6-1). The pattern of the root loci depends only on the relative separation of the
open loop poles and zeros.
Table (6-1)
11
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
Ex.(6-3): Sketch the root loci for the system shown in Fig.(6-9a). (the gain K is
assumed to be positive)
Figs.(6-9)
Sol:
1. Locate the open loop poles and zeros on the complex plane. Root loci exist on the
negative real axis between 0 and -1, and between -2 and -3.
2. The number of the open loop poles and that of finite zeros are the same. This means
there are no asymptotes in the complex region of the s plane
3. Determine the breakaway and break in points. The characteristic equation for the
system is
1
Or
K ( s  2)( s  3)
0
s( s  1)
K 
s( s  1)
( s  2)( s  3)
The breakaway and break in points are determined from
(2s  1)( s  2)( s  3)  s( s  1)(2s  5)
dK

0
2
ds
(s  2)(s  3)
The roots are
12
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
s  0.634,
s  2.366
Notice that both points are on root loci. Therefore, they are accrual breakaway or break
in points. At point s  0.634 , the value of K is
Similarly, at s  2.366 ,
K 
(0.634)(0.366)
 0.0718
(1.366)(2.366)
K 
(2.366)(1.366)
 14
(0.366)(0.634)
Because point s  0.634 lies between two poles, it is a breakaway point, and because
point s  2.366 lies between two zeros, it is a break in point.
4. Determine a sufficient number of points that satisfy the angle condition. It can be
found that the root loci involve a circle with center at -1.5 that pass through the
breakaway and break in points. The root locus plot for this system is shown in Fig.(69b). Note that the system is stable for any value of K since the entire root loci lay in
the left half s plane
Ex.(6-4): sketch the root loci of the control system shown in Fig.(6-10a).
Figs.(6-10)
13
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
Sol:
 The open loop poles are located at s  0, s  3  j 4, and s  3  j 4.
 A root locus branch exists on the real axis between the origin and -∞.
 There are three asymptotes for the root loci
 180  (2k  1)
Angle of Asymptotes 
 60  ,  60  , 180 
3
 The asymptotes are intersect at
 0  3  3
  2
3
  

 We check the breakaway and break in points, from the characteristic equation for
this system we have
K   s ( s 2  6s  25)
dK
 (3s 2  12 s  25)  0
ds
Which yields
s  2  j 2.0817 , and s  2  j 2.0817 .
Notice that at points s  2  j 2.0817 the angle condition is not satisfied. Hence,
they are neither breakaway nor break in points.
 The angle of departure from the complex pole in the upper half s plane is
  180   126 .87   90 
  36.87 
Or
 The points where root locus branches cross the imaginary axis may be found by
substituting s  j into the characteristic equation and solving the equation for
 and K as follows noting that the characteristic equation is
s 3  6s 2  25 s  K  0
We have
( j ) 3  6( j ) 2  25( j )  K  (6 2  K )  j (25   2 )  0
Which yields
  5, K  150, or   0, K  0
14
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
PROBLEMS
1) Sketch the root loci for the system shown in Fig.(6-11a)
Figs.(6-11)
1) Sketch the root loci for the system shown in Fig.(6-12a)
Figs.(6-12)
2) Plot the root loci for the closed loop control system with
G( s) 
K ( s  1)
s2
,
H ( s)  1
15
Waleed Kh. Al-Ashtari
College of Engineering
Mech. Eng. Dept.
Modern Automatic Control
7002 - 7002
3) Plot the root loci for the closed loop control system with
G (s) 
K ( s  4)
( s  1) 2
H (s)  1
,
6) Plot the root loci for the closed loop control system with
G(s) 
K
s ( s  1)( s 2  4s  5)
,
H (s)  1
7) Plot the root loci for the closed loop control system with
G(s) 
K ( s  9)
s ( s  4s  11)
2
,
H ( s)  1
16