Root Locus Techniques (Sketching Method)
DNT 354 - CONTROL PRINCIPLE
Date: 25th September 2008
Prepared by: Megat Syahirul Amin bin Megat Ali
Email: [email protected]
CONTENTS
Introduction
 Properties of Root Locus
 Sketching the Root Locus
 Refining the Sketch

INTRODUCTION




Root locus is a graphical representation of the closed-loop
poles as a system parameter is varied.
Provides a qualitative description of systems performance.
The advantage of this technique is in its ability to provide
solution for systems of order higher than two.
Also provide a graphical representation of system’s stability.
i.
ii.
iii.

Ranges of stability
Ranges of instability
Conditions that cause a system to break into oscillation
Two concepts that were discussed previously:
i.
ii.
Control system problem
Complex numbers and representation as vectors
PROPERTIES OF THE ROOT LOCUS

The properties of the root locus can be derived from the
general closed-loop transfer function T(s).
T ( s) 

KG( s)
1  KG( s) H ( s)
A pole s, exists when the characteristic polynomial becomes
zero.
KG( s ) H ( s )  1
 1(2k  1)180
Where,
k = 0, ±1, ±2, ±3, …
PROPERTIES OF THE ROOT LOCUS

Alternately, a value of s is a closed-loop pole if,
KG(s) H (s)  1
and,


KG( s) H ( s)  (2k  1)180
If the angle of the complex number is an odd multiple of 180°,
the value of s is a system pole for a particular value of K.
If the angle criterion is satisfied, the magnitude criterion is as
shown below.
K
1
G( s) H ( s)
PROPERTIES OF THE ROOT LOCUS

Example: For the following unity feedback system, determine
whether the following are system’s closed-loop poles for a gain
of 5.
KG( s) H ( s) 
i.
ii.
iii.
-9.47
-5.34
-0.53
K
s( s  10)
PROPERTIES OF THE ROOT LOCUS

Example: For the following unity feedback system, determine
whether -2 + j3 is a closed-loop pole for some value of gain K.
KG( s) H ( s) 
K ( s  3)( s  4)
( s  1)( s  2)
(a) Unity feedback system
(b) Pole-zero plot
PROPERTIES OF THE ROOT LOCUS

Example: For the unity feedback system below, determine
whether the following is a closed-loop pole for some value of
gain K.
K ( s  3)( s  4)
KG( s) H ( s) 
( s  1)( s  2)
Point,
 2  j ( 2 / 2)
Should the point be a closed-loop pole of the system,
determine the corresponding value of gain K for the particular
pole.
PROPERTIES OF THE ROOT LOCUS

Problem: Given a unity feedback system that has the forward
transfer function.
G( s) 
K ( s  2)
( s 2  4s  13)
Find the following:
i.
ii.
iii.
Calculate the angle of G(s) at the point (-3 + 0j) by finding the
algebraic sum of angles of the vectors drawn from the zeros and
poles of G(s) to the given point.
Determine if the point specified in (i) is on the root locus.
If the point specified in (i) is on the root locus, find the gain K,
using the lengths of the vectors.
SKETCHING THE ROOT LOCUS


The root locus can be obtained by sweeping through every point
in the s-plane to locate those points for which the angles, as
previously described will add up to an odd multiple of 180°.
However, the five sketching rules that will be introduced will
reduce the calculations for the plotting technique.
i.
ii.
iii.
iv.
v.
Number of branches
Symmetry over real-axis
Real-axis segments
Starting and end point
Behavior at infinity
NUMBER OF BRANCHES



Each closed-loop pole
moves as the gain is varied.
If we define a branch as the
path that one pole
traverses, then there will be
one branch for each closedloop pole.
Thus, the number of
branches of the root locus
equals the number of
closed-loop poles.
SYMMETRY OVER REAL-AXIS



If complex closed-loop poles
do not exist in conjugate
pairs, the resulting
polynomial would have
complex coefficients.
Physically reliable systems
can never have complex
coefficients in the transfer
function.
Thus, the root locus is
symmetrical about the realaxis.
REAL-AXIS SEGMENTS


At each point the angular contribution of a pair of open-loop
complex poles or zeros is zero.
The contribution of the open-loop poles and open-loop zeros to
the left of the respective point is zero.
REAL-AXIS SEGMENTS


The angles on the real-axis alternate between 0° to 180°.
The angle is 180° for regions of the real-axis that exist to the
left of an odd number of poles and/or zeros.
REAL-AXIS SEGMENTS

Thus, on the real-axis, for K > 0 the root locus exists to the left
of an odd number of real-axis, finite open-loop poles and/or
finite open-loop zeros.
STARTING AND ENDING POINTS

Consider the closed-loop transfer function,
T (s) 

KN G ( s ) DH ( s )
DG ( s ) DH ( s )  KNG ( s ) N H ( s )
As K approaches 0,
T (s) 

KNG ( s ) DH ( s )
DG ( s ) DH ( s)  
It is observed that as K approaches zero, the closed-loop
system poles approach the combined poles of G(s) and H(s).
STARTING AND ENDING POINTS

Consider the closed-loop transfer function,
T (s) 

KN G ( s ) DH ( s )
DG ( s ) DH ( s )  KNG ( s ) N H ( s )
As K approaches ∞,
T (s) 

KNG ( s ) DH ( s )
  KNG ( s) N H ( s)
It is observed that as K approaches ∞, the closed-loop system
poles approach the combined zeros of G(s) and H(s).
STARTING AND ENDING POINTS

Thus, the root locus begins at the finite and infinite open-loop
poles of G(s)H(s) and ends at the finite and infinite open-loop
zeros of G(s)H(s).
BEHAVIOR AT INFINITY

Apart from having finite poles and zeros, a function can also
have infinite poles and zeros.
If a function approaches infinity as s approaches infinity, then
the function has a pole at infinity. Example: G(s) = s.
ii.
If a function approaches zero as s approaches infinity, then the
function has a zero at infinity. Example: G(s) = 1/s.
Consider a system with three finite poles and three infinite zeros, the
location of the infinite zeros will need to be determined to show the
movement of the locus from the finite poles to infinite zeros.
i.

BEHAVIOR AT INFINITY


Thus, the root locus approaches straight lines as asymptotes as the
locus approaches infinity.
The equation of the asymptotes are given by the real-axis intercept,
σa, and angle, θa, as follows:
a
finite


poles   finite zeros
# finite poles  # finite zeros
(2k  1)
a 
# finite poles  # finite zeros

Where,
k = 0, ±1, ±2, ±3, …
The angle is given in radians with respect to the positive extension of
the real-axis.
SKETCHING THE ROOT LOCUS

Example: Sketch the root locus for the following system.
Finding the real-axis intercept, σa.
a 
(1  2  4)  (3) 4

4 1
3
SKETCHING THE ROOT LOCUS

Example: Sketch the root locus for the following system.
The angles of the lines that intersect at -4/3, are:
a   3

 5 3
for k  0
for k  1
for k  2
The running index, k, yields a multiplicity of lines that account
for the many branches of a root locus that approach infinity.
SKETCHING THE ROOT LOCUS
Root locus as
sketched from the
example.
REFINING THE SKETCH



The rules covered previously permit rapid sketching of the root
locus.
However, it is also important to find additional points on the
locus which is associated with gain.
The additional points that will be stressed on are:
i.
ii.
Real-axis breakaway and break-in points
The jω-axis crossings
BREAKAWAY & BREAK-IN POINTS


Numerous root loci
appears to break
away from the real
axis as the system
poles move from the
real axis to the
complex planes.
At other times the
loci appear to return
to the real axis as a
pair of complex
poles becomes real.
BREAKAWAY & BREAK-IN POINTS


The point where the
locus leaves the real
axis, -σ1, is called the
breakaway point.
The point where the
locus returns to the
real axis, -σ2, is called
the break-in point.
BREAKAWAY & BREAK-IN POINTS

Two methods to find breakaway and break-in points.
i.
Differential calculus:
For points along the real-axis segments, s = σ.
1
1
K 
K 
G ( s) H ( s)
G ( ) H ( )
ii.
Transition method:
Breakaway and break-in points satisfy the following relationship
m
n
1
1

1   z 1   p
i
i
where zi and pi are the negative of the zero and pole values, of
G(s)H(s).
BREAKAWAY & BREAK-IN POINTS

Example: Find the breakaway and break-in points for the
following open-loop system using differential calculus.
K ( s  3)( s  5)
KG( s) H ( s) 
( s  1)( s  2)

Solution:
KG( ) H ( ) 
( 2  3  2)
K  2
(  8  15)
K (  3)(  5)
(  1)(  2)
dK (11 2  26  61)

0
2
2
d
(  8  15)
  1.45, 3.82
(Breakaway and break-in points)
BREAKAWAY & BREAK-IN POINTS

Problem: Find the breakaway and break-in points for the
following open-loop system using the transition method.
K ( s  3)( s  5)
KG( s) H ( s) 
( s  1)( s  2)
THE jω-AXIS CROSSINGS



The jω-crossings are points on the root locus that separates the
stable operation of the system from the unstable operation.
The value of ω at the axis crossing yields the frequency of
oscillation, while the gain at the jω-axis crossing yields the
positive maximum gain for system stability.
The Routh-Hurwitz Criterion is used to determine the jω-axis
crossings by the following means:
i.
ii.
Forcing a row of zeros in the Routh table will yield the gain.
Going back one row to the even polynomial equation and solving
for the roots yields the frequency at the imaginary axis
crossings.
THE jω-AXIS CROSSINGS

Example: For the system below, find the frequency and gain, K,
for which the root locus crosses the imaginary axis. For what
range is the system stable?
T ( s) 

Routh Table:
K ( s  3)
s 4  7 s 3  14s 2  (8  K ) s  3K
THE jω-AXIS CROSSINGS

Example: For the system below, find the frequency and gain, K,
for which the root locus crosses the imaginary axis. For what
range is the system stable?
T ( s) 

K ( s  3)
s 4  7 s 3  14s 2  (8  K ) s  3K
Forcing a row of zeros at the s1 row.
 K 2  65K  720  0
K  9.65
 74.65 (Gain valu es should be positive)
THE jω-AXIS CROSSINGS

Example: For the system below, find the frequency and gain, K,
for which the root locus crosses the imaginary axis. For what
range is the system stable?
T ( s) 

K ( s  3)
s 4  7 s 3  14s 2  (8  K ) s  3K
Forming the even polynomial using the s2 row with K = 9.65.
(90  K ) s 2  21K  0
80.35s 2  202.7  0
s   j1.59 (Frequency at j - crossings for K  9.65)

The system is stable for 0 ≤ K <9.65.
FURTHER READING…

Chapter 8
i.
Nise N.S. (2004). Control System Engineering (4th Ed), John
Wiley & Sons.
“Shoot for the moon. Even if you miss it you will land among the
stars…"
THE END…