Chapter 5
THE ROOT-LOCUS DESIGN METHOD
Feedback Control System
Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
Root Locus: Illustrative Example
Examine the following closed-loop system, with unity
negative feedback.
The closed-loop transfer
function is given as:
Y (s)
K
 2
R( s) s  s  K
The roots of the characteristic
equation are:
1  1  4 K
s1,2 
2
• The characteristic equation
• The denominator of the
closed-loop transfer function
Erwin Sitompul
Feedback Control System
6/2
Root Locus: Illustrative Example
 1
1  4K
1


,
0

K



2
4
s1,2   2
 1  j 4 K  1 , K  1

 2
2
4
Imaginary Axis
2
K=∞
K=1/4
1
0
: the poles of open-loop
transfer function
K=0
K=0
-1
-2
-2
Erwin Sitompul
K=∞
-1
Real Axis
0
1
Feedback Control System
6/3
Root Locus: Illustrative Example
Where are the location of the closed-loop roots
when K=1?
n2  1  n  1
Y (s)
K
1
2n  1    0.5
 2
 2
R( s ) s  s  K s  s  1
  n  0.5
d  n 1   2  0.866
1.5
K=1
Imaginary Axis
1
0.866
(  jd )  (0.5  j 0.866)
0.5
0
-0.5
K=1
-1
-1.5
-2
Erwin Sitompul
-1
–0.5
Real Axis
–0.866
0
There is a relation
between gain K and the
position of closed-loop
poles, which also affects
the dynamic properties of
1 the system (ζ and ωd)
Feedback Control System
6/4
Root Locus of a Basic Feedback System
The closed-loop transfer function of the basic feedback
system above is:
Y ( s)
D( s)G( s)
 T ( s) 
R( s )
1  D( s)G( s) H ( s)
The characteristic equation, whose roots are the poles of this
transfer function, is:
1  D(s)G(s) H (s)  0
Erwin Sitompul
Feedback Control System
6/5
Root Locus of a Basic Feedback System
To put the characteristic equation in a form suitable for study
of the roots as a parameter changes, it is rewritten as:
1  KL(s)  0
where
KL(s)  D(s)G(s) H (s)
b( s )
L( s ) 
a( s)
 K is the gain of controller-plant-sensor combination.
 K is selected as the parameter of interest.
 W. R. Evans (in 1948, at the age of 28) suggested to plot
the locus (location) of all possible roots of the characteristic
equation as K varies from zero to infinity  root locus plot.
 The resulting plot is to be used as an aid in selecting the
best value of K.
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Feedback Control System
6/6
Root Locus of a Basic Feedback System
The root locus problem shall now be expressed in several
equivalent but useful ways.
1  KL(s)  0
b( s )
1 K
0
a( s)
a(s)  Kb(s)  0
1
L( s )  
K
 The equations above are sometimes referred to as “the
root locus form of a characteristic equation.”
 The root locus is the set of values of s for which the above
equations hold for some positive real value of K.
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Feedback Control System
6/7
Root Locus of a Basic Feedback System
 Explicit solutions are difficult to obtain for higher-order
system  General rules for the construction of a root locus
were developed by Evans.
 With the availability of MATLAB, plotting a root locus
becomes very easy, using the command “rlocus(num,den)”.
 However, in control design we are also interested in how to
modify the dynamic response so that a system can meet
the specifications for good control performance.
 For this purpose, it is very useful to be able to roughly
sketch a root locus which will be used to examine a system
and to evaluate the consequences of possible
compensation alternatives.
 Also, it is important to be able to quickly evaluate the
correctness of a MATLAB-generated locus to verify that
what is plotted is in fact what was meant to be plotted.
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Feedback Control System
6/8
Guideines for Sketching a Root Locus
Deriving using the root locus form of characteristic equation,
1  KL(s)  0
1
L( s )  
K
b( s )
1

a( s)
K
Taking the polynomial a(s) and b(s) to be monic, i.e., the
coefficient of the highest power of s equals1, they can be
factorized as:
( s  z1 )( s  z2 ) ( s  zm )
1

( s  p1 )( s  p2 ) ( s  pn )
K
If any s = s0 fulfills the equation above, then s0 is said to be
on the root locus.
Erwin Sitompul
Feedback Control System
6/9
Guideines for Sketching a Root Locus
The magnitude condition implies:
( s  z1 ) ( s  z2 ) ( s  zm )
( s  p1 ) ( s  p2 )
( s  pn )
s  s0
1
1
 
K K
Magnitude
Condition
The phase condition implies:
( s  z1 )( s  z2 ) ( s  zm )
1

    180
( s  p1 )( s  p2 ) ( s  pn ) s  s
K
Phase
Condition
0
Defining (s  zi )   i and (s  pi )  i , the phase
condition can be rewritten as:
m
n
  
i 1
i
i 1
Erwin Sitompul
i
 180  360(l  1), l  1, 2,3,
Feedback Control System
6/10
Guideines for Sketching a Root Locus
“
“
The root locus is the set of values of s for which
1 + KL(s) = 0 is satisfied as the real parameter K varies
from 0 to ∞. Typically, 1 + KL(s) = 0 is the characteristic
equation of the system, and in this case the roots on the
locus are the closed-loop poles of that system.
”
The root locus of L(s) is the set of points in the s-plane
where the phase of L(s) is 180°. If the angle to a test
point from a zero is defined as ψi and the angle to a
test point from a pole as Φi, then the root locus of L(s)
is expressed as those points in the s-plane where, for
integer l, Σψi – ΣΦi = 180° + 360°(l–1).
”
Erwin Sitompul
Feedback Control System
6/11
Guideines for Sketching a Root Locus
Consider the following example.
s 1
L( s ) 
s( s  5) ( s  2)2  4
: the poles of L(s)
: the zero of L(s)
s0  1  j 2
Test
point
 1  90
1  tan 1 (2 1)  116.6
2  0
3  tan1 (4 1)  76.0
4  tan 1 (2 4)  26.6
Erwin Sitompul
   
i
i
1
 1  2  3  4
 90  116.6  0  76.0  26.6
 129.2  180  360(l  1)
 s0 is not on the root locus
Feedback Control System
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Rules for Plotting a Root Locus
RULE 1:
The n branches of the locus start at the poles of L(s) and m of
these branches end on the zeros of L(s), while n–m branches
terminate at infinity along asymptotes.
Recollecting
(s  z1 ) (s  z2 )
b( s ) 1
b( s )
1

 

a( s ) K
a( s)
K
(s  p1 ) (s  p2 )
(s  z1 ) (s  z2 )
1
lim    lim
K 0 K
s  pi ( s  p ) ( s  p )
1
2
( s  zm )
(s  z1 ) (s  z2 )
1
lim  0  lim
K  K
s  zi ( s  p ) ( s  p )
1
2
(s  zm )
( s  zm )
1

(s  pn ) K
(s  pn )
(s  pn )
 The root locus starts at K = 0 at the poles of L(s)
and ends at K = ∞ on the zeros of L(s)
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Feedback Control System
6/13
Rules for Plotting a Root Locus
RULE 2:
On the real axis, the loci (plural of locus) are to the left of an
odd number of poles and zeros.
5
4
3
2
1
: The root locus
1
4
3
2
1
 Angles from real poles or zeros are 0° if the test point is
to the right and 180° if the test point is to the left of a
given pole or zero.
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Feedback Control System
6/14
Rules for Plotting a Root Locus
The rule is now applied to obtain the root locus of:
1
L( s ) 
s ( s  4)2  16
Erwin Sitompul
p1  0
p2,3  4  j 4
Feedback Control System
6/15
Rules for Plotting a Root Locus
For any test point s0 on the real
axis, the angles φ1 and φ2 of two
complex conjugate poles cancel
each other, as would the angles of
two complex conjugate zeros
(see figure below).
1  tan1 (2 4)  26.6
2  tan1 (2 4)  26.6
1  2  0
3  0 The pair does not
give contribution to
the phase condition
1  2  3  0
s1
 180  360(l  1)
 s0 is not on the
root locus
Now, check the phase
condition of s1!
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Feedback Control System
6/16
Rules for Plotting a Root Locus
RULE 3:
For large K and s, n–m of the loci are asymptotic to lines at
angles Φl radiating out from the point s = α on the real axis,
where:
180  360(l  1)
Angles of Asymptotes
l 
, l  1, 2, ,3
nm
pi  zi

Center of Asymptotes

nm
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Feedback Control System
6/17
Rules for Plotting a Root Locus
For L( s) 
1
, we obtain n  3, m  0
2
s ( s  4)  16
p1  0, p2,3  4  j 4
180  360(l  1)
l 
nm
180  360(l  1)

30
 60  120(l  1)
 60,180,300
60°
180°
–2.67
p  z


i
i
nm
(4  j 4)  (4  j 4)  0

30
8

 2.67
3
Erwin Sitompul
300°
Feedback Control System
6/18
Rules for Plotting a Root Locus
RULE 4:
The angle of departure of a branch of a locus from a pole is
given by:
l ,dep   i  i  180  360(l  1)
i l
and the angle of arrival of a branch of a locus to a zero is
given by:
 l ,arr  i   i  180  360(l  1)
i l
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Feedback Control System
6/19
Rules for Plotting a Root Locus
For the example, the root loci must depart with certain angles
from the complex conjugate poles at –4 ± j4, and go to the
zero at ∞ with the angles of asymptotes 60° and 300°.
Erwin Sitompul
Feedback Control System
6/20
Rules for Plotting a Root Locus
From the figure,
2   i  i  180  360(l 1)
i 2
2  1  3 180  360(l 1)
But
1  90
3  tan 1 (4 4)  135
Thus
2  90 135 180
2  405  45
By the complex conjugate symmetry
of the roots, the angle of departure
of the locus from –4 – j4 will be +45°.
Erwin Sitompul
Feedback Control System
6/21
Rules for Plotting a Root Locus
So, the root loci will start their journey from –4 ± j4 towards ∞
with the direction of 45°.
±
Erwin Sitompul
Feedback Control System
6/22
Rules for Plotting a Root Locus
RULE 5:
The locus crosses the jω axis (imaginary axis) at points where:
 The Routh criterion shows a transition from roots in the left
half-plane to roots in the right half-plane.
 This transition means that the closed-loop system is
becoming unstable.
 This fact can be tested by Routh’s stability criterion, with
K as the parameter, where an incremental change of K
will cause the sign change of an element in the first
column of Routh’s array.
 The values of s = ± jω0 are the solution of the characteristic
equation in root locus form, 1 + KL(s) = 0.
 The points ± jω0 are the points of cross-over on the
imaginary axis.
Erwin Sitompul
Feedback Control System
6/23
Rules for Plotting a Root Locus
For the example, the characteristic equation can be written as:
1  KL(s)  0
1
1 K
0
2
s (s  4)  16
s3  8s 2  32s  K  0
s3 :
s2 :
1
8
8  32  K
1
s :
8
s0 :
K
Erwin Sitompul
32
K
 The closed-loop system is stable for K > 0
and K < 256  for 0 < K < 256.
 For K > 256 there are 2 roots in the RHP
(two sign changes in the first column).
 For K = 256 the roots must be on the
imaginary axis.
Feedback Control System
6/24
Rules for Plotting a Root Locus
The characteristic equation is now solved using K = 256.
s3  8s 2  32s  256  0
s1  8
s2,3   j5.66   j0
Points of Cross-over
Another way to solve for ω0 is by simply replacing any s with jω0
without finding the value of K first.
( j0 )3  8( j0 )2  32( j0 )  K  0
 j03  802  j320  K  0
K  802  j(320  03 )  0
≡0
≡0
320  03
K  80 2
 K  8  32
 0 2  32
 K  256
 0  5.66
Same results for K and ω0
Erwin Sitompul
Feedback Control System
6/25
Rules for Plotting a Root Locus
The points of cross-over are now inserted to the plot.
5.66
–5.66
Erwin Sitompul
Feedback Control System
6/26
Rules for Plotting a Root Locus
The complete root locus plot can be shown as:
Erwin Sitompul
Feedback Control System
6/27
Rules for Plotting a Root Locus
RULE 6:
The locus will have multiple roots at points on the locus where:
da ( s )
db( s )
b( s )
 a(s)
0
ds
ds
The branches will approach and depart a point of q roots at
angles separated by:
180  360(l  1)
q
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Feedback Control System
6/28
Rules for Plotting a Root Locus
 A special case of point of multiple roots is the intersection
point of 2 roots that lies on the real axis.
 If the branches is leaving the real axis and entering the
complex plane, the point is called the break-away point.
 If the branches is leaving the complex plane and entering
the real axis, the point is called the break-in point.
Imag
axis
Imag
axis
90
90
Real
axis
Break-away
point
Erwin Sitompul
Real
axis
Break-in
point
Feedback Control System
6/29
Example 1: Plotting a Root Locus
Draw the root locus plot of the system shown below.
K
Y (s)
s ( s  1)( s  2)

K
R( s) 1 
s ( s  1)( s  2)
 1  KL(s)
Erwin Sitompul
1
 L( s ) 
s( s  1)( s  2)
RULE 1
n  3, m  0
3 zeros at infinity
p1  0, p2  1, p3  2
Feedback Control System
6/30
Example 1: Plotting a Root Locus
p1  0, p2  1, p3  2
3
Imag
axis
2
1
RULE 2
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
6/31
Example 1: Plotting a Root Locus
180  360(l  1)
l 
nm
180  360(l  1)

30
 60  120(l  1)
 60,180,300
p  z


i
nm
0 1 2

30
 1
Erwin Sitompul
i
Angles of Asymptotes
p1  0, p2  1, p3  2
Center of Asymptotes
Feedback Control System
6/32
Example 1: Plotting a Root Locus
l  60,180,300
  1
3
Imag
axis
60°
2
1
180°
–4
–3
–2
–1
0
–1
RULE 4
Not applicable. The angles
of departure or the angles of
arrival must be calculated
only if there are any complex
poles or zeros.
Erwin Sitompul
1
2
Real
axis
RULE 3
–2
300°
–3
Feedback Control System
6/33
Example 1: Plotting a Root Locus
1  KL(s)  0
1
1 K
0
s( s  1)( s  2)
s3  3s 2  2s  K  0
Replacing s with jω0,
( j0 )3  3( j0 )2  2( j0 )  K  0
 j03  302  j 20  K  0
K  302  j(20  03 )  0
≡0
≡0
20  03
K  30 2
 K  3 2
 0 2  2
K 6
 0  1.414
Points of Cross-over
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Feedback Control System
6/34
Example 1: Plotting a Root Locus
3
0  1.414
Imag
axis
2
1.414
RULE 5
1
–4
–3
–2
–1
0
1
2
Real
axis
–1
–1.414
–2
–3
Erwin Sitompul
Feedback Control System
6/35
Example 1: Plotting a Root Locus
1  KL(s)  0
b( s )
1
1
L( s ) 

 3
a( s) s( s  1)( s  2) s  3s 2  2s
The root locus must have a break-away point, which
can be found by solving:
da ( s )
db( s )
b( s )
 a(s)
0
ds
ds
1 (3s2  6s  2)  (s3  3s2  2s)  0  0
3s 2  6s  2  0
s1  1.577, s2  0.423
Erwin Sitompul
Feedback Control System
6/36
Example 1: Plotting a Root Locus
s1  1.577, s2  0.423
Not on the
root locus
3
On the root locus
The break-away
point
–0.423
–4
–3
–2
–1
Imag
axis
2
1.414
RULE 6
1
0
1
2
Real
axis
–1
–1.414
–2
–3
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Feedback Control System
6/37
Example 1: Plotting a Root Locus
After examining RULE 1 up to
RULE 6, now there is enough
information to draw the root
locus plot.
3
Imag
axis
2
–0.423
1.414
1
90
–4
–3
–2
–1
0
1
2
Real
axis
–1
–1.414
–2
–3
Erwin Sitompul
Feedback Control System
6/38
Example 1: Plotting a Root Locus
The final sketch, with direction of
root movements as K increases
from 0 to ∞ can be shown as:
3
Imag
axis
Final Result
2
–0.423
–4
–3
–2
–1
1.414
1
0
1
2
Real
axis
–1
–1.414
–2
–3
Determine the locus of
all roots when K = 6!
Erwin Sitompul
Feedback Control System
6/39
Example 2: Plotting a Root Locus
a) Draw the root locus plot of the system.
b) Define the value of K where the system is stable.
c) Find the value of K so that the system has a root at s = –2.
Erwin Sitompul
Feedback Control System
6/40
Example 2: Plotting a Root Locus
1
K
s2
Y (s)
( s  4)
RULE 1
 L( s ) 

s  4 There is one branch,
R ( s ) 1  K 1 ( s  2)
starts from the pole and
( s  4)
n  1, m  1 approaches the zero
p1  4, z1  2
 1  KL(s)
Erwin Sitompul
Feedback Control System
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Example 2: Plotting a Root Locus
p1  4, z1  2
3
Imag
axis
2
1
RULE 2
RULE 3
–4
–3
Not applicable, since n = m.
RULE 4
Not applicable. The angles
of departure or the angles of
arrival must be calculated
only if there are any complex
poles or zeros.
Erwin Sitompul
–2
–1
0
1
2
Real
axis
–1
–2
–3
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Example 2: Plotting a Root Locus
1  KL(s)  0
s2
1 K
0
s4
s  4  K (s  2)  0
Replacing s with jω0,
( j0 )  4  K ( j0  2)  0
4  2K  j0 (1  K )  0
≡0
Erwin Sitompul
≡0
4  2K
K 2
0 (1  K )
 0  0
Points of Cross-over
Feedback Control System
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Example 2: Plotting a Root Locus
RULE 5
3
The point of cross-over, as can
readily be guessed, is at s = 0.
2
K=2
–4
–3
Imag
axis
–2
–1
1
0
1
2
Real
axis
–1
RULE 6
Not applicable. There is no
break-in or break-away point.
Erwin Sitompul
–2
–3
Feedback Control System
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Example 2: Plotting a Root Locus
a) Draw the root locus plot of the system. Imag
3
The final sketch, with direction of
root movements as K increases
from 0 to ∞ can be shown as:
axis
Final Result
2
1
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
Erwin Sitompul
Feedback Control System
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Example 2: Plotting a Root Locus
b) Define the value of K where the
system is stable.
3
Imag
axis
2
K=0
–4
K=2
–3
–2
–1
K=∞
1
0
1
2
Real
axis
–1
System is stable when the root of
the characteristic equation is on
the LHP, that is when 0 ≤ K < 2.
–2
–3
Erwin Sitompul
Feedback Control System
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Example 2: Plotting a Root Locus
c) Find the value of K so that the
system has a root at s = –2.
3
2
K = 0.5
K=0
–4
K=2
–3
–2
Inserting the value of s = –2 in
the characteristic equation,
–1
K=∞
1
0
1
2
Real
axis
–1
s  4  K (s  2) s2  0
–2
2  4  K (2  2)  0
K (2  2)  2 4  0.5
–3
Erwin Sitompul
Imag
axis
Feedback Control System
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Homework 6
 No.1, FPE (5th Ed.), 5.2.
Hint: Easier way is to assign reasonable values for the zeros and poles in
each figure. Later, use MATLAB to draw the root locus.
 No.2, FPE (5th Ed.), 5.7.(b)
Hint: After completing the hand sketch, verify your result using MATLAB.
Try to play around with Data Cursor.
 No.3
Sketch the root locus diagram of the following closedloop system as accurate as possible.
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Feedback Control System
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