Outline
• Rules for sketching the root locus.
• Examples showing how the rules
are applied.
Root Locus Rules
M. Sami Fadali
Professor of electrical Engineering
University of Nevada
R(s) 

K
G(s)
C(s)
H(s)
1
Root Locus Branches
Number of Root Locus Branches
1. The number of root locus branches (closed-loop
poles) is equal to the number of open-loop poles
of the loop gain
.
Equation has
2
roots.
3
2. Branches start at the open-loop poles and end at the
open-loop zeros or at infinity.
⋯
⋯
zeros at
4
Example
Real axis root loci
How many branches? Start? End?
3- Real axis root loci have an odd number of real
poles plus zeros to their right.
• Apply the angle condition to test points.
• No contribution from complex poles and zeros.
• 3 root locus branches
• Start at: 0,2, 5
x
• One branch ends at 1 (zero), two branches
go to infinity
s2
s1
x
a
• 3  1 = 2 zeros at infinity
s1+a
x
5
6
Example
Asymptotes
Determine the real-axis loci
4- The branches going to infinity asymptotically
approach the straight lines defined by the
angle and intercept
Pole-Zero Map
1
0.8
0.6
Imaginary Axis
0.4
0.2
a 
0
-0.2
np
-0.4
-0.6
a 
-0.8
-1
-5
 2m  1180
, m  0,1, 2,
n p  nz
-4.5
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
nz
 pi   z j
i 1
j 1
n p  nz
0
Real Axis
7
8
Example
Breakaway/Break-in Points
5- Breakaway points (points of departure from the
real axis) correspond to maxima of , while breakin points (points of arrival at the real axis)
correspond to minima of .
• Find maxima (minima) using one of two
)
approaches (on real axis
Find the asymptotes for
a 
 2m  1180  2m  1180

3 1
n p  nz
 2m  190 , m  0,1, 2,
np
a 
 p z
i 1
i.
nz
i
j 1
n p  nz
j

0  2  5   1
3 1
ii. Plot
 3
9
10
Example
Graphical Approach
Find the breakaway points for
Root Locus
10
(i)
8
6
MAPLE
plot(K,s=-5..-2)
• Peak at about 3.3
4
Imaginary Axis
MAPLE
g := (s+1)/(s*(s+2)*(s+5))
K := -1/g
diff(K, s)
solve(% = 0., s)
K
(ii) Plot
2
0
-2
-4
-6
-8
-10
-6
-5
-4
-3
-2
-1
0
1
Real Axis
s
11
12
Proof of Rule
Departure & Arrival Angles
n z 1
6-The angle of departure from a complex pole
(angle of arrival at a complex zero ) is


L( s )  s  z n 
z
 s  zi 
i 1
np
 s  p 
nz
 s  z 
1

 i 1
s  p n n 1
 s  p 
p
p
j
j
j 1


L( scl )  s  z n  Gn scl  
)
z
z
i
j 1
1
 Gn scl 
s  pn
p
p
)
• Apply the angle condition to test points
close to the complex pole (zero).
    zn  Gnz  z n    zn    Gnz  z n 
)
)
13
Example
    pn  Gn p  pn    pn    Gn p  pn 
14
Root Locus for Example
Sketch the root locus for the system
j-axis
intercept
See root locus
15
16
j-axis
intercept
Example 8.5 (Nise)
=
Multiple Constraints
, Closed-loop Transfer Function
• Negative 2nd derivative of quadratic (negative
coefficient): maximum (concave)
KN ( s )
K s  3
T (s) 
 4
3
KN ( s )  D( s ) s  7 s  14s 2  8  K s  3K
14
s4 1
8 K
s3 7
21K
s 2 90 2K
 K  65 K  720
s1
90  K
s 0 21K
• Quadratic is positive for
3K
• Stable range=intersection of K ranges for positive
entries of the first column
• Auxiliary Equation:
• Third row multiplied by 7 to simplify calculations.
• For stability > 0, < 90, and the quadratic is positive
• The quadratic has the roots
= −74.646,
= 9.6465
17
Conclusion
• Apply rules to sketch root locus.
• Some rules may not be applicable.
• MATLAB: numerical solution for roots
without the use of rules for root locus
sketching.
19
• Intersections of RL with j-axis:
18