Root Locus Method
Control Systems Engineering
Christian Dave T. Rolloque
Niel Martin Codes
Orland Nuñez
Jay Mark Mellomida
Lyndon Agupo
John Christian Absalon
4.1 Introduction
Definition
A graph displaying the roots of a polynomial equation when one of the parameters in
the coefficients of the equation changes from 0 to .
Motivation
To satisfy transient performance requirements, it may be necessary to know how to
choose certain controller parameters so that the resulting closed-loop poles are in the
performance regions, which can be solved with Root Locus technique.
2
4.1 Introduction
R(s) C(s)
k G(s)
Fig. 4.1
H (s)
Consider the feedback control system shown in Fig. 4.1, the closed
loop transfer function is given by:
k G (s)
C (s)
T (s) R (s) 1 k G (s) H
Where k is a constant gain(s)parameter. The poles of the
transfer
function are the roots of the characteristic equation given by:
1 k G (s) H (s) 0
3
Definition of Root
Locus
The root locus is the locus of the characteristic equation of the closed-
loop system as a specific parameter (usually, gain K) is varied from
to infinity. For 0 ≤ 𝑘 < ∞ , if 1 k G(s) H (s) 0
zero
1
then G (s)H (s) k
and for positive k, this means that a point s which is a point on locus
must satisfy the magnitude and angle and conditions:
The magnitude condition:
1
| G ( s ) H (s) | k
The angle condition:
G (s) H (s) 180 o i360 o i 0,1,2,3,....
w here 4
4.2.1 Root Locus
Rules 1: The number of
Rule branches of the root locus is equal to
the number of closed-loop poles (or roots of the characteristic
equation).
open-loop
Rule zeros
2: Root (when
locus starts at open-loop poles (when K= 0) and ends at
K=). Q(s)
1 k G (s) = 1 + k = P(s) + kQ(s) = 0
P(s)
Rule 3: Root locus is symmetric about the real axis, which reflects the
fact that closed-loop poles appear in complex conjugate pairs.
5
4.2.1 Root Locus
Rules
Rule 4: Along the real axis, the root locus includes all segments that
are to the left of an odd number of finite real open-loop poles and
zeros.
Rule 5: If number of poles (n) exceeds the number of zeros (m), then as
K, (n - m) branches will become asymptotic to straight lines. These
straight lines intersect the real axis with angles A at A .
A p i
z sum open loop poles sum open loop
nm
i # of open loop poles # of open loop
zeros
zeros
A 180 i360 where i 0,1,2,3,....
n
m 6
4.2.1 Root Locus
Rules
Rule 6: “Breakaway and/or break-in points” where the locus between
two poles on the real axis leaves the real axis is called the breakaway
point and the point where the locus between two zeros on the real axis
returns to the real axis is called the break-in point. The loci leave or
return to the real axis at the maximum gain k of the following equation:
1
| G(s )H (s ) | k
o o
Rule 7: The departure angle for a pole pi ( the arrival angle for a zero
zi) can be calculated by slightly modifying the following equation:
7
4.2.1 Root Locus
Rules
The angle of departure p of a locus from a complex pole is given by:
p = 180 – [sum of the other GH pole angles to the pole under consideration]
+ [sum of the GH zero angles to the pole]
The angle of approach z of a locus to a complex zero is given by:
z = 180 + [sum of the other GH pole angles to the zero under consideration] –
[sum of the GH zero angles to the zero]
Rule 8: If the root locus passes through the imaginary axis (the
stability boundary), the crossing point j and the corresponding gain
K can be found using Routh-Hurwitz criterion.
8
Steps to Sketch Root Locus
(1/2)
Step#1: Transform the closed-loop characteristic equation into the
standard form for sketching root locus:
1 k G (s) H (s) 0
Step#2: Find the open-loop zeros and the open-loop poles. Mark the
open-loop poles and zeros on the complex plane. Use ‘x’ to represent
open-loop poles and ‘o’ to represent the open-loop zeros.
Step#3: Determine the real axis segments that are on the root locus by
applying Rule 4.
Step#4: (If necessary) Determine the number of asymptotes and the
corresponding intersection and angles by applying Rules 2 and 5.
9
Steps to Sketch Root Locus
(2/2)
Step#5: (If necessary) Determine the break-away and break-in points using
Rule 6.
Step#6: (If necessary) Determine the departure and arrival angles
using Rule 7.
Step#7: (If necessary) Determine the imaginary axis crossings using Rule 8.
Step#8: Use the information from Steps 1-7, sketch the root locus.
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4.2.2 Root Locus Procedure – Example
4.1
Example 4.1: Sketch the root locus for the system shown in Fig.4.2
Fig. 4.2
Solution:
Step 1: “Write the open loop transfer function of the system and find the
open loop Poles & zeros” as follows:
k(s
kG(s)H (s)
s(s 3) 2)(s
1)(s
4) # of poles = n =4
Open loop Poles = {0, -1, -2, -4}
Open loop Zeros= {-3} # of zeros = m =1
11
4.2.2 Root Locus Procedure – Example 4.1
(Cont.)
Step 2: “Draw the pole-zero plot of kG(s)H(s), then locate the segments of real axis
that are root loci”.
Root locus segments
exists
after the poles at 0, -2 and -4
Asymptotic
Beakaway
Intersection with imaginary??
12
4.2.2 Root Locus Procedure – Example 4.1
(Cont.)
Step 3. Asymptotic Angles: “As k approaches +∞, the branches of the locus become
asymptotic to straight lines with angles:
A 180 i360 where i
n 0,1,2,3,....
• m are three [n-m = 3]
The number of asymptotes
• for i=0 1 = 60, for i=1 2=180, and for i= -1 3=-60.
Step 4. Center of Asymptotes: The starting point of the asymptotes. These linear
asymptotes are centered at a point on the real axis at:
A pole v al ues of G H z e ro v al ue s of G H
nm
1 2 4 3
1 .3 3 13
4.2.2 Root Locus Procedure – Example 4.1
(Cont.)
14
4.2.2 Root Locus Procedure – Example 4.1
(Cont.)
Step 5. Breakawayand break-in or entry points:
1
1 s(s 1)(s 2)(s 4)
| G(s )H (s ) |k and k | G(s)H (s) |
o o | (s
3)
|
Therefore, the break away point lies between 0 and -1
at
maximum value for k which is calculated as follows:
s -0.3 -0.4 -0.45 -0.5
k 0.489 0.532 0.534 0.525
It is clear that the breakaway point is at s = -0.45.
15
4.2.2 Root Locus Procedure – Example 4.1
(Cont.)
Step 7. j axis crossing:
• the closed loop transfer function is given by:
G(s) k(s
T (s)
1 3)
G(s)H (s) 4 7s 14s (8 k )s
3 2
s
3k
• The c/s equation is 1 G(s)H (s) s 4 7s 3 14s 2 (8 k )s
3k
• The Routh-table for the characteristic equation is shown
• The coefficient of s1 = 0
k 2 65k 720 0
• Then k = 9.65
• The intersection points are
(90 k )s 2 21k 0
80.35s2 202.7 0
thus s1,2 j1.59
We also conclude that the system is stable for 0 < k < 9.65. 16
4.2.2 Root Locus Procedure – Example 4.1
(Cont.)
17
4.2.2 Root Locus Procedure – Example
4.2
Example 4.2: Sketch the root locus for the system shown in Fig.4.4
R(s) k(s 3)(s C(s)
5)
(s 1)(s 2) Fig. 4.4
Solution
Step 1: Write the open loop transfer function or the characteristic equation
of the system as follows:
k(s 3)(s 5)
k G( s) H ( s)
(s 1)(s 2)
18
4.2.2 Root Locus Procedure – Example 4.2
(Cont.)
Step 2: “Draw the pole-zero plot of kG(s)H(s), then locate the segments of real axis
that are root loci”.
Root locus segments
exists after the zero at 5
and after the pole at -1.
Beakaway & Break-in
Intersection with imaginary 19
4.2.2 Root Locus Procedure – Example 4.2
(Cont.)
Step 3: Breakaway and break-in or entry points
1
1 (s 1)(s
| G(s )H (s ) | k and k | G(s)H (s) | (s 3)(s
o o
5) 2)
Therefore, the breakaway point lies between -1 and -2 and the break in point lies
between 3 and 5.
at s=-1.45 breakaway
&
at s=-3.82 break-in
points
20
4.2.2 Root Locus Procedure – Example 4.2
(Cont.)
Step 7. j axis crossing:
• the closed loop transfer function is given by:
G(s)
k(s 2 8s 15)
T (s)
• The c/s equation is 1 G(s)H1(s) G(s)H (s) (1 k )s 2 (3 (1 )s (2
8k
• The coefficient of s1k=)s0 (3 15k
8k )s
) (2 15k )
2
3 8k 0
• Then k = 3/8=0.375
• The intersection points are
(1 k )s 2 (2 15k ) 0
1.375s 2 7.625 0
thus s1,2 j2.35
We also conclude that the
system is stable for 0 < k < 21
4.2.2 Root Locus Procedure – Example 4.2
(Cont.)
22
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