The document provides an overview of root locus analysis and design of control systems. It begins with an introduction to root locus including motivation, definition, and the basic feedback control system model. It then covers the key rules and steps for constructing and interpreting root loci, including determining asymptotes, breakaway/break-in points, and imaginary axis crossings. Three examples are worked through step-by-step to demonstrate how to apply the rules and steps to sketch root loci for different open-loop transfer functions. The document explains how root locus can be used to choose controller parameters to satisfy transient performance requirements.

1. ‫ر‬َ‫ـد‬ْ‫ق‬‫ِـ‬‫ن‬،،،‫لما‬‫اننا‬ ‫نصدق‬ْْ‫ق‬ِ‫ن‬‫ر‬َ‫د‬
LECTURE (4)
Analysis and Design of Control
Systems using Root Locus
Assist. Prof. Amr E. Mohamed

2. Lecture Outline
4.1 Introduction
4.2 Root Locus Construction
4.3 Root Locus Examples
4.4 Stability Analysis
4.5 Compensation and Controller Design using Root Locus
2

3. 4.1 Introduction
 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.
 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 .
3

4. 4.1 Introduction
 Consider the feedback control system shown in Fig. 4.1, the closed loop
transfer function is given by:
 Where k is a constant gain parameter. The poles of the transfer
function are the roots of the characteristic equation given by:
)()(1
)(
)(
)(
)(
sHsGk
sGk
sR
sC
sT


Fig. 4.1
)(sR )(sC
)(sG
)(sH
k
0)()(1  sHsGk
4

5. 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 zero
to infinity. For 0 ≤ 𝑘 < ∞ , if
 then
 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:
 The angle condition:
0)()(1  sHskG
k
sHsG
1
)()( 
k
sHsG
1
|)()(| 
,....3,2,1,0360180)()(  iwhereisHsG oo
5

6. 4.2.1 Root Locus Rules
 Rule 1: The number of branches of the root locus is equal to the
number of closed-loop poles (or roots of the characteristic equation).
 Rule 2: Root locus starts at open-loop poles (when K= 0) and ends at
open-loop zeros (when K=). If the number of open-loop poles is
greater than the number of open-loop zeros, some branches starting
from finite open-loop poles will terminate at zeros at infinity.
 Rule 3: Root locus is symmetric about the real axis, which reflects the
fact that closed-loop poles appear in complex conjugate pairs.
6

7. 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 .
zerosloopopenofpolesloopopenof
zerosloopopensumpolesloopopensum
mn
zp ii
A
## 





 
,....3,2,1,0
360180



 iwhere
mn
i
A
7

8. 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:
 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:
k
sHsG oo
1
|)()(| 
8

9. 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.
9

10. Steps to Sketch Root Locus (1/2)
 Step#1: Transform the closed-loop characteristic equation into the
standard form for sketching root locus:
 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.
0)()(1  sHsGk
10

11. 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.
11

12. Fig. 4.2
4.2.2 Root Locus Procedure – Example 4.1
 Example 4.1: Sketch the root locus for the system shown in Fig.4.2
 Solution:
 Step 1: “Write the open loop transfer function of the system and find the
open loop Poles & zeros” as follows:
 Open loop Poles = {0, -1, -2, -4}  # of poles = n =4
 Open loop Zeros= {-3}  # of zeros = m =1
)4)(2)(1(
)3(
)()(



ssss
sk
sHskG
12

13.  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??
4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
13

14.  Step 3. Asymptotic Angles: “As k approaches +∞, the branches of the locus become
asymptotic to straight lines with angles:
• The number of asymptotes are three [n-m = 3]
• 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:
,....3,2,1,0
360180



 iwhere
mn
i
A
33.1
3
3421




 

mn
GHofvalueszeroGHofvaluespole
A
4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
14

15. 4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
15

16.  Step 5. Breakawayand break-in or entry points:
|
)3(
)4)(2)(1(
|
|)()(|
11
|)()(|



s
ssss
sHsG
kand
k
sHsG oo
 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.
4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
16

17.  Step 7. j axis crossing:
• the closed loop transfer function is given by:
• The c/s equation is
• The Routh-table for the characteristic equation is shown
• The coefficient of s1 = 0
• Then k = 9.65
• The intersection points are
ksksss
sk
sHsG
sG
sT
3)8(147
)3(
)()(1
)(
)( 234





kskssssHsG 3)8(147)()(1 234

0720652
 kk
021)90( 2
 ksk
07.20235.80 2
s
59.12,1 jsthus 
 We also conclude that the system is stable for 0 < k < 9.65.
4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
17

18. 4.2.2 Root Locus Procedure – Example 4.1 (Cont.)
18

19. 4.2.2 Root Locus Procedure – Example 4.2
 Example 4.2: Sketch the root locus for the system shown in Fig.4.4
 Solution
 Step 1: Write the open loop transfer function or the characteristic equation
of the system as follows:
Fig. 4.4
)(sR )(sC
)2)(1(
)5)(3(


ss
ssk
)2)(1(
)5)(3(
)()(



ss
ssk
sHskG
19

20.  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
4.2.2 Root Locus Procedure – Example 4.2 (Cont.)
20

21.  Step 3: Breakaway and break-in or entry points
 Therefore, the breakaway point lies between -1 and -2 and the break in point lies
between 3 and 5.
)5)(3(
)2)(1(
|)()(|
11
|)()(|



ss
ss
sHsG
kand
k
sHsG oo
at s=-1.45 breakaway
&
at s=-3.82 break-in points
4.2.2 Root Locus Procedure – Example 4.2 (Cont.)
21

22.  Step 7. j axis crossing:
• the closed loop transfer function is given by:
• The c/s equation is
• The coefficient of s1 = 0
• Then k = 3/8=0.375
• The intersection points are
)152()83()1(
)158(
)()(1
)(
)( 2
2
ksksk
ssk
sHsG
sG
sT





)152()83()1()()(1 2
ksksksHsG 
083  k
0)152()1( 2
 ksk
0625.7375.1 2
s
35.22,1 jsthus 
 We also conclude that the system is stable for 0 < k < 0.375.
4.2.2 Root Locus Procedure – Example 4.2 (Cont.)
22

23. 4.2.2 Root Locus Procedure – Example 4.2 (Cont.)
23

24.  Example 4.3: Sketch the root locus for the system shown in Fig.4.6
 Solution
 Step 1: Write the open loop transfer function or the characteristic equation
of the system as follows:
Fig. 4.6
)(sR )(sC
)22)(3(
)2(
2


sss
sk
)22)(3(
)2(
)()( 2



sss
sk
sHskG
4.2.2 Root Locus Procedure – Example 4.3
24

25.  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 -2.
 Asymptotic
 Departure Angles
 Intersection with imaginary????
4.2.2 Root Locus Procedure – Example 4.3 (Cont.)
25

26.  Step 3. Asymptotic Angles:
• The number of asymptotes are Two [n-m = 2]
• for i=0  1 = 90, for i=, 2=-90 .
 Step 4. Center of Asymptotes:
,....3,2,1,0
13
360180



 iwhere
i
A
1
3
211113




 

jj
mn
GHofvalueszeroGHofvaluespole
A
4.2.2 Root Locus Procedure – Example 4.3 (Cont.)
26

27.  Step 5. Angles of departure: 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]
p = 180 – [2+4] + [3]
= 180 – [90+30] + [45]
= 105 degree
4.2.2 Root Locus Procedure – Example 4.3 (Cont.)
27

28. 4.2.2 Root Locus Procedure – Example 4.3 (Cont.)
28

29. 4.2.2 Root Locus Procedure – Example 4.4
 Example 4.2: Sketch the root locus for the system shown in Fig.4.8 (a) and
find the following:
a) The point and the gain on the root locus where the damping ration (ξ=0.45).
b) The range of k within which the system is stable.
 Solution
 Step 1: Write the open loop transfer function or the characteristic equation of the
system as follows:
Fig.4.8 (a)
)4)(2(
)204(
)()(
2



ss
ssk
sHskG
29

30.  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 pole at -2.
 Beakaway
 Arrival Angles
 Intersection with imaginary
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
30

31.  Step 3. Angles of Arrival: The angle of
Arrival z of a locus from a complex pole
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]
z = 180 + [1+2] - [3]
= 180 + [45+30] – [90]
= 165 degree
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
31

32.  Step 4: Breakaway points
 Therefore, the breakaway point lies between -2 and -4.
 The breakaway point lies at s=-2.88 at maximum value for k = 0.0248.
)204(
)4)(2(
|)()(|
11
|)()(| 2



ss
ss
sHsG
kand
k
sHsG oo
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
32

33.  Step 5. j axis crossing:
• the closed loop transfer function is given by:
• The c/s equation is
• The coefficient of s1 = 0
• Then k = 1.5
• The intersection points are
)208()46()1(
)204(
)()(1
)(
)( 2
2
ksksk
ssk
sHsG
sG
sT





)208()46()1()()(1 2
ksksksHsG 
046  k
0)208()1( 2
 ksk
0385.2 2
s
89.32,1 jsthus 
 We also conclude that the system is stable for 0 < k < 1.5
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
33

34. a) The angle of the desired poles
is at = cos-1(0.45) =63.3 °. A
line is drawn from the origin to
intersect the root locus at the
point 3.4 116.7° with a gain k
= 0.417.
b) Therefore, this system is stable
for 0 < k < 1.5.
4.2.2 Root Locus Procedure – Example 4.4 (Cont.)
34

35. CONTROL SYSTEMS DESIGN
USING ROOT-LOCUS APPROACH
35

36. 36

37. 37