GEE338 Linear Control Systems
Chapter 3
Stability of Control Systems in Laplace
Doman
Prof. Fawzy Ibrahim and Eng. Medhat Toubar
Electronics and Communication Department
Misr International University (MIU)
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
Chapter Contents
3.1 Stability Concept
3.2 Stability Analysis Using Routh-Hurwitz Criterion
3.3 Root Locus Techniques
3.3.1 Definition of Root Locus
3.3.2 Root Locus Construction Procedure
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.1 Stability Concept
Stability: A system is stable if every bounded input yields a bounded output. We
call this statement the Bounded-Input, Bounded-Output (BIBO) definition of
stability, so:
• A system is stable if every Bounded Input yields a Bounded Output (BIBO).
• A system is unstable if any bounded input yields an unbounded output.
• If the closed-loop system poles are in the left half of the plane and hence have
a negative real part, the system is stable. That is, stable systems have closedloop transfer functions with poles only in the Left Half-Plane (LHP).
Example 3.1 Check the stability of a control system described by closed-loop
transfer functions, T(s) as:
C (s)
32
T (s) 
 2
R ( s ) s  12s  32
The impulse response function g(t) can be described as follows:
• Knowing the system transfer function, T(s) and the input, R(s) then output, C(s)
can be determined:
C ( s )  R ( s )T ( s )
•
If the input is unit impulse, r(t) = (t) its Laplace transform R(s) = 1, therefore:
32
32
C ( s )  R ( s )T ( s )  1x 2

s  12 s  32 ( s  4)( s  8)
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GEE338 Ch.3 Stability
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3.1 Stability Concept
•
Impulse response function g(t) is the inverse Laplace transform of T(s), using
partial fraction, we get:
32
32
A
B
T (s)  2



s  12s  32 ( s  4)( s  8) ( s  4) ( s  8)
A  ( s  4)T ( s) s 4 
32
( s  8)
B  ( s  8)T ( s) s 8 
T (s) 

s  4
32
( s  4)
32
8
4

s  8
32
 8
4
8
8

( s  4) ( s  8)
g (t )  8(e 4t  e 8t )u (t )
•
•
•
The system is stable because g(t)  0 for t   or every Bounded Input yields
a Bounded Output (BIBO).
Note that all the closed-loop system poles are in the Left Half-Plane (LHP).
A system is stable if the natural response approaches zero as time approaches
infinity.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.1 Stability Concept
•
•
•
A system is unstable if the natural response approaches infinity as time
approaches infinity.
A system is unstable if any bounded input yields an unbounded output.
If some of the closed-loop system poles are in the right half of the s-plane and
hence have a positive real part, the system is unstable.
Example 3.2 Check the stability of a control system described by closed-loop
transfer functions, T(s) as:
T (s) 
C (s)
5
 2
R( s) s  s  6
The impulse response function g(t) can be described as follows:
• Knowing the system transfer function, T(s) and the input, R(s) then output, C(s)
can be determined:
•
If the input is unit impulse, r(t) = (t) its Laplace transform R(s) = 1, therefore:
C ( s )  R ( s )T ( s )
C ( s )  R ( s )T ( s )  1x
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5
5

s 2  s  6 ( s  3)( s  2)
GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.1 Stability Concept
•
Impulse response function g(t) is the inverse Laplace transform of T(s), using
partial fraction, we get:
5
5
A
B
T (s)  2



s  s  6 ( s  3)( s  2) ( s  3) ( s  2)
A  ( s  3)T ( s) s 3 
5
( s  2)
B  ( s  2)T ( s) s 2 
T (s) 
•
•
•
•

s  3
5
( s  3)
1
1

( s  2) ( s  3)
5
 1
5

s 2
5
1
5
g (t )  (e 2t  e 3t )u (t )
The system is unstable because g(t)   for t   or every bounded input
yields a unbounded output.
Note that some of the closed-loop system poles are in the Right Half-Plane
(RHP).
A system is stable if the natural response approaches infinity as time
approaches infinity.
If the closed-loop system poles are in the right half of the s-plane and hence
have a positive real part, the system is unstable.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.2 Stability Analysis Using Routh-Hurwitz Criterion
•
•
•
A stable system is a dynamic system with a bounded response to a bounded
input.
Consider the closed loop transfer function of a system as:
bm s m  bm1 s m1    b0
N ( s)
T ( s) 

n
n 1
a n s  a n 1 s    a0
( s )
Where N(s) is the numerator of the transfer function and ∆(s) or Q(s) is
denominator or the characteristic equation or polynomial of the system which is
given by:
(s)  Q( s)  an s n  an1s n1    a0
•
•
•
For the system described by T(s) to be stable, the root of the characteristic
equation must lie in the Left Half Plane (LHP).
The Routh-Hurwitz criteria or test is a numerical procedure for determining the
number of Right Half-Plane (RHP) and imaginary axis roots of the characteristic
polynomial.
The method requires two steps: (1) Generate a data table called a Routh table
and (2) interpret the Routh table to tell how many closed-loop system poles are
in the left half-plane, the right half-plane, and on the j-axis.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.2 Stability Analysis Using Routh-Hurwitz Criterion
Step #1: Generate a date table called a Routh table as follows:
• Consider the characteristic equation which is given by:
(s)  Q(s)  a4 s 4  a3 s 3  a2 s 2  a1s  a0
Rules for Routh table creation
• Any row of the Routh table can be multiplied by a positive constant without
changing the values of the rows below.
• To avoid the division by zero, an epsilon is assigned to replace the zero in the
first column.
Table 3.1 Initial layout for Routh table
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GEE338 Ch.3 Stability
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3.2 Stability Analysis Using Routh-Hurwitz Criterion
Step #1: Generate a date table called a Routh table as follows:
• Further rows of the schedule are then completed as follows:
Table 3.2 Completed Routh table
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3.2 Stability Analysis Using Routh-Hurwitz Criterion
Step #2: Interpret the Routh table to tell how many closed-loop system poles are
in the:
- left half-plane
- right half-plane.
• The number of roots of the polynomial that are in the right half-plane is equal to
the number of sign changes in the first column.
• Notes:
1. If the coefficients of the characteristic equation have differing algebraic, there is
at least one RHP root. For Example:
(s)  Q( s)  7s 5  5s 4  3s 3  2s 2  s  10
Has definitely one or more RHP roots.
2. If one or more of the coefficients of the characteristic equation have zero value,
there are imaginary or RHP roots or both. For Example:
(s)  Q( s)  s 6  3s 5  2s 4  8s 2  3s  17
has imaginary axis roots indicated by missing s3 term.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.2 Stability Analysis Using Routh-Hurwitz Criterion
Example 3.3 For the system shown in Fig. 3.1, determine T(s) and then apply the
Routh-Hurwitz Method to check the its Stability.
Solution:
The system closed loop transfer function T(s) is calculated as:
1000
G(s)
1000
( s  2)( s  3)( s  5)
T (s) 

 3
2
1000
1  G(s) H (s) 1 
s  10 s  31s  1030
( s  2)( s  3)( s  5)
Therefore, the characteristic equation is given by:
( s )  s 3  10s 2  31s  1030
(a)
(b)
Fig. 3.1 (a) Feedback system for; (b) equivalent closed loop system.
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GEE338 Ch.3 Stability
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3.2 Stability Analysis Using Routh-Hurwitz Criterion
( s )  s  10s
Example 3.3 Solution:
Step #1: Generate the Routh table as follows:
3
2
 31s  1030
Table 3.3 The Completed Routh table for Example 3.3
Divide by 10
Note: Any row of the Routh table can be multiplied by a positive constant without
changing the values of the rows below.
Step #2: Interpret the Routh table: Since there are two sign changes in the left
column (+ + - +). Therefore, the system is unstable because the characteristic
equation, (s) has two roots in the right half plane so two poles of T(s) are in
the right half plane.
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GEE338 Ch.3 Stability
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3.2 Stability Analysis Using Routh-Hurwitz Criterion
Example 3.4 Apply the Routh-Hurwitz Method to determine the stability of a the
closed loop system whose transfer function is given by:
5
4
3
2
Q
(
s
)


(
s
)

3
s

5
s

6
s

3
s
 2s  1
Solution
10
T (s)  5
3s  5s 4  6s 3  3s 2  2s  1
Step #1: Generate the Routh table as follows:
Table 3.4 The Completed Routh table
Step #2: Interpret the Routh table: Since there are two sign changes in the left
column (+ + - +). Therefore, the system is unstable because the characteristic
equation, (s) has two roots in the right half plane so two poles of T(s) are in
the right half plane.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.2 Stability Analysis Using Routh-Hurwitz Criterion
Example 3.5 Apply the Routh-Hurwitz Method to check the stability of a system
whose characteristic equation is given by:
( s)  s5  2 s 4  3 s3  6 s 2  5s  3
Solution
Step #1: Generate the Routh table as follows:
Table 3.5 (a) The Completed Routh table for Example 3.5
Note:
To avoid the division by zero,
an epsilon () is assigned to
replace the zero in the first
column.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.2 Stability Analysis Using Routh-Hurwitz Criterion
Example 3.5 Solution
Step #2: Interpret the Routh table to tell how many closed-loop system poles are
in the right half-plane. To begin the interpretation, we must assume a sign,
positive or negative, for the quantity (ε) as illustrated in Routh table.
Table 3.5(b) The Completed Routh table for Example 3.5
•
Since, there are two sign changes in the left column (+ + - +); Therefore, the
system is unstable because the characteristic equation, (s) has two roots in
the right half plane so two poles of T(s) are in the right half plane.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.2 Stability Analysis Using Routh-Hurwitz Criterion
Example 3.5 Apply the Routh-Hurwitz Method to determine the values of K that
make the system stable whose characteristic equation is given by.
s 5  13s 4  54s 3  82s 2  (60  K ) s  3K  0
Solution
Step #1: Generate the Routh table as follows:
Table 3.5(b) The Completed Routh table for Example 3.5
s5
1
54
60  K
s4
13
82
3K
s3
47.7
60  0.769K
0
3K
0
0
0
65.6  0.212K
s2
s
s
1
3940  105K  0.163K 2
65.6  0.212K
0
3K
0
0
65.6  0.212K  0
 K  309
3940  105K  0.163K 2  0
K0
 K  35
Step #2: Interpret the Routh table. Then the values of k that makes the system
stable are:
0  K  35
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3.3 Root Locus Techniques
3.3.1 Definition of Root Locus
• Consider the feedback control system shown in Fig. 3.2,
transfer function is given by:
C (s)
k G( s)

R( s) 1  k G ( s) H ( s)
T ( s) 
•
the closed loop
Where k is a constant gain parameter. The poles of the transfer function are
the roots of the characteristic equation given by:
1  k G( s) H ( s)  0
•
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 < ∞.
R (s )
G (s )
k
C (s )
H (s )
Fig. 3.2 feedback control system.
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3.3 Root Locus Techniques
3.3.1 Definition of Root Locus
• If the roots of the characteristic equation given by: 1  k G ( s ) H ( s )  0
kG( s ) H ( s )  1 or G ( s ) H ( s )  
1
k
•
Then
•
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:
| kG( s ) H ( s ) | 1 or | G ( s ) H ( s ) |
•
1
k
The angle condition:
G ( s ) H ( s )  (2i  1)180o where
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i  0, 1, 2, 3, ....
GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.6 Sketch the root locus for the system shown in Fig.3.3.
Solution Step 1: Write the open loop transfer function or the characteristic
equation of the system and find the open loop poles and zeros as follows:
kG( s ) H ( s ) 
k ( s  3)
s ( s  1)( s  2)( s  4)
Open loop Poles = {0, -1, -2, -4}  # of poles = n = 4.
Open loop Zeros= {-3}  # of zeros = m = 1. So, n – m = 3.
Root locus properties :
1) Symmetry As all roots are either real or complex conjugate pairs so that the
root locus is symmetrical about the real axis.
2) Number of branches of the root locus is equal to the number of poles of the
open-loop transfer function.
Fig. 3.3 feedback control system.
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.6 Solution
3) Locus start and end points: The locus starting points (k=0) are at the openloop poles and the locus ending points (k = ∞) are at the open-loop zeros. n -m
branches end at infinity.
Step 2: Draw the pole-zero plot of kG(s)H(s), then locate the segments of real axis
that are root loci. The root locus on the real axis always lies in a section of real
axis to the left of an odd number of poles and zeros as shown in Fig. 3.4(a).
Root locus segments
exists after the poles at 0,
-2 and -4
Fig. 3.4(a) pole-zero plot of kG(s)H(s).
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GEE338 Ch.3 Stability
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.6 Solution
Step 3: Draw the asymptotes as follows:
Asymptotic Angles: As k approaches +∞, the branches of the locus become
asymptotic to straight lines with angles:
(2i  1)1800 (2i  1)1800
A 

where i  0, 1, 2, 3 ,....
nm
3
The number of asymptotes are three [n - m = 3].
for i = 0  1 = 60o, for i = 1  2 = 180o, and for i = -1  3 = -60o.
Center of Asymptotes: The starting point of the asymptotes. These linear
asymptotes are centered at a point on the real axis at:
 pole values of GH   zero values of GH
A 
nm
1  2  4  3

 1.33
3
•
These asymptotes are shown in Fig. 3.4(b).
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.6 Solution
Step 3: Draw the asymptotes as follows:
Fig. 3.4(b) The asymptotes of the root locus.
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.6 Solution
Step 4. Breakaway and break-in or entry points: The point where the locus
leaves the real axis is called the breakaway point and the point where the locus
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. A pair of locus segments leave or enter
the real axis at angles ±90 degree and k is calculated by trial and error from the
1
1
s( s  1)( s  2)( s  4)
equation:
| G( so ) H ( so ) | and k 
|
|
k
| G( s) H ( s) |
( 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. This point is shown in Fig. 4.4(c).
1
s( s  1)( s  2)( s  4)
s 4  7 s 3  14s 2  8s
1  KG ( s) H ( s)  0 and k 


G( s) H ( s)
( s  3)
( s  3)
• For maximum K, dK/ds = 0, we get:
dk
( s  3)(4s 3  31s 2  28s  8)  ( s 4  7 s 3  14s 2  8s)

0
2
ds
( s  3)
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.6 Solution
Step 4. Breakaway and break-in or entry points:
4 s 4  26s 3  77s 2  84s  24  0
•
Solve for s, we get the
K is at s = - 0.434
and K = 0.5346.
Fig. 3.4(c) The complete root locus.
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.6 Solution
Step 5. j axis crossing: The closed loop transfer function is given by:
T ( s) 
The c/s equation is:
KG ( s )
k ( s  3)
 4
1  KG ( s ) H ( s ) s  7 s 3  14s 2  (8  k ) s  3k
1  KG (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
80.35s 2  202.7  0
thus s1, 2   j1.59
•
These points are shown in Fig. 3.4(c). We also conclude that the system is
stable for 0 < k < 9.65.
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.7 Sketch the root locus for the system shown in Fig.3.5.
Solution Step 1: Write the open loop transfer function or the characteristic
equation of the system and find the open loop poles and zeros as follows:
kG( s ) H ( s ) 
k ( s  3)( s  5)
( s  1)( s  2)
Open loop Poles = {-1, -2}  # of poles = n = 2.
Open loop Zeros= {3, 5}  # of zeros = m = 2. So, n – m = 0.
Step 2 Real Axis Segments: Draw the pole zero plot of kG(s)H(s), then locate the
segments of real axis that are root loci. The root locus on the real axis always
lies in a section of real axis to the left of an odd number of poles and zeros as
shown in Fig. 3.6(a)
R(s )
k ( s  3)( s  5)
( s  1)( s  2)
C (s )
Fig. 3.5 feedback control system.
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.7 Solution
Step 3: Draw the asymptotes, center of Asymptotes are not applied for this
example.
Root locus segments
exists after the zero
at 5 and after the
pole at -1.
Fig. 3.6(a) pole-zero plot of kG(s)H(s).
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.7 Solution
Step 4. Breakaway and break-in or entry points: A pair of locus segments leave
or enter the real axis at angles ±90 degree and k is calculated by trial and error
from the equation:
1
1
( s  1)( s  2)
| G( so ) H ( so ) |
k
and k 
| G( s) H ( s) |

( s  3)( s  5)
Therefore, the breakaway point lies between -1 and -2 and the break in point
lies between 3 and 4 at maximum value for k which is calculated as listed in
Table 3.6(a).
Table 3.6(a).
It is clear that the breakaway and
break-in points are at:
• at s = -1.45 breakaway
and
• at s = -3.82 break-in points
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.7 Solution
Step 5: Angles of departure and approach is not applied for this example.
Step 6 j axis crossing: The closed loop transfer function is given by:
G( s)
k ( s 2  8s  15)
T ( s) 

1  G ( s ) H ( s ) (1  k ) s 2  (3  8k ) s  ( 2  15k )
The c/s equation is: 1  G( s) H ( s)  (1  k ) s 2  (3  8k ) s  (2  15k )
The Routh-table for the characteristic equation is shown in Table 3.6 (b)
The coefficient of s1 = 0
3  8k  0
Table 3.6(b).
Then 3/8 = 0.375
The intersection points are:
(1  k ) s 2  (2  15k )  0
1.375s 2  7.625  0
•
thus s1, 2   j 2.35
These points are shown in Fig. 3.6(b). We also conclude that the system is
stable for 0 < k < 0.375.
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.7 Solution
Fig. 3.6(c) The complete root locus.
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.8 Sketch the root locus for the system shown in Fig.3.7.
Solution Step 1: Write the open loop transfer function or the characteristic
equation of the system and find the open loop poles and zeros as follows:
k ( s 2  4 s  20)
kG( s ) H ( s ) 
( s  2)( s  4)
Open loop Poles = {-2, -4}  # of poles = n = 2.
Open loop Zeros= {(2 + j4), (2 – j4)}  # of zeros = m = 2. So, n – m = 0.
Step 2 Real Axis Segments: Draw the pole zero plot of kG(s)H(s), then locate the
segments of real axis that are root loci. The root locus on the real axis always
lies in a section of real axis to the left of an odd number of poles and zeros as
shown in Fig. 3.6(a)
Fig. 3.7 feedback control system.
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GEE338 Ch.3 Stability
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.8 Solution
Step 3: Draw the asymptotes, center of Asymptotes are not applied for this
example.
Root locus segments
exists after the pole at -2.
Fig. 3.8(a) pole-zero plot of kG(s)H(s).
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GEE338 Ch.3 Stability
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3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.8 Solution
Step 4. Breakaway and break-in or entry points: A pair of locus segments leave
or enter the real axis at angles ±90 degree and k is calculated by trial and error
from the equation:
1
1
( s  2)( s  4)
| G( so ) H ( so ) |
k
and k 
| G( s) H ( s) |

( s 2  4s  20)
Therefore, the breakaway point lies between -2 and -4.
The breakaway point lies at s = -2.88 at maximum value for k = 0.0248.
Step 5 Angles of departure and approach: The angle of departure p of a locus
from a complex pole is given by:  p   dep  180    poles    zeros
p = 180 – [(sum of the other GH pole angles to the pole under
consideration) - (sum of the GH zero angles to the pole )]
[It is applied only when there are two complex poles]
The angle of approach z of a locus to a complex zero is given by:
 z   arv  180    zeros    poles
z = [(sum of the other GH pole angles to the zero under
consideration) - (sum of the GH zero angles to the zero )] -180
z = 180 + [1 + 2] - [3] = 180 + [26.6 + 45] – [90] = 161.6 degree
These angles are shown in Fig.3.8(b). [if there are two complex zeros]
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.8 Solution Step 5 Angles of departure (p) and approach (z):
Fig. 3.8(a) Angles of departure (p) and approach (z) of kG(s)H(s).
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.8 Solution
Step 6 j axis crossing: The closed loop transfer function is given by:
G( s)
k ( s 2  4 s  20)
T ( s) 

1  G ( s ) H ( s ) (1  k ) s 2  (6  4k ) s  (8  20k )
The c/s equation is: 1  G( s) H ( s )  (1  k ) s 2  (6  4k ) s  (8  20k )
The coefficient of s1 = 0
6  4k  0
Then k = 1.5
The intersection points are:
(1  k ) s 2  (8  20k )  0
2.5s 2  38  0
thus s1, 2   j 3.89
•
These points are shown in Fig. 3.8(c). We also conclude that the system is
stable for 0 < k < 1.5.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.8 Solution
Notes:
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.
Fig. 3.8(c) The complete root locus.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.9 Sketch the root locus for the system shown in Fig.3.9
Solution
Step 1: Write the open loop transfer function or the characteristic equation of
the system as follows:
kG( s ) H ( s ) 
R(s )
k ( s  2)
( s  3)( s 2  2 s  2)
k ( s  2)
( s  3)( s 2  2 s  2)
C (s )
Fig. 3.9
Step 2. Real Axis Segments: Draw the pole zero plot of kG(s)H(s), then locate
the segments of real axis that are root loci. The root locus on the real axis
always lies in a section of real axis to the left of an odd number of poles and
zeros as shown in Fig. 3.10
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.9 Solution
Root locus segments
exists after the zero at -2.
The are two complex
poles -1± j1
A
Fig. 3.10 The root locus
of Example 3.9.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim
3.3 Root Locus Techniques
3.3.2 Root Locus Construction Procedure
Example 3.9 Solution
Step 3. Asymptotic Angles: is not applied for this example.
Step 5. Breakaway and break-in or entry points: is not applied for this example.
Step 6. Angles of departure and approach: 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 = 180o – [(2 + 4) - 3] = 180o - [90o + 26.6o - 45o] =108.4o
This point is shown in Fig. 3.10
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GEE338 Ch.3 Stability
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3.4 Design of Compensators by Root-Locus Method
Example 3.11 Solution
Note: The solution to such a problem is not unique, there are infinitely many
solutions.
• One possible solution is shown in Fig. 3.13 and Kc is calculated as before.
Gc ( s )  K c
s 1/
s  1.5
 1.03
s  1 / 
s  3.6
Fig. 3.13 Another possible pole and zero selection of compensator.
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GEE338 Ch.3 Stability
Prof Fawzy Ibrahim