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Bahir Dar University

Bahir Dar Institute of Technology

Faculty of Electrical & Computer Engineering

Introduction to Control System

By : Abrham T.
Email: abrham2048@gmail.com

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 Root Locus

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 Introduction
 Root locus, a graphical presentation of the closed-loop
poles as a system parameter is varied, is a powerful
method of analysis and design for stability and transient
response (Evans,1948;1950)
 In the root locus diagram, we can observe the path of the
closed loop poles. Hence, we can identify the nature of
the control system. In this technique, we will use an open
loop transfer function to know the stability of the closed
loop control system
 The root locus is a graphical representation in s-domain
and it is symmetrical about the real axis. Because the
open loop poles and zeros exist in the s-domain having
the values either as real or as complex conjugate pairs.

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 Introduction
 Consider a unity feedback control system shown below.

K
R (s ) C (s )
s 1

 The open loop transfer function G(s) of the system is

 And the closed transfer function is
K
G (s) 
s 1

C (s) G (s) K
 
R( s) 1  G ( s) s  1  K

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 Introduction
 Location of closed loop Pole for different
values of K (remember K>0).
C (s) K

R( s) s  1  K
Pole-Zero Map
1
K Pole
0.5 -1.5
0.5
1 -2 Imaginary Axis

2 -3 0
3 -4
5 -6 -0.5

10 -11
15 -16 -16 -14 -12 -10 -8 -6 -4 -2
Real Axis

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.

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 Construction of Root Loci
 The root locus is the path of the roots of the characteristic
equation traced out in the s-plane as a system parameter
varies from zero to infinity.

 The roots corresponding to a particular value of this

parameter can then be located on the resulting graph.

 By using the root-locus method the designer can predict

the effects on the location of the closed-loop poles of
varying the gain value or adding open-loop poles and/or
open-loop zeros.
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 Angle & Magnitude Conditions
 In constructing the root loci angle and magnitude
conditions are important.
 Consider the system shown in following figure.

 The closed loop transfer function is

C (s) G (s)

R( s) 1  G ( s) H ( s)

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 Construction of Root Loci
 The characteristic equation is obtained by setting the
denominator polynomial equal to zero.
1  G (s) H (s)  0
Or
G ( s ) H ( s )  1

 Since G(s)H(s) is a complex quantity it can be split

into angle and magnitude part.

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 Angle & Magnitude Conditions

 Angle Condition
G ( s ) H ( s )  180 (2k  1) (k  1,2,3...)

 Magnitude Condition
G (s) H (s)  1

 The values of s that fulfill both the angle and

magnitude conditions are the roots of the
characteristic equation, or the closed-loop poles.

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 Rules for Construction of Root Locus

Follow these rules for constructing a root locus.

 Rule 1 – root locus is always symmetrical about the
real axis. The roots are either real or complex
conjugates or a combination of both.
 Rule 2 − No of loci:
Locate the open loop poles and zeros in the ‘s’ plane.
Pole-Zero Map
1

0.5
Imaginary Axis

-0.5

-1
-5 -4 -3 -2 -1 0 1 2
Real Axis
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  Rule 3 − Find the number of root locus branches.
 We know that the root locus branches start at the open
loop poles and end at open loop zeros.
 So, the number of root locus branches N is equal to the
number of finite open loop poles P or the number of finite
open loop zeros Z, whichever is greater.
 Mathematically, we can write the number of root locus
branches N as
 N=P if P≥Z
 N=Z if P<Z
Rule 4 - The root locus starts at the poles of GH where K=0
and ends at the zeros of GH where 𝐾 = ±∞.

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 Rule 5 − Real axis root loci.
 If odd number of the open loop poles and zeros exist to the right
side of a point on the real axis, then that point is on the root locus
branch.
 Therefore, the branch of points which satisfies this condition is the
real axis of the root locus branch.
Pole-Zero Map
1

0.5
Imaginary Axis

-0.5

-1
-5 -4 -3 -2 -1 0 1 2
Real Axis

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  Rule 6 − Centroid and Angle of asymptotes.
 Before we draw these asymptotes in the complex plane, we
need to find the point where they intersect the real axis.
 Since the number of poles greater than the number of zeros
n-m branches will move to infinity and these branches
move along the asymptote.
 Point of intersection of asymptotes on real axis (or centroid
of asymptotes) is
 poles   zeros

nm
 180(2k  1)
Angle of asymptotes   
nm
 where
 n-----> number of poles
 m-----> number of zeros

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  Rule 7 − Find the intersection points of root locus
branches with an imaginary axis.
 We can calculate the point at which the root locus
branch intersects the imaginary axis and the value
of K at that point by using the Routh array method and
special case (ii).
 If all elements of any row of the Routh array are zero,
then the root locus branch intersects the imaginary axis
and vice-versa.
 Identify the row in such a way that if we make the first
element as zero, then the elements of the entire row are
zero. Find the value of K for this combination.
 Substitute this K value in the auxiliary equation. You
will get the intersection point of the root locus branch
with an imaginary axis.

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  Rule 8 − Find Break-away and Break-in points.
 If there exists a real axis root locus branch between two open
loop poles, then there will be a break-away point in between
these two open loop poles.
 If there exists a real axis root locus branch between two open
loop zeros, then there will be a break-in point in between these
two open loop zeros.
 Note − Break-away and break-in points exist only on the real
axis root locus branches.
 Follow these steps to find break-away and break-in points.
 Write K in terms of s from the characteristic
equation 1+G(s)H(s)=0.
 Differentiate K with respect to s and make it equal to zero.
Substitute these values of s in the above equation.
dK
0
ds
 The values of s for which the K value is positive are the break
points.

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  The characteristic equation of the system is
K
1  G( s) H ( s)  1  0
s( s  1)(s  2)
K
 1
s( s  1)(s  2)
K  s ( s  1)(s  2)
The breakaway point can now be determined as
  s( s  1)(s  2)
dK d
ds ds

  s( s  1)(s  2)

dK d
ds ds
dK
ds

d 3
ds

s  3s 2  2s 
 3s 2  6 s  2  0
dK 3s 2  6 s  2  0
 3s 2  6s  2
ds s  0.4226
Set dK/ds=0 in order to determine breakaway point.
 1.5774
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 s  0.4226
 1.5774

 Since the breakaway point needs to be on a root locus

between 0 and –1, it is clear that s=–0.4226 corresponds
to the actual breakaway point.
 Point s=–1.5774 is not on the root locus. Hence, this
point is not an actual breakaway or break-in point.

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  Rule 9 − Find the angle of departure and the angle of
arrival.
 The Angle of departure and the angle of arrival can be
calculated at complex conjugate open loop poles and
complex conjugate open loop zeros respectively.
 The formula for the angle of departure ϕd is
 ϕd=180+arg[G(s)H(s)]

 Where arg[G(s)H(s)] is the angle of G(s)H(s) excluding the pole where the angle is to be
calculated

 The formula for the angle of arrival ϕa is

 ϕa=180-arg[(G(s)H(s)]
 Where arg[G(s)H(s)] is the angle of G(s)H(s) excluding the zero where the angle is to be calculated

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 Plot the root locus
Example 1
 G(s)H(s)=k/s(s+1+j)(s+1-j)=k/s(𝑠 2 + 2𝑠 + 2 )

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 For k=4, Marginally stable

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 For k<4, Stable

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 For k>4, Unstable

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 Example 2
 G(s)H(s)=k/s(s+3)(s+5)=k/𝑠 3 + 8𝑠 2 + 15𝑠

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 For k=120, Marginally Stable

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 For k<120, Stable

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 For k>120, Unstable

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 Stability analysis using Bode plot

 #Method I
 Gm and Pm Positive-stable System
 Gm and Pm negative-unstable System
 Gm and Pm zero- Marginally stable System

#Method II
 Wcp>Wcg-Stable System
 Wcp=Wcg-Marginally stable system
 Wcp<Wcg-Unstable System

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 #Example 1
G=tf([100],[1 10.1 1 0])
bode(G)  Gm = 0.1010
grid on
% set(gca, 'XColor', 'r')
[Gm Pm Wcp Wcg]=margin(G)
title('Bode plot of 100/(s^3+10.1s^2+s)')
Gm_db=20*log10(Gm)  Pm = -15.3186

 Wcp = 1.0000

 Wcg = 3.0902

Gm_db = -19.9136
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 #Example 1

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 #Example 1
Wcg, Wcp

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 Gm and Pm

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 Example 2  Gm = 2.4000

G=tf([100],[1 12 20 0])
bode(G)  Pm = 19.9079
grid on
set(gca, 'XColor', 'r')
[Gm Pm Wcp Wcg]=margin(G)
title('Bode plot of 100/(s^3+12s^2+20s)')
Gm_db=20*log10(Gm)  Wcp = 4.4721

 Wcg = 2.7992

 Gm_db = 7.6042

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 Example 2

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 #Example 2
Wcg and Wcp

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 Gm and Pm

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 Questions?

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