Sharanabasava C Pilli Principal, KLE Societys

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ME55 CONTROL ENGINEERING
Sharanabasava C PilliPrincipal, KLE Societys
College of Engineering and Technology,
Udyambag, Belgaum-590008Email
scpilli_at_yahoo.co.in
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ROOT LOCUS TECHNIQUE

• Root locus technique was introduced by W. R.
  Evans.
• Provides a graphical method of plotting the locus
  of roots in the s-plane as a given system
  parameter is varied over a range (generally K is
  varied from 0 to ?).

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ROOT LOCUS CONCEPT
Root locus is plotted for characteristic equation.
Characteristic equation can be rearranged as
1P(s)0 or P(s) -1 For single loop
P(s)G(s)H(s) Where, G(s)H(s) is open loop
transfer function.

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ROOT LOCUS CONCEPT
A plot of the points in the s-plane satisfying
angle criterion is the root locus. The value of
gain corresponding to a root can be determined
from the magnitude criterion.
Plotting the root locus satisfying angle
criterion is tedious. Hence, certain rules are
developed for making a quick approximate sketch
of root locus.
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RULES TO CONSTRUCT ROOT LOCUS
Rule 1The root locus is symmetrical about real
axis. The roots of the characteristic equation
are either real or complex conjugate or
combination of both. Therefore their locus must
be symmetrical about the real axis.
Rule 2 As K increases from zero to infinity,
each branch of the root locus originates from an
open loop pole with K 0 and terminates either on
an open loop zero or on infinity (zero _at_ ?) with
K?. The number of branches terminating on
infinity is equal to n-m.
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RULES TO CONSTRUCT ROOT LOCUS
Rule 3 A point on the real axis lies on the
locus if the number of open loop poles plus zeros
on the real axis to the right of this point is
odd.
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RULES TO CONSTRUCT ROOT LOCUS
Rule 6 The breakaway points or breakin points of
the root locus are the solutions of dK/ds0. All
the real roots on real axis are break points,
while complex roots are break points if they
satisfy the characteristic equation
The value of gain K at break point must be a
real positive number.
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RULES TO CONSTRUCT ROOT LOCUS
Rule 7 Angle of departure from a pole Angle of
departure from real open pole is 0? or 180?.
for complex pole ?p180? -?sum of all
angles of vectors to the complex pole in
question from other poles ?sum of
all angles of vectors to the complex pole in
question from zeros
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RULES TO CONSTRUCT ROOT LOCUS
Rule 8 Angle of arrival at a zero Angle of
arrival at a real open loop zero is 0? or 180?.
for complex zero ?p180? - ?sum of all
angles of vectors to the complex zero in
question from other zeros ?sum of all
angles of vectors to the complex zero in
question from poles
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RULES TO CONSTRUCT ROOT LOCUS
Rule 9 The intersection of root locus branches
with the imaginary axis can be determined by the
use of the Rouths criterion.
It can also, be obtained by equating the real and
imaginary parts of the characteristic equation to
zero.
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ILLUSTRATION OF ROOT LOCUS
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ILLUSTRATION OF ROOT LOCUS
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PROBLEM 1
Root locus is symmetrical about real axis.
There is no open loop zero Open loop
pole is at s 0 One branch of root locus starts
from the open loop pole when K 0 and goes to ?
asymptotically when K??.
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PROBLEM 1
The root locus departs at an angle of ?180?. The
root locus does not cross the imaginary axis.
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PROBLEM 1
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PROBLEM 2
Root locus is symmetrical about real axis.
There is open loop zero at s -2 Open
loop poles are at s -1, -1 Two branches of root
locus start from the two open loop poles when K
0 and one branch goes to ? when K ? ? and other
goes to open loop zero when K ? ? .
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PROBLEM 2
zero _at_ s -2 poles _at_ s -1, -1
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PROBLEM 2
Break points
Angle of departure at poles ? Angle of arrival
at poles?
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PROBLEM 2
The intersection with imaginary axis is obtained
from Rouths criterion.
Does not cross imaginary axis for K gt 0
System is stable for all values of K gt 0
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PROBLEM 2
zero _at_ s -2 poles _at_ s -1, -1 ?A 0.0, K
0 asymptotes at ?A ?180? break point _at_ s
-1.0, K 0 s -3.0, K
4 no j? cross over 4 gt K gt 0 ? lt 1 K
4 ? 1 K gt 4 ? gt
1
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PROBLEM 3
The open loop transfer function is G(s)
Sketch the root locus plot
Root locus is symmetrical about real axis.
There is open loop zero at s -4 Open
loop poles are at s 0, -2 Two branches of root
locus start from the two open loop poles when K
0 and one branch goes to ? when K?? and other
goes to open loop zero when K?? .
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PROBLEM 3
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PROBLEM 3
Break points
Angle of departure at poles ? Angle of arrival
at poles?
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PROBLEM 3
The intersection with imaginary axis is obtained
from Rouths criterion.
System is stable for all values of K gt 0
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PROBLEM 3
zero _at_ s -4, poles _at_ s 0, -2 ?A 2.0, K
-1.33 asymptotes at ?A ?180? break point _at_ s
-1.172, -6.828 K 0.343, 11.7 resp. no j?
cross over 0 gt K gt 0.343 ? gt 1 K
0.343 ? 1 0.343 gt K gt 11.7 ? lt
1 K 11.7 ? 1 K gt 11.7
? gt1
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PROBLEM 4
Root locus is symmetrical about real axis.
There is open loop zero at s -0.2
Open loop poles are at s 0, 0, -3.6 Three
branchs of root loci start from the open loop
poles when K 0 and two branches go to ?
asymptotically when K?? and one branch goes to
open loop zero when K?? .
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PROBLEM 4
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PROBLEM 4
Break points are given by dK/ds 0
Angle of departure at poles ? Angle of arrival
at poles?
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PROBLEM 4
The intersection with imaginary axis is obtained
from Rouths criterion.
K gt 0 the system is stable. K 0 the loci at the
origin are tangential to the imaginary axis.
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PROBLEM 4
PROBLEM 4
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PROBLEM 5
Root locus is symmetrical about real axis.
There are no open loop zeros Open loop
poles are at s 0, -3? j4 Three branches of root
locus start from the three open loop poles when K
0 and go to ? asymptotically when K??.
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PROBLEM 5
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PROBLEM 5
Break points
K -s(s26s25) -(s36s225s) Break points
are given by dK/ds 0 -(3s212s25)0 s1,2 -2
? j2.0817 K 34 ? j18.04 Hence, s1,2 are not
break points
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PROBLEM 5
Angle of departure from s0 is ?180? Angle of
departure from the complex Pole s -3 j4 in
the upper half s plane is ? 180 - 126.87 - 90
-36.87? Angle of departure from the complex pole
s-3 j4 in the lower half s plane is ? 180 -
233.13 - 270 -323 or 36.87?
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PROBLEM 5
Use Rouths criterion to find imaginary cross over.
or
The points where root locus branches cross the
imaginary axis is found by substituting s j?
into the characteristic equation and solving for
? and K by equating real and imaginary parts to
zero.
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PROBLEM 5
K gt 150 system is unstable
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RECAP

• Review of
• what is root locus
• rules to sketch the root loci
• Illustration of root loci using the rules
• pole _at_ s 0
• zero _at_ s -2, poles _at_ s -1, -1
• zero _at_ s -4, poles _at_ s 0, -2
• zero _at_ s -0.2, poles _at_ s 0, 0, -3.6
• poles _at_ s 0, -3?j4