Automatic Control 2

Lecture-2
• Construction of root loci

Dr. Mountasser Mohamed Ramadan

email: mountasser.m.r@gmail.com
 Lecture Outline

• Introduction
• The definition of a root locus
• Construction of root loci
• Closed loop stability via root locus
 Construction of root loci
• Step-1: The first step in constructing a root-locus plot
is to locate the open-loop poles and zeros in s-plane.
Pole-Zero Map
1

0.5

Imaginary Axis
0

K
G( s) H ( s)  -0.5

s( s  1)(s  2)
-1
-5 -4 -3 -2 -1 0 1 2
Real Axis
 Construction of root loci
• Step-2: Determine the root loci on the real axis.
• To determine the root loci Pole-Zero Map
on real axis we select some 1
test points.
• e.g: p1 (on positive real
0.5
axis).

Imaginary Axis
p1
0

• The angle condition is not

satisfied. -0.5

• Hence, there is no root

-1
locus on the positive real -5 -4 -3 -2 -1 0 1 2
axis. Real Axis
 Construction of root loci
• Step-2: Determine the root loci on the real axis.
• Next, select a test point on the
negative real axis between 0 and Pole-Zero Map
–1. 1

• Then
0.5

Imaginary Axis
• Thus p2
0

• The angle condition is satisfied. -0.5

Therefore, the portion of the
negative real axis between 0 and
–1 forms a portion of the root -1
locus. -5 -4 -3 -2 -1 0 1 2
Real Axis
 Construction of root loci
• Step-2: Determine the root loci on the real axis.
• Now, select a test point on the
negative real axis between -1 and Pole-Zero Map
1
–2.
• Then
0.5

Imaginary Axis
p3
• Thus 0

-0.5
• The angle condition is not
satisfied. Therefore, the negative
real axis between -1 and –2 is not -1
-5 -4 -3 -2 -1 0 1 2
a part of the root locus. Real Axis
 Construction of root loci
• Step-2: Determine the root loci on the real axis.

Pole-Zero Map
• Similarly, test point on the 1

negative real axis between -2

and – ∞ satisfies the angle 0.5
condition.

Imaginary Axis
p4
0
• Therefore, the negative real
axis between -3 and – ∞ is part
of the root locus. -0.5

-1
-5 -4 -3 -2 -1 0 1 2
Real Axis
 Construction of root loci
• Step-2: Determine the root loci onMap
Pole-Zero the real axis.
1

0.5
Imaginary Axis

-0.5

-1
-5 -4 -3 -2 -1 0 1 2
 Construction of root loci
• Step-3: Determine the asymptotes of the root loci.

Asymptote is the straight line approximation of a curve

Actual Curve
Asymptotic Approximation
 Construction of root loci
• Step-3: Determine the asymptotes of the root loci.
 180(2k  1)
Angle of asymptotes   
nm
• where
• n-----> number of poles
• m-----> number of zeros
K
• For this Transfer Function G( s) H ( s) 
s( s  1)(s  2)

 180(2k  1)

30
 Construction of root loci
• Step-3: Determine the asymptotes of the root loci.

  60 when k  0
 180 when k  1
 300 when k  2
 420 when k  3

• Since the angle repeats itself as k is varied, the distinct angles

for the asymptotes are determined as 60°, –60°, -180°and
180°.
• Thus, there are three asymptotes having angles 60°, –60°,
180°.
 Construction of root loci
• Step-3: Determine the asymptotes of the root loci.

• Before we can draw these asymptotes in the complex

plane, we must find the point where they intersect the
real axis.

• Point of intersection of asymptotes on real axis (or

centroid of asymptotes) can be find as out

 poles   zeros

nm
 Construction of root loci
• Step-3: Determine the asymptotes of the root loci.

K
• For G( s) H ( s) 
s( s  1)(s  2)

(0  1  2)  0

30

3
  1
3
 Construction of root loci
• Step-3: Determine the asymptotes of the root loci.
Pole-Zero Map
1

0.5

  60 ,60 , 180

Imaginary Axis

180 60

  1
0
  60

-0.5

-1
-5 -4 -3 -2 -1 0 1 2
Real Axis
 Home Work
• Consider following unity feedback system.

• Determine
– Root loci on real axis
– Angle of asymptotes
– Centroid of asymptotes
 Construction of root loci
• Step-4: Determine the breakaway point.
Pole-Zero Map
• The breakaway point 1

corresponds to a point
in the s plane where 0.5
multiple roots of the
characteristic equation Imaginary Axis

occur. 0

• It is the point from

which the root locus -0.5

branches leaves real

axis and enter in
-1
complex plane. -5 -4 -3 -2 -1 0 1 2
Real Axis
 Construction of root loci
• Step-4: Determine the break-in point.
Pole-Zero Map
• The break-in point 1

corresponds to a point
in the s plane where 0.5
multiple roots of the
characteristic equation Imaginary Axis

occur. 0

• It is the point where the

root locus branches -0.5

arrives at real axis.

-1
-5 -4 -3 -2 -1 0 1 2
Real Axis
 Construction of root loci
• Step-4: Determine the breakaway point or break-in point.

• The breakaway or break-in points can be determined from the

roots of
dK
0
ds
• It should be noted that not all the solutions of dK/ds=0
correspond to actual breakaway points.

• If a point at which dK/ds=0 is on a root locus, it is an actual

breakaway or break-in point.

• Stated differently, if at a point at which dK/ds=0 the value of K

takes a real positive value, then that point is an actual breakaway
or break-in point.
 Construction of root loci
• Step-4: Determine the breakaway point or break-in point.
K
G( s) H ( s) 
s( s  1)(s  2)
• 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
 Construction of root loci
• Step-4: Determine the breakaway point or break-in point.

  s( s  1)(s  2)

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

s  0.4226
 1.5774
 Construction of root loci
• Step-4: Determine the breakaway point or break-in point.
s  0.4226
 1.5774
• Since the breakaway point must lie 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.
• In fact, evaluation of the values of K corresponding to s=–
0.4226 and s=–1.5774 yields
 Construction of root loci
• Step-4: Determine the breakawayPole-Zero
point.Map
1

0.5
Imaginary Axis

s  0.4226
0

-0.5

-1
-5 -4 -3 -2 -1 0 1 2
Real Axis
 Construction of root loci
• Step-4: Determine the breakaway point.
Pole-Zero Map
1

0.5

s  0.4226
Imaginary Axis

180 60
0
  60

-0.5

-1
-5 -4 -3 -2 -1 0 1 2
Real Axis
 Example
• Determine the Breakaway and break in points
 Solution

K ( s 2  8s  15)
 1
s  3s  2
2

( s 2  3s  2)
K  2
( s  8s  15)
• Differentiating K with respect to s and setting the derivative equal to zero yields;
dK [(s 2  8s  15)(2s  3)  ( s 2  3s  2)(2s  8)]
 0
ds ( s  8s  15)
2 2

11s 2  26s  61  0

Hence, solving for s, we find the break-away and break-in points;

s = -1.45 and 3.82

 Construction of root loci
• Step-5: Determine the points where root loci cross the
imaginary axis.
Pole-Zero Map
1

0.5
Imaginary Axis

180 60
0
  60

-0.5

-1
-5 -4 -3 -2 -1 0 1 2
 Construction of root loci
• Step-5: Determine the points where root loci cross the
imaginary axis.
– These points can be found by use of Routh’s stability criterion.

– Since the characteristic equation for the present system is

– The Routh Array Becomes

 Construction of root loci
• Step-5: Determine the points where root loci cross the
imaginary axis.
• The value(s) of K that makes the system
marginally stable is 6.

• The crossing points on the imaginary

axis can then be found by solving the
auxiliary equation obtained from the
s2 row, that is,

• Which yields
 Construction of root loci
• Step-5: Determine the points where root loci cross the
imaginary axis.
• An alternative approach is to let s=jω in the characteristic
equation, equate both the real part and the imaginary part to
zero, and then solve for ω and K.

• For present system the characteristic equation is

s 3  3s 2  2s  K  0

( j )3  3( j ) 2  2 j  K  0

( K  3 2 )  j (2   3 )  0
 Construction of root loci
• Step-5: Determine the points where root loci cross the
imaginary axis.
( K  3 2 )  j (2   3 )  0
• Equating both real and imaginary parts of this equation
to zero
(2   3 )  0

( K  3 2 )  0
• Which yields
 Root Locus
5

2
Imaginary Axis

-1

-2

-3

-4

-5
-7 -6 -5 -4 -3 -2 -1 0 1 2
Real Axis
 Example
• Consider following unity feedback system.

• Determine the value of K such that the damping ratio of

a pair of dominant complex-conjugate closed-loop poles
is 0.5.
K
G (s) H (s) 
s ( s  1)( s  2)
 Example
• The damping ratio of 0.5 corresponds to
  cos 

  cos 1 

  cos 1 (0.5)  60

?
 Example
• The value of K that yields such poles is found from the
magnitude condition

K
1
s( s  1)( s  2) s 0.3337 j 0.5780
 Example
• The third closed loop pole at K=1.0383 can be obtained
as
K
1  G( s) H ( s)  1  0
s( s  1)(s  2)
1.0383
1 0
s( s  1)(s  2)

s(s  1)(s  2)  1.0383  0

 Home Work
• Consider following unity feedback system.

• Determine the value of K such that the natural

undamped frequency of dominant complex-conjugate
closed-loop poles is 1 rad/sec.
K
G( s) H ( s) 
s( s  1)(s  2)
 Root Locus

1.5

-0.2+j0.96
1

0.5
Imaginary Axis

-0.5

-1

-1.5

-2

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5

 Example
• Sketch the root locus of following system and
determine the location of dominant closed loop
poles to yield maximum overshoot in the step
response less than 30%.
 Example
• Step-1: Pole-Zero Map
Pole-Zero Map
1

0.8

0.6

0.4
Imaginary Axis

0.2

-0.2

-0.4

-0.6

-0.8

-1
-5 -4 -3 -2 -1 0 1
Real Axis
 Example
• Step-2: Root Loci on Real axis
Pole-Zero Map
1

0.8

0.6

0.4
Imaginary Axis

0.2

-0.2

-0.4

-0.6

-0.8

-1
-5 -4 -3 -2 -1 0 1
Real Axis
 Example
• Step-3: Asymptotes
Pole-Zero Map
1

0.8

0.6
  90 0.4
Imaginary Axis

0.2

  2 0

-0.2

-0.4

-0.6

-0.8

-1
-5 -4 -3 -2 -1 0 1
Real Axis
 Example
• Step-4: breakaway point
Pole-Zero Map
1

0.8

0.6

0.4
Imaginary Axis

0.2

-0.2 -1.55
-0.4

-0.6

-0.8

-1
-5 -4 -3 -2 -1 0 1
Real Axis
 Example#2
Root Locus
8

2
Imaginary Axis

-2

-4

-6

-8
-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5
Real Axis
 Example
• 𝑀𝑝 < 30% corresponds to


1 2
Mp e 100



1 2
30%  e 100

  0.35
 ExampleRoot Locus
8

6
6 0.35

2
Imaginary Axis

-2

-4

-6 0.35
6

-8
-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5
Real Axis
 Example#2
Root Locus
8

6
6 0.35

System: sys
4
Gain: 28.9
Pole: -1.96 + 5.19i
2 Damping: 0.354
Overshoot (%): 30.5
Imaginary Axis

Frequency (rad/sec): 5.55

0

-2

-4

-6 0.35
6

-8
-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5
Real Axis