Introduction

• What is root locus?

– Root locus is the locus (graphical presentation) of the
closed-loop poles as a specific parameter (usually gain, K)
is varied from 0 to infinity.

• Why do we need to use root locus?

– We use root locus to analyze the transient response
qualitatively. (E.g. the effect of varying gain upon percent
overshoot, settling time and peak time).

– We can also use root locus to check the stability of the

system.
 Introduction

 we use root locus to analyze the feedback control system.

 K in the feedback system is called a gain. Gain is used to vary the

system in order to get a different output response.
 Example

How the dynamics of the system (camera) changes as K is varied ?

 Drawing the root locus

Characteristic equation:

𝑺𝟐 + 𝟏𝟎𝑺 + 𝑲 = 𝟎
 Drawing the root locus

Next step is to plot the poles values on the s-plane by varying the gain, K,
value.
 Drawing the root locus

Join the poles with solid lines and you will get the shape of the locus (path)
 Drawing the root locus
• The process of drawing a root locus is time
consuming. If the system is complex.

• An alternative approach is to sketch the

root locus instead of drawing the root locus.
Sketching the root locus

In order to sketch the root locus we must follow

these six rules.
 Example

1) Sketch the root locus of the following

system:

2) Determine the value of K such that the

damping ratio ζ of a pair of dominant
complex conjugate closed-loop is 0.5.
 Construction of root loci
• Step-1: The first step in constructing a root-locus plot is to
locate the open-loop poles and zeros of G(s)H(s) in s-plane.
K 1
G(s)H (s)
 s(s 1)(s 
2) 0.
5

 n=3 -----> number of poles 0

p1 = 0; p2 = -1; p3 = -2

• m=0----> number of zeros

-
0.5

-1
-5 -4 -3 -2 -1 0 1 2

13
 Construction of root loci
• Step-2: Determine the root loci on the real axis.

 The loci on the real axis are

to the left of ODD 1
an poles
number REALof zeros of
and/or 0.
5
G(s)H(s) REAL
0
 Red lines on s-plane are
parts of the real-axis where
-
the root locus exists. 0.5

-1
-5 -4 -3 -2 -1 0 1 2

14
 Construction of root loci
• Step-3: Determine the asymptotes of the root loci.

 Number of asymptotes = n - m

 Intersection point of asymptotes with real axis:

 p  z
i
i (0 1 2)  0  3  1
  n m  
nm 30 3
 Angles of asymptotes with real axis:

180o 2k 1

 , k=0,1,2,…,(n-m-1)
n
m
15
  60 when k  0
180 2k
o
  180 when k 
1 3   300 1 when k
0
2
 60 ,60 ,180

16
 Construction of root loci
• Step-3: Determine the asymptotes of the root loci.
1

0.5

 60 ,60 ,180 180 

60 

  1
0
  60 

-0.5

-1
-5 -4 -3 -2 -1 0 1 2
17
 1

0.5

Break
Away
-0.5 Point

-1
-5 -4 -3 -2 -1 0 1 19
2
 Construction of root loci
• Step-4: Determine the breakaway point or break-in point.

• The breakaway or break-in points are the closed-loop poles that

satisfy:
dK
ds  0
• It should be that not all the solutions of dK/ds=0
noted
correspond to actual breakaway points.
K
G(s)H (s)
 s(s 1)(s 
2)
• The characteristic equation of the system is

K
1 G(s)H (s)  0
1 s(s 1)(s 
2) 20
 K  s(s 1)(s 
2)
K
 1
s(s 1)(s 
2)
• The breakaway point can now be determined as
dK

d
s(s 1)(s  2)
ds
ds
dK

d
s 3
 3s 2  2s 
ds
dK
ds  3s  6s 
ds 2

2
• Set dK/ds=0 in order to determine breakaway
point.
 3s 2  6s  2  0
s1  1.5774 , s2  0.4226
21
 1

0.5

Break
Away
-0.5 Point
s
0.4226
-1
-5 -4 -3 -2 -1 0 1 22
2
23
26
 Construction of root loci
• Step-5: Intersection points with 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 )
29
 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

30
31
 Root Locus
5

1
Imaginary Axis

-1

-2

-3

-4

-5
-7 -6 -5 -4 -3 -2 -1 0 1 2
Real Axis 32
33
 Example#1
• 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)
34
 Example#1
• The damping ratio of 0.5 corresponds to


 cos

  cos 1 

  cos 1 (0.5)  60

35
?

36
 Example#1
• The value of K that yields such poles is found from
the magnitude condition

K

s(s 1)(s  s0.3337 j 1
2) 0.5780

37
38
 Example#1
• The third closed loop pole at K=1.0383 can be obtained
as
K
1 G(s)H (s)  0
1 s(s 1)(s 
2)
1.0383
1 0
s(s 1)(s 
2)
s(s 1)(s  2) 1.0383 
0

39
40
41
 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)
42
 Example#2
• 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%.

43
 Example#2
• Step-1: Pole-Zero Map
Pole-Zero Map
1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8
-1
-5 -4 -3 -2 -1 0 1
Real Axis 44
 Example#2
• Step-2: Root Loci on Real axis
Pole-Zero Map
1

0.8

0.6

0.4

0.2

-0.2

-0.4

-0.6

-0.8
-1
-5 -4 -3 -2 -1 0 1
Real Axis 45
 Example#2
• Step-3: Asymptotes
Pole-Zero Map
1

0.8

0.6
 90 0.4

0.2

  0

2 -0.2

-0.4

-0.6

-0.8

-1
-5 -4 -3 -2 -1 0 1
Real Axis
46
 Example#2
• Step-4: breakaway point
Pole-Zero Map
1

0.8

0.6

0.4

0.2

-0.2 -1.55
-0.4

-0.6

-0.8

-1
-5 -4 -3 -2 -1 0 1
Real Axis 47
 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 48
 Example#2
• Mp<30% corresponds to


1 2
Mp  e 100




30%  e 1 2
100
  0.35

49
 Example#2
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
50
 Example#2
8

6
6 0.35

System: sys
4
Gain: 28.9
Pole: -1.96 +
5.19i
2
Damping:
0.354
Overshoot
0
(%): 30.5
Frequency
(rad/sec):
-2
5.55

-4

-6 0.35
6

-8
-4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5
51
 Root Locus of 1st Order System
• 1st order systems (without zero) are represented by following
transfer function.
K
G(s)H (s)
 s

• Root locus of such systems is a horizontal line starting from -α
and moves towards -∞ as K reaches infinity.

jω

- σ
-α
∞
52
 Home Work
• Draw the Root Locus of the following systems.

K
1) G(s)H (s)
 s
2
2) K
G(s)H (s)
 s
1
3) K
G(s)H (s)  s
53
 Root Locus of 1st Order System
• 1st order systems with zero are represented by
following
transfer function.
K (s   )
G(s)H (s)  s  
• Root locus of such systems is a horizontal line starting from -α
and moves towards -β as K reaches infinity.

jω

σ
-β -α

54
 Home Work
• Draw the Root Locus of the following systems.

Ks
1) G(s)H (s)
 s
2
2) K (s  5)
G(s)H (s)  s
1
3) K (s 
G(s)H (s)  s
3)
55
 Root Locus of 2nd Order System
• Second order systems (without zeros) have two poles and the
transfer function is given
K
G(s)H (s)
 (s  1 )(s 
2 )
• Root loci of such systems are vertical lines.

jω

σ
-α2 -α1

56
 Home Work
• Draw the Root Locus of the following systems.

K 4)
K
G(s)H (s) 2
1) G(s)H (s)  s(s   s  3s
2) 10

2) K
G(s)H (s) 
s2

K
3) G(s)H (s)  (s 1)(s 
3) 57
 Root Locus of 2nd Order System
• Second order systems (with one zero) have two poles and the
transfer function is given
K (s   )
G(s)H (s) (s  1 )(s   2 )

• Root loci of such systems are either horizontal lines or circular

depending upon pole-zero configuration.

jω jω jω

σ σ σ
-α2 -β -α1 -β -α2 -α1 -α2 -α1 -β

58
 Home Work
• Draw the Root Locus of the following systems.

K (s  1)
1) G(s) H (s)  s(s 
2)

2) K (s  2)
G(s)H (s) 
s2

K (s 
3) G(s)H (s)  (s5)
1)(s 
3) 59
 Example
• Sketch the root-locus plot of following
system with complex-conjugate open loop poles.

60
 Example
• Step-1: Pole-Zero Map

• Step-2: Determine the root loci on real axis

• Step-3: Asymptotes

61
 Example
• Step-4: Determine the angle of departure from the
complex-conjugate open-loop poles.

– The presence of a pair of complex-conjugate open-loop

poles requires the determination of the angle of
departure from these poles.

– Knowledge of this angle is important, since the root

locus near a complex pole yields information as to
whether the locus originating from the complex pole
migrates toward the real axis or extends toward the
asymptote.
62
 Example
• Step-4: Determine the angle
of departure from the
complex-conjugate open-loop
poles.

63
 Example
• Step-5: Break-in point

64
65
 Root Locus of Higher Order System
• Third order System without zero

K
G(s)H (s)
 (s  1 )(s   2 )(s   3
)

66
 Root Locus of Higher Order System
• Sketch the Root Loci of following unity feedback system

K (s 
G(s)H (s)
s(s 3)  2)(s 
1)(s

4)

67
• Let us begin by calculating the asymptotes. The real-axis intercept is
evaluated as;

• The angles of the lines that intersect at - 4/3, given by

68
• The Figure shows the complete root locus as well as the asymptotes
that were just calculated.

69
 Example: Sketch the root locus for the system with the characteristic equation
of;

a) Number of finite poles = n = 4.

b) Number of finite zeros = m = 1.
c) Number of asymptotes = n - m = 3.
d) Number of branches or loci equals to the number of finite poles (n) = 4.
e) The portion of the real-axis between, 0 and -2, and between, -4 and -∞, lie
on the root locus for K > 0.
• Using Eq. (v), the real-axis asymptotes intercept is evaluated as;

−2 + 2 −4 − (−1) −10 + 1
σ𝑎 = = = −3
𝑛 −𝑚 4 −1

• The angles of the asymptotes that intersect at - 3, given by Eq. (vi), are;

(2𝑘 + 1)π (2𝑘 + 1)π For K = 0, θa = 60o

θ𝑎 = = For K = θa = 180o
𝑛 −𝑚 4−1 1, For K θa = 70
= 2, 300o
• The root-locus plot of the system is shown in the figure below.
• It is noted that there are three asymptotes. Since n – m = 3.
• The root loci must begin at the poles; two loci (or branches) must leave the double pole
at s = -4.
• Using Eq. (vii), the breakaway point, σ, can be determine as;

• The solution of the above equation is 𝜎 = −2.59.

71
 Example: Sketch the root loci for the system.

• A root locus exists on the real axis between points s = –1 and s = –3.6.
• The intersection of the asymptotes and the real axis is determined as,
0 + 0 + 3.6 − 1 2.6
σ𝑎 = = −1.3
𝑛 −𝑚 3 −1
=
• The angles of the asymptotes that intersect at – 1.3, given by Eq. (vi), are;
(2𝑘 + 1)π (2𝑘 + 1)π For K = 0, θa = 90o
θ𝑎 = = For K = θa = -90o or 270o
𝑛 −𝑚 3−1
1,
• Since the characteristic equation is

• We have (a)
72
• The breakaway and break-in points are found from Eq. (a) as,

From which we get,

• Point s = 0 corresponds to the actual breakaway point. But points ar

neither breakaway nor break-in points, because the corresponding gain values K e
become complex quantities.

73
• To check the points where root-locus branches may cross the imaginary axis, substitute 𝑠
= 𝑗𝜔 into the characteristic equation, yielding.

• Notice that this equation can be satisfied only if

𝜔 = 0, 𝐾 = 0.

• Because of the presence of a double pole at the

origin, the root locus is tangent to the 𝑗𝜔axis
at
𝑘 = 0.

• The root-locus branches do not cross the 𝑗𝜔axis.

• The root loci of this system is shown in the

74
Figure.
75
 Home Work
• Consider following unity feedback system.

• Determine
– Root loci on real axis
– Angle of asymptotes
– Centroid of asymptotes
76
77
78