1) The document discusses root locus analysis for a control system. It provides definitions of root locus, direct root locus, and inverse root locus. 2) Seven rules for drawing root loci are described, including the number of loci branches, real axis loci, asymptotes, breakaway points, and angle of arrival/departure. 3) An example problem demonstrates applying the rules to determine the root locus for a given open-loop transfer function. The poles and zeros are identified and the root locus is drawn.

1. Shantilal Shah Engineering College

2. TOPIC:ROOT LOCUS
Branch : Electrical
Sem : 5th
Prepared By :
NO. NAME ENROLLMENT NO
1 RATHOD MADHAV 130430109044
2 PATEL BHAVIN 130430109039
3 SINGH SUDHIR 130430109052

3. Introduction
 The stability of any closed loop system depends on the location of poles of closed loop system.
 The characteristic equation is associated with the roots to predict the response of control system.
consider a canonical system as show in fig
= Closed loop transfer function
The closed loop transfer function poles are given by
Characteristic equation is,
1+G(s) .H(s) = 0
In this system gain is variable parameter . Therefore roots of the characteristic equation depends on system
Gain (k). This system gain is varied from - ∞ to + ∞

4.  Definition
1) Root Locus : The plot of the locus of the closed loop poles as a function of the open loop gain k, when k
is varied form 0 to + ∞ .
2) Direct Root Locus : When system gain k is varied form 0 to + ∞, the locus is called as Direct root locus.
3) Inverse Root Locus : When system gain k is varied form - ∞ to 0, the locus is called as inverse root
locus .
 If the root locus branch moving toward the left then the system is more relatively stable.
 If the root locus branch moving toward the right the system is less relatively stable.
 The best method to find relatively stability is root locus.
 The best method to find absolute stability is RH- criterion.

5.  Draw the root locus to the following system
The root locus is nothing but drawing the closed loop poles path, the closed loop poles path given by characteristic eq.
closed loop = 1 + G(s) = 0
1 + K/S = 0
K + S = 0
K = - S
S = - K

6. K
VALUE
POLE LOCATION (S
= - K)
0 0
1 -1
2 -2
3 -3
4 -4
5 -5
∞ -∞
S =0
K =0
S =∞
K =∞
S = - 3 S = - 2 S = - 1
K = 3 K = 2 K = 1

7. NOW :
G(S) = K / S² , H(S) = 1
1 + G(S) H(S) = 1 + K /S² = 0
S² = - K
S = ± j√k
K value Pole location s =
-j√k
Pole location =
j√k
0 S = 0 S = 0
1 S = -j S = +j
2 S = -j √2 S = +j √2
3 S = -j √3 S = +j √3
4 S = -j √4 S = +j √4
∞ S = -j √∞ S = +j √∞
S = 0k = 0
k = 1
k = 1
k = 2
k = ∞
k = 2
k = ∞
S = +j √1
S = -j √1
S = -j √2
S = -j √∞
S = +j √2
S = +j √∞
imagin
ary axis

8. Relation between OLTF and CLTF Poles and Zeros
• The relation between open loop transfer function poles and zero and closed loop transfer function poles .
for open loop control system we can write,
G(s) . H(s) = k ―
Where
k = Gain of OLTF
N(s) = Numerator Polynomial in s
D(s) = Denominator Polynomial in s
(1) OLTF Zeros :
the zeros of G(s) . H(s) are give by,
G(s) .H(s) = 0 K
i.e. N(s) = 0
(as k non- zero gain )
OLTF zeros are given by,
N(s) = 0
N(S)
N(S)
D(s)
= 0
D(S)

9. (2) OLTF Poles :
The poles of G(s) . H(s) are given by,
G(s) . H(s) = ∞ or k — = ∞
i.e. D(s) = 0
OLTF poles are given by, D(s) = 0
(3) CLTF Poles :
They are given by, equation (1) i.e
1 + G(s) . H(s) = 0 1 + k — = 0
D(s) + k N(s) = 0
If N(s) and D(s) are maintained constant , and k alone is change CLTF poles will change. This
Plot is nothing but root locus.
(1) K = 0
D(s) + k N(s) = 0 i.e D(s) = 0
which is OLTF poles , the CLTF poles are the same as OLTF poles for k = 0 . Thus root locus start for poles of OLTF.
N(s)
D(s)
N(s)
D(s)

10. (2) K = ∞
K = ∞ = - —
N(s) = 0
N(s)
D(s)
K = ± ∞
When
k = ∞ closed loop pole = open loop zeros
k = - ∞ closed loop pole = open loop zeros
 When k increase from 0 to ∞ the direction of the root locus branch is open loop pole to open loop zero because
open loop zero the k value is infinity
 If K is increase from - ∞ to zero the direction of the root locus branch is open loop zero to open loop pole because at
open loop zero the k value - ∞ at open loop pole the k value is 0.
K ( 0 to ∞)
K ( - ∞ to 0)
x x 0
K=0
0
K=∞ K= - ∞
K=0
CL pole
OL pole
OL pole
CL pole
»»»»»»» »»»»»»»

11. Method for Drawing Root loci
 RULE 1 : The root locus is always symmetrical about the real axis i.e x-axis
 RULE 2 : Number of loci
 RULE 3 : Real axis loci
 RULE 4 : Angle of asymptotes
 RULE 5 : Center of asymptotes / Centroid
 RULE 6 : Breakaway point/Saddle point
 RULE 7 : Angle of arrival /departure
 RULE 8 : j ɷ crossover or intersection of imaginary axis

12.  RULE 1 : The root locus is always symmetrical about the real axis i.e x-axis
The root locus diagram is symmetrical about the real axis because the location of the poles and zeros in the s plan symmetrical
about the real axis.
---------------------------------
----------------------------------
--------------------------
----
-------------
-------------
------------------
Real axis
(symmetrical)
pol
e
zero
The symmetrical not only depends on the pole and zeros location it depends on which graph the plot is constructed.
 RULE 2 : Number of loci
The no. of root locus branch depends on no. of poles and zeros.
if case (i) P > Z no. of loci = P
case (i) P < Z no. of loci = Z

13.  RULE 3 : Real axis loci
• A – point to exist on real axis root locus branches the sum of the pole and zeros to the right
hand side of that point should be odd.
• Once the pole occupy the zero than it became the completed root locus branch for that
particular pole. When k increases from 0 to ∞.
k = 0
×××××××××××××××
k = 0k = 0
Complete RL branch, (0 to ∞)
k = ∞
(P + Z) → ODD (RL
branch)
Root locus
No path
exist for CL
poles
 Problem : identify the section of real axis which belong to root locus to the given system.
for G(s) H(s) =
k.(s+1)(s+3)
s.(s+2)(s+5)
Poles = S=0, -2 , -5
Zero = S= -1 , -5

14. ««««««««««««
-5 -2 -1-3
K = 0 K = ∞
K = 0
K = ∞K = 0K = ∞ polezero
• All the poles & zeros must come on root locus because they are starting and ending of root locus.
• Never apply angle & magnitude condition at the position of poles & zeros.
 RULE 4 : Angle of asymptotes
• The asymptotes are the root locus branches which are approach to the infinity.
• The no. of asymptotes n = p – z
An angle of asymptotes ᵦ=
(2X + 1) (180°)
P – z
Where x= 0 , 1, 2 ……p – z ..
• The asymptotes are symmetrical about the real axis.
 NOTE : the asymptotes gives the direction of zero
When the no. of pole are greater then zeros only (p > z).

15.  RULE 5 : Center of asymptotes / Centroid
o The centroid is nothing but intersection point of asymptotes on the real axis.
centroid =
Ʃ real part of pole of OLTF – Ʃ real part of zeros of OLTF
P – Z
• For Ex. Real part of pole = -2, -3, 0 and zeros = -1
Centroid =
Ʃ – (0 + 2 + 3) – Ʃ (- 1)
3 – 1
=- 5 + 1
2
=- 2
• The fig is now which shows the asymptotes and the center of asymptotes
jɷ
s = 0s = - 1s = - 2s = - 3
k = 0
k = ∞
zerospole
−−−−−−−−−−−−−−−−−−−−−−−
ɵ = 90°
Asymptotes(1)
Asymptotes(2)
ɵ = 270°
90°
jɷ
s = 0s = - 1s = - 2s = - 3
k = 0
k = ∞
zerospole

16.  RULE 6 : Breakaway point/Saddle point
 The point at which two or more poles meet or directly located at any location then it is called break point.
 Break away point : The point at which the root locus branches leaves the real axis is called breakaway point.
 Break in point : The point at which the root locus branches enter into the real axis is called break in point.
 The root locus branches enter or leaves the real axis with an angle of ±180°/n ,
where n = no .of poles at that break point.
 Finding the existence of break point.
 Case : 1
 whenever their exist the two adjacent placed poles in between their exist a root locus branch then these
should be the minimum one break away point in between adjacently placed poles.
»»»»»» »»»»»»
↠↠↠↠
GH(s) =
S(s + 2)
k
Pole = 0 , -2
»»»»»»
↠↠↠↠
-2 0
»»»»»»

17.  Case : 2
 Whenever these are two adjacent placed zeros in between their exit the root locus branch then these
should be the minimum one break in point in adjacent placed zeros.
×
 Case : 3
 whenever their exist left side zeros to the left most side of that zero their exist a root locus branch then these
should be the minimum one break in point to the left side of that zeros when the no .of poles are greatest then
zeros only (p > z).
 Case : 4
 whenever there exist left most side pole to the left most side of that pole their exist a root locus branch then there
should be the left most side of that pole when the no. of pole less then zero only ( p < z) only.
Above systems is practically not exist because the control system are low pass filter .
»»»»»»»»
×−−−−−−−−−−−−
»»
»»

18.  Breakaway points:
• We work with the characteristic equation ,
1 + G(s) H(s) = 0
s(s + 1) (s + 2) + k = 0
s³ + 3s² + 2s + k = 0
k = - (s³ + 3s² + 2s )
0 = - ( 3s² + 6s + 2) i.e 0 = 3s² + 6s + 2
s = - 0.42 and s = - 1.57
Since only s = - 0.42 line on root locus , it is a valid break away point.
dk―
ds
= 0
 1 + = 0k
S(s + 1 ) (s + 2)
×××
-1-2
-jɷ
jɷ
-j1.42
J1.42
-0.42
Break away
point

19.  Procedure to find the location of break point
Step 1 : from the characteristic equation’
Step 2 : Rearing the above equation in the from of k = f(s)
Step 3 : Differentiate k with respect to s and make equation to zeros.
Step 4 : The roots of do/ds = 0 use the valid and invalid break point.
• the valid break point is the one it must be on root locus
branch .
• For valid break point the k value in step 2 is positive .

20.  RULE 7 : Angle of arrival /departure
 Angle of departure is calculated at a complex conjugate poles.
 Angle of arrival is calculated at a complex conjugate zeros.
 Angle of departure :it gives that with what angle the pole depart or leaves from the initial position is called
angle of departure.
ᶲd = 180° + arg G H’
 It give that with what angle the poles arrive near and terminate the complex zeros.
ᶲ = 180° - arg G H’
a
 Angle of arrival

21.  Calculate ᶲ at complex conjugate poles
G(s) H(s) =
d
K (s + 2) (s + 4)
(s² + 2s + 2)
Zeros = -2, -4
Pole = s² + 2s + 2 = 0
s = -1 ± j
 ∟GH (s)│s = -1 ± j1 =
∟k ∟(s +2) ∟(s + 4)
∟(s +1 – j1 ) ∟(s + 1 + j1)
= ∟k ∟(1 +j1) ∟ 3 + j1
∟0 ∟ j2
0° + 45° + 18.43°
0° + 90°
=
= 45° + 18.43° + 90° = - 26.56°
ᶲd
=180° + ∟GH (s) = 153°
−−−−−−−−−−−−−
−−−−−−−−−−−−−
−−−−−−−−−−
−−−
+j1
-j1
-1-2-4
153°
-153°

22. −−−−−−−−−−−−−
−−−−−−−−−−−−−
−−−−−−−−−−
−−−
+j1
-j1
-1-2-4
153°
-153°
a
K (s² + 2s + 2)
(s + 2) (s + 4)
Zeros = -2, -4
Pole = s² + 2s + 2 = 0
s = -1 ± j
∟k ∟(s +1 – j1) ∟(s + 1 + j1)
= ∟k ∟0 ∟ j2
∟1 + j1 ∟3 + j1
0 + 90° + 0
90° + 18.43°
=
= + 26.57°
ᶲa
=180° - 26.57° = 153.43°
(s + 2) (s + 4)
=
 Calculate ᶲ at complex conjugate poles

23.  RULE 8 : j ɷ crossover or intersection of imaginary axis
 Step 1 : Consider the characteristic equation .
1 + G (s) . H(s) = 0
 Step 2 : Apply Routh’s Array in terms of k.
 Step 3 : Determine kmarginal
 Step 4 : Make auxiliary equation.
 Step 5 : Derive the auxiliary equation roots of auxiliary equation are nothing but the j ɷ crossover points or intersection point
of the root locus with imaginary axis.

24.  Intersection of imaginary axis.
 EX. 1 + G(s) H(s) = 0
• 1 +
k
S(s + 5 ) (s + 10)
= 0
 S(s + 5 ) (s + 10) + k = 0
 s³ + 15s² + 50s + k = 0
 Routh’s array is drawn below, …
s³
s²
s¹
s°
1 50
15 K
750 - K
15
-
K
 For stability , column 1 should have no sign change.
 i.e k > 0 and > 0
 i.e k > 0 and k < 750
 Two range 0 < k < 162 as the stable range of k. Kmar = 750 .
 With kmarg = 750 get of zeros (s¹ row ). Hence the preceding row (s² row ) is the auxiliary equation . For s² row,
 15s² + kmar = 0
750 - K
15
 i.e 15s² = - kmar
15s² = - 750
s² = - 50
s = ±j7.07 = ±jɷ mar.
ɷ mar = 7 .07 rad/s
Hence the root locus intersect the imaginary at ±j 7.07.
×××
-1-2-3-5-10
-jɷ
jɷ
-j7.07
j7.07

25. prepare by:
 RATHOD MADHAV,
 PATEL BHAVIN,
 SINGH SUDHIR
THANK YOU