Chapter No. 8 Root Locus of control System

Chapter No. 8 Root Locus of control System

1. Analysis &
Design Through
Root Locus

2. Pre requisites before we start handling with
Root Locus
•Vector representation of a complex number
in s plane
•evaluation of a complex function via vectors
Example:
Given that F(s) = (s+1) / s(s+2). Find F(s) at the point
s= -3 +j4
Ans: 0.217 -114.3º

3. Vector representation of example

4. •
KG(s) H(s) = K N/D ( Open loop transfer function)
C/R = G / 1 + K GH ( closed loop transfer function)
Closed loop poles are the roots of characteristic equation
1 + K G(s) H(s) =0
The location of the poles in the s plane changes with the open loop
gain factor K. The locus of these roots plotted in s plane as a function
of K is called as root locus.

5. Defining Root Locus
•The poles of a transfer function can be displayed graphically
in the s plane by means of a pole zero map.
•root locus is a graphical / analytical method for displaying the
location of poles of the closed loop transfer function G / 1 + GH
without requiring the need for factorizing the characteristic
polynomial.
•They yield accurate computation of time domain response as
well as the frequency response information.
•it represents qualitatively the performance of a system as
various parameters are changed like effect of change in gain
with percent overshoot, settling time, and peak time.
•It also readily gives information about system stability and
relative stability, ranges of instability and conditions that may
break the system into oscillations.

6. Example of variation of pole location of a system
with K

7. Pole location as a
function of Gain
Root Locus

8. Information obtained from Root locus
•all poles are real for gain less than 25 (real poles
account for overdamped systems) ς>1
•at a gain of 25 poles are real and multiple hence are
critically damped. ς=1
•for gains above 25 system is under damped. ς<1
•as real parts of poles is same as K varies and settling
time is proportional to the real part of poles so K change
does not affect settling time. Ts = 4/ςwn
•ξdecreases with increase in K, % OS increases,
damped frequency of oscillation which is equal to the
imaginary part of pole also rises with rise in gain.
•For every value of K system is stable it never runs into
oscillations.

9. Properties of Root locus(1)
Angle & Magnitude Criterion
G(s) H(s) = ( 2k + 1 ) 180º
For k = 0, ±1, ±2 , ±3 ………
K1 + K G (s) H(s) = 0
K G(s) H(s) = -1
│K G(s) H(s) │= 1 Magnitude Equation
Angle Equation
Also K = 1 / │G(s) H(s)│
Every point that lies on Root Locus must satisfy
the magnitude and angle criterion.

10. Example
Given poles & zeros of open loop TF K G(s) H(s) a
point in the s plane is on the root locus if angle and
magnitude criterion are fulfilled.
K G(s) H(s) = -1
K = -1 / G(s) H(s) if H(s) = 1
K = -1/G(s) = (s-p1)(s-p2)….(s-pn)/ (s-z1)(s-z2)…..(s-zn)

11. Root Locus Sketch Using Root locus Properties(1)
1.Number of branches in root locus equals the number
of closed loop poles.
2.the root locus is always symmetrical about real axis.
3.on the real axis for K>0 the root locus exists to the
left of an odd number of real axis finite poles and/or
finite zeros.

12. Root Locus Sketch Using Root locus Properties(2)
4.The root Locus begins at the finite and infinite
poles G(s) H(s) and ends at finite and infinite zeros
of G(s) H(s).
The poles start out at -1 and -2 and move through the real axis
space between the two ploes, they meet somewhere between two
poles and break out into complex plane moving as complex
conjugates, they return to real axis between zeros at -3 & -4 away
from each other and end up at zeros.

13. Root Locus Sketch Using Root locus Properties(3)
5)The root locus approaches straight lines as
asymptotes as the locus approaches infinity.
Equation of asymptotes is given by
σa = ∑ finite poles -∑ finite zeros
# finite poles -# finite zeros
Θa = ( 2k + 1 ) π
# finite poles -# finite zeros
If the function approaches infinity as s becomes infinity then
there is a pole at infinity, and if the function approaches zero
when s becomes infinity then there is a zero at infinity. Discuss
G(s) = s , 1/s , 1/ s(s+2)
Every function of s has an equal number of poles and zeros if we
include the infinite poles and zeros as well as finite poles and
zeros.
Angle is
measured in
radians w.r.t
+ve extension
of real axis

14. Example No. 2
•No. of finite poles=4 No. of finite zeros = 1
•First of all calculate asymptotes, the real axis
intercept is -4/3 & angles are π/3 , πand 5π/3.
•there are more open loop poles than open loop
zeros so there must be zeros at infinity, asymptotes
tell us a way to go there.

15. •real axis segments lie to
the left of an odd number
of poles and /or zeros.
•The locus starts at open
loop poles and ends at
open loop zeros
•one open loop finite
zero and three open loop
infinite zeros.
•the three zeros at
infinity are at the ends of
the asymptotes.
•no. of branches = 4 =
no. of closed loop poles.
Solution

16. Example No. 3
Sketch the root locus and its asymptotes for a unity
feedback system that has the following forward
transfer function
G(s) = K / (s+2) (s+4) (s+6)

17. Root Locus Sketch Using Root locus Properties(4)
Real axis Breakaway & Break-in Points:
The point where locus leaves the real axis is called the breakaway
σ1 point and the point where locus returns to the real axis is called
as break-inσ2point.
At the breakaway or break-in point the branches of root locus form
an angle of 180º/n with the real axis where
n = no. of closed loop poles arriving at or departing from the single
breakaway or break –in point on the real axis.
Breakaway occurs at the point of maximum gain on the
real axis between the open loop poles.
Break-in point occurs at the point of minimum gain on
the real axis between open loop zeros.

18. How to find Breakaway & Break-in points?
Condition:Using magnitude criterion these points
can be found as K G(s) H(s) = -1 , equating the open
loop TF to -1, put s = σ, and differentiate w.r.t σ, and
equating to zero gives the value of σ1 & σ2.
K G(s) H(s) = K ( s-3) (s-5) / (s+1) (s+2)
K (σ
2
-8σ+15) / (σ
2
+ 3σ+12) = -1
Differentiating w.r.t σ, and equating to
zero gives -1.45 and 3.82

19. Root Locus Sketch Using Root locus Properties(5)
The jw axis crossingis a point on the root locus that
separates the stable operation of the system from the
unstable portion.
The value of w at the axis crossings yields the frequency of
oscillation.
The gain at jw crossing yields the minimum positive gain for
system stability.
Ruth Hurwitz criterion is used to find the jw axis crossings.
Forcing a row of zeros in the Routh table will yield the
gain(done before)
Going back one row to the even polynomial equation and
solving for the roots yields the frequency at imaginary axis

20. Example No. 4
For the following system find the frequency and gain K
for which the root locus crosses the imaginary axis. For
what range of K the system is stable.
Ans: ±j 1.59 , K = 9.65 thus the system is stable for
0 <= K < 9.65

21. Root Locus Sketch Using Root locus Properties(6)
Angle of Departure & Arrival:
As root locus starts from the poles and ends at zeros
thus we need to calculate the departure angle from
complex poles and arrival angles at complex zeros.
Angle Criterion is used here i.e ∟G(s) H(s) = ( 2k + 1 ) 180º
Calculation of Angle
of Departure
Sum of angles from zeros –sum of
angles from poles = ( 2k +1 ) 180

22. Calculation of
Angle of Arrivals
Sum of angles from zeros –sum of
angles from poles = ( 2k +1 ) 180

23. Example No. 5
Calculation of angle of Departure from the complex
poles and sketch the root locus
Using the poles and zeros of G(s) = (s+2) /( s+3) ( s2 + 2s +2)
Calculate the sum of angles drawn to a point close to the
complex pole -1 + j1 in the second quadrant. Thus
-Θ1-Θ2+ Θ3-Θ4 = -Θ1 -90º + tan
-1
(1/1) –tan
-1
( ½) = 180º
Θ1 = -251.6 º = 108 .4º

24. Solution

25. How to find the gain K at particular points on the root
locus?
Find exact point at which root locus
crosses ξ= 0.45 line and 20% OS.
As ξ= cos
-1
Θ& OS = e
-(ξπ/ √ 1-ξ2)
Choose some test points on the radial line
corresponding to 20% OS , ξ= 0.45 , apply
angle criterion for legal points and then find
K
For r =2
K = πpole lengths
πZero lengths
= 1.71

26. To find a point on root locus according to a value of
given ξand % OS, and gain at that point we
“ search a given line for the point yielding a
summation of angles ( zero angles-pole angles )
equal to an odd multiple of 180º then the point lies on
root locus.
The gain at that point is then found by multiplying the
pole lengths drawn to that point and dividing by the
product of zero lengths drawn to that point.”

27. Example No.5
Given the unity feedback system that has the forward
transfer function as G(s) = K( s +2) / ( s2 –4s +13)

29. Solution

30. Example No.5
Sketch the root locus for the closed loop system
GH = K / s(s+2) ( s+4)
Ans: center of asymptotes = -2
angles are : 60, 180 and 300
Breakaway point : -0.845

31. Solution

32. Example No.6
Sketch the root locus for the closed loop system
GH = K / (s+1) ( s+2-j) (s+2+j)
Ans: No. of asymptotes: 3
angles are : 60, 180 and 300
Departure Angle from a complex pole at s = -2 + j is -45º

33. Solution(2)

34. Example No.7
Sketch the root locus for the closed loop system
GH = K(s+2) / (s+1) ( s+3-j) (s+3+j)
Ans: No. of asymptotes: 2
angles are : 90 , 270
Departure Angle from a complex pole at s = -3 + j is 72º
and using symmetry the other departure angle is -72
from s = -3-j

35. Solution

36. Example No.8
Sketch the root locus for the closed loop system
GH = K / s(s+1) ( s+3) (s+4)
Ans: No. of asymptotes: 4
center of asymptotes: -2
angles are : 45, 135, 225, 315
Breakaway point : -0.424, -0.3576

37. Solution