Root locus compensation

Design Via Root Locus
Dr.R.Subasri
Professor, Kongu Engineering College,
Perundurai, Erode, Tamilnadu, INDIA
Dr.R.Subasri.KEC,INDIA

•The root locus typically allows us to choose the
proper loop gain to meet a transient response
specification.
•As the gain is varied, we move through different
regions of response.
•Setting the gain at a particular value yields the
transient response dictated by the poles at that
point on the root locus.
•Thus, we are limited to those responses that
exist along the root locus.
Design Via Root locus

Design K such that the settling time is 2.5 sec and
damping ratio is 0.6 in closed loop
k
G(s)H(s)
s(s 2)(s 4)

 
s
n
n
n
2
n n
4
t 2.5
δω
4
δω 1.6
2.5
1.6
ω 2.667
0.6
Required closed loop pole :-δω jω 1 δ
 
 
 
 
-1.6j2128
This closed loop pole should lie on the root locus

But , the closed loop pole lie on root locus along 0.6 damping ratio line is
-0.714 +0.926i
The settling time is 5.5 sec. The equivalent K value cannot be obtained

One way to solve our problem is to replace the existing
system with a system whose root locus intersects the
desired design point, B. Unfortunately, this replacement is
expensive and counterproductive.
Rather than change the existing system, we augment, or
compensate, the system with additional poles and zeros,
so that the compensated system has a root locus that
goes through the desired pole location for some value of
gain.

Both methods change the open-loop poles and zeros,
thereby creating a new root locus that goes through the
desired closed-loop pole location.
Compensators that use pure integration for improving
steady-state error or pure differentiation for improving
transient response are defined as ideal compensators
Ideal compensators must be implemented with active
networks, which, in the case of electric networks, require
the use of active amplifiers and possible additional power
sources

An advantage of ideal integral compensators is that
steady-state error is reduced to zero.
Compensators implemented with passive elements such
as resistors and capacitors are not ideal compensators.
Advantages of passive networks are that they are less
expensive and do not require additional power sources for
their operation.
Their disadvantage is that the steady-state error is not
driven to zero

Systems that feed the integral of the error to the plant
are called integral control systems.
Thus, we use the name PI controller interchangeably with
ideal integral compensator, and we use the name lag
compensator when the cascade compensator does not
employ pure integration.
Improving Steady-State Error
via Cascade Compensation

Lag Compensation
Ideal integral compensation-with its pole on the origin-
active network
passive networks- the pole and zero are moved to the left,
close to the origin, as shown
This placement of the pole and zero although it does
not increase the system type, does yield an
improvement in the static error constant over an
uncompensated system.

In order to reduce the steady state error,
the gain is increased.
The compensated steady state gain Kpc =
p times the uncompensated gain kpu
The factor p = Zc/Pc
For the lag compensator pole should be
nearer to origin compared to zero.
Also both compensator pole and zero should
be close to origin to retain the transient
response specifications.Dr.R.Subasri.KEC,INDIA

1
G(s)H(s)
(s 1)(s 2)(s 10)

  
Compensate the system to improve the steady-
state error by a factor of 10 if the system is
operating with a damping ratio of 0.174.
From the root locus, the corresponding gain value for
damping ratio of 0.174 is K =162.
The error constant Kpu = 8.1. The steady state error is =
1/1+kpu = 0.109.
s 0 s 0
162
kpu Lt G(s) Lt
(s 1)(s 2)(s 10) 
 
  

pc
pu
K 91.59
11.3
K 8.1
 
c
c
Z
11.3 β
P
 
To improve the steady state error requires to reduce the
error by 10 times = 0.0109.
The equivalent Kpc = 91.59
Construct a line at an angle of 10 (or less) with  line
from sd. Intersection with real axis is the compensator
zero(zc)
The compensator pole (pc) = zc/
The open loop transfer function of the compensated
system is
c
c
c
s z
G (s) K G(s)
s p




Improving Transient Response via Cascade
Compensation- Lead Compensator
The objective is to design a response that has a desirable
percent overshoot and a shorter settling time than the
uncompensated system.
For this, ideal derivative compensation, a pure
differentiator (a proportional-plus-derivative (PD)
controller) is added to the forward path of the feedback
control system.
Adding differentiation is the addition of a zero to the
forward-path transfer function (s+Zc). This type of
compensation requires an active network for its
realization. Dr.R.Subasri.KEC,INDIA

The second technique does not use pure differentiation.
Instead, it approximates differentiation with a passive
network by adding a zero and a more distant pole (a lead
compensator). to the forward-path transfer function
The advantages of a passive lead network over an active
PD controller are that
(1) no additional power supplies are required and
(2) noise due to differentiation is reduced.

Lead Compensator
Select a desired dominant, second-order pole on the s-
plane from the given specifications
The sum of the angles from the uncompensated
system's poles and zeros to the design point can be
found
The difference between 180° and the sum of the
angles must be the angular contribution required
of the compensator.

Design a lead compensator for the given system that will
reduce the settling time ts by a factor of 2 while maintaining
the overshoot as 30%.
k
G(s)
s(s 4)(s 6)

 
1. Draw root locus for the given system
2. Translate the transient response specifications into a pair of
complex dominant root
Given overshoot = 0.30.
Determine the equivalent damping ratio  =0.36
find = cos-1()
Draw a  line with an angle of cos-1() from the negative real
axis. The point of intersection of  line on the root locus, is the
complex dominant root (sd).
Sd =-1.01+j2.6
2
d n ns δω jω 1 δ   

3. Calculate settling time ts= 4/ n =4/1.01 =3.96 sec
required settling time = tsn =ts/2 =3.96/2 =1.98sec
Determine the desired dominant root with the revised settling
time tsn :
Real part n = 4/1.98 = 2.02
desired dominant root -2.02+j6Dr.R.Subasri.KEC,INDIA

Place the compensating zero on the real axis in the region
below the desired dominant root sd. If the
uncompensated system has an open loop pole on the real
axis in the region below sd, then the compensating zero
should lie to the left of the pole.
Let the compensator zero (zc) be at -2.5
Find the angle contribution at sd by all the poles and zeros (c )
s 2 j5
k(s 2.5)
G(s)
s(s 4)(s 6)
k( 2 j5 2.5)
( 2 j5)( 2 j5 4)( 2 j5 6)
5.02 84
(5.3 111 )(5.3 68 )(6.4 51 )
0.027 150
 


 
  

       
 

     
    Dr.R.Subasri.KEC,INDIA

The angle should be contributed by the compensator
pole to get -180 is d = -180 – (c)
= -180 – (-150) = 30
6. Determine the compensator pole (pc) using the formula
d
d
c n
ω
tanθ
p δω


c
6
tan15 0.26
p 2.6
 

Pc = -24.9
The open loop transfer function of the compensated system is
c
c new
c
s z
G (s) K G(s)
s p



where the gain (Knew) at that point sd is determined by
d
new
c s s
1
K
G (s) 

K new= 155Dr.R.Subasri.KEC,INDIA

Thank you

Root locus compensation

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Root locus compensation