Design with Root Locus
Lecture 9
Objectives for desired response
1. Improving transient response
Percent overshoot, damping ratio,
settling time,
peak time
2. Improving steady-state error
Steady state error
Gain adjustment
• Higher gain, smaller steady stead error,
larger percent overshoot
• Reducing gain, smaller percent
overshoot, higher steady state error
Compensator
• Allows us to meet transient and steady
state error.
• Composed of poles and zeros.
• Increased an order of the system.
• The system can be approx. to 2nd order
using some techniques.
Improving transient response
• Point A and B have the
same damping ratio.
• Starting from point A,
cannot reach a faster
response at point B by
adjusting K.
• We have pole at A on
the root locus, but we
want response like at B.
• Compensator is
preferred.
Compensator configulations
Cascade
Compensator
Feedback
Compensator
The added compensator can change a pattern of root locus
compensator
Method of implementing compensator:
1. Proportional control systems: feed the error forward to the plant.
2. Integral control systems: feed the integral of the error to the plant.
3. Derivative control systems: feed the derivative of the error to the plant.
Types of compensator
1. Active compensator
– PI, PD, PID use of active components, i.e., OP-AMP
– Require power source
– ss error converge to zero
– Expensive
2. Passive compensator
– Lag, Lead use of passive components, i.e., R L C
– No need of power source
– ss error nearly reaches zero
– Less expensive
Improving steady-state error (PI)
Placing a pole at the origin to increase system order;
decreasing ss error as a result!!
(a) Original system without compensation
(b) Add a pole at the origin but angular contribution at point A
is no longer 180
Improving Steady-State Error (PI)
Also add a zero close to the
pole at the origin. As angular
contribution of the
compensator zero and pole
cancels out, point A is still
on the root locus, and the
system type has been
increased.
Improving Steady-State Error (PI):Example
Choose zero at -1
Damping ratio = 0.174 in both uncompensated and PI cases
Improving Steady-State Error (PI)
>> z=[1];
>> n=conv([1 3 2],[1 10]);
>> sys=tf(z,n)
>> sgrid(0.174,[2,10])
>> [k p]=rlocfind(sys)
Improving Steady-State Error (PI)
Improving Steady-State Error (PI)
As shown in the figure, the step response of the PI
compensated system approaches unity in the steady-state,
while the uncompensated system response approaches
1−0.108 = 0.892. The simulation shows that it takes 18
seconds for the compensated system to reach and stay
within ±2% of the final value of unity, while the
uncompensated system takes about 6 seconds to settle to
within ±2% of its final value of 0.892. This is because there is
no pole-zero cancelation and the pole not canceled is very
close to the origin.
Improving Steady-State Error (PI)
158.2(s+0.1)
s(s+1)(s+2)(s+10)
Step
Zero-Pole
Scope
164.6
(s+1)(s+2)(s+10)
Step1
Zero-Pole2
Finding an intersection between
damping ratio line and root locus
• Damping ratio line has an equation:
b   ma
where a = real part, b = imaginary part of the
1
intersection point, m  tan(cos ( ))
• Summation of angle from open-loop poles and
zeros to the point is 180 degrees
tan
1 


  tan
1 a 
b
1 


  tan
2a
b
1 


  180
 10  a 
b
Arctan formula
tan
tan
1
( A )  tan
1
( B )  tan
1
( A )  tan
1
( B )  tan
AB 


 1  AB 
1 
AB 


 1  AB 
1 
• Use the formula to get the real and
imaginary part of the intersection point
and get
a  -1.5893, 0.6936
b  - 3.9255
• Magnitude of open loop system is 1
K 
p1 p 2 p 3
No open loop zero
1
K 
1  a  2
b
2
2  a 2
1
b
2
10
 a  b
2
2
 164 . 53
• Draw root locus with
compensator (system order is up
by 1--from 3rd to 4th)
• Needs complex poles
corresponding to damping ratio of
0.174 (K=158.2)
• From K, find the 3rd and 4th poles
(at -11.55 and -0.0902)
• Pole at -0.0902 can do phase
cacellation with zero at -1 (3th
order approx.)
• Compensated system and
uncompensated system have
similar transient response (closed
loop poles and K are aprrox. The
same)
PI Controller
A compensator with a pole at the origin and a zero close to
the pole is called an ideal integral compensator, or PI
controller
Gc (s)  K1 
K2
s


K1

K1 s 
K2

s




Lag Compensator
Ideal integral compensation: pole is in the origin, requires
active network (costly).
Real (passive) integral compensation: pole is close to origin
(not in the origin), cheaper.
(a) Type 1 uncompensated
system
(b) Type 1 compensated
system
Type is not increased.
What about steadystate error
Example
With damping ratio of 0.174, add lag
Compensator to improve steady-state error
by a factor of 10
Step I: find an intersection of root locus and
damping ratio line (-0.694+j3.926 with K=164.56)
Step II: find Kp = lim G(s) as s0 (Kp=8.228)
Step III: steady-state error = 1/(1+Kp)= 0.108
Step IV: want to decrease error down to 0.0108
[Kp = (1 – 0.0108)/0.0108 = 91.593]
Step V: require a ratio of compensator zero to pole
as 91.593/8.228 = 11.132
Step VI: choose a pole at 0.01, the corresponding


Zero will be at 11.132*0.01 = 0.111  ss00.111
. 01


3rd order approx. for lag compensator
(= uncompensated system)  making
Same transient response but 10 times
Improvement in ss response!!!
If we choose a compensator pole at 0.001 (10 times
closer to the origin), we’ll get a compensator zero
at 0.0111 (Kp=91.593)
New compensator:
s  0 . 0111
s  0 . 001
4th pole is at -0.01
(compared to -0.101)
producing a longer
transient response.
SS response improvement conclusions
• Can be done either by PI controller (pole
at origin) or lag compensator (pole closed
to origin).
• Improving ss error without affecting the
transient response.
• Next step is to improve the transient
response itself.
Improving Transient Response
• Objective is to
– Decrease settling time
– Get a response with a desired %OS
(damping ratio)
• Techniques can be used:
– PD controller (ideal derivative compensation)
– Lead compensator
PD controller: Improving transient response
System above controlled by a pure gain (P controller) in the
forward path has its root locus going through point A for some
value of gain K.
• Our goal is to speed up the response at A to that at B, while
keeping the percent overshoot unchanged.
• The above root locus with a P controller cannot go through point
B (sum of angles from the open-loop finite poles and zeros to point
B is not an odd multiple of 180◦). A solution is to add a (nonzero)
zero to the forward path (e.g., PD controller).
PD controller: Improving transient response
• Transfer function of the PD controller Gc(s) = K2 s + K1 = K2(s+K1/K2) =
K(s+zc) introduces a zero at −zc Into the forward path.
• Effect of the added zero: The added zero will contribute to make the sum
of angles from the open-loop finite poles and zeros to the desired point
(point B) be an odd multiple of 180◦.
Note: an added zero has the effect of pushing the root locus to the left
while an added pole has the effect of pushing it to the right.
• The new root locus can meet the specific transient response (with
shorter settling time) by going through point B for some value of gain K.
Ideal Derivative Compensator
• So called PD controller
• Compensator adds a zero to the system
at –Zc to keep a damping ratio constant
with a faster response
GC  s  zc
(a) Uncompensated system, (b) compensator zero at -2
(d) compensator zero at -3, (d) compensator zero at -4
Indicate peak time
Indicate settling time
• Settling time & peak time: (b)<(c)<(d)<(a)
• %OS: (b)=(c)=(d)=(a)
• ss error: compensated systems has lower value than
uncompensated one cause improvement in transient
response always yields an improvement in ss error
Example
design a PD controller to yield 16%
overshoot with a threefold reduction in
settling time
• Step I: calculate a corresponding damping
ration (16% overshoot = 0.504 damping ratio)
• Step II: search along the damping ratio line for
an odd multiple of 180 (at -1.205±j2.064) and
corresponding K (43.35)
• Step III: find the 3rd pole (at -7.59) which is far
away from the dominant poles  2nd order
approx. works!!!
More details in step II and III
Characteristic equation:
3
2
s  10 s  24 s  K  0
( s  1 . 2  j 2 . 06 )( s  1 . 2  j 2 . 06 )( s  c )  0
3
2
2
2
2
2
s  ( 2 a  c ) s  ( a  2 ac  b ) s  c ( a  b )  0
 2 a  c  10
 c(a
2


 solve to get the third pole and gain K
2
b ) K

• Step IV: evaluate a desired settling time:
uncompensa ted system : T s 
compensate d system : T s 
4

3 . 320

n
4
 3 . 320 sec
1 . 205
 1 . 107 sec
3
• Step V: get corresponding real and
imagine number of the dominant poles
(-3.613 and -6.193)
 
4
Ts

4
 3 . 613
1 . 107
 d  3 . 613 tan(cos
1
( 0 . 504 ))  6 . 193
Location of poles
as desired is at
-3.613±j6.192
• Step VI: summation of angles at the desired
pole location, -275.6, is not an odd multiple of
180 (not on the root locus) need to add a
zero to make the sum of 180.
• Step VII: the angular contribution for the point
to be on root locus is +275.6-180=95.6  put a
zero to create the desired angle
6 . 192
3 . 613  
  3 . 006
 tan( 180  95 . 607 )


Compensator: (s+3.006)
Might not have a pole-zero cancellation for
compensated system
PD Compensator
Gc  K 2s  K1  K 2 (s 
K1
K2
)
Lead Compensation
Zeta2-zeta1=angular contribution
• A PD controller can be approximated with a lead
compensator, which is implemented with a passive network.
• If the lead compensator pole is farther from the imaginary
axis than the compensator zero, the angular contribution of
the compensator is still positive and thus approximates an
equivalent single zero.
• The advantages of a passive lead compensator over an
active PD controller are that (1) no additional power supplies
are required and (2) noise due to differentiation is reduced.
Lead Compensation
Zeta2-zeta1=angular contribution
• The concept behind lead compensation: the difference between 180◦ and
the sum of the angles from the uncompensated system’s poles and zeros
to the design point (desired pole location) must be the angular
contribution required of the compensator.
That is,
q2 −q1 −q3 −q4 +q5 = (2k+1)180◦
where q2 −q1 = qc is the angular contribution of the lead compensator.
• The angular contribution qc can be determined from the rays originating
from the desired closed-loop pole and terminating at the compensator
pole and zero. These rays can be rotated about the desired closed-loop
pole and thus different pairs of compensator pole and zero can be used to
meet the transient response requirement.
• Different possible lead compensators: differences are in the values of
the static error constants, the static gain, the difficulty in justifying a
second-order approximation when the design is complete, and the
ensuing transient response.
• For design we arbitrarily select either a lead compensator pole and zero
and find the angular contribution at the design point of this pole or zero
with the uncompensated system’s open-loop poles and zeros. The
difference between this angle and 180◦ is the required contribution of the
remaining compensator pole or zero.
Example
Design three lead compensators for the system
that has 30% OS and will reduce settling time down
by a factor of 2.
• Step I: %OS = 30% equaivalent to damping ratio =
0.358, Ѳ= 69.02
• Step II: Search along the line to find a point that gives
180 degree (-1.007±j2.627)
• Step III: Find a corresponding K ( K  63 .212 )
• Step IV: calculate settling time of uncompensated
system
Ts 
4


4
 3 . 972 sec
1 . 007
n
• Step V: twofold reduction in settling time (Ts=3.972/2 =
1.986), correspoding real and imaginary parts are:
 
4
Ts

4
 2 . 014
1 . 986
 d  2 . 014 tan(cos
1
( 0 . 358 ))  5 . 253
• Step VI: let’s put a zero at -5 and find the
net angle to the test point (-172.69)
• Step VII: need a pole at the location
giving 7.31 degree to the test point.
5 . 252
p c  2 . 014
p c  42 . 96
 tan 7 . 31

lead compensato r 
( s  5)
( s  42 . 96 )
Note: check if the 2nd order approx. is
valid for justify our estimates of percent
overshoot and settling time
– Search for 3rd and 4th closed-loop poles
(-43.8, -5.134)
– -43.8 is more than 20 times the real part of
the dominant pole
– -5.134 is close to the zero at -5
The approx. is then valid!!!