INC341
Design with Root Locus
Lecture 9
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2 objectives for desired response
1. Improving transient response
Percent overshoot, damping ratio, settling
time, peak time
2. Improving steady-state error
Steady state error
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Gain adjustment
• Higher gain, smaller steady stead error,
larger percent overshoot
• Reducing gain, smaller percent
overshoot, higher steady state error
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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.
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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.
• Compensator is
preferred.
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Compensator configulations
Cascade
Compensator
Feedback
Compensator
The added compensator can change a pattern of root locus
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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
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Improving steady-state error
Placing a pole at the origin to increase system order;
decreasing ss error as a result!!
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The pole at origin
affects the transeint
response  adds a
zero close to the pole
to get an ideal
integral compensator
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Example
Choose zero at -1
Damping ratio = 0.174 in both uncompensated and PI cases
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• Draw root locus
without compensator
• Draw a straight line
of damping ratio
• Evaluate K from the
intersection point
• From K, find the last
pole (at -11.61)
• Calculate steadystate error
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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
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1 


  tan
1 a 
b
1 


  tan
2a
b
1 


  180
 10  a 
b
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Arctan formula
tan
tan
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1
( A )  tan
1
( B )  tan
1
( A )  tan
1
( B )  tan
AB 


 1  AB 
1 
AB 


 1  AB 
1 
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• 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
b
2
10
 a  b
2
2
 164 . 53
1
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• 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)
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Comparason of step response
of the 2 systems
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PI Controller
Gc (s)  K1 
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K2
s


K1
K 1  s 
K2





s
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Lag Compensator
•Build from
passive elements
•Improve ss error
by a factor of
Zc/Pc
•To improve both
transient and ss
responses, put
pole and zero
close to the
origin
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Uncompensated system
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With lag compensation
(root locus remains the same)
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Example
With damping ratio of 0.174, add lag
Compensator to improve steady-state error
by a factor of 10
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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

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3rd order approx. for lag compensator
(= uncompensated system)  making
Same transient response but 10 times
Improvement in ss response!!!
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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.
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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.
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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
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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
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(a) Uncompensated system, (b) compensator zero at -2
(d) compensator zero at -3, (d) compensator zero at -4
Indicate peak time
Indicate settling time
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• 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
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Example
design a PD controller to yield 16%
overshoot with a threefold reduction in
settling time
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• 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!!!
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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

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• 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
1 . 107
 d  3 . 613 tan(cos
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 3 . 613
1
( 0 . 504 ))  6 . 193
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Location of poles
as desired is at
-3.613±j6.192
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• 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  
 tan( 180  95 . 607 )


  3 . 006
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Compensator: (s+3.006)
Might not have a pole-zero cancellation for
compensated system
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PD Compensator
Gc  K 2s  K1  K 2 (s 
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K1
)
K2
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Lead Compensation
Zeta2-zeta1=angular contribution
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Can put pairs of poles/zeros to get a desired θc
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Example
Design three lead compensators for the system
that has 30% OS and will reduce settling time down
by a factor of 2.
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• 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
1 . 986
 d  2 . 014 tan(cos
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 2 . 014
1
( 0 . 358 ))  5 . 253
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• 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
 tan 7 . 31

p c  42 . 96
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lead compensato r 
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( s  5)
( s  42 . 96 )
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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!!!
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