Outline
• PD compensation.
• PI compensation.
• PID compensation.
PD, PI, PID Compensation
M. Sami Fadali
Professor of Electrical Engineering
University of Nevada
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2
PD Control
Remarks
= loop gain
= desired closed-loop pole location.
• Find a controller s.t.
or
its odd multiple (angle condition).
• Controller Angle for
• Graphical procedures are no longer
needed.
• CAD procedure to obtain the design
parameters for specified
.
» evalfr(L,scl) % Evaluate L at scl
» polyval( num,scl) % Evaluate num at scl
• The complex value and its angle can also
be evaluated using any hand calculator.
• Zero location
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Procedure:MATLAB or Calculator
Procedure 5.2: Given
1. Calculate
»theta = pi  angle(evalfr(l, scl) )
2. Calculate zero location using
3. Calculate new loop gain (with zero)
» Lc = tf( [1, a],1)*L
4. Calculate gain (magnitude condition)
» K = 1/abs( evalfr( Lc, scl))
5. Check time response of PD-compensated
system. Modify the design to meet the desired
specifications if necessary (MATLAB).
&
1. Obtain error constant
from
and
determine a system parameter
that
is fixed for the
remains free after
system with PD control.
2- Rewrite the closed-loop characteristic
equation of the PD controlled system as
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Procedure 5.2 (Cont.)
Example 5.5
3. Obtain corresponding to the desired closedcan
loop pole location. As in Procedure 5.1,
be obtained by clicking on the MATLAB root
locus plot or applying the magnitude condition
using MATLAB or a calculator.
4. Calculate the free parameter from the gain .
Use a CAD package to design a PD controller
for the type I system
5. Check the time response of the PD
compensated system and modify the design to
meet the desired specifications if necessary.
7
to meet the following specifications
(a) = 0.7 &
rad/s.
(b) = 0.7 &
due to a unit ramp.
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(a)
&
(a) Design (cont.)
MATLAB
Calculate pole location & corresponding gain
>> scl = 10*exp( j*( pi-acos(0.7) ) )
scl =
7.0000 + 7.1414I
>> g=zpk([ ],[0,-4],1);
>> theta=pi( angle( evalfr( g, scl) ) )
theta =
1.1731
» a = 10 * sqrt(1-0.7^2)/ tan(theta) + 7
a=
10.0000
» k =1/abs(evalfr(tf( [1, a],1)*g,scl)) %
k=
10.0000
Gain at scl:
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RL of Uncompensated System
Root Locus
8
0.7
0.7
System: gcomp
Gain: 10
Pole: -7.01 + 7.13i
Damping: 0.701
Overshoot (%): 4.56
Frequency (rad/sec): 10
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6
4
5
2
Imaginary Axis
Imaginary Axis
RL of PD-compensated System
Root Locus
8
7
10
4
3
10
0
-2
2
-4
1
-6
0
-8
-7
-6
-5
-4
-3
-2
-1
0
-8
-20
Real Axis
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0.7
-18
-16
-14
-12
-10
Real Axis
-8
-6
-4
-2
0
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(b)
for unit ramp
L
(b) Design (cont.)
L
→
→
• Assume that K varies with K a fixed, then
• Closed-loop characteristic equation with
PD
• RL= circle centered at the origin
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RL of PD-compensated System
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(b) Design Cont.
fixed
Desired location: intersection of root locus with
radial line.
K = 10, a =10 (same values as in Example 5.3).
MATLAB command to obtain the gain
» k = 1/abs( evalfr( tf([1,10],[ 1, 4, 0]), scl) )
k=
10.0000
• Time responses of two designs are identical.
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PI Control
•
•
•
PI Remarks
Integral control to improve
(increases type by one)  worse
transient response or instability.
Add proportional control  controller has
a pole and a zero.
Transfer function of proportional-plusintegral (PI) controller
• Used in cascade compensation (integral term in
the feedback path is equivalent to a differentiator
in the forward path)
• PI design for a plant transfer function
is the
same as PD design of
.
• A better design is often possible by "almost
canceling" the controller zero and the controller
pole (negligible effect on time response).
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Procedure 5.3
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Comments
1. Design a proportional controller for the
system to meet the transient response
specifications, i.e. place the dominant
.
closed-loop poles at
2. Add a PI controller with the zero location
such that
(small angle)
• Use PI control only if P-control meets the
transient response but not the steady-state
error specifications. Otherwise, use
another control.
• Use pole-zero diagram to prove zero
location formula.
or
3. Tune the gain of the system to moved the
closed-loop pole closer to
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Proof
Pole-zero Diagram of PI Controller
scl
j
tan 180°
d
 z  scl  a 
• From Figure
 p  scl


a
• Controller angle at
n
tan 180°
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Controller Angle at
Proof (Cont.)
• Trig. Identity
3.5
3
2.5
2
1.5
• Solve for
• Multiplying by
1
0.5
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
gives
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Solution
Example 5.6
• To use the PD procedure for PI design, consider
Design a controller for the position control
system to perfectly track a ramp input with a
dominant pair with
.
• Procedure 5.1 with modified transfer function.
• Unstable for all gains (see root locus plot).
• Controller must provide an angle  249 at the
desired closed-loop pole location.
• Zero at 1.732.
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RL of System With Integrator
• Cursor at desired pole location on compensated
system RL gives a gain of about 40.6.
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RL of PI-compensated System
Root Locus
6
5
Imaginary Axis
4
3
0.7
2
1
0
-10
-8
4
-4
-6
Real Axis
-2
0
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Analytical Design
MATLAB
» scl = 4*exp(j*(pi  acos( 0.7)) )
scl =
-2.8000 + 2.8566i
» theta = pi + angle( polyval( [ 1, 10, 0, 0], scl ) )
theta =
1.9285
» a = 4*sqrt(1-.7^2)/tan(theta)+ 4*.7
a=
1.7323
» k = abs(polyval( [ 1, 10, 0,0], scl )/ polyval([1,a], scl) )
k=
40.6400
• Closed-loop characteristic polynomial
• Values obtained earlier, approximately.
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Design I Results
Closed-loop transfer function for Design I
• Zero close to the closed-loop poles 
excessive PO
(see step response together with the response for a
later design: Design II).
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Step Response of PI-compensated
System
Design I (dotted), Design II (solid).
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Design II: Procedure 5.3
For
zero location
Compare Designs I & II
rad/s.
7.143
0.7
1
0.49/ tan 3°
• Design II dominant poles:
0.5
Closed-loop transfer function
rad/s
• Time response for Designs I and II
• Percentage overshoot for Design II (almost
pole-zero cancellation)
<< Percentage overshoot for Design I (II better)
Gain slightly reduced to improve response.
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PID (proportional-plus-integralplus-derivative) Control
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PID Design Procedures
• Improves both transient & steady-state
response.
proportional, integral, derivative
gain, resp.
• Controller zeros: real or complex conjugate.
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1. Cancel real or complex conjugate LHP poles.
2. Cancel pole closest to (not on) the imaginary
axis then add a PD controller by applying
Procedures 5.1 or 5.2 to the reduced transfer
function with an added pole at the origin.
3. Follow Procedure 5.3 with the proportional
control design step modified to PD design. The
PD design meets the transient response
specifications and PI control is then added to
improve the steady-state response.
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Example 5.7
Design I
• Cancel pole at 1 with a zero & add an
integrator
Design a PID controller for the transfer
function to obtain zero
due to
and
rad/s.
ramp,
• Same as transfer function of Example 5.6
• Add PD control of Example 5.6
• PID controller
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Design II
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Design II (Cont.)
• Design PD control to meet transient response
specs.
• Select
rad/s (anticipate effect of adding PI).
Design PD controller using MATLAB commands
» [k,a,scl] = pdcon(0.7, 5, tf(1, [1,11,10,0]))
k =
43.0000
a =
2.3256
scl =
-3.5000 + 3.5707i
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• Obtain PI zero (approx. same as
)
» b =5 /(0.7 + sqrt(1-.49)/tan(3*pi/ 180) ) %
b=
0.3490
Transfer function for Design II (reduce gain to
40 to correct for PI)
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Time response for Design I(dotted)
and Design II (solid)
Conclusions
• Compare step responses for Designs I and II:
Design I is superior because the zeros in Design II
result in excessive overshoot.
• Plant transfer function favors pole cancellation in
this case because the remaining real axis pole is
far in the LHP. If the remaining pole is at 3, say,
the second design procedure would give better
results.
• MORAL: No easy solutions in design, only
recipes to experiment with until satisfactory results
are obtained.
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Example 5.8
Solution
Design a PID controller for the transfer
function to obtain zero
due to step,
and
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• Complex conjugate pair slows down the
time response.
• Third pole is far in the LHP.
• Cancel the complex conjugate poles with
zeros and add an integrator
rad/s
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Step Response of PIDcompensated System
Solution (cont.)
• Stable for all gains.
• Closed-loop characteristic polynomial
• Equate coefficients with
(meets design specs)
• PID controller
In practice, pole-zero cancellation may not occur but near
cancellation is sufficient to obtain a satisfactory time response
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