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clg;
echo on
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University of California, Santa Barbara.
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ECE 147 A:
Feedback Control Systems
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module2:
PID control
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% This module will show graphically the varying effects of
% proportional (P), proportionalintegral (PI), proportional
% derivative (PD) and proportionalintegralderivative (PID)
% controllers on several typical systems.
% The plants, we will discuss here are a motor with a voltage input
% and a velocity output, the spring mas dashpot (SMD) system from
% module 1, and a % third order Butterworth filter.
The
% controllers we will look at are of the form:
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C1(s) = Kp
(proportional only)
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C2(s) = Kp + Ki/s
(proportional and integral)
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C3(s) = Kp + Kd*s/(s/beta + 1)
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(prop. and derviative)
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NOTE: The (s/beta + 1) term is added to the
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derivative term to make it proper.
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beta is chosen as a frequency well
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above the bandwidth of the system so
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that the effect of 1/(s/beta +) is
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approximately equal to 1 in the frequency
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range of interest.
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C4(s) = Kp + Ki/s + Kd*s/(s/beta + 1)
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(prop., integ., deriv.)
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pause
% press any key to continue
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% The plants we will use have the following form:
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P1(s) = A/(s+B)
(voltage > velocity)
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P2(s) = C/(S^2 + 2*z*wn*s + wn^2)
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(force > position)
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P3(s) = 1/((s + 1)(s^2 + 1.4142*s + 1))
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(Butterworth filter)
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% We will be graphing time response curves to illustrate
% the effect of each gain on the closed loop time response.
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View Full Document % Typically, PID controllers are tuned by control engineers
% that understands which way to adjust specific gains in
% order to achieve the desired response.
For simple systems
% there are some rules of thumb about tuning PID controllers
% (look up ZeiglerNichols rules for example).
Often it is
% a trial and error process with the control engineer
% eventually getting a feel for the effect of the various gains.
% In each case we will consider the standard unity gain negative
% feedback setup:
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_________
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y




error
ref.
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<o plant <
C(s)
<(+)<
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________
^ 
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% The objective is to design C(s) so that the output, y, is equal
% to the reference (this is the ideal  we will never get them
% exactly equal).
pause
% press any key to continue.
% For this control example, we must first build the models
% of the plants and controllers.
For the motor model, we
% use values of A = 5 and B = 15.
Thus, P1(s) = 5/(s+15).
num = [0 5];
den = [1 15];
P1 = nd2sys(num,den);
% The SMD model values are the same as in module 1;
% C = 90, wn = 10, and z = .1.
So,
% P2(s) = 90/(s^2 + 2s + 100).
num = [0 0 90];
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This note was uploaded on 04/06/2010 for the course ECE 145 taught by Professor Rodwell during the Spring '07 term at UCSB.
 Spring '07
 RODWELL

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