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PID Control

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usually only add together the last n' error terms ... if it stays negative for a long time, the contribution to f(t) can become significant ... – PowerPoint PPT presentation

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Title: PID Control


1
PID Control
  • spring-mass-damper problem

2
In an experiment, the following output was
obtained and the system parameters were found to
be ? c/(2(k1m1)1/2)0.1203, ?n (k1/m1)1/2
18.81 rad/sec m1 0.5773 kg, k1 204.25 N/m, c
2.6135 N sec/m x1(0) ? 0.025 m 2.5 cm, from
rest.
3
  • lets model this system using Simulink
  • governing differential equation
  • for our case, f(t) 0, (0) 0, x1(0) 0.025
    m
  • thus

4
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5
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6
new case ? 0.938, ?n 10.43 rad/sec 1.66
cycles/sec
7
  • suppose the mass is initially positioned at the
    free length and is at rest
  • we want to move the mass to the position x 0.3
    m
  • how do we determine f(t)?

8
  • proportional controller for the underdamped case
  • f(t) (1 N/m) e(t)

large steady state error large settling
time never gets close to the desired position
x, m
Simulink
time, sec
9
Kp 10,000 N/m
x, m
Kp 1000 N/m
Kp 100 N/m
Kp 10
time, sec
as proportional gain Kp increases steady state
error improves rise time improves settling time
worsens overshoot and frequency of oscillation
worsens
10
free response
overdamped system new case ? 10.43, ?n
10.43 rad/sec 1.66 cycles/sec
was 4.5 N sec/m
underdamped case
this overdamped case
11
  • proportional controller for the overdamped case

Kp 1 N/m
underdamped case
x, m
Kp 1 N/m
this overdamped case
Simulink
time, sec
12
Kp 10,000 N/m
Kp 1000 N/m
Kp 100 N/m
Kp 10
proportional controller looks pretty good for
overdamped case when gain value is high
13
Kp 10,000,000 N/m
Kp 1,000,000 N/m
Kp 100,000 N/m
response can overshoot for overdamped case if
gain value Kp becomes too large
14
  • suppose we have a critically damped system (?1)

15
critically damped case, Kp 1
Simulink
16
critically damped case for various Kp
Kp 100,000 N/m
Kp 10,000 N/m
Kp 1000 N/m
Kp 100 N/m
Kp 10
17
Kp 1 N/m for all 3 cases
c 4.5 N sec/m, ? 0.094
c 47.9583 N sec/m, ? 1
c 500 N sec/m, ? 10.43
all cases (underdamped, critically damped, and
overdamped) have the same steady state error
18
  • apparent rules of thumb for a 2nd order system
    using only a proportional controller
  • for all three cases, increasing Kp reduces the
    steady state error
  • if Kp gets too large you will getoscillations
  • proportional controller can workpretty well,
    unless the system isunderdamped

underdamped
overdamped
critically damped
19
underdamped
overdamped
critically damped
20
  • how can we do better?
  • lets try to reduce the steady state error
  • next try ? PI controller
  • we will add a new term to f(t)
  • this will be proportional to the time integral of
    the error
  • what about the limits of the integration?
  • do we integrate from tdinosaur age to current
    time?
  • - usually only add together the last n error
    terms
  • note that e(t) may be positive or negative
  • if it stays negative for a long time, the
    contribution to f(t) can become significant
  • so, if there is a steady state error, eventually
    this error will add up and give a boost to f(t)

21
  • lets look at the underdamped system again

22
underdamped system
Kp 1 N/m, Ki 1 N sec/m
Kp 1 N/m, Ki 0
23
underdamped system
Kp 1 N/m, Ki 100 N sec/m
Kp 1 N/m, Ki 300 N sec/m
Kp 1 N/m, Ki 500 N sec/m
24
underdamped system
Kp 0.1 N/m, Ki 250 N sec/m
Kp 5 N/m, Ki 250 N sec/m
Kp 50 N/m, Ki 250 N sec/m
Kp 500 N/m, Ki 250 N sec/m
25
overdamped system
Kp 1 N/m, Ki 1 N sec/m
Kp 1 N/m, Ki 0
26
overdamped system
Kp 1 N/m, Ki 100 N sec/m
Kp 1 N/m, Ki 300 N sec/m
Kp 1 N/m, Ki 500 N sec/m
27
critically damped system
Kp 1 N/m, Ki 1 N sec/m
Kp 1 N/m, Ki 0
28
critically damped system
Kp 1 N/m, Ki 100 N sec/m
Kp 1 N/m, Ki 300 N sec/m
Kp 1 N/m, Ki 500 N sec/m
29
critically damped system
Kp 1 N/m, Ki 1000 N sec/m
Kp 1 N/m, Ki 2000 N sec/m
Kp 1 N/m, Ki 5000 N sec/m
30
overdamped system
underdamped system
critically damped system
critically damped system w/ larger Ki values
Kp 1 N/m, Ki 100 N sec/m
Kp 1 N/m, Ki 300 N sec/m
Kp 1 N/m, Ki 500 N sec/m
31
  • Can we decrease the overshoot and the settling
    time?

32
  • lets look at the underdamped case againKp 100
    N/m

33
underdamped case, Kp 100 N/m
x (position)
m
error
time, sec
how can we reduce the oscillations ?
34
underdamped case, Kp 100 N/m
x (position)
x (position)
error
m
derivative of error
error
derivative of the error
time, sec
lets look at the derivative of the error if we
have a positive error, and is positive, then
the error is getting worse ! so, whenever we
have a positive error and is positive, lets
add more to f(t) lets add Kd ? , where Kd is
some constant
35
e(t) xdesired - x
lets look at all the possible cases
I
II
III
IV
36
underdamped case, Kp 100 N/m
xdesired 0.3 m
x (position)
m
derivative of error / 3
I
II
I
II
time, sec
for this example, the error is always positive,
but goes positive and negative lets look at
a proportional and derivative controller (PD)
37
underdamped case, Kp 100 N/m
xdesired 0.3 m
Kd 0
Kd 10 N/m-sec
m
Kd 100
time, sec
the Kd term helped tremendously with the
oscillations, but didnt change the steady state
error
38
In Summary
39
PID Controller
40
PID Controller
underdamped case, Kp 100 N/m, Ki 250 N sec/m
xdesired 0.3 m
Kd 0
m
Kd 10 N/m-sec
Kd 100
time, sec
41
General Tips for Designing a PID Controller
?
  • Obtain an open-loop response and determine what
    needs to be improved
  • Add a proportional control to improve the rise
    time
  • Add a derivative control to improve the overshoot
  • Add an integral control to eliminate the
    steady-state error
  • Adjust each of Kp, Ki, and Kd until you obtain a
    desired overall response. (TUNING!)

?
?
?
?
42
PI Control with Velocity Feedback
43
underdamped system PI controller
44
underdamped system PI controller with velocity
feedback
45
underdamped system PI controller with velocity
feedback
Kp 5 Ki 250
Kv 0
Kv 1
m
Kv 5
Kv 10 N-sec/m
time, sec
46
underdamped system
Kp 5 Ki 250
Kv 0
Kv 1
m
Kv 5
Kv 10 N-sec/m
time, sec
47
  • WHY DESIGN CONTROLLER?
  • Feedback Control System Has to Satisfy Several
    Conditions in Order to Perform in Some Prescribed
    Optimal Sense.
  • Most Systems Need Some Adjustment in Order to
    Satisfy all Conditions (or Comprise Among
    Conflicting and Demanding Specifications).
  • May Need to Change the Structure of the System
    and Redesign to Meet Performance Requirements.
  • Suitable Control System
  • Stable
  • Acceptable response to input commands
  • Less sensitive to system parameter changes
  • Minimal steady-state error for input command
  • Able to eliminate the effect of undesirable
    disturbances

48
Compensationtechniquesa. cascadeb.
feedbackOther CompensationOutput or Load
Input
Design via gain adjustment. Design via cascaded
or feedback filters.
49
Types of cascade compensators (continued on next
slide)
50
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51
Ziegler Nichols PID Tuning (closed loop method)
  • Start with a pure proportional controller and run
    a step test with a low value of Kp.
  • Repeat the test by increasing (or decreasing) Kp
    until a stable oscillation begins to occur
    (oscillations with a constant amplitude). This
    value of Kp is called the ultimate gain, Ku.
  • Determine Ku and Pc ? Kp , Ki and Kd
  • Then touch up
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