I need help whith control pid

On Wed, 30 May 2007 22:52:16 -0700, julio.maragano@gmail.com wrote:


This almost has to be homework -- few real-world problems are so tidy, nor
is one often called upon to control systems that resonate at 4kHz.
Moreover, people with real-world problems usually start by saying
something like "I'm trying to control this itty bitty piezo actuator with
PID, and I'm having a problem...".

So tell us more!  If it's homework go ahead and fess up -- most newsgroup
populations will help with homework if you say so, and if you'll accept
_help_ and not _answers_.  If you're doing something real, please give us
more details like your required performance, any interesting limitations
(like hysteresis if it's a piezo actuator), and what you're trying to do
it with (op-amps? DSP? FPGA?).

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html

julio.maragano@gmail.com wrote:

There are an infinite number of solutions.  Which one do you want?

Peter Nachtwey

Peter Nachtwey wrote:

The third one from the end if the list?

Jerry
--
Engineering is the art of making what you want from things you can get.
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¯AF¯AF¯AF¯AF¯AF¯AF¯AF¯AF¯AF¯AF¯AF
Jerry, what is your solution? I have one that works for 3 real poles
at -5.163978E4.  I just arbitrarily picked 3 poles at 2 times the
natural frequency.

Peter Nachtwey

Peter Nachtwey wrote:

   ...


I'm sorry, Peter. I didn't even look at the problem. "Third from the
end" of an infinite list was an attempt at humor.

Jerry
--
Engineering is the art of making what you want from things you can get.
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G(s) is undamped. Damp it!

Insert a damper in your system and your control works.

T^2 = 1.5e-9

G(s)_1 = 400/(T^2s^2 + 2*delta*T*s + 1)

E.g. delta = 1.1

See therefore
http://home.arcor.de/janch/janch/_control/20070531-controldoc/

2*delta*T*s must be added!


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Jan C. Hoffmann  eMail aktuell: janch@nospam.arcornews.de
                 Microsoft-kompatibel/optimiert für IE7+OE7

That isn't a PID. It isn't even closed loop control.

Peter Nachtwey

On Fri, 01 Jun 2007 08:15:36 -0700, Peter Nachtwey wrote:


Yet it's realistic, in an odd sort of way -- if you have a really nasty
plant that you need to control, you should first ask if there's any chance
of changing the plant.

The OP hasn't come back with sufficient detail to solve his problem.  :(

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html

: On Fri, 01 Jun 2007 08:15:36 -0700, Peter Nachtwey wrote:
:
: Yet it's realistic, in an odd sort of way -- if you have a really nasty
: plant that you need to control, you should first ask if there's any chance
: of changing the plant.

That wouldn't help the OP with his tuning problem.
:
: The OP hasn't come back with sufficient detail to solve his problem.  :(

The home work probably needed to be turned in by now.  In any case the
problem can be solved symbolically.  Then one doesn't care if there is
damping or not.

Peter Nachtwey

Once I had to care about such a problem in practice.

It may be of some interest. I moved the set point according to the defined
benchmark scheme for comparing with standard PID. The difference can be seen
at page 3. As discussed before the controller range of 0.2...1 may not be
exceeded.

If someone is interested: page 3
http://home.arcor.de/janch/janch/_control/20070531-controldoc/


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Jan C. Hoffmann  eMail aktuell: janch@nospam.arcornews.de
                 Microsoft-kompatibel/optimiert für IE7+OE7

You are still trying to make the PID look bad. This just proves you
still can't calculate PID gains or know how to properly implement a
PID.  If have provided a few example in previous threads but you
obviously keep ignoring them.  This is also a non-solution to the
wrong problem.

This one of an infinite number of solutions to julio's non-damped
problem.
ftp://ftp.deltacompsys.com/public/NG/Mathcad%20-%20t0p2%20simple%20OSC%20NG.pdf

In this example the closed loop transfer function has 3 poles so I
chose a generic desired characteristic equation of ((s+a)^2+b^2) *(s
+c). This means there is a real pole at -c and complex pole at -a-jb
and -a+jb.  I set a=c=3*natural frequency.  I want the response to be
faster than JCH's example just to show that the fault isn't with the
PID, even if it wasn't realistically possible.  I set b=0 because I
wanted all three poles to be real. The PID gains are all calculated
symbollically in terms of the system parameters and the desired
characeristic equation.  It is the desired response that julio was
missing.  By chosing where I want the poles I can get any closed loop
response I want, in theory.  The reality of sampling times and
feedback resolution often limits how quick the response can be.

To get my example to work required an update time of 5 microseconds
which is very fast.

Peter Nachtwey

Ok, you could have 3rd order system. But the 'given system' is

G(s) @0/(T^2*s^2 + 1)

T^2=1.5e-9

Damped directly keeping 2nd order

G(s) @0/(T^2*s^2 + 2*delta*T*s + 1)

delta >= 1

Then the process can't oscillate for sure.

If you won't completely rebuild the system you have just to 'repair it
mathematically' and add a damper like 2*1.1*T*s.

I was repairing a given system and you have completely build a new system!
Our results cannot be compared.


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Jan C. Hoffmann  eMail aktuell: janch@nospam.arcornews.de
                 Microsoft-kompatibel/optimiert für IE7+OE7

JCH wrote:

ftp://ftp.deltacompsys.com/public/NG/Mathcad%20-%20t0p2%20simple%20OSC%20NG.pdf

Surprise, surprise, that is what the closed loop transfer function is
when you add a PID to a second order plant and don't cancel any poles
with zeros. I can place a few zeros on top of the poles and make the
closed loop transfer function 1st or 2nd order. That would make the
little bobble in the Bode plot response go away. I wasn't trying to be
fancy and I didn't need to be to beat your comparison PID.  So what kind
of closed loop transfer function do you have for your poor example? You
don't show your work.


So is mine. G(s)@0*omega*2/(s^2+2*zeta*omega*s+omega^2)


I set omega = 1/T.  I would think you would be able to figure that out.
  I include zeta which is the damping factor but I set it to 0.0 to be
compatible with the julio's plant.


Yes, but that wasn't the problem.  The problem is to find PID gains. You
have a PID that you compare your over damped system to but you posted no
gains or show how you got your results.

This works well if you are adding a few resistors to a circuit.  It
doesn't work on large mechanical systems.  Think of the wasted energy.
I have just added a PID as julio requested.


Again, you have modified the problem to fit you math.  I have just added
a PID to julio's system.


The PIDs can be compared.  You can see your PID gets into position
slowly and only after overshooting the set points. My complaint about
your posts is that you make the PID look bad to prove what ever point
you want to make. I have shown you how I calculate PID gains in previous
threads but you ignore my examples and continue to make your PID
examples look bad to what ever you are comparing them to. You also
change the problem to fit the your math.  Finally, you don't show your
work or supply PID gains even though you had to calculate some for your
comparison PID.  You have not presented a solution to julio's problem.

Peter Nachtwey

ftp://ftp.deltacompsys.com/public/NG/Mathcad%20-%20t0p2%20simple%20OSC%20NG.pdf

What if it would be a F-18 air fighter 'mechanical system' on all planes?


--
Regards/Grüße    http://home.arcor.de/janch/janch/menue.htm
Jan C. Hoffmann  eMail aktuell: janch@nospam.arcornews.de
                 Microsoft-kompatibel/optimiert für IE7+OE7

On Fri, 01 Jun 2007 23:23:33 -0700, Peter Nachtwey wrote:


No it won't.  But one of the saddest realities I eventually had to accept
was that there are some closed-loop control problems that simply cannot be
solved by a more-clever application of theory.  Ultimately, every plant
has it's limitations, and often the only way to get a system to come up
to snuff is to change the plant, with more sensors, more actuators, or
just plain old modifications.

True.  The biggest reason I didn't want to reply substantively to his post
is that without knowing what the expected range of good behavior of the
plant is, you really can't go designing a controller to it.  I know of at
least four different ways of designing a controller for a highly resonant
plant (not including changing the plant); each one applies to a different
combination of plant characteristics and system requirements, and you
cannot recommend any one of them without knowing a lot more about the
plant than just a 2nd-order s-domain model.

--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com

Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html