Automatic optimization of high order control loops

What happens if the initial error changes?   If have found the optimal
gains exist only for the current conditions.    The optimizer should
find gains that keep the control output in saturation much of the time
when responding to a step jump.  I didn't see the control output going
into saturation during a response to a step change.   I think you must
be using the target filter.   You shouldn't use the target filter when
optimizing PID gains.

To keep the control output from wanting to saturate all the time you
can try minimuizing the control output too.

There are many minimization functions.  SSE, IAE, ITAE.


There are better ways of findimg the minimum.  They involve finding a
gradient at a point and using that to go down hill.   See Levenberg
Marquardt or BFGS.  Numerical Recipes in C has these routines.


First you must optimize PID gains WITHOUT the target filter and feed
forwards.
Tuning your 5 repeated poles without a feed forwards is not easy.

Peter Nachtwey

See Page 4
http://home.arcor.de/janch/janch/_control/20071104-pd5 (pid)z1z2/

The initial error e = 0

That's usually done by changing from manual to automatic operation.



v2 is desinged to get fully 'open' but not more. The red line covers partly
the blue line. That's why you can't see it.



It's an all over evaluation that also optimizes. Compare to Page 7
http://home.arcor.de/janch/janch/_control/20071104-pd5 (pid)z1z2/



Yes, but it is in this case not necessary. Maybe in special cases.



Yes, but it will be more difficult. One has 3 parameters (PID) to optimize.



I don't tune it. I all calculated. If designed Ai=Bi=Ci then it's best. If
differently designed then the computer will also do the work in the same
manner. It's just a
different solution.


Example

F3(s) = (0,3000031 s + 1)^3 equivalent to
0,02700085 y''' + 0,2700056 y'' + 0,9000094 y' + y = 0,3993063

F3(s) is chosen with slower performance. Otherwise it would exceed the
ranges (<0,2 and >1).

Optimisation is calculated automatically.

See Page 6
http://home.arcor.de/janch/janch/_control/20071104-pd5 (pid)z1z2/


Note:
Be aware that all is very realistic because the process transfer function is
based on real data, made from measured data points not physically derived.
It would be hard work to derive a 5th order process tranfer function without
few assumptions.


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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

3 parameters is easy but what if you need more gains?  Using a grid
limits your gain resolution


Only after figure out the difference between a type 0 and type 1
plant.  What you call positin should be velocity.


Ah, this points out my complaint about IAE, ITAE or SSE tuning.  The
optimal gains are only optimal for one set of conditions.


What do you do if you come across a real system that is faster or
slower?


How can that be?  The relationship beween the control and the velocity
is not right.

For the past year you have confused positions with velocity.   When
the control signal goes to 0 the velocity will go to 0 not the
position.  In your examples the position goes back to 0 when the
control output goes to 0.  That isn't right

Peter Nachtwey

http://home.arcor.de/janch/janch/_control/20071104-pd5 (pid)z1z2/
Where did I say that? Page ?



http://home.arcor.de/janch/janch/_control/20071104-pd5 (pid)z1z2/
I defined it clearly in Page 2.



http://home.arcor.de/janch/janch/_control/20071104-pd5 (pid)z1z2/
The process (transfer function) is defined in Pages 2,3,4,5,6,7

And the Target Filter is defined as constant:
Example 1: Page 4
Example 2: Page 6

Another process will have another process transfer function.



http://home.arcor.de/janch/janch/_control/20071104-pd5 (pid)z1z2/
v1 and v2 could refer to any physical value.


--
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

Do you understand the difference between a type 0 ( non integrating )
and type 1 ( integrating ) system?

Peter Nachtwey

http://home.arcor.de/janch/janch/_control/20071104-pd5 (pid)z1z2/
and all Pages before:

v1 can be 'any' physical value (process value PV).

Examples

v1''''' = d^5 v1/dt^5, the 5th derivative of v1

v1''''  = Integral v1''''' dt + C4
v1'''   = Integral v1''''  dt + C3
...
v1      = Integral v1'     dt + C0

Ci = initial values

These are the definitions I work with.

Refering to Page 5 (y=v1)
http://home.arcor.de/janch/janch/_control/20071104-pd5 (pid)z1z2/

2,786423E-05 y''''' + 0,001135031 y'''' + ... + y = 1,00095
or equivalent
F(s) = (0,1227466 s + 1)^5


--
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