Question about PI control integration time

Andrew wrote:

The integrator has two main effects: it eliminates steady-state error
and it reduces stability. The linger you can comfortably wait for a
steady-state error to be removed, the stabler your system will be and
the less overshoot it will experience. The more overshoot you can
tolerate, the shorter the has to last. Ideal solutions exist on paper.
The real world is a compromise.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

http://www.controlguru.com/wp/p71.html
This is the closest example I could find on the net quickly but it
should be all you need.  The heat exchanger example is different
because its gain is negative whereas you gain will be positive but the
rest should apply.  Is your pressure proportional to the control
output when you are in manual? If so follow the example on the page.
There are other pages that explain how to do the different steps.  It
look like you found Tp on your own.  If you look down the page you
will see that Ti=Tp=3.75.  If you look some more you will see there is
a formula for the controller gain.

Peter Nachtwey

pnachtwey@gmail.com wrote:


The example Peter N. referred to above is good, and practical.  The
information you provided for your open loop test was incomplete.  You
must always provide four figures (as documented in my Control Engineer's
Training Manual, p96):
   1) steady-state gain (Kp) of your process = delta(measured
variable)/delta(manipulated variable) in any suitable engineering units
   2) theta = approximation of the dead time
   3) t25 = time when 25% of the response has been achieved
   4) t75 = time when 75% of the response has been achieved

Assuming that your plant is a first-order plus dead time (FOpDT)
process, we can use the following simple formulas to estimate the
plant's transfer function:
   a) tau = 0.91(t75-t25) (first-order time constant)
   b) theta = t75-1.386*tau  (first-order dead time)

You did *not* provide the steady state gain.  Hence, it is not possible
to calculate the controller gain Kc.

Using a typical plant response curve that shows steady-state after 15
seconds give a first-order time constant of 5.0 to 5.5 seconds.

Assuming that you obtained the Kp of the plant, then the following
simple formulas give you a starting point for your PI controller tuning
constants:
   i) Kc = 1/Kp  (say, Kp = 5, then Kc = 0.18)
  ii) Ti = tau   (say, tau = 5.5, then Ti = 5.5)

This is also the tuning that results from the "moderate response tuning"
in reference http://www.controlguru.com/wp/p71.html .

All the above assumes that you are working with an electronic (or
pneumatic) analogue controller.  The equations hold also if you are
using a digital controller with a loop sample time that is at least 10
times faster than the first-order time constant, or in your case 5.5/10
= 0.55 seconds or faster.

The process you need to go through to obtain a plant response is the
same for any plant, regardless of time constants and steady-state gains.
  In the process industry we see time constants that vary from seconds
to hours (large reactors, distillation trains, etc).  Once you get that
skill down pat, and you can tune controllers well, you will be in demand.
--
Regards / John C H (not JCH in Germany)

jch wrote:

Above should read
   i) Kc = 1/Kp  (say, Kp = 5, then Kc = 0.2)

When you ask questions about control loop tuning it would most helpful
if you describe the process in simple terms.  All we know is that your
plant has to do with a pump system.  The major first-order time constant
is quite small; in the order of seconds.  Perhaps you are measuring and
controlling a fast pressure process?

It is standard (and best) practice to show the plant as a "black box",
clearly identifying the controlled variable (Y), and the manipulated
variable (M) [in engineering units] like this:

          ----------          Where G(s) is a first-order plus dead time
   M(s)   |        |  Y(s)    process
-------->|  G(s)  |--------> G(s)=Y(s)/M(s)=Kp[1/(1+tau*s)]e^(-theta*s)
          |        |
          ----------

The process gain Kp is thus delta(Y)/delta(M).  It is customary to keep
the delta(M) as small as possible to avoid large process upsets, but big
enough to get an open-loop response curve that can be charted easily.  A
common starting value for delta(M) is in the 10% to 15% of range.

There are other tuning methods around.  Ziegler-Nichols and others
developed methods that rely on getting the plant into a sustained
oscillation.  I personally dislike that method.  The plant operator(s)
may not like you doing the Z-N method.

I often use a graphical method.  The graphs were developed (via
simulations) for "setpoint tuning" as compared to "load tuning" because
the tuning for these two cases differ.  From my graphs i get Kc = 0.2,
and Ti = 4.2.
--
Regards / John C H (not JCH in Germany)

See page 1 and page 2
http://home.arcor.de/janch/janch/_control/20070603-controldoc/


--
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

Where did this come from?  It isn't right.
Kp*(1+3.67*s) = Kc*[1+1/(Ti*s)]

Where did Ti=3.3 come from?  It works but it looks like you really
'tuned' this system instead of calculating the gains  You certainly
didn't use
Kp*(1+3.67*s) = Kc*[1+1/(Ti*s)]
You are confusing the rookies again.

see
http://lorien.ncl.ac.uk/ming/digicont/control/digital2.htm
I know I have posted the link above before in the last month or two.
http://www.jechura.com/ChEN403/15_DirectSynthesis.pdf
Both links will show you how to calculate the gains using direct
synthesis which is the method used on the www.controlguru.com site.

What does
ITAEcriterion = t*|e|*dt = 1.273 MIN = optimum
mean?  You certainly didn't use the minimum ITAE technique to tune
your system.  The minimum ITAE tuning is under damped so the response
would over shoot the set point.  Your response didn't noticeably over
shoot.  It is also doubtful that the minimum ITAE would return nice
numbers such as 3 or 3.3.

Peter Nachtwey

Peter:

Where did this come from?  It isn't right.
Kp*(1+3.67*s) = Kc*[1+1/(Ti*s)]


Jan:

Text interpretation. That is PI controlled 1st order process system. Any
other interpretation is welcomed.

Citation:

Peter:

Where did Ti=3.3 come from?  It works but it looks like you really
'tuned' this system instead of calculating the gains


Jan:

It is ITAE tuned/calculated. You can do it manually or automatically.


Peter:

You certainly didn't use Kp*(1+3.67*s) = Kc*[1+1/(Ti*s)]


Jan:

I did it exactly.

Citation:

Peter:

You are confusing the rookies again.


Jan:

If rookies don't know this then they have spent too much money for
education.
http://home.arcor.de/janch/janch/_control/20070603-controldoc/
Page 1.


Peter:

What does ITAEcriterion = t*|e|*dt = 1.273 MIN = optimum mean?


Jan:

http://sts.bwk.tue.nl/7y500/readers/.%5CInstellingenRegelaars_ErrorCriteria.pdf
Eq. 2, 3rd one


Peter:

You certainly didn't use the minimum ITAE technique to tune
your system.


Jan:

Why not? It is appropriate for the task and is just one more integration.


Peter:

The minimum ITAE tuning is under damped so the response
would over shoot the set point.  Your response didn't noticeably over
shoot.


Jan:

ITAE shows time related quality under defined conditions: All
simulation/tuning must be done such that anything is in a controlled range
(0.2...1). I did not and you should not exceed the limits and suppress
overshoots if necessary! I suppressed it.

See v2 in:
http://home.arcor.de/janch/janch/_control/20070603-controldoc/
Page 2 (v2<=1 !!!)

ITAE definition:
http://sts.bwk.tue.nl/7y500/readers/.%5CInstellingenRegelaars_ErrorCriteria.pdf

If using ITAE you reduce 'indefinite solutions' to just 'defined solution',
the optimum! I used e=w-v1. If I also want to reduce/control forces I could
use

Integration [t*|w-v1|+|v1''|]*dt =>MIN

e is time weighted, acceleration is weigthed but not time weighted. Too high
acceleration can damage a plant. If much money is involved then anything
must be defined/considered in advance. It is not only the process variable
that counts.

Remember the jerk filter used in PD3(PI) controller. Don't let PD3(PI) work
without (jerk) filter. If the actuator had enough power then the forces
would be very high.


Peter:

It is also doubtful that the minimum ITAE would return nice
numbers such as 3 or 3.3.


Jan:

I rounded them. Changes of Kc and Ti have less influence at optimum.


--
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

This is not ITAE optimal soulution!

Your ITAE index is J=1.27.
Your controller output is 300psi. The plant is destroyed (RIP).
Congratulations.

Use this settings to get ITAE index J = 0.32.
Kc=10, Ti=0.6
You can see your solution is not optimal in ITAE sense.

I can get even better ITAE index but completely unrealistic in practise
just like yours is.

You use software to get results but you don't understand them.
You don't know control theory nomenclature so you are misleading people.=


You haven't shown us not even once your calculation
so my conclusion is that you are having fun with some software and than
you are trying to find on the internet what does your results mean.

You are not serious oponent to debate,
you must be kid or self-taught person or some very mentally-ill former  =

engineer.
You are wasting my and others time and you will go to troll room (PLONK)=
.

-- =

Mikolaj

This is not ITAE optimal soulution!

Your ITAE index is J=1.27.
Your controller output is 300psi. The plant is destroyed (RIP).
Congratulations.


My ITAE is based on 0.2...1 and is "relativ".

My "process value" is 60 psi in a range of 0...100 psi:
http://home.arcor.de/janch/janch/_control/20070603-controldoc/
Controller output is max. 1 (note max 1!). Look at it.

Thanks for the PLONK! For the next 300 years, I hope.

Mikolaj is saying your indexd is 1.27 and is not optimal.  You
misunderstood what Mikolaj was saying.

What do you mean by that?  Your should read this thread.
http://groups.google.com/group/sci.engr.control/browse_frm/thread/64d35b9b6b8c3768/c368d3e56217b8d3?lnk=gst&q=Dave+Y+ITAE&rnum=1#c368d3e56217b8d3
Dave Y and I basically proved that the origanal coefficients for ITAE
were wrong.  I don't know if someone has a better set of coefficient
other than those that Dave Y and I posted.
JCH, by now you should realize that I wasn't asking about minumum ITAE
because I don't know what it is.  I was asking to get you to explain
discrepencies that Mikolaj has pointed out.In most case I questions
just to get you to think about what you are doing wrong.  I could be
far more blunt.  I think every know this except you.

psi:http://home.arcor.de/janch/janch/_control/20070603-controldoc/

I agree with Mikolaj, you never show your calculations so it is hard
to explain exactly where your are wrong.   I do know that you don't
know how to tune PID because you always make them look bad.   I know
you really don't understand feed forwards.  Your feed forward
coefficients for velocity, acceleration and jerk feed forwards are not
right.   JCH must always use a target filter.  I dont know why.   I
don't need one for simple applications.

Were is Jerry, Tim and PaulM?  I know Tim and Jerry have challenged
the fractor, why not JCH?.  There are many more people that just lurk
or only participate in the non-technical threads.  Study some so you
can challenge what doesn't seem right.    I have thought for years
that there is no intelligent life on this newgroup.  Don't let JCH
prove me right.

JCH, be sensible and show your calculations and answer the questions
asked about your posts.  If you don't want to then don't post links to
your site that you are unwilling to back up.  You still didn't answer
why your response isn't underdamped like a minimum ITAE response
should be.  WELL?

Peter Nachtwey

[...]


Subject: Question about PI control integration time

Did Mikolaj show calculations (plots)!?
Did YOU show calculations (plots)!?

I DID! All the plots are '100% calculated' according to the task defined.

The formulae and start conditions are input and the calculation is done by a
program. You see the well defined results (v1, v2 plots) in time domain.

See again: definitions and results (plots)!
http://home.arcor.de/janch/janch/_control/20070603-controldoc/

Page 1 definitions!
Page 2 are results! (plots)

v1 = process value
v2 = controller output

If you solve this very easy task under the same conditions you will have the
same result (plots) because it's mathematics.

Internal calculation ranges:

v1 = 0.2...1 (for 0...100 psi)
v2 = 0.2...1 (for 4... 20 mA)

See again: definitions and results (plots)!
http://home.arcor.de/janch/janch/_control/20070603-controldoc/

Do not exceed (note: not exceed!) 0.2...1 and make the best of it and show
it.

I LIMITED the proportional step to max. v2 = 1
                                   ^^^^^^^^^^^
Set Kc = 3 and Ti = 3.3 seconds as example. The primary task is not to
exceed limits if testing.

See definitions for Kp, Kc, Ti

See again: definitions and results (plots)!
http://home.arcor.de/janch/janch/_control/20070603-controldoc/

Just use these data and we will have the same results. Show your plot! And
you can prove that I am right, not wrong.

What are you thinking I am doing? Just painting the plots?


--
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