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steve sulis
science forum beginner
Joined: 30 Sep 2005
Posts: 3
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Carl
science forum Guru Wannabe
Joined: 02 May 2005
Posts: 113
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 Posted: Thu Oct 13, 2005 5:16 am Post subject:
Re: Ever Feel Like This When talking to your Equipment Rep?
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"Bob" <robert.rice@gmail.com> wrote in message
news:1128102788.389236.48930@g14g2000cwa.googlegroups.com...
ROFLMAO... Tears in my eyes.
Thanks,
Carl |
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Tim Wescott
science forum Guru Wannabe
Joined: 03 May 2005
Posts: 292
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| Posted: Fri Dec 30, 2005 11:32 pm Post subject:
Re: PID control
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Varun Aggarwal wrote:
| Quote: | Dear Sir
I saw the Ziegler-Nichols PID tuning method on your site. I have a
question regarding the same.
If I am controlling a second order delay-less plant, it will never
give sustained oscillations with P control. Then how do I find the
ultimate gain?
Does ZN method not work for second order plants? Is there some other
method to tune the PID for second order plants?
Kindly let me know
Thanks
Varun
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I'm posting this to the sci.engr.control newsgroup. Please direct any
further comments there -- not only will your interesting question be
posted for the benefit of all, but you'll get a richer variety of
opinions than just mine.
I don't have any information on Z-N tuning on my site.
If your plant were truly 2nd order with no delay then yes, you could
increase P to infinite values and you'd never see oscillation. In the
real world you don't have to worry about this, because when you push it
fast enough anything will show higher order behaviors.
Ignoring that issue for the moment, you can extract plant information
using the plant's step response, then doing some curve fitting to get
the various parameters you need for Z-N tuning. It's called "open loop"
Z-N tuning.
If you're serious about this, check out Astrom-Haggerlund tuning -- Karl
Astrom is one of the giants of control theory (at least practical
control theory). He was unsatisfied with the tunings he got using Z-N
because it often results in an under damped system. He and Haggerlund
came up with a similar method that they claim works better than Z-N, and
at least gives more conservative results. If I had to use one or the
other I'd use Astrom-Haggerlund. Before I used either I'd want to use a
system identification/structured design approach.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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Peter Nachtwey
science forum addict
Joined: 06 Nov 2005
Posts: 68
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| Posted: Sat Dec 31, 2005 1:11 am Post subject:
Re: PID control
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If the system is truely second order lile this Gain/((tau1*s+1)(tau2*s+1)),
then one can easily calculate the gains K, Ti and Td to get the desired
response if one know Gain, tau1 and tau2.
Peter Nachtwey |
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Curtis
science forum beginner
Joined: 18 Jun 2005
Posts: 10
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| Posted: Sat Dec 31, 2005 1:53 am Post subject:
Re: PID control
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I don't have near the training you guys have, but I thought that
anything over a first order process showed some "inherent deadtime",
and therefore would oscillate. I thought that only the first order
process with no deadtime would not oscillate...
curtis |
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Tim Wescott
science forum Guru Wannabe
Joined: 03 May 2005
Posts: 292
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| Posted: Sat Dec 31, 2005 6:10 am Post subject:
Re: PID control
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Curtis wrote:
| Quote: | I don't have near the training you guys have, but I thought that
anything over a first order process showed some "inherent deadtime",
and therefore would oscillate. I thought that only the first order
process with no deadtime would not oscillate...
curtis
Even a first-order system has some delay. In a linear system the delay |
shows up as a certain phase shift vs. frequency relation. A 2nd-order
non-minimum phase system approaches the requisite 180 degrees of phase
shift asymptotically at the same time that it's gain approaches zero.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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John Shaw
science forum beginner
Joined: 25 Mar 2005
Posts: 29
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| Posted: Mon Jan 02, 2006 7:45 pm Post subject:
Re: PID control
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Assuming that a control system is digital (a good assumption these
days) with a finite scan rate (usually 4 times/sec or slower) there is
an inherent time delay of 1/2 the scan time.
A first order process will have sustained oscillations if the gain is
high enough. If the single lag time constant is not much different from
the scan time, the gain does not have to be very large.
A second order system with two equal lags and the small scan time will
often have sustained oscillations at a moderate gain.
There are certainly loops that will not oscillate at even a high gain.
Then the ZN closed loop is not useful. The ZN open loop method may be
used if the "pseudo dead time" (the small lag before the point of
inflection in the response curve) can be measured. If that cannot be
measured, ZN (and other tuning methods) cannot be of much help. A high
gain may be used, with the limitation on the gain being the effect of
any noise or sepoint adjustments on the valve (not wanting the valve to
be always moving back and forth).
John Shaw
www.jashaw.com/pid
Tim Wescott wrote:
| Quote: | Curtis wrote:
I don't have near the training you guys have, but I thought that
anything over a first order process showed some "inherent deadtime",
and therefore would oscillate. I thought that only the first order
process with no deadtime would not oscillate...
curtis
Even a first-order system has some delay. In a linear system the delay
shows up as a certain phase shift vs. frequency relation. A 2nd-order
non-minimum phase system approaches the requisite 180 degrees of phase
shift asymptotically at the same time that it's gain approaches zero.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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Jerry Avins
science forum Guru
Joined: 03 May 2005
Posts: 534
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| Posted: Mon Jan 02, 2006 11:00 pm Post subject:
Re: PID control
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John Shaw wrote:
| Quote: | Assuming that a control system is digital (a good assumption these
days) with a finite scan rate (usually 4 times/sec or slower) there is
an inherent time delay of 1/2 the scan time.
|
That's a good measure of the average delay is processing time is small
(and at 4 scans/sec it will be). The peak delay can be as much as a full
scan time, and sometimes that matters.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Peter Nachtwey
science forum addict
Joined: 06 Nov 2005
Posts: 68
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| Posted: Tue Jan 03, 2006 12:23 am Post subject:
Re: PID control
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"Tim Wescott" <tim@seemywebsite.com> wrote in message
news:X_udndG66JdCvSveRVn-tw@web-ster.com...
Define delay. The last time I look at first order response it was something
like 1-exp(-a*t).
Even at time t=0+ there is some response. I suppose that even an ideal RC
circuit has some dead time if you consider the dead time due to the speed of
electric current.
Peter Nachtwey |
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Tim Wescott
science forum Guru Wannabe
Joined: 03 May 2005
Posts: 292
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| Posted: Tue Jan 03, 2006 12:50 am Post subject:
Re: PID control
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Peter Nachtwey wrote:
| Quote: | "Tim Wescott" <tim@seemywebsite.com> wrote in message
news:X_udndG66JdCvSveRVn-tw@web-ster.com...
Even a first-order system has some delay.
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Define delay. The last time I look at first order response it was something
like 1-exp(-a*t).
Even at time t=0+ there is some response. I suppose that even an ideal RC
circuit has some dead time if you consider the dead time due to the speed of
electric current.
Peter Nachtwey
Classic phase delay. I'm surprised at your example, because it's not |
restricted to 1st-order systems: any continuous-time, linear system with
a finite number of states has some response at time t = 0+; in a strict
sense "dead time" only has meaning in sampled-time systems and systems
with an infinite number of states (thermal, fluid flow and long strings
of coax all come to mind).
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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Peter Nachtwey
science forum addict
Joined: 06 Nov 2005
Posts: 68
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| Posted: Tue Jan 03, 2006 12:55 am Post subject:
Re: PID control
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"John Shaw" <john@jashaw.com> wrote in message
news:1136231116.179573.211680@f14g2000cwb.googlegroups.com...
| Quote: | Assuming that a control system is digital (a good assumption these
days) with a finite scan rate (usually 4 times/sec or slower) there is
an inherent time delay of 1/2 the scan time.
A first order process will have sustained oscillations if the gain is
high enough. If the single lag time constant is not much different from
the scan time, the gain does not have to be very large.
A second order system with two equal lags and the small scan time will
often have sustained oscillations at a moderate gain.
|
You should make it clear that it is the dead times and long sample times
that are the cause of the sustained oscillations, not the fact the system is
a first or second order system.
| Quote: |
There are certainly loops that will not oscillate at even a high gain.
Then the ZN closed loop is not useful. The ZN open loop method may be
used if the "pseudo dead time" (the small lag before the point of
inflection in the response curve) can be measured. If that cannot be
measured, ZN (and other tuning methods) cannot be of much help. A high
gain may be used, with the limitation on the gain being the effect of
any noise or sepoint adjustments on the valve (not wanting the valve to
be always moving back and forth).
John Shaw
www.jashaw.com/pid
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There are other forms of tuning that work very well without dead time.
Peter Nachtwey |
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Jerry Avins
science forum Guru
Joined: 03 May 2005
Posts: 534
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| Posted: Tue Jan 03, 2006 3:13 am Post subject:
Re: PID control
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Peter Nachtwey wrote:
| Quote: | "Tim Wescott" <tim@seemywebsite.com> wrote in message
news:X_udndG66JdCvSveRVn-tw@web-ster.com...
Even a first-order system has some delay.
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Define delay.
|
For a sampled system, or continuous-time?
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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John Shaw
science forum beginner
Joined: 25 Mar 2005
Posts: 29
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| Posted: Tue Jan 03, 2006 12:36 pm Post subject:
Re: PID control
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In some times of control this may make dead time unusual. However, in
process control there is almost always some amount of dead time,
although, fortunately, it is usually small. It comes from the transport
delay between where the action happens (mixing, heating, etc.) and the
measurement is made (analyzer, temperature, etc. in a pipe). Frequently
the dead time is too small to have any serious consequence other than
limiting the gain that can be used on an otherwise almost 1st or 2nd
order system.
Since control systems are usually digital, the scan time is also a
factor if the scan rate is slow. Scan rates should be as fast as
possible. It is rare for a scan rate faster than 1 sec. to produce a
dead time that has any serious effect, except on some fast loops such
as flow loops. Scan rates of 4 seconds or longer (I have seen people
use 10 seconds) will often limit the allowable gain.
My suggestion on configuration of a DCS is to buy many controllers,
putting only a few loops on each, and run all the loops fast. But I
used to work for a company that made DCSs ;-}
Tim Wescott wrote:
<snip>
| Quote: | restricted to 1st-order systems: any continuous-time, linear system with
in a strict
sense "dead time" only has meaning in sampled-time systems and systems
with an infinite number of states (thermal, fluid flow and long strings
of coax all come to mind). |
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John Shaw
science forum beginner
Joined: 25 Mar 2005
Posts: 29
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| Posted: Tue Jan 03, 2006 12:51 pm Post subject:
Re: PID control
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Peter:
| Quote: | You should make it clear that it is the dead times and long sample times
that are the cause of the sustained oscillations, not the fact the system is
a first or second order system.
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True. It is often the dead time, sample time, and other small lags in
the measurement and control system, and final control element (such as
variable speed drive) that has the greatest affect on the tuning.
| Quote: | There are other forms of tuning that work very well without dead time.
|
Sometimes the ZN open loop (and related methods) will work even when
sustained oscillations cannot be produced for the ZN closed loop. For a
simple two lag process with some oscillation that is not sustained at a
high gain, the gain can be reduced until 1/4 wave damping is achieved,
reset added based on the period of the decaying oscillations, and the
gain then further reduced.
Also, for simple one or two lag processes the tuning may be based on
criteria other than best control of the controlled variable. For
example, not long ago I reduced the gain of a well controlled pressure
loop (N2 header supplying gas to the top of a vessel) because the high
gain on the controller was causing upsets on the header, affecting
other parts of the process. |
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Tim Wescott
science forum Guru Wannabe
Joined: 03 May 2005
Posts: 292
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| Posted: Tue Jan 03, 2006 4:56 pm Post subject:
Re: PID control
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John Shaw wrote:
| Quote: | In some times of control this may make dead time unusual. However, in
process control there is almost always some amount of dead time,
although, fortunately, it is usually small. It comes from the transport
delay between where the action happens (mixing, heating, etc.) and the
measurement is made (analyzer, temperature, etc. in a pipe). Frequently
the dead time is too small to have any serious consequence other than
limiting the gain that can be used on an otherwise almost 1st or 2nd
order system.
|
I do high-performance motion control stuff, generally with full custom
hardware. There are enough other contributors to phase delay that "dead
time" per se is just not an issue.
| Quote: |
Since control systems are usually digital, the scan time is also a
factor if the scan rate is slow. Scan rates should be as fast as
possible. It is rare for a scan rate faster than 1 sec. to produce a
dead time that has any serious effect, except on some fast loops such
as flow loops. Scan rates of 4 seconds or longer (I have seen people
use 10 seconds) will often limit the allowable gain.
|
I think the slowest loop I've ever closed was at 10Hz, and that was a
"park" mode for a controller that used a few too many processor cycles
during an active move for the communications queue to get serviced
correctly. Aside from that the lowest one is a 60Hz loop on a
temperature controller (with deep integrators -- it closes at around 1
or 1/2 Hz).
| Quote: |
My suggestion on configuration of a DCS is to buy many controllers,
putting only a few loops on each, and run all the loops fast. But I
used to work for a company that made DCSs ;-}
That may not be bad advice -- in low production environments (like the |
one-off mill that you have to keep running) it's often better to solve
problems by throwing equipment at them rather than by spending lots of time.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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