Curious about some links posted for "Understand PID control"

Peter Nachtwey wrote:

Yes.  I would rather word it that damping is determined by where the
poles happen to fall, because pure pole-placement design will lead you
down the garden path -- successful pole-placement design is done when
you have a pretty good grasp of where the poles would go in a robust
system; forcing them to that safe place works, but attempting to place
them arbitrarily usually results in pain.


Yes, and there are good ways to implement integrator saturation,
including ones that will give you a controlled overshoot for fast
overall settling (if overshoot is allowable).

If you only know how to fix integrator windup by simple integrator
limiting then I-PD will be better with regards to overshoot.


I'll defer _that_ question to Jerry.

I will comment, though, that sometimes I'll put derivative into my
forward path and sometimes I won't, depending on the problem at hand and
the expected nature of the command (i.e. if I know the command won't
have any steps I'll be more likely to include the derivative).


For equal closed-loop poles the linear response to a _disturbance_
should be exactly the same; it's only the response to a _command_ that
will differ because of the zeros in the forward path.

Usually if I'm worried about integrator windup I limit the integrator
state based on the proportional gain and error, so the integrator state
is no larger than that which would cause the output to saturate with
just the proportional term thrown in -- even if that means having the
integrator state oppose the proportional term.

I have used an I-PD-like approach once, however, when I needed to
control a plant to hold four different outputs to four different maxima
(one "real" controlled variable and three things that needed to be kept
in their safe regions).  In this case I had four PD filters followed by
a minimum function, so any of the four could pull the output integrator
down but they all had to be below target for the output to rise.  The
system worked very well at transitioning control from one output
variable to another, although for other reasons it went into production
with only two output variables under such control.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Yes, I opt for over damped or critically damped.  The point is the poles for
both
PID and I-PD are placed the same way.  Therefore the advantage #1 is not
valid.


Yes, but a PID that is well designed will not use simple intetgrator
limiting.
I claim that advantage is only partially valid because well designed PIDs
will
handle saturation just as well


You have brought up a good point that Jerry's link doesn't.  It depends.
Jerry's link just says I-PD is better.  One must use the right PID form
depending on the application.


I agree and that contridicts the advantage in Jerry's link.

 it's only the response to a _command_ that

Again I agree.


The example your article integrates error.  Do you really integrate error
in your applications?

 so the integrator state

That is another fine point that isn't listed in the links.  Not even in your
link.


That is a good example of using the right tool for the application.  I worry
about the affect those.
that say there is only one best form of PID, have on rookies.
Those that say there is only one best form of PID or the derivative gain
doesn't
need to be used make me cringe.  It perpetuates the ignorance.

I think Jerry needs to update or replace that Phelan document.  I-PD is
good.
I-PD just isn't best in all cases.  I-PD is just one of many tools.

Peter Nachtwey

Tim Wescott wrote:

   ...


We do many things with software that aren't feasible with hardware. What
I dubbed I-PD -- not Phelan's term -- is easy to implement with op-amps
(and diodes for the integrator limiter).


An unadulterated PID follows a ramp with constant error after the
start-up transient, The error term is has the form of the step response
of an R-C low pass. [1 - exp(-xt)] An I-PD responds to a ramp with the
same initial form as a PID, but the error decays asymptotically to zero
even while the ramp command continues, as if the error were capacitively
coupled.


An I-PD controller with integrator limiting converts a step input (which
no physical load can follow) to a ramp at the load that uses the maximum
available power (or allowed mechanical load, etc.).

   ...

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

 but the error decays asymptotically to zero

No. What would make the error decay asymptotically to zero during the ramp?
The type (0,1,2) of the CLTF doesn't change just because one changes the
controller from a PID to I-PD.


? I don't believe it, at least not a linear ramp.  As the error decreases
the rate at which the integrator term increases slow down so the response is
as you pointed out above.  The response is exponential.  If you want to call
that a ramp then fine.


?  I don't believe it.  Doesn't it depend on the integrator gain and the
error?  If the error isn't very big the output will not use all the
available power.

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Peter Nachtwey

Peter Nachtwey wrote:

Let's assume a velocity servo, which is what I blocked out in an earlier
post. (A position servo would require an additional feedback term,
raising the order of the system.) The integrate time constant is much
less than a PID's slow-acting "reset". When a step is commanded, the
integrator quickly ramps up to the limit. During integrator ramp-up, the
motor's acceleration ramps up too, but it will have not picked up much
speed by the time the power is limited. That means that there will be
very little feedback at the pseudo-derivative summing node, and almost
all of the integrator's output is effective drive. As the motor picks up
speed, the feedback at the PD node cancels some of the drive, but the
bounds network compensates by allowing the integrator output to
increase. Ignoring the details of the linear region, By the time the
motor reaches the commanded speed, the output of the integrator is
exactly what is needed to keep it there at whatever load is on it. The
integrator stops ramping, and there is no overshoot. With a narrow
proportional range, the system behaves like a bang-bang servo, but no
element of it saturates.


The integrator has a very short time constant compared to PID. Even a
small error would drive the output close to saturation in a short time
if the bound arrangement allowed it. Remember: the bound measures what
happens at the actuator and adjusts the integrator accordingly. Because
of that, in response to a step, the feedback signal and the integrator
output rise together to maintain maximum drive at the actuator.

Classical servo equations are normalized, usually to the integrator's
time constant. With such curves, PID and I-PD look similar. The
difference becomes evident in real time. A critically damped I-PD system
reaches its setpoint in about 1/3 the time it takes a 4% overshoot PID
to settle.

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

<snip>

I understand that but Phelan's point 3 says response to a ramp.  Not a step
input.


What is a classical servo equation?  What is normalized?


NO! Not possible because the PID has a higher bandwidth.  A properly
designed PID will handle saturation with out overshoot just fine and ramp
into the setpoint just a few milliseconds faster than the I-PD.  WHY,
because the PID's control output will jump directly to saturation if the
error and gain is large enough and a I-PD's control output  will ramp up
given the same gains.  Therefore the PID will respond just a little faster.

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Peter Nachtwey wrote:

Not Phelan's point: my paraphrase. I may have had in mind the action
when a higher derivative is included, but I don't think so. I don't have
hos book or my notes handy.


Response time measured in units of integrator time constant.


What do you base that on?


That's hard to do with a strictly analog approach. With digital tweaks,
approaching I-PD performance isn't too hard. but I find it /ad hoc/.


I think you're confused. An I-PD's integrator typically had a shorter
time constant than a PID's. As for saturation, what matters is that the
output saturate rapidly. What happens to the integrators is (as the
Chinese often say to us) an internal matter.


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

I measure response in terms of a desired characteristic equation.   The link
I posted
shows what I mean.
ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20t1p1%20i-pv%200.pdf
Tim Wescott has Mathcad and can verify the .pdf.   I will send him the .mcd
file if he wants.


I was asking you what you based the advantages in your link on and I haven't
got an
answer that I like yet, but I will go first anyway.  See the link.   Notice
that the I-PV or I-PD does not follow a ramp without error.  Notice also
that the response to a step is faster with a PID than the I-PV.
Compare the ITAE values.

DO YOU SEE ANY OVER SHOOT?  Notice the implementation is very simple, not ad
hoc.
Doesn't it make sense that the control system with highest bandwidth will
respond to high frequency moves faster?  I don't see why this is so
difficult to see even for a rookie.


Who said the control is limited to analog only?   I have never used an
analog controller that I remember.


What is ad hoc?   Check out the link  Notice the PID and I-PV implementation
is very simple and very similar except for necessary differences.   One can
see the PID out performs the I-PV making the same change in the set point
using the same gains.   Isn't it obvious that a PID with zeros will have a
higher bandwidth than the I-PV so the PID will respond faster to a step
which has very high frequency components?


I am not confused.  The link should show that.


Why?  Why not the same as in my example.  If you make the integrator time
constant faster in the I-PV
than in the PID it is no wonder you think the I-PV is faster but that is not
a apples to apples comparison.


The PID saturates faster and gives it a few millisecond head start to the
set point.  I hope everyone can see that.


I know the example in the link above is a position system an not a
temperature system.  If you want a temperature example then see this:
ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20TempPID.pdf

Peter Nachtwey





¯¯¯¯¯¯¯¯¯¯¯¯

Peter Nachtwey wrote:

   ...


I don't see any PID overshoot. How is that possible? When a PID has
settled on the setpoint, its integrator, in the absence of a load, is
zero. The integrator charges -- to saturation if the command is large --
  before the setpoint is reached, and must be discharged by an error
signal of the opposite sense (i.e. overshoot) before steady state is
achieved. The overshoot is always the same fraction of the linear
region. Is the linear region in your simulation zero?

   ...

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

Jerry Avins wrote:

More thoughts on the issue. As far as I can see, you have chosen the
same integrator gain for both the PID and I-PD simulations. I-PD's
typically run with 3 to 10 times the K1 that would make a PID unstable.

For an interesting comparison, simulate the effects of step and ramp
loads on both systems.

I was wrong about following a ramp load. What I wrote applies to a
system a degree higher, one with a second pseudo derivative (needing one
differentiation) added.

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

I know.  Customers don't like overshoot.  The point is that that the PID
 and I-PV will not overshoot if the gains are chosen properly and one
 makes just the smallest effort to manage the integrator windup.


You are thinking within the narrow confines of analog circuits.  This is
 a digital implementation that 'discharges' the integrator when the set
 point jumps and then the integrator windup up to 0.  I know it is
 backwards from what you are familiar.  You can see the integrator limit
 is simple.  It uses a technique that Tim Wescott mentioned above.
I called it a fine point.


Again you are thinking analog.  If I implemented a PID, like Tim
 Westcott does in his article, it would overshoot as you expect.  The
 point I am making is that PID doesn't need to overshoot.  This is just a
 implementation detail.  My example is very simple and requires no more
 code that Tim's PID implementation in his article


I added more to the .pdf
ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20t1p1%20i-pv%201.pdf
I added the PID response to a ramp and both PID and I-PD response to a sine
wave.

Also,  NO ONE CAUGHT THE NON-MINIMUM PHASE PROBLEM WITH MY PID IN THE MY
PREVIOUS .PDF!!!  Notice the second feedback gain, the velocity feedback or
derivative gain in the K array,  is negative.   One can see from the
equation for the velocity feed back gain that it will be negative if lambda,
the desired corner frequency of the 3 poles of  the desired characteristic
equation, is too low.   I increased the desired pole locations in the second
link above.  The I-PD will work well with negative feedback gains a PID will
not.  Sometimes negative feedback gains are required for critically damped
response.  Odd but it works.  I asked if anyone has tried negative feed back
gains a couple months ago and no one responded.

Peter Nachtwey

Peter Nachtwey wrote:

-- snip --


We were afraid (well, I was at least).

I recently had to convince a client that they _really_ didn't want to
fix a problem with a control rule change; they _really_ wanted to change
their plant, instead.  I wasn't making much headway until I pointed out
that the compensator had unstable zeros in it -- then they listened.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

proclaimed to the world:


I worked with negative gains once but it was a fix for a bad control
design to take out instability. In this case the system needed to have
a fast response but there was a lot of hysteresis in the final control
element. Negative feedback was used at the element to the main
controller to dampen oscillation. It worked.

Be well,

HoP

The preceding message represents personal opinions
and/or advice that may prove incorrect or harmful. But then maybe not.
Feel free to disregard.

 ------- Words have no Warranty ------
 ------- No View without Merit  ------

I responded to most of Peter's letter from home a few days ago. This is
a copy:

Peter Nachtwey wrote:

 > Newsgroups: sci.engr.control
 > Sent: Wednesday, November 09, 2005 6:44 AM
 > Subject: Re: Curious about some links posted for "Understand PID control"
 >
 >> Peter Nachtwey wrote:
 >>
 >>   ...
 >>> I was asking you what you based the advantages in your link on and
I haven't got an
 >>> answer that I like yet, but I will go first anyway.  See the link.
Notice that the I-PV or I-PD does not follow a ramp without error.
Notice also that the response to a step is faster with a PID than the I-PV.
 >>> Compare the ITAE values.
 >>>
 >>> DO YOU SEE ANY OVER SHOOT?  Notice the implementation is very
simple, not ad hoc.
 >>> Doesn't it make sense that the control system with highest
bandwidth will respond to high frequency moves faster?  I don't see why
this is so difficult to see even for a rookie.
 >>
 >> I don't see any PID overshoot.
 >
 >
 > I know.  Customers don't like overshoot.  The point is that that the
PID and I-PV will not overshoot if the gains are chosen properly and one
makes just the smallest effort to manage the integrator windup.
 >  >How is that possible?
 >
 > I think the formula is simple enough.   The integrator is limited to
 > 10 volts - P and D terms on the high side and -10 volts - P and D
 > terms on the low sides.  Simple.
 >> When a PID has settled on the setpoint, its integrator, in the
 >> absence of a load, is zero. The integrator charges -- to saturation
 >> if the command is large --  before the setpoint is reached, and must
 >> be discharged by an error signal of the opposite sense (i.e.
 >> overshoot) before steady state is achieved.
 >
 >
 > You are thinking within the narrow confines of analog circuits.  This
 > is a digital implementation that 'discharges' the integrator when the
 > set point jumps and then the integrator windup up to 0.  I know it is
 > backwards from what you are familiar.  You can see the integrator
 > limit is simple.  It uses a technique that Tim Wescott mentioned
 > above.
 > I called it a fine point.


I call it an ad hoc solution, the kind I implemented also with analog
servos. That was easier to do once transmission gates became available.
With bounded I-PD, it is unnecessary.

 >> The overshoot is always the same fraction of the linear region. Is
 >> the linear region in your simulation zero?
 >
 > Again you are thinking analog.  If I implemented a PID, like Tim
 > Westcott does in his article, it would overshoot as you expect.  The
 > point I am making is that PID doesn't need to overshoot.  This is just
 > a implementation detail.  My example is very simple and requires no
 > more code that Tim's PID implementation in his article.


A classical PID, whether implemented as an analog or digital circuit
must overshoot. If it doesn't, then it's something else and should have
another name. "Encrusted PID" sounds derogatory. Do you have another
suggestion. (For God's sake, don't suggest "enhanced"!) Abraham Lincoln
once asked his cabinet, "If you call a tail a leg, how many legs does a
sheep have?" He replied to the answer "five", "No; four. Calling a tail
a leg doesn't make it one." Calling a controller with no overshoot a PID
doesn't make it one. Don't confuse the rookies. :-)

 > The point that I am trying to make is that people on their sites and
 > statements are made without really knowing if they are true.   Rookies
 > that read your document will think I-PD is better than PID which is
 > depending on the application.   Even without the simulation one should
 > know by intuition that PIDs have a higher bandwidth and therefore will
 > respond faster because of the zeros.

Intuition is wrong, then. The I-PD integrator gain is much higher.

I will fix the ramp-following statement. I don't see how you come to
your bandwidth conclusion. Try a velocity servo with a real motor/tach
combination both ways, using the highest practical Ki in each case.

 > I see the same old myths repeated over and over again.  I see people
 > saying this or that form of PID is best when they are just tools.
 > The control engineer needs to know which tool works best in a
 > particular application. Our products use both PID and I-PV or I-PD.
 > Both have their advantages.


I coined I-PD to mean integral/pseudo derivative. What does I-PV mean?

 > BTW, I design digital motion controllers.   I have seen a motion
controller that uses time constants to express integrator or derivative
gains, but the ones I have seen are all digital.


I agree that TC is an odd way to express it. If you post a copy of this,
I'll add this response.

Jerry

P.S. There is another way to look at the "pseudo derivative" inner loop.
Consider a simple velocity servo. A good PM or separately excited motor
is quite good on its own. (It's nearly perfect with negative
compounding.) Reducing the excitation actually brakes the motor. Driven
by a current source with velocity feedback, it retains that
characteristic. Driven by a voltage source with velocity feedback, it
becomes so highly damped that the inertia is nearly swamped by the
electrically created viscous damping, reducing the order of the system,
as seen at the summing point into the power amplifier, by one. Such a
system is stable with an integral-only controller for small signals, and
also stable for large signals if the integrator is bounded by conditions
at the output.

Jerry
--
Never ascribe to malice what might be be ignorance, stupidity, or sloth.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯





My own P.S.: I don't see the integrator waveform in the I-PD plots. The
power amplifier saturates quickly, but the integrator stops charging as
soon as it does. As the feedback at the PD node increases with the plant
ramps up, the integrator output rises just enough to keep the power
amplifier at the edge of saturation (or wherever the output limit is
set). Integrator and plant ramp up together until the system enters the
(usually narrow) linear band. I'm not sure of the block diagram of the
systems you're modeling. What is I-PV? Are we discussing the same devices?

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

I don't care what you call it.  You can see it works and that is what is
important to me and my customers. The limiting technique is simple and only
a little different from Tim Wescott's in his article.  Is Tim's PID
integrator limiting ad hoc too?  I have other schemes that are far more
complicated and don't work as well.


Yes, in position or final value form.  Not in velocity or incremental form.
(OT, I hate the terms position and velocity form because it has nothing to
do with position or velocity.  This too confuses the rookies.)

 If it doesn't, then it's something else and should have

PID with dynamic integrator limiting.   Tim's example uses fixed integrator
limiting.
I think that is simple enough and easy to remember.


WAIT, I am using the same integrator gain for both the PID and I-PD in my
.PDFs.
I am initializing Ki once and using the same Ki through out the .PDF
I am doing a apples to apples comparison so the CEs ( characteristic
equations ) have the poles in the same place.  I can easily change the
lambda so the poles are far to the left ( higher corner frequency ) and
compare them.  THE ONLY THING THAT IS DIFFERENT IS THE PID HAS ZEROs.   This
calls for another .PDF of bode plots for PID, PI-D and I-PD.  This will show
the difference in the response.

ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20T0C1%20Bode.pdf

Many PLCs use the PI-D form.  AB has for a while

This shows the only difference between these forms is where the P and D
gains are applied in the loop.  If the P and the D gains are in the forward
path the bandwidth will be higher.  It has NOTHING to do with integrator
time constants or gains.  The difference is in the zeros.
Even the rookies can see that.


Just check out the link above.  I feel like I should write a book.  "Peter
Ponders Practical PID or what every ad hoc name you call it"


The problem you are seeing is the noise and quantizing errors being
amplified by the zeros.  If one needs faster response be prepared to pay for
better, higher resolution and less lag, feedback devices.  It is these two
problem that discourage people from using the derivative and higher order
gains.


I consider them to be almost the same.  I-PV is just the state feedback
term.  Each state, position, velocity, acceleration and jerk, can be used
for feedback and have its own gain.   The simple case I-PV or I-PD just
means the derivative or velocity feedback gain in feedback path.  The
derivative in the feedback path means one is actually taking the derivative
of the PV.   In some case one could read the position from a take rather
than differentiate positions.  It is a fine difference.  From a control
theory point of view I don't see where there is a difference between I-PD
and I-PV.  It is a implementation detail.

I am also surprised no one has commented on the way I calculate gains
symbolically.
Matlab is good for getting answer quickly with out understanding the
process.  Mathcad is better when used symbolically.  The symbols make it
much easy to see the relationship between the different variables.

So what do the equations say about the relationship between the integrator
and proportional gain?

Peter Nachtwey
Delta Computer Systems, Inc.

Peter Nachtwey wrote:



To compare like things, use, say, half the integrate gain that just
induces instability. When comparing the performance of a Ferrari and an
18 wheeler, would you have them both carry 50 tons?


What I dubbed I-PD has only an integrator in the forward path and
proportional and pseudoderivative feedback in the reverse path within
the proportional band. Outside that, the feedback is primarily via the
bounds circuit.

Recall the block diagram I posted on 11/6:



                        +<<<<<<-bounds circuit-<<<<<<-+
                        |                             |
           |¯¯¯¯¯|  |¯¯¯¯¯¯¯¯|  |¯¯¯¯¯|  |¯¯¯¯¯¯¯¯¯|  |
  Command->|  -  |->|integral|->|  -  |->|power amp|->+->>> actuator
           |_____|  |________|  |_____|  |_________|  |
              |                    |                  |
              |                    +<<<<<<<-Kd-<<<<<<-+
              |                                       |
              +<<<<<<<<<<<<<<<-Kp-<<<<<<<<<<<<<<<<<<<-+





   ...


Did I mention a problem? There is no additional lag when using a
pseudoderivative because there is no differentiator. To augment the
block diagram for a position servo, replace the proportional velocity
with proportional position feedback and increase the order of the
circuit by feeding both position and velocity to the pseudoderivative
pseudoderivative node.

   ...


How about a block diagram? I believe we are discussing different
circuits as if there were no difference.


I'll have to hunt up my notes.

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

When the goal is to move 50 tons, yes.  The Ferrari loses.


Yes, that is why your ciruit has no zeros and it doesn't have the band width
of a PID
Can't you see that.  I do think your circuit is a little different from
mine.

My I - PD

                                                                      G*alpha
SP ------>+-----Ki/s -----+-------+------   ---------  -----+------- PV
                 |                        |             |             s +
alpha          |
              -PV                   -Kp     -Kd*s
|
                 |                        |             |
|
                 +--------------+--------+------------------------+
My PID
                            +------Ki/s------+
                            |                          |           G*alpha
SP-------+-------+----- Kp-------+----  -----------   -----+---------PV
               |            |                          |          s + alpha
|
            -PV         +------Kd*s----+
|
              |
|
              +--------------------------------------------------+

In the denominator my transfer functions you see ( Ki/s + Kp + Kd*s)
Your I-PD would have (Kp*Ki/s+Kd*s) I think, It is hard to figure out what
you really meant by your diagram.

What is the transfer function?  I can tell by that and it is easier to
right.
You should have been able to compare your I-PD with my version of I-PD
to check for a difference.

What do you expect to see?



A differentiator doesn't cause lag.
   ...

Yes, I think so but doesn't make any difference to the statement about PID
overshooting.
 It doesn't need to if you limit the integrator correctly.


Check this site out.  Mike Borrello has two technotes that apply.  One is
about the pseudo derivative and the other is about integrator windup.  In
the pseudo derivative article he say the characteristic equations of the PI
and what he calls PDF (Pseudo Derivative Control)  One can see the only
difference is in the numerator which is what I am saying.  Now he, like you,
says the PI overshoots which it will unless using only a simple integrator
limit..   See his technote about  Control windup.  There he mentions
limiting the integrator the same way I do.
http://users.adelphia.net/~maborrello/

Peter Nachtwey

Jerry Avins wrote:


- lots of back and forthing snipped -


I think the real item of interest here is that Peter's system is using a
form of integrator bounding that ends up working the same as Jerry's
circuit.  The difference is that Jerry's way is easier to implement but
more obscure in its operation for a beginner.  Peter's way looks, with
the exception of a rather complicated integrator limit, like a PID that
you may see in a textbook.

I suspect that the two forms are essentially, if not exactly, the same
behaviorally because of the way that Peter's method limits the integrator.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Tim Wescott wrote:

I'm tempted to believe that, but if Peter's PID-style integrator has the
same time constant as a classical PID integrator, then it's not
equivalent to Phelan's. By time constant, I mean the time constant of an
equivalently performing op-amp integrator; namely, the R-C consisting of
the input resistor and the feedback capacitor.

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