On Saturday, June 27, 2020 at 1:10:00 PM UTC-4, Rick C wrote:
> On Saturday, June 27, 2020 at 11:31:58 AM UTC-4, David Brown wrote:
> > On 27/06/2020 16:53, pozz wrote:
> > > Il 27/06/2020 09:07, Rick C ha scritto:
> > >> On Saturday, June 27, 2020 at 2:12:21 AM UTC-4, Paul Rubin wrote:
> > >>> DJ Delorie <dj@delorie.com> writes:
> > >>>> Typical systems will limit the I term to some range so it won't go out
> > >>>> of bounds (or "wind up" as they say).&nbsp; It doesn't decay on its own, but
> > >>>> the system can cause its value to reduce to zero due to the Ki math.
> > >>>
> > >>> Do you mean it saturates (controller stops increasing it once it reaches
> > >>> some maximum)?&nbsp; Ok, but won't it reach the maximum and stay there?&nbsp; The
> > >>> Ki math I think was:
> > >>>
> > >>> &nbsp;&nbsp; I += Ki * err&nbsp;&nbsp; (note += not = )
> > >>> &nbsp;&nbsp; Vout = P + I
> > >>>
> > >>> So if the error is almost always positive, I will keep increasing.
> > >>>
> > >>>>> E.g. in a thermal system, you have to keep adding heat
> > >>>>
> > >>>> Yes, but the amount of extra energy you put in is constant in that
> > >>>> case,
> > >>>> and when the extra energy balances so that you stay at your set point,
> > >>>> the I term stops increasing, because the error is now zero.&nbsp; Drain more
> > >>>> hot water, the I term goes up.&nbsp; Insulate the tank, the I term goes
> > >>>> down..
> > >>>
> > >>> I'm imagining a tank starting with 20 degree water and you want to heat
> > >>> it to 50 degrees.&nbsp; Now the error is 30 degrees, or am I interpreting the
> > >>> wrong thing as the error?&nbsp; You turn on the heat and the error decreases
> > >>> as the water gets hotter.&nbsp; Eventually the water reaches 50 degrees and
> > >>> the error is 0, but I is some positive number (the integral of the error
> > >>> over the time it took to reach 50 degrees).
> > >>>
> > >>> Now you draw out some hot water, say to wash dishes; you replace it with
> > >>> more 20 degree water, so the system is say at 40 degrees now (error is
> > >>> 10).&nbsp; You turn the heat back on, but again, I keeps increasing.&nbsp; You
> > >>> might overshoot a little (say to 51 degrees, error is -1) in which case
> > >>> I might decrease a little until the tank cools back down.&nbsp; But really,
> > >>> the error will always be either positive or near 0.&nbsp; So I keeps
> > >>> increasing.
> > >>>
> > >>> I have to be missing something.&nbsp; Thanks.
> > >>
> > >> Yes, you are missing that the error variable can be both positive and
> > >> negative and by definition will balance the I variable to a value that
> > >> makes error return to zero.
> > >>
> > >> Once the water in your heater reaches the set temperature the error
> > >> will be zero but the heat sent to the water heater won't be zero.&nbsp; If
> > >> water is drawn the temperature will drop and the error term will go
> > >> positive again, the integral term will increase to restore the error
> > >> to zero, but at that point the heat into the tank will be higher than
> > >> before so that the temperature will rise above the set point and the
> > >> error will go negative.
> > >>
> > >> It is a mistake to think there will be more positive error than
> > >> negative error.&nbsp; Once the water is warmed to the appropriate
> > >> temperature the integral value will be back at the value it had before
> > >> cold water was added.
> > > 
> > > I think the wrong assumption was: as soon as the error goes to zero, P
> > > and I term should be zero as well, because we reached the set-point.
> > 
> > Yes - P goes to zero, and I /stabilises/ to a value of output that
> > results in a steady state.
> > 
> > If you overshoot - perhaps I has built up too much - the error term will
> > be negative, so the I term will decrease (but remain positive for a
> > while), while the P term will be negative.
> 
> Not "if" but just "how much" I will cause overshoot.  The I term is an oscillator with P as a damping factor.  
> 
> 
> > > For example, when I need to move the optical bar of a scanner to a
> > > certain position.
> > > As soon as the bar reached the wanted position, the power from the
> > > motors must be cut off immediately (because the bar doesn't move if the
> > > motors are off).
> > > This means that when the error is zero, P+I must be zero as well.
> > 
> > That works for some kinds of systems.  In other systems, you might have
> > gravity or a spring that means keeping the same position requires a
> > constant small force - you get that from the I term.
> > 
> > In this case, the I term will be zero at the perfect stable spot.  But
> > it could well be non-zero while moving - a P term alone rarely gets you
> > to the target.  (Imagine you are near the target.  The P term might give
> > too low a force to overcome the static friction and move the bar - then
> > the I term will build up until things are moving.)
> > 
> > > However this is impossible. P is really zero, but I isn't zero, because
> > > it's the sum of all the previous errors that are all positive. So the
> > > power doesn't drop to zero and we have an overshoot (the bar moves after
> > > the wanted position). Now the error starts being negative and I starts
> > > decreasing.
> > 
> > Yes.
> > 
> > In any regulation system, you have a balance between reacting quickly to
> > changes and minimising overshoot and oscillations.  The D term can be
> > helpful in reducing these unwanted effects.
> > 
> > > 
> > > I think that systems similar to optical bars (where the controlled
> > > signal must be stopped as soon as the set-point is reached) is best
> > > controlled by a P-only algorithm.
> > 
> > You'll never get to the target that way.  
> 
> Not correct.  This design incorporates the integrator into the mechanism.  Position is the integral of velocity which is proportional to the output of the controller.  So the additional integrator is not required in the controller.  But the P term will get to the end position because there is non-ideal behavior in moving the bar because of it's mass.  So velocity is not perfectly proportional to the control output. 
> 
> This is why the theory is so hard to manage in real world situations.  There are many, many nonidealities.  While many of them won't have a significant impact on the controller, some may and it can be hard to know which need to be factored in. 
> 
> I'm working on a PID simulation in LTspice.  I thought I'd use the "universal op amp" as an ideal device, but I guess I should have read the data sheet first... lol.  I should have something working later in the day. 
> 
> Working on lunch at the moment. 

I didn't like the op amp circuits too many components.  Here is an LTspice simulation with voltage sources as amplifiers, simpler. 

The ringing of the controlled quantity can be managed by "tuning" the ratio of the proportional section to the integral section.  Too little P term and ringing starts.  Too much P term and the response is slow.  I tried it with no P term and the ringing is all you see practically.  Ramping down both together rounds the corners of the response.  Ramping up both together sharpens the corners crisply.  

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-- 

  Rick C.

  --- Get 1,000 miles of free Supercharging
  --- Tesla referral code - https://ts.la/richard11209

On 6/27/20 12:41 AM, Paul Rubin wrote:
> Richard Damon <Richard@Damon-Family.org> writes:
>> 2) It can go to some finite, non-zero state. This indicated neither a
>> pole or zero near zero. A proportional controller will have a finite
>> error result, so you may want an integral term in the system.
> 
> Slightly tangential but can someone explain if the integral term is
> supposed to decay?  Otherwise, can't it potentially keep getting bigger
> and bigger without bound?  E.g. in a thermal system, you have to keep
> adding heat to account for losses, or for (say) hot water being drawn
> out of a water heater, and being replaced by incoming cold water.
> 

The integral term is supposed to be a pure integral, no decay. It will
only keep getting bigger if you keep having error of the same sign.

Often, limits are put on the integration, to limit the overshoot from
'windup' as the system tries to get close or special rules to hold the
integrator when the P-D part of the loop is driving hard.

On Saturday, June 27, 2020 at 11:31:58 AM UTC-4, David Brown wrote:
> On 27/06/2020 16:53, pozz wrote:
> > Il 27/06/2020 09:07, Rick C ha scritto:
> >> On Saturday, June 27, 2020 at 2:12:21 AM UTC-4, Paul Rubin wrote:
> >>> DJ Delorie <dj@delorie.com> writes:
> >>>> Typical systems will limit the I term to some range so it won't go out
> >>>> of bounds (or "wind up" as they say).&nbsp; It doesn't decay on its own, but
> >>>> the system can cause its value to reduce to zero due to the Ki math.
> >>>
> >>> Do you mean it saturates (controller stops increasing it once it reaches
> >>> some maximum)?&nbsp; Ok, but won't it reach the maximum and stay there?&nbsp; The
> >>> Ki math I think was:
> >>>
> >>> &nbsp;&nbsp; I += Ki * err&nbsp;&nbsp; (note += not = )
> >>> &nbsp;&nbsp; Vout = P + I
> >>>
> >>> So if the error is almost always positive, I will keep increasing.
> >>>
> >>>>> E.g. in a thermal system, you have to keep adding heat
> >>>>
> >>>> Yes, but the amount of extra energy you put in is constant in that
> >>>> case,
> >>>> and when the extra energy balances so that you stay at your set point,
> >>>> the I term stops increasing, because the error is now zero.&nbsp; Drain more
> >>>> hot water, the I term goes up.&nbsp; Insulate the tank, the I term goes
> >>>> down..
> >>>
> >>> I'm imagining a tank starting with 20 degree water and you want to heat
> >>> it to 50 degrees.&nbsp; Now the error is 30 degrees, or am I interpreting the
> >>> wrong thing as the error?&nbsp; You turn on the heat and the error decreases
> >>> as the water gets hotter.&nbsp; Eventually the water reaches 50 degrees and
> >>> the error is 0, but I is some positive number (the integral of the error
> >>> over the time it took to reach 50 degrees).
> >>>
> >>> Now you draw out some hot water, say to wash dishes; you replace it with
> >>> more 20 degree water, so the system is say at 40 degrees now (error is
> >>> 10).&nbsp; You turn the heat back on, but again, I keeps increasing.&nbsp; You
> >>> might overshoot a little (say to 51 degrees, error is -1) in which case
> >>> I might decrease a little until the tank cools back down.&nbsp; But really,
> >>> the error will always be either positive or near 0.&nbsp; So I keeps
> >>> increasing.
> >>>
> >>> I have to be missing something.&nbsp; Thanks.
> >>
> >> Yes, you are missing that the error variable can be both positive and
> >> negative and by definition will balance the I variable to a value that
> >> makes error return to zero.
> >>
> >> Once the water in your heater reaches the set temperature the error
> >> will be zero but the heat sent to the water heater won't be zero.&nbsp; If
> >> water is drawn the temperature will drop and the error term will go
> >> positive again, the integral term will increase to restore the error
> >> to zero, but at that point the heat into the tank will be higher than
> >> before so that the temperature will rise above the set point and the
> >> error will go negative.
> >>
> >> It is a mistake to think there will be more positive error than
> >> negative error.&nbsp; Once the water is warmed to the appropriate
> >> temperature the integral value will be back at the value it had before
> >> cold water was added.
> > 
> > I think the wrong assumption was: as soon as the error goes to zero, P
> > and I term should be zero as well, because we reached the set-point.
> 
> Yes - P goes to zero, and I /stabilises/ to a value of output that
> results in a steady state.
> 
> If you overshoot - perhaps I has built up too much - the error term will
> be negative, so the I term will decrease (but remain positive for a
> while), while the P term will be negative.

Not "if" but just "how much" I will cause overshoot.  The I term is an oscillator with P as a damping factor.  


> > For example, when I need to move the optical bar of a scanner to a
> > certain position.
> > As soon as the bar reached the wanted position, the power from the
> > motors must be cut off immediately (because the bar doesn't move if the
> > motors are off).
> > This means that when the error is zero, P+I must be zero as well.
> 
> That works for some kinds of systems.  In other systems, you might have
> gravity or a spring that means keeping the same position requires a
> constant small force - you get that from the I term.
> 
> In this case, the I term will be zero at the perfect stable spot.  But
> it could well be non-zero while moving - a P term alone rarely gets you
> to the target.  (Imagine you are near the target.  The P term might give
> too low a force to overcome the static friction and move the bar - then
> the I term will build up until things are moving.)
> 
> > However this is impossible. P is really zero, but I isn't zero, because
> > it's the sum of all the previous errors that are all positive. So the
> > power doesn't drop to zero and we have an overshoot (the bar moves after
> > the wanted position). Now the error starts being negative and I starts
> > decreasing.
> 
> Yes.
> 
> In any regulation system, you have a balance between reacting quickly to
> changes and minimising overshoot and oscillations.  The D term can be
> helpful in reducing these unwanted effects.
> 
> > 
> > I think that systems similar to optical bars (where the controlled
> > signal must be stopped as soon as the set-point is reached) is best
> > controlled by a P-only algorithm.
> 
> You'll never get to the target that way.  

Not correct.  This design incorporates the integrator into the mechanism.  Position is the integral of velocity which is proportional to the output of the controller.  So the additional integrator is not required in the controller.  But the P term will get to the end position because there is non-ideal behavior in moving the bar because of it's mass.  So velocity is not perfectly proportional to the control output. 

This is why the theory is so hard to manage in real world situations.  There are many, many nonidealities.  While many of them won't have a significant impact on the controller, some may and it can be hard to know which need to be factored in. 

I'm working on a PID simulation in LTspice.  I thought I'd use the "universal op amp" as an ideal device, but I guess I should have read the data sheet first... lol.  I should have something working later in the day. 

Working on lunch at the moment. 

-- 

  Rick C.

  ++ Get 1,000 miles of free Supercharging
  ++ Tesla referral code - https://ts.la/richard11209

On 6/27/20 10:53 AM, pozz wrote:
> Il 27/06/2020 09:07, Rick C ha scritto:
>> On Saturday, June 27, 2020 at 2:12:21 AM UTC-4, Paul Rubin wrote:
>>> DJ Delorie <dj@delorie.com> writes:
>>>> Typical systems will limit the I term to some range so it won't go out
>>>> of bounds (or "wind up" as they say).&nbsp; It doesn't decay on its own, but
>>>> the system can cause its value to reduce to zero due to the Ki math.
>>>
>>> Do you mean it saturates (controller stops increasing it once it reaches
>>> some maximum)?&nbsp; Ok, but won't it reach the maximum and stay there?&nbsp; The
>>> Ki math I think was:
>>>
>>> &nbsp;&nbsp; I += Ki * err&nbsp;&nbsp; (note += not = )
>>> &nbsp;&nbsp; Vout = P + I
>>>
>>> So if the error is almost always positive, I will keep increasing.
>>>
>>>>> E.g. in a thermal system, you have to keep adding heat
>>>>
>>>> Yes, but the amount of extra energy you put in is constant in that
>>>> case,
>>>> and when the extra energy balances so that you stay at your set point,
>>>> the I term stops increasing, because the error is now zero.&nbsp; Drain more
>>>> hot water, the I term goes up.&nbsp; Insulate the tank, the I term goes
>>>> down..
>>>
>>> I'm imagining a tank starting with 20 degree water and you want to heat
>>> it to 50 degrees.&nbsp; Now the error is 30 degrees, or am I interpreting the
>>> wrong thing as the error?&nbsp; You turn on the heat and the error decreases
>>> as the water gets hotter.&nbsp; Eventually the water reaches 50 degrees and
>>> the error is 0, but I is some positive number (the integral of the error
>>> over the time it took to reach 50 degrees).
>>>
>>> Now you draw out some hot water, say to wash dishes; you replace it with
>>> more 20 degree water, so the system is say at 40 degrees now (error is
>>> 10).&nbsp; You turn the heat back on, but again, I keeps increasing.&nbsp; You
>>> might overshoot a little (say to 51 degrees, error is -1) in which case
>>> I might decrease a little until the tank cools back down.&nbsp; But really,
>>> the error will always be either positive or near 0.&nbsp; So I keeps
>>> increasing.
>>>
>>> I have to be missing something.&nbsp; Thanks.
>>
>> Yes, you are missing that the error variable can be both positive and
>> negative and by definition will balance the I variable to a value that
>> makes error return to zero.
>>
>> Once the water in your heater reaches the set temperature the error
>> will be zero but the heat sent to the water heater won't be zero.&nbsp; If
>> water is drawn the temperature will drop and the error term will go
>> positive again, the integral term will increase to restore the error
>> to zero, but at that point the heat into the tank will be higher than
>> before so that the temperature will rise above the set point and the
>> error will go negative.
>>
>> It is a mistake to think there will be more positive error than
>> negative error.&nbsp; Once the water is warmed to the appropriate
>> temperature the integral value will be back at the value it had before
>> cold water was added.
> 
> I think the wrong assumption was: as soon as the error goes to zero, P
> and I term should be zero as well, because we reached the set-point.
> 
> For example, when I need to move the optical bar of a scanner to a
> certain position.
> As soon as the bar reached the wanted position, the power from the
> motors must be cut off immediately (because the bar doesn't move if the
> motors are off).
> This means that when the error is zero, P+I must be zero as well.
> However this is impossible. P is really zero, but I isn't zero, because
> it's the sum of all the previous errors that are all positive. So the
> power doesn't drop to zero and we have an overshoot (the bar moves after
> the wanted position). Now the error starts being negative and I starts
> decreasing.
> 
> I think that systems similar to optical bars (where the controlled
> signal must be stopped as soon as the set-point is reached) is best
> controlled by a P-only algorithm.
> 
> Now come to mind another scenario. Suppose I need to transfer heat to a
> liquid to reach a set-point temperature, for example 80&deg;C starting from
> 20&deg;C. It's very import to avoid overshoots, because some strange and
> very dangerous chemical reactions happen when the temperature is over 82&deg;C.
> I know I can tune the PID controller (i.e., find an optimal Kp, Ki, Kd)
> to avoid overshoots, but I'm wondering if some other techniques can be
> implemented too.
> 
> For example, changing dynamically PID coefficients when the error is
> negative.

The motor situation where you want 0 signal to hold at the desired set
point, basically implies that the system has an integrator in its
response. (A pole at zero). With the motor that is because the input
Voltage determines the Speed of the motor, and the motor mechanics
integrates that speed to become a position, which is the thing that is
being controlled. Such a system may not need an integral term for 0
error (because the plant provides that integral term).

On the other hand, if there is a bit of force on the motor (maybe
gravity) and with zero input, the system slides of the control point, so
we need to add a bit of voltage to provide some torque to counter that
force, that brings back the need for the integral loop to provide that
balancing voltage.

One issue with your P only controller for position, is that as you get
close, you slow down, so you close slower, and slower, so it takes a
while to get to where you want (theoretically, forever, as you will
close exponentially). You can make the loop faster by raising P, but at
some point, the gain gets too high, and other dynamics or time delays
get in the way and the system may mal-perform.

PID loops, when properly tuned, can get you closer faster when you tune
the system to be critically damped, the key being that you let the PI
terms give you a higher starting gain, but then the damping term backs
that off giving you more gain at slow speeds, which handles the base
error, and less gain at high frequencies where the instabilities live.

As to changing parameters, yes that can be done, and that gets you into
the domain of non-linear control. Such system can be harder to think
about or analyze, and it can introduce strange phenomenon like limit
cycles. I would probably not have a sharp change of values at 0 error,
as that would likely be the nominal operating point, so very apt to
create limit cycles. You might either gradually adjust the parameters as
you go negative, or have a break-point at a certain amount of error that
you don't normally expect to reach.

Paul Rubin <no.email@nospam.invalid> writes:
> Do you mean it saturates (controller stops increasing it once it reaches
> some maximum)?  Ok, but won't it reach the maximum and stay there?  The
> Ki math I think was:
>
>   I += Ki * err   (note += not = )
>   Vout = P + I
>
> So if the error is almost always positive, I will keep increasing.

The error is *after* all three PID factors are added.  The I factor
reduces the error.  If it reduces the error to zero, the I term stops
increasing.

In real life terms, consider this: you want to heat a pot of water to a
slow boil.  You turn on the stove, at first a lot as a "guess" but as
the pot approaches the boiling you turn the heat down (P term).
Eventually you have to start tweaking the heat up a bit to bring the
temperature up, this is the I term.  As you approach the right temp, you
tweak in smaller increments, until you're there.

> I'm imagining a tank starting with 20 degree water and you want to heat
> it to 50 degrees.  Now the error is 30 degrees, or am I interpreting the
> wrong thing as the error?

Error is 30.

> You turn on the heat and the error decreases
> as the water gets hotter.  Eventually the water reaches 50 degrees and
> the error is 0, but I is some positive number (the integral of the error
> over the time it took to reach 50 degrees).

Better example: when the heat gets to 45 degrees (error=5) the P term
has been reduced enough that the water doesn't get any warmer.  If the
water were to warm up, the heat would be reduced, and it would cool
again.  You've reached steady state, and this is as close as the P term
can get you.  This happens relatively quickly.  Let's say the P term is
providing 500 watts at this point.

Now, you need a little more heat to make up the difference.

The I term sees the small remaining error, and slowly adds more heat,
until the temperature slowly climbs to the set point.  The I term might
be adding another 5 watts or so

> Now you draw out some hot water,

And the error quickly jumps up, the P term immediately reacts.  The I
term slowly starts rising again.  More heat pours in.  The temperature
might rise past the set point, but then the error is *negative* and the
I term slowly goes down until it's back where we started.

> But really, the error will always be either positive or near 0.  So I
> keeps increasing.
> I have to be missing something.  Thanks.

You're missing that if the I term keeps increasing, eventually the water
will be too hot, the error negative, and that reduces the I term again.

It's like the pot on the stove.  You turn the heat up until the water is
the right temperature, but if you *keep* turning the knob up, the water
becomes too hot, and you have to turn the knob back down again.

On 27/06/2020 16:53, pozz wrote:
> Il 27/06/2020 09:07, Rick C ha scritto:
>> On Saturday, June 27, 2020 at 2:12:21 AM UTC-4, Paul Rubin wrote:
>>> DJ Delorie <dj@delorie.com> writes:
>>>> Typical systems will limit the I term to some range so it won't go out
>>>> of bounds (or "wind up" as they say).&nbsp; It doesn't decay on its own, but
>>>> the system can cause its value to reduce to zero due to the Ki math.
>>>
>>> Do you mean it saturates (controller stops increasing it once it reaches
>>> some maximum)?&nbsp; Ok, but won't it reach the maximum and stay there?&nbsp; The
>>> Ki math I think was:
>>>
>>> &nbsp;&nbsp; I += Ki * err&nbsp;&nbsp; (note += not = )
>>> &nbsp;&nbsp; Vout = P + I
>>>
>>> So if the error is almost always positive, I will keep increasing.
>>>
>>>>> E.g. in a thermal system, you have to keep adding heat
>>>>
>>>> Yes, but the amount of extra energy you put in is constant in that
>>>> case,
>>>> and when the extra energy balances so that you stay at your set point,
>>>> the I term stops increasing, because the error is now zero.&nbsp; Drain more
>>>> hot water, the I term goes up.&nbsp; Insulate the tank, the I term goes
>>>> down..
>>>
>>> I'm imagining a tank starting with 20 degree water and you want to heat
>>> it to 50 degrees.&nbsp; Now the error is 30 degrees, or am I interpreting the
>>> wrong thing as the error?&nbsp; You turn on the heat and the error decreases
>>> as the water gets hotter.&nbsp; Eventually the water reaches 50 degrees and
>>> the error is 0, but I is some positive number (the integral of the error
>>> over the time it took to reach 50 degrees).
>>>
>>> Now you draw out some hot water, say to wash dishes; you replace it with
>>> more 20 degree water, so the system is say at 40 degrees now (error is
>>> 10).&nbsp; You turn the heat back on, but again, I keeps increasing.&nbsp; You
>>> might overshoot a little (say to 51 degrees, error is -1) in which case
>>> I might decrease a little until the tank cools back down.&nbsp; But really,
>>> the error will always be either positive or near 0.&nbsp; So I keeps
>>> increasing.
>>>
>>> I have to be missing something.&nbsp; Thanks.
>>
>> Yes, you are missing that the error variable can be both positive and
>> negative and by definition will balance the I variable to a value that
>> makes error return to zero.
>>
>> Once the water in your heater reaches the set temperature the error
>> will be zero but the heat sent to the water heater won't be zero.&nbsp; If
>> water is drawn the temperature will drop and the error term will go
>> positive again, the integral term will increase to restore the error
>> to zero, but at that point the heat into the tank will be higher than
>> before so that the temperature will rise above the set point and the
>> error will go negative.
>>
>> It is a mistake to think there will be more positive error than
>> negative error.&nbsp; Once the water is warmed to the appropriate
>> temperature the integral value will be back at the value it had before
>> cold water was added.
> 
> I think the wrong assumption was: as soon as the error goes to zero, P
> and I term should be zero as well, because we reached the set-point.

Yes - P goes to zero, and I /stabilises/ to a value of output that
results in a steady state.

If you overshoot - perhaps I has built up too much - the error term will
be negative, so the I term will decrease (but remain positive for a
while), while the P term will be negative.

> 
> For example, when I need to move the optical bar of a scanner to a
> certain position.
> As soon as the bar reached the wanted position, the power from the
> motors must be cut off immediately (because the bar doesn't move if the
> motors are off).
> This means that when the error is zero, P+I must be zero as well.

That works for some kinds of systems.  In other systems, you might have
gravity or a spring that means keeping the same position requires a
constant small force - you get that from the I term.

In this case, the I term will be zero at the perfect stable spot.  But
it could well be non-zero while moving - a P term alone rarely gets you
to the target.  (Imagine you are near the target.  The P term might give
too low a force to overcome the static friction and move the bar - then
the I term will build up until things are moving.)

> However this is impossible. P is really zero, but I isn't zero, because
> it's the sum of all the previous errors that are all positive. So the
> power doesn't drop to zero and we have an overshoot (the bar moves after
> the wanted position). Now the error starts being negative and I starts
> decreasing.

Yes.

In any regulation system, you have a balance between reacting quickly to
changes and minimising overshoot and oscillations.  The D term can be
helpful in reducing these unwanted effects.

> 
> I think that systems similar to optical bars (where the controlled
> signal must be stopped as soon as the set-point is reached) is best
> controlled by a P-only algorithm.

You'll never get to the target that way.  Maybe that's okay, and you'll
get close enough.  Sometimes overshoot must be avoided at all costs,
other times it is merely a mild inconvenience - there is no single
answer to all regulation problems.

And pure PID is just one way to regulate a system.  You can use
variations or ad-hoc methods (such as P regulation, but using an
artificial setpoint beyond the real target, or applying a break when you
hit the target), or you can use completely different regulation techniques.

> 
> Now come to mind another scenario. Suppose I need to transfer heat to a
> liquid to reach a set-point temperature, for example 80&deg;C starting from
> 20&deg;C. It's very import to avoid overshoots, because some strange and
> very dangerous chemical reactions happen when the temperature is over 82&deg;C.
> I know I can tune the PID controller (i.e., find an optimal Kp, Ki, Kd)
> to avoid overshoots, but I'm wondering if some other techniques can be
> implemented too.
> 
> For example, changing dynamically PID coefficients when the error is
> negative.

Il 27/06/2020 09:07, Rick C ha scritto:
> On Saturday, June 27, 2020 at 2:12:21 AM UTC-4, Paul Rubin wrote:
>> DJ Delorie <dj@delorie.com> writes:
>>> Typical systems will limit the I term to some range so it won't go out
>>> of bounds (or "wind up" as they say).  It doesn't decay on its own, but
>>> the system can cause its value to reduce to zero due to the Ki math.
>>
>> Do you mean it saturates (controller stops increasing it once it reaches
>> some maximum)?  Ok, but won't it reach the maximum and stay there?  The
>> Ki math I think was:
>>
>>    I += Ki * err   (note += not = )
>>    Vout = P + I
>>
>> So if the error is almost always positive, I will keep increasing.
>>
>>>> E.g. in a thermal system, you have to keep adding heat
>>>
>>> Yes, but the amount of extra energy you put in is constant in that case,
>>> and when the extra energy balances so that you stay at your set point,
>>> the I term stops increasing, because the error is now zero.  Drain more
>>> hot water, the I term goes up.  Insulate the tank, the I term goes down..
>>
>> I'm imagining a tank starting with 20 degree water and you want to heat
>> it to 50 degrees.  Now the error is 30 degrees, or am I interpreting the
>> wrong thing as the error?  You turn on the heat and the error decreases
>> as the water gets hotter.  Eventually the water reaches 50 degrees and
>> the error is 0, but I is some positive number (the integral of the error
>> over the time it took to reach 50 degrees).
>>
>> Now you draw out some hot water, say to wash dishes; you replace it with
>> more 20 degree water, so the system is say at 40 degrees now (error is
>> 10).  You turn the heat back on, but again, I keeps increasing.  You
>> might overshoot a little (say to 51 degrees, error is -1) in which case
>> I might decrease a little until the tank cools back down.  But really,
>> the error will always be either positive or near 0.  So I keeps increasing.
>>
>> I have to be missing something.  Thanks.
> 
> Yes, you are missing that the error variable can be both positive and negative and by definition will balance the I variable to a value that makes error return to zero.
> 
> Once the water in your heater reaches the set temperature the error will be zero but the heat sent to the water heater won't be zero.  If water is drawn the temperature will drop and the error term will go positive again, the integral term will increase to restore the error to zero, but at that point the heat into the tank will be higher than before so that the temperature will rise above the set point and the error will go negative.
> 
> It is a mistake to think there will be more positive error than negative error.  Once the water is warmed to the appropriate temperature the integral value will be back at the value it had before cold water was added.

I think the wrong assumption was: as soon as the error goes to zero, P 
and I term should be zero as well, because we reached the set-point.

For example, when I need to move the optical bar of a scanner to a 
certain position.
As soon as the bar reached the wanted position, the power from the 
motors must be cut off immediately (because the bar doesn't move if the 
motors are off).
This means that when the error is zero, P+I must be zero as well. 
However this is impossible. P is really zero, but I isn't zero, because 
it's the sum of all the previous errors that are all positive. So the 
power doesn't drop to zero and we have an overshoot (the bar moves after 
the wanted position). Now the error starts being negative and I starts 
decreasing.

I think that systems similar to optical bars (where the controlled 
signal must be stopped as soon as the set-point is reached) is best 
controlled by a P-only algorithm.

Now come to mind another scenario. Suppose I need to transfer heat to a 
liquid to reach a set-point temperature, for example 80&deg;C starting from 
20&deg;C. It's very import to avoid overshoots, because some strange and 
very dangerous chemical reactions happen when the temperature is over 82&deg;C.
I know I can tune the PID controller (i.e., find an optimal Kp, Ki, Kd) 
to avoid overshoots, but I'm wondering if some other techniques can be 
implemented too.

For example, changing dynamically PID coefficients when the error is 
negative.

On 27.6.20 07:41, Paul Rubin wrote:
> Richard Damon <Richard@Damon-Family.org> writes:
>> 2) It can go to some finite, non-zero state. This indicated neither a
>> pole or zero near zero. A proportional controller will have a finite
>> error result, so you may want an integral term in the system.
> 
> Slightly tangential but can someone explain if the integral term is
> supposed to decay?  Otherwise, can't it potentially keep getting bigger
> and bigger without bound?  E.g. in a thermal system, you have to keep
> adding heat to account for losses, or for (say) hot water being drawn
> out of a water heater, and being replaced by incoming cold water.


The integral term is supposed to approach a stable value, to
compensate for the steady-state error of pure proportional
control. What is supposed to decay is the total error term.

-- 

-TV

On Saturday, June 27, 2020 at 2:18:44 AM UTC-4, Paul Rubin wrote:
> Rick C <gnuarm.deletethisbit@gmail.com> writes:
> > In the water heater example the integral term will be ramping up the
> > entire time the heater is not up to temperature.  The rate of increase
> > of the heat flowing in will lower until the set point is achieved at
> > which point the integral term will be at a maximum, but too high with
> > the water temperature continuing to rise.
> 
> Let's say the water heater is insulated, but not perfectly (which is
> impossible).  That means heat steadily escapes from the system and you
> have to keep adding energy to maintain the temperature.  That is, the
> error (difference between measured temperature and goal temperature) is
> always positive or zero, except maybe for a few moments when you
> overshoot slightly.  Therefore, the integral of the error should keep
> increasing.
> 
> I tried reading the Wikipedia article about PID and thought I understood
> it, but still come away from it with the same impression.  I'm clearly
> missing something.

Your mistake is thinking the integral term will only overshoot "slightly".  Once the integral term rises to heat the added water it is too high to maintain the temperature of the take once it reaches the set point.  So the error will shoot past the zero point and become negative quickly.  The integrator doesn't make the temperature rise to gently reach the set point.  At the point the temperature crosses the set point it increasing the fastest it will move.  So now the error has to be negative for a while to bring it back to zero from the other direction.  

Increasing the gain on the proportional term and keeping the gain on the integral term low will minimize the ripples, but never eliminate them.  So the error will always wing positive and negative.  

You need to keep in mind that if the integral term is high, the heat output will be high.  Thinking the set point will only be passed a little bit is not accurate.  You need to stop "feeling" the design and try looking at some real numbers.  

If LTspice is too complicated for you, you can use a spread sheet to make calculations in a stepwise manner.  Write the equations for the three controller terms and one for the combined heat output based on the previous values.  Write the equations for the heat output (losses) based on present temperature and the intermittent water draw.  Write the equations for the temperature of the tank in terms of the previous temperature and the heat inputs and outputs.  

Copy each of these from one row to the next for many rows and watch the temperature change across the rows.  I evaluate DSP algorithms this way.  You can even chart the columns to see temperature and the heat flows change and watch the oscillations of temperature and heat output. 

-- 

  Rick C.

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