Re: Curious about some links posted for "Understand PID control"


"Jerry Avins" <jya@xxxxxxxx> wrote in message
news:47ednT0eJId_I-7eRVn-tQ@xxxxxxxxxx
> Peter Nachtwey wrote:
>
>
>> 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.
>
> 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?

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

>
>> ftp://ftp.deltacompsys.com/public/PDF/Mathcad%20-%20T0C1%20Bode.pdf
>>
>> Many PLCs use the PI-D form. AB has for a while.
>
> 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:

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

>
>> 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.
>
> ...
>
>Try a velocity servo with a real motor/tach combination both ways, using
>the highest practical Ki in each case.

What do you expect to see?


> Did I mention a problem? There is no additional lag when using a
> pseudoderivative because there is no differentiator.

A differentiator doesn't cause lag.
...
>
>>>I coined I-PD to mean integral/pseudo derivative. What does I-PV mean?
>>
>>
>> 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.
>
> How about a block diagram? I believe we are discussing different circuits
> as if there were no difference.

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.

>
>> 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?
>
> I'll have to hunt up my notes.
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
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





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