[time-nuts] PI Math question
Ulrich Bangert
df6jb at ulrich-bangert.de
Thu Apr 17 08:14:39 UTC 2014
Warren,
the job of a controller, regardless of P, PI od PID, is to minimize the
error between a process value and its setpoint. Since I see no setpoint
value in any of your versions my 50 ct is that none of them really
incorporates a controller at all and that for this reason the question
whether they produce the same output is close to being irrelevant.
Best regards
Ulrich
> -----Ursprungliche Nachricht-----
> Von: time-nuts-bounces at febo.com
> [mailto:time-nuts-bounces at febo.com] Im Auftrag von WarrenS
> Gesendet: Mittwoch, 16. April 2014 18:50
> An: Discussion of precise time and frequency measurement
> Betreff: [time-nuts] PI Math question
>
>
>
> A question to the math time-nuts
>
> With the values of K1, K2 & K3 constant,
> and the initial state of I#1, I#2 and Last_Input all zero
> assuming there is no rounding, clipping or overflow in the
> math and that if I've made any obvious dumb typo errors that
> they are corrected,
>
> Given this PID type of controller;
> D = (Input - Last_Input))
> Last_Input = Input
> I#1 = I#1 + (K1 * Input)
> I#2 = I#2 + (K2 * D)
> Output = I#1 + I#2 + (K3 * Input)
>
> Is the above Input to Output's transfer function any
> different than any of
> the following more simplified versions of PI controllers?
> Or asked another way, if each of the four codes are given the
> exact same
> input string and same K Gains, will the difference between
> any of their
> outputs ever be non zero?
>
>
> a)
> D = Input - Last_Input
> Last_Input = Input
> I#1 = I#1 + (K1 * Input) + (K2 * D)
> Output = (K3 * Input) + I#1
>
> b)
> D = (Input - Last_Input)
> Last_Input = Input
> Output = Output + (K1 * Input) + (K2 + K3) * D
>
> a)
> I#1 = I#1 + (K1 * Input)
> Output = I#1 + ((K2 + K3) * Input)
>
> ws
>
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