[time-nuts] PI Math question

[Ulrich Bangert]

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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