10-18-2019 03:02 PM
I found the block diagram that governs the inner working of the PID.vi in LabVIEW from here.
So, output=Kc⁄Ti&int e(t)dt + Kc e(t).
To my knowledge PID, in discrete domain, is governed by the equation:
, where .
The problem is I am not sure how to use just Kc to tune the PID controller in LabVIEW (without Ki). And I don't know what Ti is.
Solved! Go to Solution.
10-18-2019 04:01 PM
10-23-2019 09:37 AM - edited 10-23-2019 09:49 AM
Hi, thanks. I was looking more for the significance of Ti. Like if my PID is continuous sampling and but I know that due to the limit of the system I hook up to LabVIEW a new process variable only gets to the PID once every 9 sec. Then, should I put 9 for Ti?
Then, there is the question of why the block diagram is dividing the sum of error by the sampling period Ti. At first, I thought that is taking the average. But upon closer examination, if I run the PID from time 0 to t_0, the average should be the integral from 0 to t_0 divided by t_0 not Ti.
10-23-2019 10:00 AM
@MyLord wrote:
1. Hi, thanks. I was looking more for the significance of Ti. Like if my PID is continuous sampling and but I know that due to the limit of the system I hook up to LabVIEW a new process variable only gets to the PID once every 9 sec. Then, should I put 9 for Ti?
2. Then, there is the question of why the block diagram is dividing the sum of error by the sampling period Ti. At first, I thought that is taking the average. But upon closer examination, if I run the PID from time 0 to t_0, the average should be the integral from 0 to t_0 divided by t_0 not Ti.
1. Usually you want the Ti to be several samples, let's say 10. In that case Ti is 90 secs, but if you read the help it's measured in minutes, so 1.5 (90/60).
2. You're right about the average, but the PID does not look at the full integral, but the PID-window, which is Ti. Older samples are ignored.
/Y
10-29-2019 10:26 AM
I actually found the formula:
In PID.vi's help page, under "implementing the PID algorithm with the PID VIs",