04-15-2010 05:46 AM
Hi,
I am trying to use the Lead-Lag.vi block from the tool kit but I don't understand how it works. When you open its from panel it looks like this:
The calculation shown is not what I expected, im pretty sure that this is not how lead lag should work. Should there not be two poles and zeros? I guess the attenuation term is included in the gain?
If anyone knows and can let me know that would be great.
Thanks
Adam
Solved! Go to Solution.
04-16-2010 09:18 AM
04-16-2010 03:23 PM
Hello Adamkse,
I believe that PID Lead-Lag VI is really a Lead/Lag VI. As stated in the context help, it implements a PID controller with a lead/lag function. This corresponds to the expression
Output = Gain * ((s*Tlead +1)/(s*Tlag+1))*Input
with only 1 pole and 1 zero
04-16-2010 03:44 PM
Hi,
Thanks for the reply.
I still don't really understand. Is it for lead or lag, not both? If so I don't get how the time constants are called lead and lag and if not then I don't understand how lead-lag can be done with one pole and zero.
Sorry if this is obvious.
Adam
04-18-2010 11:14 AM
The VI is either Lead OR Lag depending on the pole/zero location. You can think of the lead and lag time in term of frequency response.
Tlead = a*T
Tlag = T
so P = 1/T
Z = 1/aT
04-18-2010 12:38 PM
Thank you, I seen now.
Adam
06-04-2015 01:11 AM - edited 06-04-2015 01:12 AM
Hi there,
I believe there is a bug in Lead-Lag VI because it sends out the exact input at time t=0 and, also whenever the block is re-initialized. Can somebody double-check this please?
Thanks.
06-05-2015 05:54 AM
Hi Bij2012,
Could you please post this as a new question? This will help us keep track of questions and potential bugs a bit easier as well as boost the visibility of your post so other community members can help answer it as well as NI staff.
This particular post is 5 years old so I doubt anybody is going to see it.
Best wishes,
Charlotte