LabVIEW Derivative x(t).vi

[unc98]

Hello All,

I'm trying to interpret the output from the Derivative x(t).vi and want to make sure I'm doing so appropriately.  The default value for the "dt" input is 1.0.  I have force plate data that are sampled at 1200Hz.  I interpret this to mean that the value of 1.0 for "dt" represents a real time interval of 0.833ms [(1200pts/1s) x (1s/1000ms) = 0.833ms/pt] and that the units of the derivative (i.e. loading rate = change in force/unit time) will be in N/0.833 ms.  I would then need to convert this to N/s by multiplying by 1000ms/1s. 

can anyone confirm (or deny!) this?  Thanks for the help.    

[Alexander_Sobolev]

Yes, it is correct.

Interval is constant so it simply divides differences (dx = dN array) by interval (dt = 0.833 ms = 0.000833 s)

BTW, you can verify it if you input x = [1 2 3] and check how result depends on dt... 

[unc98]

Excellent - thanks so much for your help!