LabVIEW

cancel
Showing results for 
Search instead for 
Did you mean: 

Problem with signal integral

Hello everybody,

 

I'm novice in the labVIEW developement and I have a probably simple question to suggest. I want to realise the integral of a signal delivered by a NI ADC converter (represented by the box "simulate signal") in real time. So, for a sin(wt) signal, the result must be -cos(wt) but we can see on the enclosed picture that it is not what it is given by the integral box. Indeed, the magnitude of the integrated signal is not the same as the input sinus signal and it is always positive, what is more, the magnitude of the integrated signal is frequency dependant...

 

So my question is : How can I do to obtain the result -1.cos(wt) when in put the 1.sin(wt) signal on the input? I think it's a simple question but I do not have enough experience to solve that problem.

 

Thank you Smiley Happy,

 

HRicochet

 

 

Download All
0 Kudos
Message 1 of 4
(2,344 Views)

A code picture does now show the configuration of the express VIs.

 

What is the "dt" of your data? That will determine the scaling, of course.

0 Kudos
Message 2 of 4
(2,332 Views)

HRicochet,

 

I hope your calculus teacher does not monitor this Forum.

 

Integral[sin(wt)dt] = -(1/w)*cos(wt)

 

The numerical integration VIs calculate a definite integral, which essentially measures the area under the curve from zero to x.  Look at the detailed help files for the Integral VIs for more information.

 

Lynn

0 Kudos
Message 3 of 4
(2,316 Views)

Thank you Johnsold for your useful answer! I'm ashamed Smiley Surprised . Right now, I now that the result given by the integrator box is correct and I will just remove the mean value to obtain the result I initially wanted. 

 

Sincerely,

 

Hricochet.

 

 

0 Kudos
Message 4 of 4
(2,304 Views)