Numeric Integration "dt"

[dumbest_dummie]

We are doing data acquisition with PCB accelerometers using Easy AI (Acquire Waveforms). In order to have displacement the signal from accelerometers should be integrated twice. I'm trying to use Numeric Integration box wich seems to give better results than Integral x(t) box.
But I don't know what should I use for the "dt"? Say I have a scan rate of 10kHz and Number of Samples = 1k.
dt=0.001?
dt=0.0001?
neither?
It really makes a big difference and I'm kinda lost.

Thanks

Erick

[JoshuaP]

The dt is the sampling interval and must be greater than zero. When performing an analog input you want to wire the sampling rate (in time) to dt. In your example the scan rate is 10 kHz so the dt (time between samples) would be 0.0000001 or 1.0E-7. Think of it as the dt when you perform a discrete integration.

I hope this helps.

Joshua

[ali_masoudi]

Hi Joshua,

I have a question. The "dt" should be greater that the sampling rate for sure. but if dt=0.0001 and sampling rate is 1kHz, it means that each sample is count 10 times? i.e. we have 1000 data points in 1 second, but we are sampling 10,000 times.

Cheers,

Ali

[Dennis_Knutson]

The dt is the inverse of the sample rate. If dt is .0001, then your sample rate 10kHz. If your sample rate is 1kHz, then the dt is .001. Your numbers are not correct.

[ali_masoudi]

I'm pretty sure my numbers are correct. My question was that if I choose dt=0.0001 AND my sampling rate be 1kHz what is happening. 

My qustion was that in the integrator each point contributes 10 times.

[Dennis_Knutson]

You choose one or the other. You simply cannot have a dt of .0001 and a sample rate of 1kHz. I will repeat - the dt is simply the inverse of the sample rate and vice versa. It's basic arithmetic.