9239 number of samples versus rate

[SeanJ]

Hi All,

I am running a VI for the 9239. 

In it I set the sample clcok to 2048.

I then set the the number of samples to 20480, i.e run for ten seconds.

However instead of the final point being measured at 10s it is measure at 9.82992

If it is set for 2048 samples the final time is 0.98256

I realise that say if I had a rate of 5 samples per second and a took five samples, I would get the following 0, .2, .4, .6, .8

but if I was to increase this to 10 samples, I would get

0, .2, .4, .6, .8, 1, 1.2, 1.3, 1.4, 1.6, 1.8

The important point here is that the offset remains the same

In addiiton 1 - 1/2048 =.99951171875  not .98256.

Does anyone know what is going on here?

Sean

[John_P1]

Hi Sean,

The behavior comes from the fact that the 9239 only has a certain number of fixed sample rates available.  They are defined by the following equation:

        

If using the internal Master Timebase, then f_M would be 12.8 MHz  (±100 ppm).

So, if you specify a rate of 2048, the driver will actually coerce this up to 2083.3333 Hz (n=24).  20480 samples divided by 2083.333 Hz gives 9.8304 seconds, which is approximately the time you measured.

Best Regards,

[John_P1 · Accepted solution]

To add to that,

You made a very good point about the first sample being taken at t=0.  Accounting for this we get:

20480 samples

    (20480 / 2083.33333) * (1-1/20480) = 9.82992

2048 Samples

    (2048 / 2083.33333) * (1-1/2048) = 0.98256

Which gives exactly the numbers that you reported.

Best Regards,

[SeanJ]

Thanks John, mystery solved!