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9237 &9205 with chassis 9178 sampling rate problem

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      I have two 9237 and one 9205 installed on chassis 9178. In order to acquire data simultaneously, all the 9237 and 9205 use the onboardclock of chassis 9178. But when the sample rate is 20000 Hz, getting 20000 points just takes less than 1 second. When the sample rate is 10000Hz, getting 10000 points takes 1 second.   Why does this happen ?

      One 9237 use one channel, another 9237 use 4 channels, the 9205 use 3 channels (differential).  And the 9178 manual make me a little confused, is the onboardclock frequency 80 MHz ?

 

Look forward your help. Thanks.

 

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Message 1 of 12
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The time to acquire is just basic arithmetic - number of samples/Samples per second so both should take exactly 1 second.
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Message 2 of 12
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Sorry, but I really find that acquiring 20000 points takes less than 1s, And it is apparent.

Nobody come accross this problem ?

 

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Message 3 of 12
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Solution
Accepted by topic author dffs

By default the 9237 derives its clock from dividing down its own 12.8 MHz timebase (see the specifications).

 

12.8 MHz / 256 / N

 

Where N is any integer 1 to 31.

 

 

10 kHz is achievable (12.8 MHz / 256 / 5), but 20 kHz is not (12.8 MHz / 256 / 2.5).

 

The driver will round up to the next available frequency, which happens to be 25 kHz (12.8 MHz / 256 / 2).  At 25 kHz we would expect 20k samples to be acquired in 0.8 seconds.

 

 

Best Regards,

John Passiak
Message 4 of 12
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Sorry, I don't think so.

The timebase means the biggest sample rate should less than this.

I mean sample rate <= timebase

But thank you all the same.

 

Best regards.

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Message 5 of 12
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John's post is correct.  You can read back the coerced sample rate to see what rate your task is running at.  The 9237 cannot run at 20kHz so it coerces the sample rate of the task up to 25kHz. 

 

See section 2-3 of the cDAQ-917x manual (Sigma-delta modules).  The 9237 uses a 12.8MHz internal timebase instead of the chassis 80MHz.  And as John mentioned, the 9237 manual has the list of valid sample rates (12.8MHz/256/1..31)

 

coercion.png

Message 6 of 12
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Thank you and John.

Your posts are correct.

So I can not use the sample rate (20kHz) with 9237?

 

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Message 7 of 12
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@dffs wrote:

 

So I can not use the sample rate (20kHz) with 9237?

 


You can only sample at exactly 20 kHz if you use an external timebase.  You would need to provide a clock at either of the following frequencies:

 

5.12 MHz ( /256 /1 = 20 kHz)

10.24 MHz (/256 /2 = 20 kHz)

 

Both of these frequencies are too high to pass through the PFI lines on the 9178, so you would need to have a digital module with enough bandwidth to accomodate the signal (namely, a 9401 or 9402).

 

 

 

Without providing your own timebase, you could get a little bit closer to 20 kHz by using a counter to generate a timebase instead of the onboard 12.8 MHz timebase.  The counter's output frequency is divided from 80 MHz, and the maximum timebase on the 9237 is 13.15 MHz.  This essentially gives you three options:

 

(80 MHz /8) /256 /2 = ~19.5 kHz

(80 MHz /15) /256 /1 = ~20.8 kHz

(80 MHz /7) /256 /2 = ~22.3 kHz

 

 

 

Best Regards,

John Passiak
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Message 8 of 12
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Hi John,

 

I synchronize the two device 9237 and 9205 use the 12.8MHz timebase in one program.

Now I want to know if I have 8 channels analog signals need be acquired by 9205, what could the sample rate range be ? 

I mean is the sample rate also calculated according to the formula: 12.8 MHz / 256 / N  ?

 

Thanks.

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Message 9 of 12
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If they are in the same task, then yes, that is correct, with the added constraint that the chosen sample rate cannot be greater than the 9205's maximum rate.  The 9205 is a multiplexed module with a max rate of 250kHz.  With 8 channels that gives you a max rate of 31.25kHz.  The 9237 cannot go at that rate, however, it can run at 25kHz with N = 2.

 

If the 9205 is in its own task, then that task's maximum rate is 250kHz / N where N is the number of channels in the task.

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Message 10 of 12
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