What happens when you replace samples in mode analog output buffer?

I have a sequence of output voltages that I need to update all the 1/60th of a second.  Each sequence must remain synchronized to a trigger external and controlled by an external clock.  The program is in c ++.

These are my steps to configure the task:

(1) DAQmxCreateAOVoltageChan

(2) DAQmxSetWriteAttribute: DAQmx_Val_DoNotAllowRegen

(3) DAQmxCfgSampClkTiming: DAQmx_Val_FiniteSamp (and fix the external clock signal)

(4) DAQmxCfgDigEdgeStartTrig: The front of my external trigger

(5) DAQmxSetTrigAttribute: DAQmx_StartTrig_Retriggerable

If I write too slow samples, I get the following message:

"The production has stopped to prevent the regeneration of old samples. Your application could not write samples in the back buffer fast enough to prevent the former samples of regenerated. »

I understand this message and this is the behavior I want.

My question is: what happens if I write the new samples to fast, i.e. before the previous one, who have been posted on the output channel?  This will give me an error or software allows me to crush the samples in my output buffer?

I'll echo what others have said here.  DAQmx won't let you write samples in the buffer until there are enough empty space to make it--and that's why the entry on the function DAQmx Write time-out.

There are two ways to go about writing only when the buffer is ready for the next set of data:

  1. Use the timeout to write DAQmx directly.  Do not forget that this blocks however.

  2. Use the DAQmx software event "Every N samples transferred from the buffer".  This tutorial gives a good introduction to the DAQmx software events and this forum thread goes through a few considerations to keep in mind.

PS - If you use the C API, use the DAQmxRegisterEveryNSamplesEvent function to register this callback.

Tags: NI Hardware

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