Implicit counter operation

Question

Hello,
     I have a question on how the implicit timing on a frequency input works.  I am inputting a straight frequency into a counter input on a cDaq unit.  I set it up using the 20Mhz Timebase. and counter 0, However If for example I input a 10Khz frequency and read data from the read counter vi every 10ms I would expect to see 10,000 * .010 samples or ~100 samples per read right?  I am only seeing about 10 samples per read though.  Am I not figuring something right?  I am setting up the sample clock as implicit using parameters (continuous, 200000 buffer size).

Answer 1

You should read the section in the 917x User Manual entitled Choosing a Method for Measuring Frequency.  In summary:
The one counter method loses accuracy as frequency increases.
The two counter (large range) method is the same as the one counter method but it takes the measurement over N consecutive periods which allows you to increase accuracy at the expense of measurement time.  To get a reasonable accuracy at high frequencies however would require your low frequency measurements to take a very long time.
The two counter (high frequency) method loses accuracy as frequency decreases.
The sample clocked method has good accuracy at all frequency ranges and maintains consistent measurement time.
The sample clocked method should be the obvious choice for measuring a large range of input signals.  However, it comes with the caveat that your input signal must be faster than the sample clock or you will receive an error and have to restart the task (which might not even be an issue depending on synchronization requirements).  You'll probably end up using a second counter to generate the sample clock.
Best Regards,
John Passiak