Minimal Sample Duration for FFT
Hi,
We are using the V6-2000 to continuously record IQ samples with the Raw IQ File Writer block.
With the information contained in the .xml file (sample start time & clock unit) we calculate the timestamps for each IQ value recorded to allow timely sychronisation with other GPS time stamped data during data postprocessing.
By that we intend to crop the IQ values such, that we can perform FFTs at the same times of the data of the other sensors, which e.g. deliver data only every 1.0 second.
My question is how to calculate the minimum sample duration I have to consider to comply with the requirement that a sample duration shall at least cover one full period or one cycle of the (highest) frequency?
Is it the duration of one cycle of the clock unit (sampling frequency) of (in our case) of 180 kHz (92 MHz/512) resulting in a minimum sample duration of 5.56 µsec?
If not, how to calculate else the minimum sample duration required?
In case of interest: The center frequency in our case would be 331.7 MHz.
Thanks for your support.
You might want to check this posting: https://v6-forum.aaronia.de/forum/topic/what-is-the-lowest-100-poi-the-spetran-v6-can-offer/
Thank you for the link.
But I just don't get it.
What is the relevance of POI, when there is continuous IQ recoding and FFT performed during postprocessing?
Can you pls. explain?
As mentioned: "The POI is calculated as FFT size/sampling rate".
Thank you very much!
In our case above, a sample duration or POI for an FFT size of 8 samples without overlap (no RT requirements due to post-processing) would be 8 / (92Mio/512) = 44.444 µsec. Since we are counting the IQ values anyway, we simply consider as many IQ values as the desired FFT size.
However, my post is on a different topic, but this may be "old school", saying that the sample duration should be at least one complete cycle of the frequency of interest. If I am right, the RF front end of the V6 transforms the incoming centre frequency (331.7 MHz in our case) down to 0 Hz, which would then have an infinite cycle duration. Or even a frequency very close to the centre frequency, say 90 Hz in our case, would require a minimum sampling time of 1/90 = 11 msec then. With the POI calculation mentioned above, the sample time of at least one cycle does not seem to be necessary. Could you please confirm this?
