In message
<CALBs-TipQ2tWb71L_jO+d3CysuyxmNHd1cAx+162RVC-OCHd9g@...>,
dated Mon, 16 Dec 2013, Andy <Andrew.Ingraham@...> writes:
You need at least two time samples at the highest frequency of the FFT.
1/90MHz * 0.5 = 5.56 ns
Helmut's example simulates for 20 ms, which actually gives you 50 Hz
resolution, enough to see the "trough" between alternate components
that are 100 Hz apart. ?Given that, you need at least 3600000 for the
"Number of data point samples in time" to have time samples no greater
than 5.56 ns apart. ?Using Helmut's recommended 4194304 (= 2^22) gives
you that. ?That simulation's FFT goes up to (1/2) * 1/(20ms/4194304) =
104.858 MHz, which is what I see when I run it.
Using the bare minimum (3600000) probably stops exactly AT 90 MHz and
isn't enough to show you the 90 MHz component. ?So always go higher if
there is any doubt.
Thanks. I hope that makes it all clear:
To get x Hz resolution, you should, in practice, simulate for 2/x
seconds. You can simulate for longer to get a clearer spectrum display,
say N/x seconds.
To get a spectrum up to X Hz, you then need more than 2*N*X/x samples,
preferably the next higher power of 2.
I do think that, if it's correct, is a bit more lucid than the Help
text.
--
OOO - Own Opinions Only. With best wishes. See www.jmwa.demon.co.uk
Nondum ex silvis sumus
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
