Yes! I realised that right after sending the message. I was confused because the results were not making sense to me (I was computing the bins resolution in the wrong way, using the reduced sample rate due to FFT overlapping instead of the original one), and I ended up asking a stupid question... Sorry about that :)
I don't know if you asked because you are familiarised with the function's code or because you just saw my mistake, but if it's the first case I want to ask something else. My real problem was that in this line: freq = center_freq - (tb.usrp_rate / 2) + (tb.channel_bandwidth * i_bin) I think that the "tb.channel_bandwidth" should be the bins resolution, because otherwise the value of "freq" is far away from the values it can take when centered on "center_freq" (like happens in my case where the channel bandwidth is much larger than the resolution). For example, if I am centered in 928.8MHz with samp_rate=10MHz, FFT_size=1024 and channelBW=200kHz, according to the formula the frequency of the last bin would be: freq = 928.8M - (10M/2) + 1023*200k = 1128.4MHz This value is far away from the frequency of the last bin that you get when doing an FFT with those parameters (considering that the resolution is 9765.625Hz/bin the last bin should go from 938.7902344MHz to 938.8MHz). Am I missing something? Thank you both for your answer, Marc On 25/01/2017 9:38:01, Sylvain Munaut <246...@gmail.com> wrote: > Does anyone know why the first and last 25% of FFT bins are discarded? I am > talking about the following lines of code: > > line 267: bin_start = int(tb.fft_size * ((1 - 0.75) / 2)) > line 277: bin_stop = int(tb.fft_size - bin_start) Read that code again ... it discards the first and last 12.5% ... so that it discards 25% in total. Cheers, Sylvain
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