I recorded this signal at 2.45GHz, it is coming from a camera.
Is this QAM or OFDM ?
Based on those side-lobes I would guess OFDM but I am not sure.
Also there seems to be an interesting preamble at the top.
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1 Answer
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The sharp rectangular shape of the spectrum, with the immensely flat top definitely says "OFDM, with suitable whitening/PAPR reduction"; what constellation is used on the individual carriers is a bit hard to tell.
Can you, with a no-samples-left-out spectrogram (e.g. gr-fosphor sink, or by playing back your recorded file slower and using a suitably high update rate and suitable low number of FFT points in your QT waterfall sink, determine how long the OFDM "burst" is?
Take a bunch of samples from the end, and a bunch of samples from the start (like, one fourth of the burst duration), and calculate a cross-correlation. The maximum correlation should give you the time when what is used as the cyclic prefix at the end of the burst starts; that way, you'll be able to make an educated guess at the symbol number that makes up the CP, and by that the OFDM symbol length/IFFT size. Take frame, and try to manually resample it, so that both CP starts are exactly that (guessed) IFFT size apart. FFT the hell out of the frame; because your resampling had the purpose of time-stretching your signal so that the length of it actually (hopefully) matches the transmitter's sampling, you'd have a timing-synced demodulator.
Look at the spectrum and the constellation plot of the resulting signal; maybe you can already guess the constellation used to modulate the OFDM subcarriers from that.
answered Feb 15, 2016 at 9:19
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$\begingroup$ Many thanks for the answer. I need to read up a bit on how OFDM works. $\endgroup$– jhegedusFeb 15, 2016 at 10:27
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1$\begingroup$ OFDM is simple, really. Just take a bunch of symbols that you generated with the constellation of your choice, and do an IFFT on them, to get the time-domain OFDM symbol. Send that. And because synchronization is hard, you prepend the end of your time-domain signal to its beginning (that's the cyclic prefix, CP), so when the receiver takes the FFT of the received signal to get the original symbol vector, starting "too early" will lead to nothing but a complex phase rotation of the symbol vector. $\endgroup$ Feb 15, 2016 at 11:02
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$\begingroup$ Hmm, this cyclic prefix is interesting, thanks for explanation. $\endgroup$– jhegedusFeb 16, 2016 at 6:55