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I use a demodulator that outputs X and Y quadratures, with the R and Phase which can be calculated.

For SNR measurements, when plotting the FFT to find the SNR a signal at a certain frequency, should I be plotting the FFT of |X+iY| or the FFT of R?

My understanding would be to do the FFT of R so that the phase information can be completely canceled out, and R being the amplitude (the signal strength) is what matters to the SNR calculations.

Thank you

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    $\begingroup$ It's not clear what kind of SNR calculation you're doing – SNR depends on your signal and noise levels – one man's noise is another man's signal. Also, I assume "R" is the magnitude of the signal? I'm not quite sure what useful information you'd get from discarding the phase of the signal and just transforming the magnitude... $\endgroup$ – Marcus Müller Jan 19 at 17:24
  • $\begingroup$ I'm sending a signal into a device and then demodulating the signal that comes out, and I want to calculate the SNR at a certain frequency I am modulating this device at. So are you saying that choosing FFT of R (magnitude) or |X+iY| really depends the device and measurements wanting to be taken? $\endgroup$ – digeridoo Jan 19 at 17:37
  • $\begingroup$ I'm certain that you taking FFT(R) makes no sense, in no application I can think of. Whether taking an FFT of the complex values is helpful is a different question. You assume there "one way" to estimate SNR; there isn't. It fully depends on how you define S and N. $\endgroup$ – Marcus Müller Jan 20 at 12:53
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To calculate the SNR you need to get the signal's power. One way to do this is in the time domain, but Parseval's theorem says that the power in the time domain is equal to the power in the frequency domain as it should be..it is the same signal after all. The DFT version of the theorem is $\sum_{n=1}^N|x[n]|^2=\frac{1}{N}\sum_{k=1}^N|X[k]|^2$, where $X[k]$ is the DFT of $x[n]$, which means that the power will be determined by its magnitude squared and the phase has no part in the calculation.

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    $\begingroup$ yes, but as you said: Parseval's theorem says that you can get the energy from time domain just as well as from frequency domain, so if you're just after the power, you don't do a DFT, but directly calculate the sum of squares in time domain. $\endgroup$ – Marcus Müller Jan 20 at 6:56
  • $\begingroup$ That is true. I had assumed, for some reason, that the question needed to be done in frequency domain..shouldn't have done that! $\endgroup$ – Engineer Jan 20 at 12:47

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