# Tag Info

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This question is specific to smoothing samples in the frequency domain (given by FFT and spectral density) and asking about the impact to the resulting noise floor in the same domain. The answer depends on the characteristics of the noise and any time domain windowing that is applied. For white noise, with no windowing beyond the rectangular window, each bin ...

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To fully understand interdependence of a function and its Fourier transform, be it in continuous-time or discreet flavor, you need a basic background in calculus. It may seem off-topic in the context of your question, but pay attention that a Fast Fourier Transform (FFT) algorithm is a tool, it does not define the intrinsic properties of the Fourier ...

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The OP's question requires further details to provide a definitive answer, but the following will give the considerations involved. If the noise is white and stationary then the answer is clear in that we can simply use a power spectral density and work with the SNR as an SNR/Hz quantity. If the noise is not white (meaning the average power across all bins ...

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SUMMARY The equivalent noise bandwidth (ENBW) for a window function is the bandwidth in bins of a brickwall filter that would result in the same noise power from a white noise source as the DFT "filter" (when viewing, appropriately, each bin of the DFT as a bandpass filter). The ENBW for the rectangular window (no further windowing) is 1 bin as ...

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By preference, you should calculate the noise bandwidth. The noise bandwidth is the bandwidth of the sharp-cutoff filter that has the same output power as your filter when the inputs are white Gaussian noise. So to calculate the ENBW, you integrate the filter's amplitude response squared, from frequency = zero to infinity. I'm not sure what the best ...

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I was doing quite a bit wrong, but the key thing that I was missing was the fact that the SNR needs to be calculated over the whole Nyquist spectrum instead of only looking at the peaks. This article explains everything very well: Taking the Mystery out of the Infamous Formula, "SNR = 6.02N + 1.76dB," and Why You Should Care. Another issue was that ...

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Some issues here: Your SNR formula only applies to full scale sine waves, your sine wave has -6dB amplitude so your SNR will be 6 dB lower The formula also implies rounding, not truncation, that's another 6 dB You use a frequency that's a small integer divider of the sample rate, that means you are just repeating the same samples over and over again and don'...

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Before getting to the answer, several corrections are necessary. First, the units of the Boltzmann constant are J/K, so kT is energy, not power. Thus P is not equal to 4kT. Second, the expected value of the thermal noise voltage, $\mu$, is zero, as required by the second law of thermodynamics. So $V^2/R$ is not equal to 4kT. When I do computer simulations ...

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Emphasis mine: I want to plot the expected voltage vs time trace The expectation (that's a term from probability theory and statistics) of a white process is zero. So, your expected voltage trace is a constant 0. You might mean that you want to plot an exemplary realization of a random noise process - in that case: For a discrete-time noise process, draw ...

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If you have no prior data on the signal of interest there is nothing to do actually. The more prior you have the better you can do. For instance, if the only information you have is the Bandwidth of your signal, the best you can do is apply an Band Pass / Low Pass Filter. If you know a sparse representation of your signal it will be great, as you can remove ...

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Your question is vague for me. I understand that you want to make a convolution on an received OFDM signal to synchronize right? OFDM signal is demodulated through an FFT. Usually FFT size is a power of 2 (128,256,512,etc...). To demodulate FFT symbols cleanly, they must be synchronized (beginning symbol is found) Your OFDM waveform has probably a preamble, ...

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Yes, when we can recover OFDM in fading scenarios along with gaussian noise, then yes we can also recover it in just gaussian noise. You can think of fading coefficients having a magnitude of 1 and phase zero

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Following the comments by AlexTP (thank you), there is one clear mistake in my reasoning. Using a roll-off factor indeed increases the bandwith but it colours noise too. The mistake is $P_n = W'N_0$ because that equation is assuming flat filter response. The intregration of the squared reception filter (RRC, and so, raised cosine) over all bandwidth is 1, so ...

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