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I see quite a few misstatements and possible misconceptions both in the answers and in the question, so let’s break down what is meant by “noise.” Analog noise within the same bandwidth as the signal. Analog noise outside of the bandwidth of the signal. Quantization noise introduced by the ADC. I interpreted your question as implying option 1. ...


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I'll try to answer this, although I don't have much knowledge of the area beyond the first paper you linked to. This would be a better fit on https://dsp.stackexchange.com/. If you use energy detection, the signal you are looking for has to be strong enough, such that you are able to detect the difference between noise and the signal. Yes. An energy ...


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No closed form expression exists for the capacities. The best you can do is numerical approximation. The capacity is the mutual information between Y and X, where Y=X+N, X is the constellation point (in your case a PAM), and N is Gaussian. Then, the mutual information is I(Y;x) = H(Y) - H(Y|X). The second term is easy (working it you yourself if you want). ...


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The reference level for an ADC is the maximum signal that meets some quality criterion, for instance distortion. In an ordinary ADC, exceeding the rails results in a distorted signal. In a sigma delta ADC, exceeding the stability threshold results in a distorted signal.


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As you reduce Vref the quantization size reduces but the input signal has to be reduced (the max input is probably equal to Vref) so the effect of analog noise in the input circuitry is increased and so SNR is reduced. Reducing Vref is not the same as increasing the number of bits where you could keep the input signal at the same level.


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Decimation basically means downsampling or re-sampling the signal using a lower sample rate, see the article on Wikipedia. There are several ways to do the decimation and it depends on the chosen method and the type of signal what effect decimation has. When as in your example, the 1 out of 3 samples is kept and that is combined with lowpass filtering (the ...


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When you average AC signals, the answer is zero, unless EVERYTHING is phase-locked. This means every one of your microphones must be equidistant from the sound source, or errors will be introduced, dependent on the difference in distance and the frequency of the sound. For 16kHz, the wavelength is 2.1cm, so this is a real problem. For this reason, I would ...


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Taking samples of a digital stream of data is subject to the same impositions as when sampling an analogue signal. To effectively sample an analogue signal, you need an anti-alias filter to remove unwanted higher frequency artifacts that can bounce down into the digital domain's restricted bandwidth and cause aliasing. The same applies when converting to a ...


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You need to subtract the squares of the RMS voltages then take the square root to obtain the noise of the source. However, this is only accurate if there is no coherence between sources.


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Noise performance depends on the difference between levels. Since both your signals have a difference of 1 volt, they have the same resistance to noise. However, they have different energy. Let's say that your data signal has a period of 1 second. Then, the pulses you transmit have the following energy: The 0V pulse has energy 0 J. The 1V pulse has energy ...


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