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In the case of uniform quantization, and under some light hypothesis for the signal, the error can be modeled as an additive IID signal, independent of the signal, and with uniform distribution between +/- half LSB. The power of such error is then $\Delta^2/12$, where $\Delta$ is the amplitude of one LSB. Taking the square root and calling it standard ...


4

The author is modeling quantization noise as being white (i.e., each sample is independent of previous or following samples) with each sample being a zero-mean, uniformly distributed random number with a span of one: $$p(x) = \begin{cases}1 & -\frac{1}{2} < x < \frac{1}{2} \\ 0 & \mathrm{otherwise}\end{cases}. \tag 1$$ Do the math (or look it ...


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If the signal is complex, then $1/delta > f_{max}$ with consideration to the bins of the Discrete Fourier Transform (DFT) extending from $k=0 \ldots N-1$ since the Discrete Fourier Transform will be unique in that case from bin $0$ to bin $N-1$. If the signal is real, then the DFT will be Hermitian symmetric, so we may only consider bins $0$ to $(N-1)/2$ ...


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whenever we feed N time-samples of a periodic, continuous, signal into a FFT algorithm You can't feed a continuous signal into an FFT. You need to sample it first at a specific sample rate. Sampling makes the time domain signal discrete and something that's discrete in one domain MUST be periodic in the other domain. The FFT implements the Digital Fourier ...


1

Note that your statement of the Nyquist sampling theorem only works for infinite length signals. If N is finite, then you need to sample above twice the highest spectrum frequency present by some amount. That's because any finite length window has infinite support in the frequency domain. Same with any finite length reconstruction or anti-aliasing filter. ...


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