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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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The use of $rad/\sqrt{\text{Hz}}$ suggests that this is phase noise specifically (a spectral density due to phase fluctuations), and typically in my use this has been described as a power spectral density (units of $rad^2/\text{Hz}$), so this is just the square root of that quantity. The reason the DFT (of which the FFT computes) is divided by $N$ is to ...


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I'm a big believer in this form of the continuous Fourier Transform and inverse: $$ X(f) \triangleq \mathscr{F}\Big\{x(t)\Big\} = \int\limits_{-\infty}^{+\infty} x(t) \ e^{-j 2 \pi f t} \ \mathrm{d}t $$ $$ x(t) \triangleq \mathscr{F}^{-1}\Big\{X(f)\Big\} = \int\limits_{-\infty}^{+\infty} X(f) \ e^{+j 2 \pi f t} \ \mathrm{d}f $$ because I like the symmetry ...


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According to this, the units for PSD from a DFT should be volts^2/bin: https://www.mathworks.com/matlabcentral/answers/47633-what-is-the-relation-between-dft-and-psd-of-a-signal $$ \mathrm{PSD} = [ X[k] \cdot \operatorname{conj}(X[k]) ] / N $$ $$ \mathrm{units} = \mathrm{volts} \cdot \mathrm{volts} /\mathrm{bins} = \mathrm{volt}^2/\mathrm{bin} $$ Which makes ...


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You do this by doing two things: First establish what the relationship of scale exists between your A/D converter and the numerical values that go into your DFT. Find out how many volts corresponds to a value of $1.0$ in the samples that go into the DFT. The next thing you gotta do is express your integral above without that $\frac{1}{T}$ factor (which is ...


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