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Each bin in the FFT is effectively a bandpass filter (and for the unwindowed FFT the filter response in the frequency domain is the Dirichlet Kernel which is essentially an aliased Sinc Function). Conveniently for white noise, the equivalent noise bandwidth of the Dirichlet Kernel is that of a brickwall filter 1 bin wide. The scaling of the DFT by $1/\sqrt{N}$ is consistent with Parseval’s theorem, while the scaling of the DFT by $1/N$ is convenient for tones as the DFT tone magnitude matches the time domain magnitude independent of what $N$ is used):

$$X[k]= \frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-j 2\pi nk/N}$$

the variance of the total noise applied (which is the total noise power in the signal) will go down as $1/N$ when the noise is white (the noise is evenly spread over all frequencies, and we have $N$ bins, so the noise in one bin is $1/N$ in power). Similarly, the standard deviation will be $1/\sqrt{N}$. In the bin where a tone exists (and to simplify the explanation without involving spectral leakage, let's assume the tone is exactly at a bin center), the FFT result will be the perfect tone +/- the standard deviation as $\sigma/\sqrt{N}$ where $\sigma$ is the standard deviation of the noise. This could be confirmed by repeating the experiment many times with independent and identically distributed zero-mean stationary white noise, as we would do for other experimental measurements to establish a noise statistic. Ultimately the noise variation we see on the peak is identical to the noise variation we see on every other bin (without the offset due to the tone present on bin center). So we can also confirm the noise on the peak by using the statistics from the variation we see for every other bin (again and importantly under this contrived case of not inducing spectral leakage from the tone to other bins by ensuring the tone is exactly on bin center).

From this we see that the noise is not white, but bandlimited around the tone of interest. If we assume that the noise extends evenly over $M$ bins out of the $N$ bins total and has a total variance of $\sigma^2$, then the variance in each of the bins where the noise is present (including the tone bin) would be $\sigma^2/M$ instead of $\sigma^2/N$ (assuming we scale the DFT by $1/N$).

Each bin in the FFT is effectively a bandpass filter (and for the unwindowed FFT the filter response in the frequency domain is the Dirichlet Kernel which is essentially an aliased Sinc Function). Conveniently for white noise, the equivalent noise bandwidth of the Dirichlet Kernel is that of a brickwall filter 1 bin wide. The scaling of the DFT by $1/\sqrt{N}$ is consistent with Parseval’s theorem:

$$X[k]= \frac{1}{\sqrt{N}}\sum_{n=0}^{N-1}x[n]e^{-j 2\pi nk/N}$$

For this scaling, the variance for each bin (which is the total noise power in the signal divided by the total number of bins) will go down as $1/N$ when the noise is white (the noise is evenly spread over all frequencies, and we have $N$ bins, so the noise in one bin is $1/N$ in power). Similarly, the standard deviation will be $1/\sqrt{N}$. In the bin where a tone exists (and to simplify the explanation without involving spectral leakage, let's assume the tone is exactly at a bin center), the FFT result will be the perfect tone +/- the standard deviation as $\sigma/\sqrt{N}$ where $\sigma$ is the standard deviation of the noise. This could be confirmed by repeating the experiment many times with independent and identically distributed zero-mean stationary white noise, as we would do for other experimental measurements to establish a noise statistic. Ultimately the noise variation we see on the peak is identical to the noise variation we see on every other bin (without the offset due to the tone present on bin center). So we can also confirm the noise on the peak by using the statistics from the variation we see for every other bin (again and importantly under this contrived case of not inducing spectral leakage from the tone to other bins by ensuring the tone is exactly on bin center).

From this we see that the noise is not white, but bandlimited around the tone of interest. If we assume that the noise extends evenly over $M$ bins out of the $N$ bins total and has a total variance of $\sigma^2$, then the variance in each of the bins where the noise is present (including the tone bin) would be $\sigma^2/M$ instead of $\sigma^2/N$ (assuming we scale the DFT by $1/\sqrt{N}$).

Each bin in the FFT is effectively a bandpass filter (and for the unwindowed FFT the filter response in the frequency domain is the Dirichlet Kernel which is essentially an aliased Sinc Function). Conveniently for white noise, the equivalent noise bandwidth of the Dirichlet Kernel is that of a brickwall filter 1 bin wide. Since I work mostly with Signal to Noise ratios, I prefer the scaling of the DFT by $1/N$ such that Parseval’s theorem applies (and conveniently the DFT tone magnitude matches the time domain magnitude independent of what $N$ is used):

Each bin in the FFT is effectively a bandpass filter (and for the unwindowed FFT the filter response in the frequency domain is the Dirichlet Kernel which is essentially an aliased Sinc Function). Conveniently for white noise, the equivalent noise bandwidth of the Dirichlet Kernel is that of a brickwall filter 1 bin wide. The scaling of the DFT by $1/\sqrt{N}$ is consistent with Parseval’s theorem, while the scaling of the DFT by $1/N$ is convenient for tones as the DFT tone magnitude matches the time domain magnitude independent of what $N$ is used):

Magnitude of tone in DFT (at bins $k$ and $N-k$): $NA/2$
Variance of each bin: $\sqrt{N}\sigma^2$

Magnitude of tone in DFT (at bins $k$ and $N-k$): $A/2$
Variance of each bin: $\sigma^2/\sqrt{N}$

From this we see that the noise is not white, but bandlimited around the tone of interest. If we assume that the noise extends evenly over $M$ bins out of the $N$ bins total and has a total variance of $\sigma^2$, then the variance in each of the bins where the noise is present (including the tone bin) would be $\sigma^2/\sqrt{M}$ instead of $\sigma^2/\sqrt{N}$ (assuming we scale the DFT by $1/N$).

Magnitude of tone in DFT (at bins $k$ and $N-k$): $NA/2$
Variance of each bin: $N\sigma^2$

Magnitude of tone in DFT (at bins $k$ and $N-k$): $A/2$
Variance of each bin: $\sigma^2/N$

From this we see that the noise is not white, but bandlimited around the tone of interest. If we assume that the noise extends evenly over $M$ bins out of the $N$ bins total and has a total variance of $\sigma^2$, then the variance in each of the bins where the noise is present (including the tone bin) would be $\sigma^2/M$ instead of $\sigma^2/N$ (assuming we scale the DFT by $1/N$).