= Equivalent noise bandwidth
The <equivalent noise bandwidth> of a finite window is the width of an ideal rectangular <frequency> response that passes the same white-noise power when both responses have the same peak gain. It quantifies noise collected by a spectral amplitude estimate, rather than the separation needed to distinguish two tones. For real coefficients $w_n$ with nonzero sum and sampling <frequency> $f_s$,
$$
\boxed{B_e=f_s\frac{\sum_{n=0}^{N-1}w_n^2}{(\sum_{n=0}^{N-1}w_n)^2},\qquad
b_e=\frac{B_e}{f_s/N}=N\frac{\sum_nw_n^2}{(\sum_nw_n)^2}.}
$$
$B_e$ is in hertz; $b_e$ is in bins of an $N$-sample record. The primary software documentation gives $B_e$ as the exact conversion factor between its squared spectrum and power-spectral-density scalings.<SciPy 1.16.2 documentation>[scipy.signal.periodogram, Notes: spectrum/density ratio]{ref}
An original <variance> derivation explains the formula. Let independent zero-mean sample noise $\epsilon_n$ have <variance> $\sigma^2$, and form the complex amplitude estimate at <frequency> $\nu_0$,
$$
\widehat a=\frac{\sum_n w_n\epsilon_ne^{-2\pi in\nu_0}}{\sum_nw_n}.
$$
Cross terms vanish by the independence and zero means, so
$$
\mathbb E|\widehat a|^2
=\sigma^2\frac{\sum_nw_n^2}{(\sum_nw_n)^2}
=\frac{\sigma^2}{f_s}B_e.
$$
The flat two-sided input noise spectral density is $\sigma^2/f_s$. Integrating it over the peak-normalized window response gives the same result by <Parseval identity>. A one-sided real-signal convention folds negative-<frequency> power into positive <frequencies>; factors of two must be handled consistently, rather than changing the window’s bandwidth ratio.
For a <rectangular window>, $\sum w_n=N$ and $\sum w_n^2=N$, giving $b_e=1$. For the periodic <Hann window> with $N>2$, the cosine sums give $\sum w_n=N/2$ and $\sum w_n^2=3N/8$, hence $b_e=1.5$. These agree with the instrument manual’s tabulated values.<LabVIEW Analysis Concepts by National Instruments>[printed pp. 5-23–5-24, Table 5-3, coherent gain and equivalent noise bandwidth]{ref}
For an original numerical comparison, take $f_s=4096\,\mathrm{Hz}$ and $N=256$, so the original <frequency> grid spacing is $16\,\mathrm{Hz}$. Rectangular and periodic Hann bandwidths are respectively $16$ and $24\,\mathrm{Hz}$. With independent sample noise of standard deviation $0.20\,\mathrm V$, the RMS magnitude of the zero-mean complex noise estimate is $0.0125\,\mathrm V$ for rectangular weighting and about $0.01531\,\mathrm V$ for Hann weighting. The latter <variance> is 1.5 times greater, despite Hann’s lower distant sidelobes. It is the RMS complex-error magnitude, not automatically the standard deviation of a positive spectral-magnitude display.
<Zero padding> leaves $f_s$, the actual window coefficients and $B_e$ unchanged. If a 256-sample Hann record is padded to 1024 points, the display spacing becomes $4\,\mathrm{Hz}$ and the same $24\,\mathrm{Hz}$ bandwidth spans six displayed bins. Replacing $N$ by the padded FFT length in a formula while leaving the coefficient sum unchanged confuses these two bin units.
The <noise covariance between windowed Fourier bins> also matters when combining several spectral values: a taper can correlate neighboring estimates. Bandwidth alone does not specify their joint statistics, and colored noise requires weighting its actual spectral density. A proposed check repeats the same estimator on independent synthetic noise records with known <variance>, then on measured noise with a characterized acquisition response, comparing empirical mean squared complex error with the formula. Report window convention, detrending, one-sided/two-sided scaling and record duration so the test can be reproduced.
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