With FFT, I sometimes take advantage of the fact that I can pre-calculate a signals Fourier Transform, and then you can add the noise in the spectral domain:

$$ \mathscr{F} (x[n] + h[n]) = \mathscr{F}(x[n]) + \mathscr{F}(h[n]) $$

This is useful when we want to discard of the original signal, but we still want to noise it with a spectrum. If we know the spectrum of the noise beforehand, this is also faster computationally (just adding together the two spectra).

Does the same relation hold for constant Q transforms?


1 Answer 1


It’s equivalent to a bank of linear filters, so yes it is linearly additive because each component filter is linearly additive with respect to its inputs/outputs.


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