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Damian Marek

Hi Jaffar,

I’m not sure what the convention is in electronics, but I believe in optics usually the FWHM is determined by the intensity (proportional to power) You are correct to say that the electrical signal is not power as A.U. could be considered as current or voltage depending on the component, but we use the same convention. The Width parameter in the electrical gaussian pulse generator will determine the FWHM of the power signal, so in A.U. that corresponds to the square root, which results in the full width at 1/sqrt(2).

Hope this clears it up.