There are two band-pass filters both with the same bandwidth, 200 Hz to 1.6 kHz (3 octaves). The output of a white noise generator is at input to a first filter, and the output of a pink noise generator is at input to the second filter. The true RMS voltage taken at the output of each filter is the same.

Given all of the above, is it possible to conclude what the voltage ratio at a given frequency in the pass-band, say 500 Hz, is of the filtered white noise and pink noise? That is, which has a greater amplitude and by how much?

Thanks in advance,
Pete

 

White noise has equal power per unit frequency, pink noise has equal power per octave.
White noise only applies to a limited range of frequency, because that definition would imply infinite power at infinite frequency, so it usually only applies to audio frequencies.
First you would need to know the generator output in W/Hz or W/ocvate, or the total power in a given bandwidth, for the two generators.

 

White noise has equal power per unit frequency, pink noise has equal power per octave.
White noise only applies to a limited range of frequency, because that definition would imply infinite power at infinite frequency, so it usually only applies to audio frequencies.
First you would need to know the generator output in W/Hz or W/ocvate, or the total power in a given bandwidth, for the two generators.

To make the analysis easier, change the setup to equal output voltage of the generators. Assume an equal load to each of the two generators. Then white noise in the first octave from 200 Hz- 400 Hz equals W / 200 Hz and pink noise in the first octave equals W / 3. How to proceed after that I don't know.