If I reduce the number of samples used for an FFT (increasing the width of the frequency bins), is it correct to say that the energy in two bins with smaller frequency bins (eg. 10kHz-20kHz and 20kHz-30kHz) is equivalent to the energy in a single, larger bin (e.g. 10kHz-30kHz)?
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1\$\begingroup\$ If you are correctly representing the original continuous time signal with your samples and all subsequent manipulations thereof, then yes, the total energy has to remain the same in any legitimate representation. But if you forget that voltage readings are not impulses, carelessness in mathematical manipulation can lead to violating conservation of energy... Another way of putting it would be to stay that DSP operations can (and typically do) have inherent gain or loss, and if you want to have meaningful power measurements you need to keep that gain or loss in mind. \$\endgroup\$– Chris StrattonCommented Sep 1, 2020 at 15:36
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2 Answers
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One of the most useful ways of handling this sort of question with FFTs is to use Parseval's Theorem. Paraphrasing loosely, it says that if you have a signal, it doesn't matter whether you compute its energy by summing the power in each time sample, or the energy in each frequency bin, you necessarily get the same answer - because it's the same signal.
If you keep the sample rate the same, and halve the number of samples you take, then you've halved the length and so halved the total energy in the signal.
Halving the number of samples will halve the number of frequency bins, while doubling their width.
Half the total energy, half the number of frequency bins, therefore the energy in each bin stays the same.
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\$\begingroup\$ Taking math out of context can lead to saying funny things. If a discrete time signal accurately represents a continuous time one, then the samples are not simple voltage readings but rather true impulses containing the integrated energy of the time periods they represent, hence the energy of a signal is the same irrespective of sample rate . Like many DSP operations, sample rate conversion can have inherent gain or loss; maintaining an accurate representation of the original requires tracking that gain or loss just as when passing a signal through physical circuitry. \$\endgroup\$ Commented Sep 1, 2020 at 15:47
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\$\begingroup\$ @ChrisStratton Hmm, 'reduce the number of samples used' for the FFT can have two different meanings. I took it as reduce the length, but it could be reduce the sample rate. I've clarified my answer. \$\endgroup\$– Neil_UKCommented Sep 1, 2020 at 18:06
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Usually FFT algorithms are normalized such that the total energy is constant regardless of the number of bins, in which case yes that will be true. This is not necessarily the case though (the transform could omit the normalization step for computational efficiency), so a quick check of the documentation for your specific FFT library is worthwhile.
answered Sep 1, 2020 at 16:41