0
$\begingroup$

This question already has an answer here:

For example, for an aperiodic gate pulse, the Fourier Transforms for the continuous time case is a sinc function, while the discrete time case gives a sine over sine periodic kind of a function. In either of these cases, we get a spectra that goes below the frequency axis, i.e., acquires a "negative amplitude" (namesake) on many occasions in its excursions. I realize that the spectra gives a measure of "how much" a frequency component is present. But how this negative amplitude satisfies this criteria is what I am missing. Is it kinda like.. "How much less that frequency component is present"..? I know I might not be making any sense. Please help. Thanks!!

$\endgroup$

marked as duplicate by Matt L., lennon310, MBaz, hotpaw2, Peter K. Sep 9 at 15:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

Note that in general the Fourier transform of a function is a complex-valued function, so in general it is not only positive or negative.

Roughly speaking, the magnitude of the Fourier transform says something about the presence of certain frequencies components in a signal, regardless of the phase (or sign, in the real-valued case). The phase determines the temporal alignment of the different frequency components. So a real-valued negative value of the Fourier transform just implies a 180 degrees phase shift of the respective components.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.