Fourier transform -> power spectrum

[juef]

Fourier transform --> power spectrum

Hey all!

I've been learning about the discrete Fourier transform (and FFT too) recently. What I don't understand is why applying it to a signal gives its power spectrum. I am not really good in physics, so to me it just seems like a magical formulae, one might say. :)

I don't need a deep explanation; the basics would be more than enough to please me. It would already be much more than what I can find on math sites and books about the discrete Fourier transform.

Thank you!

 

[Hash]

The inverse Fourier transform represents a signal as a sum (continuous or discrete) of frequency, each of one having an amplitude.

The (direct) Fourier transform represents this repartition of frequency from the signal. It describes how a signal is distributed along frequency.

The power spectral density (PSD) (or spectral power distribution (SPD) of the signal) are in fact the square of the FFT (magnitude). It describes how the power of a signal is distributed with frequency.

 

[antonantal]

A periodical signal can be written as a sum of harmonics (using the harmonic form of the Fourier series), each harmonic having it's own intensity (which decreases as the frequency of the harmonic increases). This harmonics are not abstract mathematical stuff. You can observe them while trying to set a tv channel and you see that you can catch that channel on several frequencies (these are the harmonics). The best quality will have the first harmonic (the fundamental) because it has the highest intensity. If you set the channel on the 2nd harmonic you will see that it will be noisy, and if you set it on the 3rd it will be even more noisy (because it's intensity is lower).
Now if you think of the inverse Fourier transform (which is an integral) as of a sum of harmonics of very close frequencies you can find an analogy: the direct Fourier transform of the signal (found in the formula of the inverse transform) plays the same role as the amplitude of the harmonics in the Fourier series. That's why it is called "spectral density of the complex amplitude". As you know, the energy is proportional to the square of the amplitude(intensity) so, the energy spectral density will be the square of the modulus of the Fourier transform (the Fourier transform has complex value). If you consider that the signal is applied on an ideal resistor of 1 ohm you can say that the square of the modulus of the Fourier transform is the spectral density of the power.

 

[juef]

Thank you both for this enlightment! I believe I understand much better now.

As you know, the energy is proportional to the square of the amplitude(intensity)[...]

Ah, well, no I didn't :shy: I should have mentioned that I'm studying maths and not physics. Thanks again!