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I have been recently working with the FFT and realized that each frequency within a bin corresponds to a phase. This concept I have no problem with because if you wanted to construct the summation of sinusoidal waves back into the original signal, you would need the phases of those sinusoidal waves.

My Questions:

  1. Why must there be an imaginary component to each of the bins corresponding phases? It makes no sense to me why a phase would need an imaginary component.

  2. In reference to DSP with the FFT, how does this imagery component come into play when characterizing a signal?

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It simple trigonometry. A vector (FFT result bin) can have a magnitude and a phase. Or you can specify the same vector identically with X,Y coordinates. The complex representation simply uses X for the real part and Y for the imaginary component. Without an imaginary component, there’s no place to put the Y, and your vector would be under specified.

The phase ends up being atan2(I,Q) == atan2(Y,X) == atan2(im(z),re(z))

As for why use a complex representation rather than 2 strictly real quantities? The complex math is more elegant, and requires less writing on the chalkboard to demonstrate similar things.

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  • $\begingroup$ It appears I misunderstood what a bin is. I thought it was just a sin wave with a specific frequency and with a specific phase. My understand was the summation of these different sine waves was a approximation of the original signal. $\endgroup$
    – Cabbage Champion
    Jul 16, 2019 at 22:44
  • $\begingroup$ Also, since I wording with a signal shouldn't the imaginary competent(y component) be 0? $\endgroup$
    – Cabbage Champion
    Jul 16, 2019 at 23:52
  • $\begingroup$ Yes, you are exactly right, the bin result is a sinewave with a magnitude and phase is a vector requiring 2 components to specify. If the imaginary component is zero, the phase is also stuck at zero. A sinewave basis vector is strictly odd. Therefore you need a phase or cosine component if your FFT input consists of any even symmetry (e.g. isn’t strictly anti-symmetric). $\endgroup$
    – hotpaw2
    Jul 17, 2019 at 0:46

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