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screenshot of a Khan Academy video link to the video: Negative frequency

Must a cosine function be expressed as two exp functions with opposite frequencies/directions of rotation and half the magnitude/amplitude(1/2), therefore resulting in two magnitude spectrum components?

Why can't there be one single component at w?

I know this is very basic but there are only videos about the "how" not the "why".Thank you in advance!

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    $\begingroup$ Hi Ming! You can check this: dsp.stackexchange.com/questions/431/… $\endgroup$
    – GKH
    Jan 19, 2020 at 19:13
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    $\begingroup$ That's just the way reality is..... Wait, is math part of reality? Or just our description of it? Anyway, negative frequencies are only meaningful for complex tones, where it defines the chirality of the spiral. When two opposing spirals are added to make a real tone, the distinction becomes meaningless. As in $\cos(\theta)=\cos(-\theta)$. $\endgroup$
    – Cedron Dawg
    Jan 19, 2020 at 19:15
  • $\begingroup$ Thanks for the swift replies. So in very simple words, I suppose any frequency on the "right" or "left" side of the spectrum represents a complex time-domain signal. Therefore, a real cosine or sine signal needs a negative component to cancel out the complex(spirals) part? $\endgroup$
    – Ming Pang
    Jan 19, 2020 at 19:49
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    $\begingroup$ You can also try reading about Analytic signal, this might help to get a different perspective. $\endgroup$
    – axk
    Jan 19, 2020 at 19:50
  • $\begingroup$ @axk yes I depend too much on visual explanations, definitely need some reading $\endgroup$
    – Ming Pang
    Jan 19, 2020 at 19:57

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Regarding your first comment, I have a figure that might help, in terms of intuition.enter image description here It's a bit badly scaled but it might do the work.

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  • $\begingroup$ I kept thinking about "a real signal requires the addition of two complex numbers" and that really confused me. This graph beautifully explains everything. Thanks $\endgroup$
    – Ming Pang
    Jan 19, 2020 at 20:32
  • $\begingroup$ You're welcome. :) Can you "accept" the answer to close the question? $\endgroup$
    – GKH
    Jan 19, 2020 at 20:37

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