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If the Fourier transform shows the magnitude of the frequency, in a $\sin(x)$ graph there is exactly one frequency.

So then why is there some "area" to the two spikes? Why are they not exactly straight lines? Is this just because of the limitation in resolution of numerical method to solve them?

Anton Vrdoljak
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Damian Games
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    What area are you referring to? The transform is (2) spikes. – operatorerror Sep 27 '23 at 15:42
  • Please see this question. The answer is in the technical aspects, not practical numerical ones. The "spike" ($\delta$, the dirac delta distribution) isn't really a real-valued function of real numbers ($f:\mathbb{R}\to\mathbb{R}$). (No real valued function can have a value of $\infty$). Instead, it's a distribution or generalized function that is, in a certain technical sense, like a limit of ordinary functions that do have area and approach $\delta$. – Jam Sep 27 '23 at 15:46
  • So the delta distribution $\delta$ is actually neither a regular function with area nor a geometric straight line/spike. But, while it is a different type of object, it does resemble those, in some convenient ways. Another loose way to think of it is like a normal distribution where we take $\sigma\to0$. We want it to have an area under its graph of $1$ (for total probability) and would expect it to look like an infinitely tall, perfectly thin spike. But see the question I linked for what it actually is, rather than how we want it to behave. – Jam Sep 27 '23 at 15:54
  • @KurtG. It is actually an artefact of discretization that was causing the effect I was confused about. Although the question regarding dirac delta is interesting tangent it does not. – Damian Games Sep 29 '23 at 10:22

1 Answers1

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Technically speaking, there are two spikes, since $\sin \omega x = \frac 1 {2i} \left ( e^{i\omega x}-e^{-i \omega x}\right)$, and thus there 2 frequencies ($\omega$ and $-\omega$).

Also, technically, those spikes are really Dirac delta distributions.

Finally, if you think you're seeing some "area" instead of pure spikes, I'm guessing you are looking at a tool that discretized the $\sin$ and computed its Discrete Fourier Transform (so not the "integral" Fourier transform). That discretization has an effect of transforming the true delta into a finite spike with "some area" underneath.

Finally, if you reduce the discretization step of your $\sin$, then you'll start seeing the bump with an area look more and more like a spike. In the limit, this would become a true Dirac delta.

Stefan Lafon
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    This is very helpful as the distinction between the result of integral and discrete fourier transform was not clear as I have just plotted and seen diagrams of "fourier transform". As well as pointing to the dirac delta function, very useful connection for someone looking to understand, thank you! – Damian Games Sep 27 '23 at 21:26