There have been many questions about the fourier transform, but as far as I know, not specifically the one I want to ask.

The fourier transform converts the time domain to a frequency domain. Let's say we take a simply sine function:

enter image description here

Then the frequency domain may look something like this (please ignore the fact that this one is centered at $0$ which it shouldn't be):

enter image description here

I understand how to interpret the big bump: this is the frequency of the sine function.

However, given that we can describe the "frequency" of the sine function with one number (e.g. $2\pi$), how do we interpret all the other smaller bumps on the frequency domain?

In other words, if the fourier transform would map a sine function to 1 number, this would make sense to me. But how do we interpret the fact that the fourier transform gives non-zero values for all kinds of frequencies, even those that are far removed from the sine function's actual frequency?

  • $\begingroup$ For a single sine wave there will only be a spike at two points (which are complex conjugates) with no other bumps. $\endgroup$
    – Ian
    Jan 30, 2018 at 21:10
  • $\begingroup$ You are rediscovering the Dirac delta "function" (generalized function, actually). en.wikipedia.org/wiki/Dirac_delta_function $\endgroup$ Jan 30, 2018 at 22:40

1 Answer 1


Your second picture appears to be the sinc function, which is the transform of a square wave - not a sine wave. You say it would make sense if the transform maps a sine function to 1 number... I understand your thinking here, but you need to remember that the frequency domain is essentially the frequency of complex sinusoids, not sine or cosine. Remember that:

$$ sin(\omega x) =\frac{i}{2\pi}(e^{i\omega x} -e^{-i\omega x}) $$

So the resulting Fourier transform has spikes at two points (as someone commented to your question), specifically at $\omega$ and $-\omega$.

To explain the smaller bumps, you can think of the Fourier transform as trying to force the time domain function into a representation as an infinite integral of complex sinusoids of the form $e^{i\omega x}$. So when you're approximating a function like a square wave, which obviously cannot be accurately created by adding sine and cosines together, you need infinitely many to form an approximation using complex sinusoids, and the "bumps" occur at frequencies that correspond to the frequency of the square wave. As an example, if the frequency of the time-domain square wave is increased, the frequency-domain image you linked would spread out to "line up" the frequencies.

  • $\begingroup$ That makes total sense. Does this interpretation still somehow work if we don't take the fourier transform as the integral from $-\infty$ to $\infty$, but on some bounded interval $(a,b)$? Because I think the reason I was confused, is that if you take the fourier transform on a bounded interval of a sine wave, then you don't get the dirac delta function. $\endgroup$
    – user56834
    Jan 31, 2018 at 6:35
  • $\begingroup$ The interpretation should still work - The transform is defined on $-\infty$ to $\infty$ though. The time-domain function could be a sine wave bounded to an interval and zero outside of that interval, and then it isn't a direct sum of complex sinusoids like the full sine wave was, so its transform won't be a set of dirac delta function(s). $\endgroup$
    – Paul
    Jan 31, 2018 at 15:30

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