# Role of i in Fourier transform

I've seen several derivations of the Fourier transform, but most don't cover the conversion to the form

# $$S(f) = \int_{\infty}^{-\infty} s(t)e^{-i2\pi ft} \;\mathrm{d}t$$

What is the role of each of the parts of the exponent to $e$ ($i, 2\pi, f,$ and $t$) in breaking up one composite wave into component sine waves? $2\pi$ is most likely something to do with circles I take it, but what is its specific role in mapping one to the other, and how is its presence seen in the result of Fourier transform in producing the component sine waves?

An ideal answer also step that introduces -i to the expression.

• When $x$ is real, the curve of the function $f(x) = e^{i x}$ describes a circle in the complex plane. Is this the kind of answer you are looking for? I'm not completely sure what you are asking for...
– Eff
Jul 6, 2015 at 20:28
• @Eff, clearer now? If not, I'm happy to hear what you're confused about and try fix it Jul 6, 2015 at 20:31
• Because $e^{ix}=\cos x+i\sin x$ you can easily rewrite $\int_{-\infty}^\infty s(t)\sin 2\pi ft\,dt$ and $\int_{-\infty}^\infty s(t)\cos 2\pi ft\,dt$ as linear combinations of $S(f)$ and $S(-f)$. The reverse also holds. Jul 6, 2015 at 20:34

The Fourier transform maps the function s(t) in the time domain to a function S(f) in the frequency domain. We often call the function s(t) a 'waveform' and usually is some periodic function representing a wave (sound, radio, etc). What this means is that the function s(t) is broken down into a number (infinite possibly) of separate simple waves which have a single frequency. Physically, we can think of a sound from a musical instrument consistings of a base note or frequency with a number of harmonics, so we have a 'superpositioning' of a finite number of frequencies, perhaps all with different amplitudes which combine into a distinct 'tone' .

To extract the individual frequencies, we do a kind of 'correlation' of the original waveform with the simplest wave components, that is simple sine/cosine waves at each frequency - this is what is being expressed in the Fourier transform integral above. This is expressed by the term exp-i2*pi*f which is the complex representation of sin and cos. So, by integrating the waveform with the exponential, this gives us the component of the waveform for a particular frequency 'f'.

For a waveform based upon a single sine wave at frequency 'F', you should be able to see that the result of the integration will be a function zero everywhere except for F. As the operation is linear, if the waveform is a linear superposition of multiple sin waves of different frequencies, the result of integration will be a function which is zero everywhere except for the points corresponding to the individual frequencies. Hence the transform function S(f) represents the distribution of frequencies of the underlying waveform.

• The confusing thing about this perspective, at least for newcomers, is that the Fourier transform doesn't actually converge in the usual sense for a periodic function. It makes sense in the sense of distribution theory, but plenty of people want to study the Fourier transform and don't want to have to also study distribution theory.
– Ian
Jul 6, 2015 at 21:55

The $2 \pi$ is not essential; that just converts between angular frequency and ordinary frequency. There are definitions of the Fourier transform which don't use this factor at all. These are common in PDE theory.

The gist of what's going on is very much like the situation in Fourier series: if you have a complex-valued square integrable function $f$ and a complex-valued square integrable function $g$, then the orthogonal projection of $f$ onto the span of $g$ is $g \frac{\int_{-\infty}^\infty f(x) \overline{g(x)} dx}{\int_{-\infty}^\infty |g(x)|^2 dx}$.

If $F$ is the Fourier transform of $f$ then $F(\xi)$ is "morally" the coefficient in the projection of $f$ onto $e^{i 2 \pi \xi t}$. (The minus sign comes from the conjugate above, since in general $\overline{e^{ix}}=e^{-ix}$.) But note that we do not divide by $\int_{-\infty}^\infty |e^{i 2 \pi \xi t}|^2 dt$. This is because this is actually $+\infty$, which would be a problem. But this is the idea. The inversion formula makes this idea precise.

You might be asking "why do we want to project onto the span of $e^{i 2 \pi \xi t}$?" One answer is that, because of Euler's identity, it is a compact way of projecting onto sinusoids, which plays nicer with derivatives. Sinusoids are what we want to use to think about frequency, which is what the Fourier transform is all about.