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Given a sinusoidal signal with a frequency $\omega$, the Fourier transform ought to simply be a delta function at $\omega$. However, what happens to multiples of these? MATLAB's FFT shows nothing at $\omega/2$ for example but I'm wondering how this distinction is made.

EDIT: Clarifying a little, perhaps talking about MATLAB is causing confusion. What I mean is this: If I take a wave with period $T$, it is also true that it has a period 2$T$, $3T$ and so on. That is, it has frequencies of $\frac{2\pi}{T}$, but also $\frac{2\pi}{2T}$ and so on. How is it that in Fourier analysis, we don't see these smaller frequencies and instead only see the largest one?

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  • $\begingroup$ I'm not sure I understand; if I read this right, why does $e^{i\,\omega\,t}$ not show a spectral component the higher harmonics? Is that your question? Actually, in the discrete Fourier transform it sometimes can owing to aliasing. So if your question's about this kind of thing, then please tell us the sampling interval, and number of points in the FFT. This is the minimum information needed to know what an FFT will do. $\endgroup$
    – Selene Routley
    Commented Jun 17, 2016 at 12:21
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    $\begingroup$ Consider a signal $x(t) = \exp(i \Omega t)$. Now Fourier transform: $\tilde{x}(\omega) \equiv \int dt x(t) \exp(-i \omega t) = \int dt \, \exp(i (\Omega - \omega) t ) = 2 \pi \delta(\Omega - \omega)$. So that's it, you get a delta function at $\Omega$, not at $2 \Omega$. In other words, $\exp(i \Omega t)$ has frequency $\Omega$; it does not have frequency $2 \Omega$. $\endgroup$
    – DanielSank
    Commented Jun 18, 2016 at 7:12

3 Answers 3

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To keep things mathematically both precise and simple, let's stick with the discrete Fourier transform. Signals are vectors of $N$ complex points, where $N$ is the dataset's length. The dimension of this vector space is $N$.

In this setting, the Fourier transform is simply a resolution of a signal, thought of as a vector, into components with respect to a certain basis of unit vectors. The fact that these basis vectors are periodic is secondary.

The "simplest" and most obvious basis for our vector space is the set of signals

$$\begin{array}{cccccc} (1,&0,&0,&\cdots,&0,&0)\\ (0,&1,&0,&\cdots,&0,&0)\\ (0,&0,&1,&\cdots,&0,&0)\\ &&\vdots&&&\\ (0,&0,&0,&\cdots,&1,&0)\\ (0,&0,&0,&\cdots,&0,&1)\\ \end{array}$$

These vectors span the vector space i.e. you can write any vector as a superposition of them, and they are linearly independent, which means that none of them is the superposition of the others. So they are a minimal set needed to build all possible signals by superposition out of.

The Fourier transform is simply the change of basis of a vector so that it is now written as a superposition of of signals of the form $\left\{\frac{1}{\sqrt{N}}\,\exp\left(i\,\frac{2\,\pi\,j}{N}\right)\right\}_{j=0}^{N-1}$.

These new basis vectors also span the vector space, and are linearly independent. So in this basis, $\exp\left(i\,\frac{2\,\pi\times 6}{N}\right)$ has only one component: namely $\exp\left(i\,\frac{2\,\pi\times 6}{N}\right)$; it cannot, as a superposition, contain any other components because the basis is linearly independent. So it cannot contain a component of the distinct $\exp\left(i\,\frac{2\,\pi\times 3}{N}\right)$, even though this one has twice the former's period. The periods are irrelevant given the linear independence of the basis.

So think of the Fourier transform as a change of basis in a vector space rather than primarily defined by periodicity. Indeed the Fourier transform is a unitary change of basis insofar that it conserves the power of signals. So it's kind of like a generalized rotation.

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A sine wave at $\omega$ is periodic in $\omega /2$, but that does not mean that a sine at $\omega $ has frequency components at $\omega /2$. Another way of looking at it is noting that you cannot synthesize a sine at $\omega $ as the sum of sines of other frequencies.

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  • $\begingroup$ I like it: much more concise than my answer, which basically amounts to the same thing as your second sentence. $\endgroup$
    – Selene Routley
    Commented Jun 18, 2016 at 22:54
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I don't know what is happening there. It always worked for me. If you have large number of cycles with smooth variation i.e. large time scale with small time interval you will definitely see the frequency components. First of all matlab stores its frequency components like. 0 to $\omega_ {max}$ then $-\omega_{max}$ to 0. Hence if you want your zero frequencies in the middle use function fftshift. Take the abs() of the Fourier transform and you get the spectrum. You will see frequencies at $\pm\omega$. If you can not resolve them try to increase the time length(decrease frequency step) and increase time step (decrease max frequency).

If you want single frequency use $\exp(i\omega t) $

If you still cannot see the frequency components there might be aliasing effects. To work around them multiply your high frequency signal with a slowly varying envelope (may be gaussian) which makes the amplitude near end of array small.

EDIT: I think the gist of original question is changed.

In my opinion now the question is about the understanding of the frequency of a wave. Frequency in a wave (in the case of pure sine wave) is just peak counting, if the time period of a wave is T then you will count 1/T peaks of the wave every second which corresponds to frequency $\frac{\omega}{2\pi}$, now if you close your eyes at every alternative peak you will get the frequency $\frac{\omega}{2.2\pi}$, but it will not present the correct picture.

However if the wave is non-symmetric or non sinusoidal you can decompose it into its Fourier components and then you will see many frequency components.

Hope this will help

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