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We all know that for any, suitable, kind of signal $f(t)$, there corresponds a Fourier transform function $F(j\omega)$ such that $$ f(t) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} F(j\omega) e^{i\omega t} d\omega $$

So, $F(j\omega)$ can be recognized as the component of $f(t)$ along $e^{i \omega t}$. This is similar to orthogonal analysis. Like a cartesian vector $\vec{a} = (1,2,3)$ can be represented as $$ \vec{a}= 1\hat{x}+2\hat{y}+3\hat{z} $$ where $\hat{x}$, $\hat{y}$, and $\hat{z}$ are the orthonormal basis vectors (unit vectors along x-y-z directions).

However, these two things are different in one perspective.Among $x$,$y$,$z$, there is no concept as 'distance'. For example, in Fourier transform. we can say that the distance between $\omega_1 = 1$ and $\omega_2=2$ is smaller than the distance between $\omega_2$ and $\omega_3 =4$.

This algebra structure in Fourier transform definitely has some extra meaning, but what is it? Does it mean that the Fourier transform makes "more sense" compared with other orthonormal basis analyses where there is no easily-definable distance between basis functions?

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closed as primarily opinion-based by Peter K. Jun 20 '16 at 19:02

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ what's "normal spectrum analysis"? $\endgroup$ – Marcus Müller Jun 20 '16 at 10:09
  • $\begingroup$ @MarcusMüller I mean some orthonormal basis without 'distance' structure $\endgroup$ – user1297181 Jun 20 '16 at 10:12
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    $\begingroup$ a basis is what gives you the ability to represent signals as sets of coefficients with respect to that basis. Having a basis and a length function defined thereupon hence directly implies measurability of the space spanned by that basis. $\endgroup$ – Marcus Müller Jun 20 '16 at 10:55
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    $\begingroup$ well, the fact that you can assign a quality "frequency" to Fourier basis vectors is what is missing for your cartesian coordinate system, indeed. But you want to do frequency analysis, so for every sensible basis to do so, your basis will have some frequency-related property, and hence, you could declare the frequency differences to be "distances". $\endgroup$ – Marcus Müller Jun 20 '16 at 11:19
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    $\begingroup$ I'm voting to close unless we can get a clear definition of what you mean by "special" or "more sense" (my interpretation in the edit to your question). $\endgroup$ – Peter K. Jun 20 '16 at 19:03
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Let $v(x, k)$ be an orthonormal set of basis functions such that $$ \int_{-\infty}^{+\infty} v(x,a) v^*(x,b) dx = \delta_{a,b} $$ where $\delta_{a,b}$ s the Kronecker delta.

I believe your question boils down to: for the Fourier Transform, we have $$ v(x,a) = e^{jax} $$ and we interpret $a$ as the frequency and $x$ as time; is there anything special about this orthonormal basis because we can interpret $a$ as a "distance" between these functions?

An example of another set of functions that are orthonormal (but over a finite interval) are the Legendre Polynomials: $$ v(x,a) = P_a(x), \mbox{where} -1\le x \le +1 $$ where now $a$ is the order (degree) of the polynomial.

For these (and other polynomial bases), the "distance" is now the difference in polynomial degree.

Whether this means such functions are "less" special than the Fourier bases is a matter of opinion.

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Great question, and a really tricky one to get your head around. I think your dual use of the word distance, and the comparison of vectors in an infinite function space to a vector in a finite-support sequence is a bit confusing.

The distance you're describing is the difference between the frequencies $w_1$, and $w_2$ of two sinusoids. In the time-domain it's easy to understand this difference because the two signals are completely localised in frequency (but specified for infinite time).

Once you've performed the Fourier transform, these signals are represented in a infinite space where sinusoids of any frequency are orthogonal to sinusoids of any other frequency. At this point, as you rightly describe, the sinusoids form an infinite orthogonal (basis) set. It feels strange, but in this space the inner product of 1Hz and 2Hz is the same as that of 1Hz and 4Hz. They're both zero! I think this is your other interpretation of distance - where a short distance would be a large inner product and a long distance a small inner product.

The reason this feels wrong is that we can still intuitively arrange the components of this infinite space in order of increasing $w$, so one component can still be 2Hz away from another. This is not the same as the distances represented by the inner products in the space generated by the Fourier transform.

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