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Fourier transform of a continuous signal is just the projection of the signal on the sinusoidal family for imaginary part and the same family with phase offseted by a quarter of period for the real part.
Beside the magical Euler's formula $e^{ix} = \cos x + i\sin x$ that ease the analysis of the transform and the fact that sinusoidal functions happen to be the solution of a prefect harmonic oscillator, I always wondered where the Fourier supremacy came from. Sinusoidal signals aren't so common in everyday's life. Even with human speech, vocal tract function isn't sinusoisdal.
For example, one transform basis that would seams interesting to me would that with a saw tooth function as signals with fast attack and slow release aren't uncommon...
I guess that the transform need to be inversible and that might not be the case for every basis, but aren't there any alternative basis that provide a bijective forward and inverse transform ? (And if so, what would be the conditions ?)

I will elaborate a little. I'm looking for a basis family of periodicals functions named $B(t)$ for example with B of unitary periods such as : $$ S(f) = T\{s(t)\}(f) = \int s(t) B\left( t * f\right) dt + i \int s(t) B\left((t - \frac{1}{4})*f\right)dt$$ with T inversible in some way...

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    $\begingroup$ The Fourier basis is the eigenbasis for time invariant linear systems. That's why it's so useful, not the fact that signals are sinusoidal. $\endgroup$ – Jazzmaniac Jul 5 '14 at 12:39
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The most important reason for the "Fourier supremacy", as you called it, has been pointed out by Jazzmaniac in his comment. The complex exponential $e^{st}$ with $s=\sigma+j\omega$ is an eigenfunction of the convolution operator:

$$e^{st}*f(t)=\int_{-\infty}^{\infty}e^{s(t-\tau)}f(\tau)d\tau= e^{st}\int_{-\infty}^{\infty}e^{-s\tau}f(\tau)d\tau=e^{st}F(s)$$

where the eigenvalue $F(s)$, if it exists, is the Laplace transform of $f(t)$, and for $\sigma=0$, i.e. for $s=j\omega$, $F(j\omega)$ is the Fourier transform of $f(t)$ (again, if it exists). Even though there are several important differences between the Fourier and the Laplace transform, they are deeply related, and for answering your question their differences are not important. By analogy, the same is true for the discrete-time Fourier transform and for the $\mathcal{Z}$-transform in the discrete domain. So when in the remainder of my answer I mention the Fourier transform, you might as well replace it by Laplace transform or $\mathcal{Z}$-transform, and everything will remain valid.

So why should we be concerned with eigenfunctions of the convolution operator? Because convolution describes the input-output relation of linear time-invariant (LTI) systems, and these systems are often very good approximations of practically important systems such as frequency selective filters, integrators, differentiators, etc.

Apart from being an invaluable tool for the analysis of LTI systems, the Fourier transform also has an important physical interpretation as the spectrum, i.e. the frequency content, of a signal. Furthermore, the FFT is a highly efficient algorithm for the computation of its discrete version for finite length signals.

So why don't we just say we're doing fine with the Fourier transform and its siblings? Because there are applications where we are looking for other properties of a transform. One example would be time localization in signal analysis, since the basis functions of the Fourier transform have no time localization at all because of their infinite extent. In such cases other transforms have become widely used, such as the short-time Fourier transform and the wavelet transform. For discrete-time signals, filter banks provide very general signal expansions. They can be designed to match the desired properties of the signal expansion.

Since you mentioned the saw-tooth function as a possible basis function I would like to point out to you the Slant transform (mentioned in this answer), which is used in image coding. Furthermore, there is the related Slantlet transform, which is a special type of discrete wavelet transform with piecewise linear basis functions. You can see the basis function in Fig. 5 of this paper.

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  • $\begingroup$ See comments on adel bibi's answer, if not evaluated on unit circle, z transform and fourier transform are not equivalents, aren't they ? Is this then an appropriate start for establishing new basis? $\endgroup$ – user9020 Jul 9 '14 at 6:51
  • $\begingroup$ @user9020: I wrote that there are important differences between the Fourier and the Laplace transforms (or - equivalently - between the discrete Fourier transform and the $\mathcal{Z}$-transform), but these differences have no bearing on your question. The basis functions are in all of these cases complex exponentials, either with or without a real-valued component in the argument of the exponential. But still, it's one class of basis functions and there's nothing new to be discovered here. $\endgroup$ – Matt L. Jul 9 '14 at 7:07
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I'll try to derive the necessary and sufficient conditions for finding a basis of periodic functions for a discrete signal space, i.e. the set $S_N:=\{f:{\mathbb{Z}}/{N\mathbb{Z}}\to\mathbb{C} :\sum_n |f(n)|^2 <\infty\}$ of all square magnitude summable $N$-periodic function from the integers to the complex numbers. The continuous case is interesting too, but out of the scope of a simple answer.

There are two statements about the basis we can immediately see to be true:

1) The number of basis functions is identical to the number of fourier basis functions (assuming complex basis functions).

2) The periods of the basis functions must divide $N$, i.e an integer number of basis periods must fit into the period $N$ of the signal.

Note that the second condition does not always lead to integer sample periods for the basis functions. The periodicity of the basis functions is therefore realized as discretely sampled function of periodic continuous functions, and the discrete representation itself is not necessarily strictly periodic.

The two statements above allow us to enumerate the basis functions. The number of integer divisors of $N$ is surely not greater than $N$, which implies that all integer divisor periods are realised as base function periods. Let's assume $N$ is even ( I leave the odd case to you ), then $N/2$ is an integer divisor and realised as a basis function period. This basis function is even exactly periodic on the discrete sample grid and can be written as the number sequence $(a,b,a,b,\dots)\in\mathbb{C}^N$ with $N/2$ repetitions.

We can write this basis vector as a complex Fourier series, and we find that only the DC and the Nyquist coefficients do not vanish. That means the basis function can be written as a linear superposition of the periodic continuations of $(1,1)$ and $(1,-1)$.

At this point, we can assume that your basis functions have zero mean, just to make things a little simpler. In this case the DC component of the Fourier decomposition also vanishes and the basis function looks just like the corresponding Fourier basis function. There is also no other linear independent basis function of that same period.

From here on, things get simpler. The next basis function candidate has fits one more period into $N$, that means $N/2+1$ repetitions and a period of $N/(N/2+1)$. The basis functions candidates for this period have a Fourier decomposition (again, zero-mean is assumed) that is only non-zero at Nyquist and Nyquist-1. This is only linearly independent from the basis function we already know if the coefficient at Nyquist-1 does not vanish. That also means we can have two independent basis functions for this period (This will typically be one basis function and its complex conjugate).

We can continue with this construction by adding another basis period, looking at the Fourier series, enforce linear independence and finding two independent basis functions. This procedure can be repeated $N/2-1$ times, and together with the constant basis function responsible for the DC component we get exactly the $N$ basis functions that we require for invertibility.

The constructed basis is precisely then a "transformation" (i.e. a bijective map) if the assumptions we made in the process are true. The necessary conditions were, that the lowest order Fourier coefficient of your periodic basis, expanded in its own intrinsic period, doesn't vanish. We also assumed, that the basis was zero-mean (apart from the constant basis function) and that $N$ was even. These are not necessary but just convenient. You can drop them if you want.

So let's summarise. A periodic function gives rise to a basis on a discrete space iff the Fourier coefficient of the base period is non-zero. Then the basis implies a discrete Fourier-like transformation, i.e. it is invertible.

Cheers,

Jazz

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There's plenty. Laplace transform is just one example. There's uncountably many bijective transformations. So the question would be if there are other useful transforms. Turns out, there are some.

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  • $\begingroup$ I am afraid that OP is asking for different basis functions for Fourier Transform, i.e. square waves. $\endgroup$ – jojek Jul 5 '14 at 20:01
  • $\begingroup$ So how would you characterize these bases suitable for a "fourier" transform in your sense? A square wave basis is not exactly trivial either - it appears to be ill-defined in the continuous case, and in the discrete case, it's plainly hopeless. Maybe the OP is interested in wavelet bases, which are a different breed, though. $\endgroup$ – user7358 Jul 5 '14 at 22:12
  • $\begingroup$ You catch my point - it's not about the Laplace transform. $\endgroup$ – jojek Jul 6 '14 at 8:26

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