# Whether the job of Fourier Transform is just to convert signals from time domain to frequency domain only or more than it?

I am a beginner . We convert a signal in time domain to frequency domain by applying Fourier transform on the signal to obtain frequency and phase spectrum.

So,whether the job of Fourier transform is just to convert signal from time domain to frequency domain only (and Whose importance is limited by Heisenberg's uncertainty principle)?

Is it used just to compute only phase and magnitude spectrum in which both spectrums are localised only in frequency domain ?

• Well, yes, that is what it does, but this has nice "side-effects": Fourier transforming derivatives gives the F.T. of the function times $(\pm i \omega)^n$ where $n$ is the order of the derivative, turning differential equations into algebraic ones, and with the only problem being the inverse, but this can be solved using powerful complex analysis methods. Propagators aka Green functions, of incredible importance to physics, can also be found this way. F.T.-ing a convolution gives just a product of the transforms, so convolution is some form of a product of functions, very important. May 7, 2015 at 8:20
• Well... yes. Does it look too little to you? May 7, 2015 at 8:21
• @devraj: Jokes aside, it does not seem too little to me. Besides the engineering and physics applications (about which I know nothing, frankly), there are huge applications to pure mathematics, such as the theory of linear differential equations (see krvolok answer). It is impossible to answer your question in finite time! May 7, 2015 at 8:54
• Have a look at this book May 7, 2015 at 8:54
• @devraj: It's "The Fourier Transform and its Applications" by Bracewell. Wait for the link to be fixed or look for it directly May 7, 2015 at 9:09

The Fourier transform is the diagonalization of the differential operator $L=\frac{1}{i}\frac{d}{dt}$. If $e_{s}(t) = \frac{1}{\sqrt{2\pi}}e^{ist}$, then $$\{ e_{s} \}_{s=-\infty}^{\infty}$$ are the eigenfunctions $Le_{s}=se_{s}$ with real eigenvalue $s$, and these form a type of "continuous" orthonormal basis for diagonalizing $L$. That is, $$f=\int_{-\infty}^{\infty}(f,e_s)e_s\,ds,$$ which is analogous to the discrete Fourier series of orthonormal eigenfunctions $\{ e_n \}_{n=-\infty}^{\infty}$ of $L$ with $Le_n=ne_n$: $$f = \sum_{n=-\infty}^{\infty}(f,e_n)e_n.$$ Both of these expansions diagonalize the differentiation operator $L$ on their respect intervals. Because of this diagonalization, $$Lf = \int_{-\infty}^{\infty}(f,e_s)se_s\,ds,\\ Lf = \sum_{n=-\infty}^{\infty}(f,e_n)ne_n.$$ Parseval's equality holds, which is equivalent to the completeness of the set of eigenfunctions: $$\|f\|^{2}_{L^{2}(\mathbb{R})} = \int_{-\infty}^{\infty}|(f,e_s)|^{2}ds,\\ \|f\|^{2}_{L^{2}[-\pi,\pi]} = \sum_{n=-\infty}^{\infty}|(f,e_n)|^{2}.$$ All of this is made precise in spectral theory, and it agrees with how these expansions were originally conceived by Fourier. The transform is really a coefficient function for the expansions of $f$ in terms of the eigenfunctions. As such, it is a type of "orthogonal" projection of $f$ onto $e_s$ given through the coefficient function $\hat{f}(s)=(f,e_s)$. So, yes, it may be interpreted as a frequency; it's the coefficient of $e_s$ needed to reconstruct $f$ in terms of the basis functions $\{ e_s \}$.
Technically $e_s$ is not in the space, but any continuous sum of such elements, say $\int_{\lambda-\epsilon}^{\lambda+\epsilon}e_{s}ds$ is in $L^{2}(\mathbb{R})$, and is very nearly an eigenfunction with eigenvalue $\lambda$, which still leads to a discrete approximation in terms of near eigenfunctions, that would be interpreted as a wave packet by those who want to view this in terms of time functions $e_{s}(t)$. A Mathematician does not need to view this in terms of time; it's a function expansion in the continuous spectral elements of $L$, and is a diagonal representation of $L$ as multiplication by $s$, which is typical of selfadjoint operators on a Hilbert space, and is analogous to the diagonalization of selfajdoint matrices in its eigenvectors.
The comments already explained a lot, but I want to add that the Fourier Transform itself is exact to the fullest, i.e. the transform $F(\omega)$ of some $f(t), t \in \mathbb{R}$ is not subject to any uncertainty. Although in practice or physics, one does not have access or is not interested in $f(t)$ from $t = -\infty$ to $t = +\infty$ and wants to know the frequency spectrum at present time. But of course you cannot assign a frequency spectrum to a single instantaneous value $f(t_\text{now})$, that'd be an absurd endeavour (without dynamics, there are now frequencies). Therefore, you have to look at a certain past time window, which then leads to the concept of Short-time Fourier transform and the trade-off between resolution in the frequency and time domains (i.e., choosing a small time window only allows for a rough computation of spectrum and vice versa), which is very much related to the Heisenberg uncertainty principle (where you are interested in both the present location and velocity).