I have some queries about the Fourier transform

  1. In most of the cases, the Fourier transform of a signal is symmetric with respect to positive and negative frequency. I think the computational complexity increases because only half of the symmetric spectrum (i.e. spectrum for only positive frequencies) is useful. Also, while working in the frequency domain we could get wrong value of energy or power due to the spectrum on negative axis.

  2. In the Fourier transform formula the limits of integration are from $-\infty$ to $\infty$. However, for a signal which is continuously or exponentially increasing with time, one can't compute it's Fourier transform.

  3. After computation of Fourier transform of a signal, we get phase and frequency spectrum of the whole signal which is localized in frequency domain only. But from both of these spectrum, we don't get any spatial component features like which frequency component is present at which time (and same with the phase value).

  4. After computation of Fourier transform of signal, with dc and positive frequencies we also get unnecessary negative frequency components. I think concept of negative frequency doesn't exists practically.

So I have observed above shortcomings within the Fourier Transform but don't know whether they are right or wrong .So can anybody give explanation on any of above doubts?

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    $\begingroup$ ...none of these read like questions, they're statements. $\endgroup$ – ACuriousMind May 7 '15 at 20:14
  • $\begingroup$ Re I think concept of negative frequency doestn't exists... Negative numbers don't "exist" either, but we can assign meaning to them. $\endgroup$ – Solomon Slow May 7 '15 at 21:39
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    $\begingroup$ @jameslarge the Fourier transform is most certainly defined for non-periodic functions. You can, for example, trivially compute the Fourier transform of a Gaussian function. I think you're confusing Fourier transform and Fourier series. I strongly recommend deleting the comment in which you say the Fourier transform is defined only for periodic functions. $\endgroup$ – DanielSank May 8 '15 at 0:08
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    $\begingroup$ Welcome to Physics Stack Exchange! These are interesting questions. However, it is much preferred that each post contain one question. It would be better for you to remove all but one question from here and put the others in their own post. If you do this you are much more likely to get good answers. $\endgroup$ – DanielSank May 8 '15 at 0:13
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    $\begingroup$ This also sounds like a matter for Signal Processing or Mathematics. I don't really see the physics in it. $\endgroup$ – David Z May 8 '15 at 12:49

I assume from your question, you are concerned with the Fourier transform of a scalar function of time (no space dependence): $$\tilde f(\omega) = \int_{-\infty}^\infty dt\, e^{-i\omega t} f(t)$$ With the inverse: $$f(t) = \frac{1}{2\pi} \int_{\infty}^\infty d\omega\, e^{i\omega t} \tilde f(\omega).$$

  1. This symmetry is due to the fact, that your signal is real ($f^*(t) = f(t)$); for a real signal you get: $$\tilde f^*(-\omega) = \int_{-\infty}^\infty dt\, e^{i\omega t} f^*(t) = \int_{-\infty}^\infty dt\, e^{-i\omega t} f(t) = \tilde f(\omega).$$ I do not understand the remarks about computational complexity. Assuming you are referring to discrete Fourier transformation (what a computer does), then the answer is yes, when you know your signal is real, you only have to do half the work (and most FFT packages supply an extra function to calculate the Fourier transform of real inputs). Due to the discussed symmetry you get the power wrong at most by a factor of 2 (and the physical spectral power density is in any case just proportional to $|f(\omega)|^2$), for complex signals you will indeed have to consider all frequencies (positive and negative) for useful results.
  2. Sure you cannot compute the Fourier transform if the involved integrals are not converging. But usually you can add a convergence generating factor by considering a modified function $g(t) = f(t)e^{-\varepsilon |t|}$ and extract your results from this (letting $\varepsilon \to 0^+$ in the end of the calculation), this scheme does obviously not work for an exponentially increasing signal (but such is a signal is clearly unphysical).
  3. This is absolutely true. And there is a fundamental limit on the sharpness with which you can extract temporal vs. frequency features of a signal (known as uncertainty relation): $$\Delta t \Delta \omega \ge \frac 1 2.$$ The exact factor depends on the precise definition of $\Delta\omega$ and $\Delta t$ (my formula assumes $\Delta x = \sqrt{\left<(x - \left<x\right>)^2\right>}$). That is, you can not clearly define the frequency of a very short signal!
  4. The concept of negative frequency makes sense in various settings, usually related to some kind of forward vs. backward propagation, negative frequencies arise even more naturally in quantum mechanics. In the Fourier transform it comes about, because you consider the decomposition of the signal into it's $e^{i\omega t}$-components. If you would only consider positive frequencies (that is forward propagating), you could not represent real signals with this formalism (but as discussed above for a real signal the negative frequency values only deliver redundant information, whereas for complex ones they are relevant).
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    $\begingroup$ Mentioning quantum mechanics here is possibly misleading. The negative frequency part of a Fourier transform is meaningful if the time domain function is complex. Reference to quantum is not needed (although I agree it's interesting!). Might be better to instead give an example of where negative frequency is meaningful in a vanilla signal processing context. $\endgroup$ – DanielSank May 8 '15 at 0:11

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