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Forward Fourier Transform $\hat f(x)$ is defined such as $$ \hat f(x) = \int_{-\infty}^{\infty}{f(t) e^{-2\pi t x i} dt} $$

but I am wondering what happens if I define it $$ \hat f(x) = \int_{-\infty}^{\infty}{f(t) e^{2\pi t x i} dt} $$

Note the missing minus sign from the exponential. Obviously, if I do this I will have to define Inverse Fourier Transform with a minus sign in order to get the original signal back.

My guess is that this will provide negative frequencies but nothing else would change; am I right?

And then, why does this happen? I would expect that $e^{ai}$ to create a positive frequency while $e^{-ai}$ create a negative but the Fourier Transform seems to use them the other way around.

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This is just a matter of choice. The theory and applications of the Fourier transform can be carried out with any of the signs. Something similar occurs with the $\pi$ in the exponent.

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  • $\begingroup$ Would you please like to detail more on what you mean with "something similar occurs with the $\pi$"? $\endgroup$ – Memleak Feb 10 '14 at 16:32
  • $\begingroup$ Depending on the field of application, the Fourier transform is defined with $e^{-ix}$, $e^{-i\pi x}$ or $e^{-i2\pi x}$. $\endgroup$ – Julián Aguirre Feb 10 '14 at 16:39
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    $\begingroup$ From my understanding Fourier Transform is using $2\pi x$ in order to use a different period. If you already know the period of your signal you can just replace the whole thing with that value and it should work the same. $\endgroup$ – Memleak Feb 10 '14 at 16:45
  • $\begingroup$ No, it is really just different conventions. The $\exp(-i ωt)$ variant is for angular velocity in radians per sec, the $\exp(-2πi\,ft)$ variant for transforms relating to frequency in cycles per sec. The minus sign relates to the fact that the original aim is to represent the function as sine waves, i.e., that the inverse Fourier transform is the primary inspiration. $\endgroup$ – LutzL Feb 11 '14 at 17:12

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