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I have seen both the formula of Fourier transform with positive and negative sign on exponential as $$ X(\omega)=\int_{-\infty}^{\infty} x(t)e^{-j\omega t}dt$$ and $$ X(\omega)=\int_{-\infty}^{\infty} x(t)e^{j\omega t}dt$$ I am confused which one is the correct formula. I also solved for Fourier transform by taking the following example $$x(t)=\begin{cases} 1, \hspace{5mm} \text{for} \hspace{2mm} |t|<1 \\0, \hspace{5mm} \text{for} \hspace{2mm} |t|>1 \end{cases}$$ and got the same result as $$ X(\omega)=\begin{cases} 2\frac{\text{sin}\omega}{\omega}, \hspace{5mm} \text{when} \hspace{2mm} \omega \neq 0 \\2, \hspace{13mm} \text{when} \hspace{2mm} \omega = 0\end{cases}$$ Can anyone explain whether both the formula for Fourier transform are correct or not?


marked as duplicate by Matt L., jojek, Peter K. Oct 5 '15 at 20:11

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The definition with the negative in the exponent is the accepted definition of the Fourier transform... however, this is an arbitrary choice. It could just as easily be defined with $e^{jw}$ and the inverse transform with $e^{-jw}$.

  • $\begingroup$ in fact, while they are negatives of each other, there is no other difference between $-j$ and $+j$. both have equal claim to being the $\sqrt{-1}$. $\endgroup$ – robert bristow-johnson Oct 5 '15 at 0:06
  • $\begingroup$ what will happen if we consider only the positive value of $t$? i.e. to say range of integration is consider from $0$ to $\infty$. $\endgroup$ – J Cian Oct 5 '15 at 1:45
  • $\begingroup$ What was the historical reason for this arbitrary choice to be the one used in the common definition? $\endgroup$ – hotpaw2 Oct 5 '15 at 4:15

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