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When you get a DFT of a signal, you use the basis functions as:

$e^{-j2\pi kn/N}$

Why is it so? Why don't we use the conjugate, $e^{j2\pi kn/N}$, or any other function?

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    $\begingroup$ Because it's the definition of the Fourier's transform : represent any function as a sum of complex exponentials. see en.wikipedia.org/wiki/Fourier_analysis $\endgroup$ – MaximGi Mar 30 '16 at 8:11
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    $\begingroup$ Whether we use the conjugate or not is essentially a convention. As far as other functions, we do use all kinds of other functions. The discrete cosine transform, the Z transform, wavelet transforms, number theoretic transforms, etc. The Fourier transform is just the basis that most resolves frequency (at the cost of temporal locality). $\endgroup$ – Derek Elkins Mar 30 '16 at 8:50
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    $\begingroup$ In fact, your basis functions are $\exp(+2\pi i kn/N)$, the minus sign stems from the sesquilinear product on complex vector spaces: It is antilinear in the first argument and linear in the second. So the basis you expand into is conjugated. $\endgroup$ – Jazzmaniac Mar 30 '16 at 15:18
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    $\begingroup$ Jazz is right, the basis functions are $e^{j2\pi kn/N}$ and they could have just as well been $e^{-j2\pi kn/N}$. by convention, we call the imaginary unit "$j$" (except everyone else in the world, other than EEs call it "$i$"), but we could just as well as declared $-j$ to be the imaginary unit. there is no difference, in any of their properties, between "$j$" and "$-j$". both have equal claim to squaring to be $-1$. if we went to all of our textbooks and replaced every instance of $j$ with $-j$ (and vise versa), every theorem would continue to be just as valid. $\endgroup$ – robert bristow-johnson Apr 1 '16 at 4:26
  • $\begingroup$ You may also want to look at this question and this one. They got some nice info on this as well. $\endgroup$ – Gilles Apr 8 '16 at 9:20
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If you look at the idea of Continuous Time Fourier Series (CTFT), it says any periodic signal can be constructed by summation of infinite number of complex exponentials. But in descrete case , only 'N' different complex exponentials are enough , because there are only 'N' distinct complex exponentials exists (N is the period of discrete signal).

DFT is nothing but DFS (Discrete Fourier Series).So You can extend the same idea to get the answer.

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  • $\begingroup$ But in descrete case , only 'N' different complex exponentials are enough , because there are only 'N' distinct complex exponentials exists (N is the period of discrete signal). Why? $\endgroup$ – GrowinMan Apr 2 '16 at 21:45
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    $\begingroup$ consider the signal $e^(j2$pi$kn/N)$ . If u replace 'k' by 'k+N' u will get the same signal,which means there are only 'N' distinct harmonics in descrete exponential signals.But this is not true in continuous case. $\endgroup$ – spectre Apr 4 '16 at 8:23
  • $\begingroup$ That makes sense. So that helps me understand this question. But I still don't have a strong intuitive grasp of why those exponentials form the basis functions exactly. $\endgroup$ – GrowinMan Oct 13 '16 at 7:24
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Developing a bit more the previous answer, note that the definition of continuous FT is:

$X(f) = \int_{-\infty}^\infty x(t)\cdot e^{- i 2\pi f t}\,dt$

now, as stated in the previous answer, we can calculate the DFT from the previous equation as:

$X_T(f)\ \stackrel{\mathrm{def}}{=} \sum_{k=-\infty}^{\infty} X\left(f - k f_s\right) \equiv T \sum_{n=-\infty}^{\infty} x(nT)\ e^{-i 2\pi f T n}$

Now, reciprocally, we can come back to time domain by doing:

$x(t) = \int_{-\infty}^\infty X(f)\cdot e^{ i 2\pi f t}\,df$

Note that in the last equation, the exponential's phase is positive. Actually, that equation is also named the inverse Fourier Transform.

Now, the negative sign in the exponent indicates the integrated juxtaposed supplements' transpolation. These supplements can be analysed through the application of variance for each function.

Also, the FT can be understood as the scalar product between the function x (t) and the complex exponential $e ^ {i2 \pi\,ft}$ evaluated over the entire frequency range f. From the usual interpretation of the scalar product, in those frequencies at which the transform has a higher value, x (t) will look more similar to a complex exponential.

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  • $\begingroup$ Thanks for taking the time to answer this, but I'm afraid I'm probably lagging in my math. You lost me at integrated juxtaposed supplements' transpolation $\endgroup$ – GrowinMan Oct 13 '16 at 7:25

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