Why are the basis functions for DFT so?

Question

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?

Answer 1

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.

Answer 2

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.