What is the DFT of DFT of discrete signal

Question

What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?

Answer 1

let

$$\begin{align} X[k] &= \mathcal{DFT} \Big\{ x[n] \Big\} \\ &\triangleq \sum\limits_{n=0}^{N-1} x[n] \, e^{-j2\pi nk/N} \end{align} $$

and

$$ y[n] \triangleq X[n] $$

(note the substitution of $n$ in for $k$.) then

$$ Y[k] = \mathcal{DFT} \Big\{ y[n] \Big\} $$

then, if the DFT is defined the most common way (as above):

$$ Y[n] = N \cdot x[-n] $$

where periodicity is implied: $x[n+N]=x[n]$ for all $n$.

Answer 2

Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).

Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).