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This question already has an answer here:

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

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marked as duplicate by Dilip Sarwate, MBaz, Matt L., lennon310, A_A Dec 11 '18 at 13:02

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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 $n$.

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  • $\begingroup$ Thanks for the clarification but can we use any property of DFT to find this solution? $\endgroup$ – Mert Ege Dec 8 '18 at 21:13
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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).

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    $\begingroup$ shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot. $\endgroup$ – robert bristow-johnson Dec 6 '18 at 20:05

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