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


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

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


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

(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 $n$.

  • $\begingroup$ 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

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).

  • 1
    $\begingroup$ 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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