# What is the DFT of DFT of discrete signal [duplicate]

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

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 $$n$$.

• Thanks for the clarification but can we use any property of DFT to find this solution? – 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).

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