I am new to DSP & Radar domain. I have one question. Say I have a discrete time signal $x[n]$ and I take an FFT of it say the output is $X[k]$. The result what i get is frequencies present in $x[n]$.

Now what will happen if I take fft on $X[k]$ ? How would the spectrum look and what would you interpret from it?

thanks in advance !

  • $\begingroup$ You probably know the IDFT (let's talk about DFT instead of FFT. FFT is just a method of calculating the DFT); now, compare the DFT and IDFT formula. Done! $\endgroup$ – Marcus Müller Apr 12 '17 at 9:09

Essentially, what you do is performing the DFT twice. Performing DFT twice amounts to time-reversing the signal:

N = 16; sN= np.sqrt(N)
x = np.random.randn(N)

X = np.fft.fft(x) / sN

x2 = np.fft.fft(X) / sN


enter image description here

blue is original signal, green is two-times DFT. As you can see, the signals are time-reversed in the sense of $y[n]=x[-n \mod N]=x[N-n], n=0...N-1$. (y is green, x is blue). Here, I used the unitary version of the DFT, to keep the signals in the same scale. Otherwise, with the "normal" definition of DFT, the green signal would be scaled by 16.

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Maximilian Matthé's answer has all the necessary information, viz., $$y[n]=x[-n \mod N]=x[N-n], ~ n=0, \ldots, N-1.$$ This corresponds to the continuous-time result which says that taking the Fourier Transform of $x(t)$ twice gives us $x(-t)$ -- the time-reversal of $x(t)$ -- a result that I used to illustrate in the "good old days" when presentations used viewgraphs by taking a viewgraph showing $x(t)$ off the projector, flipping it over, and putting it back on the projector while saying "This is what $x(-t)$ looks like". But this brings up a subtle difference between the continuous-time case and the discrete-time case, namely, that if we look at the DFT as a linear transformation mapping a vector $\mathbf x$ to a vector $\mathbf X$:

\begin{align}\operatorname{DFT}(\mathbf x) &= \mathbf X\\ &\text{i.e.}\\ \operatorname{DFT}\bigr(\left(x[0], x[1], \cdots, x[N-1]\right)\bigr) &= \left(X[0], X[1], \cdots, X[N-1]\right),\end{align} then $$\operatorname{DFT}\bigr(\operatorname{DFT} \big(\left(x[0], x[1], \cdots, x[N-1]\right)\big) \bigr) = \left(x[0], x[N-1], x[N-2] \cdots, x[2], x[1]\right)$$ and not $\left(x[N-1], x[N-2] \cdots, x[2], x[1], x[0]\right)$ which is what one would get by writing the entries of $\mathbf x$ in reverse order. Just as well, because flipping over a viewgraph on which is written $\left(x[0], x[1], \cdots, x[N-1]\right)$ results in an unreadable mess that is also wrong to boot!

The $0$-th bin is a fixed point of the transformation $\mathbf x \to \operatorname{DFT}\bigr(\operatorname{DFT} \big(\mathbf x\big)\bigr)$; it has the same value after taking the DFT of the DFT of $\mathbf x$. When $N$ is an even number, $N/2$ is also a fixed point of the transformation, and has the same value after taking the DFT twice.

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  • 2
    $\begingroup$ Good that you point out this special property that time-reversal in discrete is not just reading the values backwards. This took me some time to understand in my undergraduate times :-) $\endgroup$ – Maximilian Matthé Apr 13 '17 at 17:22

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