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I generate a Gaussian noise and then I filter it with a passband FIR Kaiser window filter. When I perform the Fourier transform of the output of the filter and plot its magnitude spectrum, it is filtered in the correct frequency band. Now, I perform a second Fourier transform on the output of the first Fourier transform (I'm performing it on a complex vector this time and in the frequency domain), and when I plot the magnitude spectrum I see something very similar to the original filtered noise. I know this isn't time domain, I know this spectrum is not even because the input signal is complex. But as the time plot of the filtered noise and the absolute value plot of the Fourier transform (of the Fourier transform of the noise) are visually similar a doubt arised on me.

  1. Why is the amplitude spectrum of the complex vector obtained after performing a Fourier tranform on the filtered noise once again occupying the whole vector (the whole frequency band) as if they were time samples spread in the time vector?
  2. Is the negative symbol in the inverse Fourier transform of the Fourier transform of the noise having no effect so that it is visually similar performing a Fourier transform of Fourier transform of noise and performing an inverse Fourier transform of Fourier transform of noise?

This doubt arised to me when I was trying to filter on a matched filter implemented in Matlab using IFFT(FFT(signal_1).*FFT(signal_1_inverted)) and I accidentally filtered the output of an already Fourier transformed signal and coudn't notice the difference (not on the output of the filter but on the output of the first FFT stage). And this leads me to another question.

  1. Suppose I would like to detect the likelihood of 2 signals by comparing their spectrums and I want to measure the likelihood by using a matched filter. Knowing that the convolution in time is the multiplication in frequency, is the convolution in frequency the multiplication in time and I get the same result if I use a matched filter (sort of convolution operation) on the spectrums than if I perform a sample by sample multiplication in time domain of the signals whose spectrums likelihood I want to determine?
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  • $\begingroup$ Related: this answer. $\endgroup$ – Matt L. Apr 3 '16 at 9:40
  • $\begingroup$ If F(F(signal(n))) = signal(-n) is what wonders you, I suggest you to look at Fractional Fourier. $\endgroup$ – Valentin Tihomirov Apr 3 '16 at 16:17
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I don't know what you've done wrong but applying the Discrete Fourier Transform (via FFT) twice to a time domain signal, should produce the original time domain signal in a circularly time reversed and linearly scaled by N form, as the Inverse Discrete Fourier transform is very similar to Forward Discrete Fourier transform except in a simple weight by N and the sign of the complex exponential, which reflects itself in the time reversal of the original signal...which is known as the Duality Property of DFT

Looking at the Inverse Discrete Fourier Transform of $X[k]$ we see: $$ x[n] = \frac {1}{N} \sum_{k=0}^{N-1} {X[k]e^{j \frac{2\pi}{N} k n}}$$

and by just changing the sign of the exponent $e^{j \frac{2\pi}{N} k n}$ into negative by a change in the sign of the $n$ of $x[n]$ what you get is:

$$ N x[(-n)_N] = \sum_{k=0}^{N-1} {X[k]e^{-j \frac{2\pi}{N} k n}} $$

which is N times the circularly reversed (obtained via modulus operator denoted by $(.)_N$) version of the original signal $x[n]$ that is obtained by an apparently forward but effectively inverse Discrete Fourier Transform applied to $X[k]$ treated as a time sequence, which is what you do at your second stage of applying forward FFT to $X[k]$ of $x[n]$

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  • $\begingroup$ What about question number 3? $\endgroup$ – VMMF Apr 5 '16 at 12:45
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    $\begingroup$ good question! If your intention is to perform a matched filtering on the "frequency domain signal" which I havent heard of much, then you would perform a convolution in frequeny domain and this is indeed equivalent to a multiplication in time domain...Select signals length properly to avoid circular aliasing. $\endgroup$ – Fat32 Apr 5 '16 at 14:15
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    $\begingroup$ Fat, if you understand the DFT and the DFS to be the same exact thing (which they are), then the DFT maps a discrete and periodic function of length $N$ to another discrete and periodic function of length $N$ and the iDFT maps it back. so $x[n+N] = x[n] \quad \forall n \in \mathbb{Z}$ and the same for $X[k]$. the "circularly reversed" is just a manifestation of that circularity. we can just say that the DFT of the DFT is $N \cdot x[-n]$ this is why some folks like the $\frac{1}{\sqrt{N}}$ scaling on both the DFT and iDFT. then applying the DFT twice simply reverses your input around $n$=0 $\endgroup$ – robert bristow-johnson May 23 '16 at 20:10
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    $\begingroup$ @robert bristow-johnson In one book the author describes DFT, X[k], as an aperiodic sequence defined for k=0 to k=N-1 for which a negative index of -k or -n is mapped into a valid ranged index by the use of modulus operator $(-n)_N$ so that the user can mathematically stay in the DFT domain without referring back to the original DFS as you stated. So if one always invokes the inherent relation between DFS and DFT (as he should), then $x[-n]$ is the choice (as I did in my original answer) but soon I realised that the reader may misinterpret it for an aperodic signal and then added the modulus $\endgroup$ – Fat32 May 23 '16 at 20:29
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    $\begingroup$ i think we're in agreement. i just sorta wanna repeat that when using the DFT, one is essentially periodically extending the finite-length (and as such, aperiodc) input to the DFT. in this question i get a little more fascistic about my insistence about formally modeling this periodic extension. $\endgroup$ – robert bristow-johnson May 23 '16 at 20:47
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The difference between an IFFT and an FFT is (merely) in the direction of the result vector (indexed forwards or backwards/reversed in time), and usually a scaling factor of N or 1/N. So you should see nearly the same thing if the input is so close to symmetrical looking that it doesn't look too different when reversed, and your plot auto-scales.

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  • $\begingroup$ it's really just the difference between "$+j$" and "$-j$". although they are negatives of each other, so they add to zero, qualitatively they are exactly the same. both have equal claim to squaring to be $-1$. $\endgroup$ – robert bristow-johnson Apr 3 '16 at 5:14
  • $\begingroup$ What about question number 3? $\endgroup$ – VMMF Apr 5 '16 at 12:45
  • $\begingroup$ (3) is unclear as stated exactly what you are asking. If clarified, it likely should be separate question, not mixed in one about IFFTs. $\endgroup$ – hotpaw2 Apr 5 '16 at 13:25

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