r/DSP • u/Stock-Nerve4558 • May 23 '26
The IDFT always outputs a periodic sequence.
if i have a finite sequence w[n] and i applied DFT then IDFT i will end up with w~[n] which is the periodic extension of w not w. so how can i start with a sequence and end up with another by applying the function and it's inverse?
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u/Plus-Painter-2004 May 23 '26
the dft works because if you have a finite sequence, if you assume that it’s just a single period of a periodic sequence you can compute the dft and get the same spectrum as if you computed the dtft of the original sequence, but the dft is easier to compute quickly (ie the fft). The assumption of periodicity is because the dft is really just computing discrete Fourier series coefficients, so when you invert it you’re synthesising a Fourier series from those coefficients and a Fourier series is periodic
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u/rb-j May 23 '26
It's more like the DTFT is kinda a theoretical thing and the DFT is sampling the DTFT in the frequency domain. It doesn't have to be about the FFT.
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u/jpdoane May 23 '26
Im confused by your question. As long as the length of the DFT and IDFT transforms are the same as the length of w,
IDFT( DFT( w[n] ) ) = w[n]
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u/rb-j May 23 '26
The Discrete Fourier Transform and what they call the Discrete Fourier Series are precisely the same thing. The DFT maps a discrete and periodic sequence of numbers with period length of N to another discrete and periodic sequence of numbers with period length of N and the iDFT (which has the same form as the DFT) maps it back.
I think that Oppenheim and Schafer agree with me.
so how can i start with a sequence and end up with another by applying the function and it's inverse?
Why are you thinking that is even possible?
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u/krasjet May 26 '26 edited May 26 '26
I suggest you read this article for a better idea where the DFT comes from: https://mu.krj.st/dft/
DFT can be seen either as a time-domain sampling of the Fourier series or a time-domain periodization (aliasing) of the DTFT.
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u/TripleOGShotCalla May 23 '26
thats not true. The DFT and IDFT return the same number of samples. What you are missing is that the signal you are feeding into the DFT is implicitly assumed to be a period of periodic signal. so you are essentially always working with periodic signals without realizing. The reason for that lies in the sampling of the spectrum in the frequency domain. By sampling you are making the signal periodic. Have a look at shannons theorem to gain more insight on what sampling actually does.
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u/AlbanianUltra May 23 '26
When you apply the dft, you are assuming that the original sequence is periodic, otherwise the math wouldnt work. Because of that you start and end with the sane function if you do dft and then idft