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I am a total beginner in this topic of DFT. I get that the series must be finite for DFT calculation. But everywhere we are assuming that this series must be limited from $0$ to $N-1$. How to evaluate DFT for a sequence extending from $n=2$ to say $n=N+1$?

I am following this video

https://www.youtube.com/watch?v=GDFTb-BwA0o

Here in the minute 21:20 to 21:50, Professor seems to be really strict about this particular range. Why is this so?

robert bristow-johnson
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Souhardya Mondal
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  • it's not limited as such. there is no operational difference between the DFT and the Discrete Fourier Series. the DFT maps a discrete and periodic sequence of numbers having period $N$ in some domain (e.g. "time domain"), to another discrete and periodic sequence of numbers having the same period $N$ in the reciprocal domain (e.g. "frequency domain"). – robert bristow-johnson Mar 23 '20 at 16:56
  • @robertbristow-johnson I had a doubt regarding your comment. You said that DFT maps N-point discrete time domain sequence to N-point discrete frequency domain sequence but we see examples of 128 point DFT of a 16 point sequence. Is that DFT after zero padding the time domain sequence? – DSP Novice Mar 24 '20 at 04:00
  • there is nothing about the DFT that zero pads samples. it's not in the natural definition, but only in the contrived definition that explicitly pads zeros. but you have to remove the zero padding, then periodically extend the $N$ samples if you are going to use any DFT theorems that cause (circular) translation or convolution. read my other answer that i pointed to. it's quite simple. the DFT maps a discrete and periodic sequence of numbers having period $N$ in some domain to another discrete and periodic sequence of numbers having the same period $N$ in the reciprocal domain. – robert bristow-johnson Mar 24 '20 at 14:50

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How to evaluate DFT for a sequence extending from n=2 to say n=N+1?

Formularistic: trivial variable substitution of the running variable $n$ by $\tilde n = n+2$, done.

Looking at it as a signal: When doing an DFT, it helps to think of it simply as mapping of $\mathbb C^N\mapsto\mathbb C^N$, i.e. a complex vector goes in, same size complex vector comes out.

Your "shifted" input hence is simply a different input value for the DFT.

Now, when we interpret the DFT as a mapping between discrete time- and discrete frequency-domain, we have to assume that the time signal is periodic with the DFT length. That means the two "new" values are just the oldest two values. I.e., your reindexing just leads to a circular shift of the values in the vector.

Marcus Müller
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