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From: Rune Allnor <allnor@tele.ntnu.no>
Newsgroups: comp.soft-sys.matlab,comp.dsp
Subject: Re: DFT the same as sampled Foureir transform?
Date: 24 May 2007 10:15:33 -0700
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On 24 May, 18:56, "Mike" <meathea...@gmail.com> wrote:
> "Rune Allnor" <all...@tele.ntnu.no> wrote in message
>
> news:1180021783.330512.237760@w5g2000hsg.googlegroups.com...
>
>
>
>
>
> > On 24 May, 09:45, "Mike" <meathea...@gmail.com> wrote:
> >> Hi all,
>
> >> I have the following question regarding the relation between DFT and
> >> Foureir
> >> Transform.
>
> > Fourier. After Jean Baptiste Joseph Fourier. Read his biography
>
> >http://www-history.mcs.st-andrews.ac.uk/Mathematicians/Fourier.html
>
> > and maybe you get enough respect for him to spell his name correctly.
>
> >> Suppose I have a sequence of discrete time signal x0, x1, x2, ... xn, ...
> >> (possibly infinite length), uniformly spaced in time, with spacing T;
> >> that's
> >> to say, x0 is the signal value at time 0, x1 is the signal value at time
> >> 1*T, x2 is the signal value at time 2*T, ...
>
> >> and the DFT of this sequence is F1(v).
>
> > Wrong. The DFT is not defined for a sequence of "possibly
> > infinite length." The DFT is defined for discrete-time sequences
> > of *finite* length.
>
> >> Also, for this sequence of signal,
>
> > What is a "sequence of signal"?
>
> >> I have an ordinary Foureir Transform
> >> F2(v), I guess it's called DTFT.
>
> > No, you don't. I have no idea what a "sequence of signal" is.
> > The Dirscrete-Time Fourier transform is defined for a discrete-
> > time sequence of *infinite* length.
>
> >> I plan to sample the F2(v) to obtain the discrete version of the F2(v)
> >> and
> >> call it F3(v).
>
> > No need to do that, the discrete-time signals are already
> > "sampled". Sampling is a way to convert from a contionuous-time
> > signal to a discrete-time signal. This can be done for signals
> > of either finite of (formally) infinite duration in time.
>
> >> My question is:
>
> >> Under what condition and for what kind of signal x's do the DFT F1(v) and
> >> sampled version of ordinary FT F3(v) equate? I want F1(v) and F3(v) to be
> >> exactly the same... what conditions shall I impose?
>
> > There is an answer to such questions. Not the one you expect
> > or will be happy to hear, but an answer to questions such as
> > yours exists. Now, I took very great care to avoid "your
> > question" in the past sentence, because you don't have
> > the necessary basis to formulate the proper question.
> > Before asking again, take your time to read up on, and
> > contemplate, the different variations of the Fourier transform.
>
> > You will have four cases to consider:
>
> > 1) Countinuos time, infinite duration
> > 2) Continuous time, finite duration
> > 3) Discrete time, infinite duration
> > 4) Discrete time, finite duration
>
> > Once you have done that, you will be able to formulate
> > a question which makes sense and, consequently, can be
> > answered in a meaningful way.
>
> > Rune
>
> Thanks Rune! It's midnight so I was too sleepy. Yes, it should be "Fourier".
> Thanks for pointing it out!
>
> I agree my question is not well-posed. Here is a reformulation:
>
> Given a continuous time signal x(t), infinitely long. Sample it to obtain
> discrete time sequence x0, x1, x2, ..., xn, ..., infinitely long, with
> uniform samples spaced at T apart.
>
> Now I do two things:
>
> (1) Truncate the above sequence to make it finite, x0, x1, ..., xn, and take
> the DFT of the truncated sequence. Call the DFT F1(v). (Capitalized letters
> denote spectrum domain)
>
> (2) Without truncation, taking the DTFT of the infinitely long sequence x0,
> x1, ..., xn, .... Call the DTFT F2(v). And then take one period of F2(v),
> since it is periodic, and then sample F2(v) in the frequency domain to
> discretize it. Call the result F3(v), which is the discretized version of
> the one period of F2(v).
>
> ---------------------
>
> Both (1) and (2) yield vectors of length n in the spectrum domain,
> representing the discretized version of the spectrum.
>
> My question is: under what conditions do these two vectors of discretized
> spectrum equate?

IFF your infinitely long sequence is truly periodic AND
you happen to truncate it so that you have exactly one
period worth of data, then -- and only then -- your DFT
and your DTFT are equal.

There are only three problems with this:

1) Real-life data series from infinite-duration processes
   are never truly priodic.
2) Even if they are *almost* periodic, you can't expect
   to sample exactly one period's worth of data.
3) You are stuck with the DFT as a computational tool
   anyway, since it is highly impragtical to implement
   infinite summations and sample infinite amounts
   of data.

Rune