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From: "Mike" <meatheadIV@gmail.com>
Newsgroups: comp.soft-sys.matlab,comp.dsp
Subject: Re: DFT the same as sampled Foureir transform?
Date: Thu, 24 May 2007 09:56:14 -0700
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"Rune Allnor" <allnor@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?

Thanks again!