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Hi group, This is an academic question, so please ignore practical considerations. What I would like to do (as a first step in some reasoning) is to take the (continuous time) Fourier transform (CTFT) of a discrete 'signal/function'. I think (I am trying to proof that to myself) that this should be equal to the DTFT of that signal, but I am not sure if that is 100% correct. When working in the discrete domain and using the DTFT my signal is just a bunch of numbers I think. (or not?) However it 'feels' to me that to be able to take the CTFT of that discrete signal I should represent the numbers by scaled diracs. I wonder if that is the right thing to do. Any insights of yours would be very appreciated! NL

"NewLine" <u...@skynet.be> writes: > Hi group, > > This is an academic question, so please ignore practical considerations. > > What I would like to do (as a first step in some reasoning) is to take the > (continuous time) Fourier transform (CTFT) of a discrete 'signal/function'. > I think (I am trying to proof that to myself) that this should be equal to > the DTFT of that signal, but I am not sure if that is 100% correct. > When working in the discrete domain and using the DTFT my signal is just a > bunch of numbers I think. (or not?) > However it 'feels' to me that to be able to take the CTFT of that discrete > signal I should represent the numbers by scaled diracs. > I wonder if that is the right thing to do. > > Any insights of yours would be very appreciated! > > NL You're on the right track. When you model the discrete signal x(nT) as a series of scaled diracs, then plug that into the CTFT, the integral from the CTFT turns them into a sum of numbers ala DTFT. Remember the sifting property of the Dirac? \int_{-infty}^{+\infty} x(t)\delta(t - tau) dt = x(tau). -- % Randy Yates % "With time with what you've learned, %% Fuquay-Varina, NC % they'll kiss the ground you walk %%% 919-577-9882 % upon." %%%% <y...@ieee.org> % '21st Century Man', *Time*, ELO http://home.earthlink.net/~yatescr

On 14 Aug, 00:55, "NewLine" <umts_remove_this_and_t...@skynet.be> wrote: > Hi group, > > This is an academic question, so please ignore practical considerations. > > What I would like to do (as a first step in some reasoning) is to take the > (continuous time) Fourier transform (CTFT) of a discrete 'signal/function'. Why do you think that's possible? The discrete signal is just that -- discrete. There is no continuous axis to integrate over. > I think (I am trying to proof that to myself) that this should be equal to > the DTFT of that signal, but I am not sure if that is 100% correct. No, it isn't. Remember Nyquist? The theorem named after him says that the sepctrum of a discrete time sequence is periodic with period Fs. There is no such theorem for continuous-time signals, meaning the DTFT and the CTFT are not related. > When working in the discrete domain and using the DTFT my signal is just a > bunch of numbers I think. (or not?) Almost. They are an *ordered* bunch of numbers, but a bunch of numbers nonetheless. > However it 'feels' to me that to be able to take the CTFT of that discrete > signal I should represent the numbers by scaled diracs. > I wonder if that is the right thing to do. Depends on what you want. If you want to troll and start a war on comp.dsp you are almost certain to succeed. If you want to get on to a side track whic will cause you huge greaf and provide no understanding about either DSP or continuous-time signal processing, you are well on your way. If you want to sort out one or two misunderstandings get a good mathematical DSP book and read about Dirac's delta as a distribution, not a function. The most important aspect of a distribution is that it, according to the mathematicans, *always* appears inside an integral. Rune

On Tue, 14 Aug 2007 00:55:34 +0200, NewLine wrote: > Hi group, > > This is an academic question, so please ignore practical considerations. > > What I would like to do (as a first step in some reasoning) is to take the > (continuous time) Fourier transform (CTFT) of a discrete 'signal/function'. > I think (I am trying to proof that to myself) that this should be equal to > the DTFT of that signal, but I am not sure if that is 100% correct. > When working in the discrete domain and using the DTFT my signal is just a > bunch of numbers I think. (or not?) > However it 'feels' to me that to be able to take the CTFT of that discrete > signal I should represent the numbers by scaled diracs. > I wonder if that is the right thing to do. > > Any insights of yours would be very appreciated! > You can unify the discrete-time and continuous-time by representing a discrete-time signal as a chain of weighted dirac impulses, yes. In some texts this is done to the extent that people are left thinking that a discrete-time signal _is_ a chain of weighted dirac impulses, which is not true. But you don't have to, unless you want them unified. In my book (see below) I chose to treat the two domains as separate entities, with rules for crossing the boundaries. This was done entirely for pedagogical reasons -- it's not as tidy when you're done, but you don't have to go into such depth with the Laplace transform either. I felt it was a good trade. -- Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html

```
NewLine wrote:
> What I would like to do (as a first step in some reasoning) is to take the
> (continuous time) Fourier transform (CTFT) of a discrete 'signal/function'.
> I think (I am trying to proof that to myself) that this should be equal to
> the DTFT of that signal, but I am not sure if that is 100% correct.
> When working in the discrete domain and using the DTFT my signal is just a
> bunch of numbers I think. (or not?)
> However it 'feels' to me that to be able to take the CTFT of that discrete
> signal I should represent the numbers by scaled diracs.
Among the transform pairs there is
discrete <--> periodic
Applying that twice, you get
discrete periodic <--> discrete periodic
which is the DTFT.
The transform will be discrete (or scaled delta functions)
if the original is periodic, discrete or not.
-- glen
```

```
>
> Depends on what you want. If you want to troll and start a war
> on comp.dsp you are almost certain to succeed. If you want to
> get on to a side track whic will cause you huge greaf and provide
> no understanding about either DSP or continuous-time signal
> processing, you are well on your way.
>
This is not at all my goal, I am just trying to get a clearer view on both
transforms...or like some others have replied in some way unify both.
Maybe I am making it difficult for myself, that can be true....so be it...
Also like others have mentioned, often in text books discrete signals are
obtained by multiplication with a dirac comb. So if it possible to go from
continuous to discrete in that way, is it such a stupid question trying to
go the other way?
```

```
>>
> You can unify the discrete-time and continuous-time by representing a
> discrete-time signal as a chain of weighted dirac impulses, yes. In some
> texts this is done to the extent that people are left thinking that a
> discrete-time signal _is_ a chain of weighted dirac impulses, which is not
> true.
>
>
> But you don't have to, unless you want them unified. In my book (see
> below) I chose to treat the two domains as separate entities, with rules
> for crossing the boundaries. This was done entirely for pedagogical
> reasons -- it's not as tidy when you're done, but you don't have to go
> into
> such depth with the Laplace transform either. I felt it was a good trade.
>
Thanks Tim, I will certainly have a look in your book.
Probably I have too often been reading books where weighted diracs are used
in a too large extent like you said.
NL
```

```
NewLine wrote:
>> You can unify the discrete-time and continuous-time by representing a
>> discrete-time signal as a chain of weighted dirac impulses, yes. In some
>> texts this is done to the extent that people are left thinking that a
>> discrete-time signal _is_ a chain of weighted dirac impulses, which is not
>> true.
>>
>>
>> But you don't have to, unless you want them unified. In my book (see
>> below) I chose to treat the two domains as separate entities, with rules
>> for crossing the boundaries. This was done entirely for pedagogical
>> reasons -- it's not as tidy when you're done, but you don't have to go
>> into
>> such depth with the Laplace transform either. I felt it was a good trade.
>>
>
> Thanks Tim, I will certainly have a look in your book.
>
> Probably I have too often been reading books where weighted diracs are used
> in a too large extent like you said.
It's perfectly reasonable to sample a continuous /signal/ and a Dirac
comb is a reasonable way to think or write about that. We can
reconstruct a samples signal with a filter and all that. You walk in
shaky qround when you sample a /transform/ because there's no going back
without a lot of IFs, BUTs, HEMs and HAWs. The continuous transform was
probably not periodic, but the sampled transform certainly is.
Jerry
--
Engineering is the art of making what you want from things you can get.
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```

On Aug 14, 4:06 pm, Jerry Avins <j...@ieee.org> wisely wrote: > > It's perfectly reasonable to sample a continuous /signal/ and a Dirac > comb is a reasonable way to think or write about that. i want to affirm that. in fact, i do not see how Tim establishes the rules for going between the discrete-time and continuous-time domain without the use of the weighted dirac comb. > We can > reconstruct a samples signal with a filter and all that. You walk in > shaky ground when you sample a /transform/ because there's no going back > without a lot of IFs, BUTs, HEMs and HAWs. sampling *anything* essentially throws away information (the data in between samples). the question is whether or not that data was somehow redundantly dependent upon the data that you're keeping (in the samples). in that case, then the data you throw away isn't anything you'll miss. the end result is that (uniform) sampling in one domain causes one to duplicate, offset, and overlap-add the data in the other domain (potentially causing aliasing when the multiple offset copies are added, unless you *know* that you're adding non-zero data to zero, you do not know what it was before adding all of these images). that is always the case. i think it's easiest to do this in the continuous- time domain when you use Hz-like frequency rather than radian frequency. then the duality-theorem of the continuous Fourier Transform is extremely simple and clear. in fact you can look at the F.T. as a sorta isomorphism (the forward transform and inverse transform are essentially the same thing) and the dirac comb is an eigenfunction (as well as the gaussian bell function, but there are an infinite number of eigenfunctions). > The continuous transform was > probably not periodic, but the sampled transform certainly is. yup, and if the transform of the continuous-time signal was not sufficiently bandlimited, that periodic transform of the sampled signal will have aliases due to the overlap and adding of the repeated "little" spectra. by thinking about the Nyquist/Shannon Sampling (and reconstruction) theorem, these relationships between the CFT and DTFT and DFT and Laplace and Z transforms all become clear and well defined. take a look at http://groups.google.com/group/comp.dsp/msg/cb32af865a6a8a30?dmode=source for my spin of the Sampling Theorem. and then ask yourself what happens if i take the continuous Fourier Transform of the dirac comb sampled signal and relate that to the spectrum before it was sampled (that is the relationship between the DTFT and CFT). then ask the same question about the Laplace Transform (of the dirac sampled signal) and the Z-Transform of the samples. i think Randy and Jerry had it right. r b-j r b-j