# CTFT of a discrete signal

Started by August 13, 2007
Hi group,

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" <umts_remove_this_and_this@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." %%%% <yates@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. &macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
On Aug 14, 1:06 pm, Jerry Avins <j...@ieee.org> wrote:
> 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 s=
ome
> >> 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 rul=
es
> >> 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 tr=