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Discussion Groups | Comp.DSP | CTFT of a discrete signal

There are 39 messages in this thread.

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CTFT of a discrete signal - NewLine - 2007-08-13 18:55:00

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


Re: CTFT of a discrete signal - Randy Yates - 2007-08-14 08:30:00

"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


Re: CTFT of a discrete signal - Rune Allnor - 2007-08-14 12:02:00

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


Re: CTFT of a discrete signal - Tim Wescott - 2007-08-14 12:26:00

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


Re: CTFT of a discrete signal - glen herrmannsfeldt - 2007-08-14 14:10:00

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


Re: CTFT of a discrete signal - NewLine - 2007-08-14 14:54:00

>
> 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?


Re: CTFT of a discrete signal - NewLine - 2007-08-14 14:58:00

>>
> 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


Re: CTFT of a discrete signal - Jerry Avins - 2007-08-14 16:06:00

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.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯


Re: CTFT of a discrete signal - Azazello - 2007-08-14 16:46:00

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 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.
> ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯- Hide quoted text -
>
> - Show quoted text -

A good book will certainly help but when I was starting DSP a
professor (or someone like a professor), was an absolute must.  As
mentioned above, be very careful with equating different transforms to
fundamentally different sets of information.  A lot of the classes
that I've taken in DSP functioned on a nearly intuitive level.
Understanding fundamentals is priority numero uno.


Re: CTFT of a discrete signal - robert bristow-johnson - 2007-08-14 17:12:00

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


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