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

# CTFT of a discrete signal

Started by ●August 13, 2007

Posted by ●August 14, 2007

"NewLine"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! > > NLYou'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." %%%%% '21st Century Man', *Time*, ELO http://home.earthlink.net/~yatescr

Posted by ●August 14, 2007

On 14 Aug, 00:55, "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'.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

Posted by ●August 14, 2007

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

Posted by ●August 14, 2007

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

Posted by ●August 14, 2007

> > 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?

Posted by ●August 14, 2007

>> > 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

Posted by ●August 14, 2007

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

Posted by ●August 14, 2007

On Aug 14, 1:06 pm, Jerry Avinswrote: > 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=ade.> > > 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. > =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF==AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF= =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF- Hide qu= oted 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.

Posted by ●August 14, 2007

On Aug 14, 4:06 pm, Jerry Avinswisely 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

Posted by ●August 20, 2007

On Aug 18, 2:28 pm, Eric Jacobsenwrote: > On Sat, 18 Aug 2007 01:21:48 -0700, robert bristow-johnson > >you see, strictly speaking, a function f(x) is just that: a mapping of x to f(x). it is not allowed to worry about meta-data like *how* f(x) came to be (regarding the limit of nascent delta functions). all we are allowed to think about f(x) = delta(x) is that for every x != 0, then what comes outa the mapping is f(x)=0. then it doesn't matter what definable number f(0) is, the integral is zero. if f(0) is not defined, then f(x) is not integrable for any region containing x=0. that's the problem.wrote: > > if any strict mathematician > >picks on you, tell them that d(t) is not strictly a dirac delta > >function (which ain't a real function, as far as they permit) but a > >simple rectangular function with 1 Planck Time in width and a > >(dimensionless) unit area (so the height is the reciprocal of the > >Planck Time). that's a true function, as far as the mathematicans are > >concerned, and it's close enough to the Neanderthal engineer's use of > >the dirac, that there is no measurable difference in effect. > > I guess during most of these discussions > I'd always just kind of built that sort of thing into the assumptions > so that it all works out, and consequently never really understood > what all the hubbub was about. > Thanks to this thread I'm finally > seeing the distinction that has bundled a few folks panties over the > years. My main reaction is that there are bigger things to get > twisted over than this...yikes.it's a problem when they say if f(x)=delta(x) and g(x)=0, we cannot say that the integral of f(x) is 1 while the integral of g(x) is zero. i want to be able to say it is so, even if my Real Analysis prof would give me an F.> On a somewhat related note (well, the 1 Planck Time bit reminds me of > it)...somebody pointed out that the precision needed in Pi to be able > to discern a single Angstrom at the radius of the known limit of the > universe is only about twenty decimal places or so (I don't remember > the exact number, but it was surprisingly low from what I thought it > would be).geez, 10^10 Ang/m * 299792458 m/s * 31557600 s/yr * 13.7x10^9 yr what does that come out to be? i think lot more than 20 digits. more like 40.> So using some small finite pulse width like you've > suggested may actually be quite practical in several ways.it's just to get the mathematicians offa my back. i remember taking "Real Analysis" and learning about the Lebesgue Integral for the first time. i knew right away i had a conceptual problem when we we told that if f(x) = g(x) for all but a countable number of discrete x points (that's what "almost everywhere" means) that the integrals of f() and g() must be the same. it's been an icky bugger ever since. so i decided to adopt the bug as a house pet. r b-j

Posted by ●August 18, 2007

On Aug 18, 1:21 am, robert bristow-johnsonwrote: > On Aug 15, 5:55 am, "NewLine" > > > >Given the uncertainty of measuring such small units of time mapped to the actual universe, your rectangular Planck-width function may actually become a probability distribution of some sort, thus still needing integration to show equivalence with the use of the ideal Platonic rectangle. IMHO. YMMV. -- rhn A.T nicholson d.0.t C-o-Mwrote: > > > 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). > > > -- > > > first of all, thanks to all the people that have responded, all your inputs > > were very useful to me...also rbj I liked a lot the link to your > > mathematical description of sampling & reconstruction. > > > Below I tried to do continue with what Randy introduced above. Are there any > > flaws in this? > > > ---------------- > > > my discrete signal, only k= 0 : N-1 values are non-zero (spaced T) > > > s[k] > > > discrete signal represented as series of scaled dirac's as if it were a > > 'continuous signal' > > > s(t) = SUM_{k=0}^_{k=N-1} s[k] * d(t-k*T) > > > where d(t) is the dirac delta > > > FT of s(t): > > > S(f) = INT_{-inf}_{+inf} SUM_{k=0}^_{k=N-1} s[k] * d(t-k*T) > > exp(-i*2*pi*f*t) dt > > > S(f) = SUM_{k=0}^_{k=N-1} s[k] INT_{-inf}_{+inf} d(t-k*T) > > exp(-i*2*pi*f*t) dt > > > S(f) = SUM_{k=0}^_{k=N-1} s[k] exp(-i*2*pi*f*k*T) > > > S(f) = SUM_{k=-inf}^_{k=+inf} s[k] exp(-i*2*pi*f*k*T) > > > replacing 2*pi*f*T with the normalized radial frequency w (radians per > > sample) this becomes: > > > SUM_{k=-inf}^_{k=+inf} s[k] exp(-i*k*w) > > > which is if I am not mistaken the DTFT. > > other than the naked delta functions which are objected to by most > strict mathematicians (who insist on clothing the naked natives with > integrals), it's fine. being a Neanderthal engineer, i generally > approve of naked dirac delta functions. if any strict mathematician > picks on you, tell them that d(t) is not strictly a dirac delta > function (which ain't a real function, as far as they permit) but a > simple rectangular function with 1 Planck Time in width and a > (dimensionless) unit area (so the height is the reciprocal of the > Planck Time). that's a true function, as far as the mathematicans are > concerned, and it's close enough to the Neanderthal engineer's use of > the dirac, that there is no measurable difference in effect.

Posted by ●August 18, 2007

On Sat, 18 Aug 2007 01:21:48 -0700, robert bristow-johnsonwrote: > if any strict mathematician >picks on you, tell them that d(t) is not strictly a dirac delta >function (which ain't a real function, as far as they permit) but a >simple rectangular function with 1 Planck Time in width and a >(dimensionless) unit area (so the height is the reciprocal of the >Planck Time). that's a true function, as far as the mathematicans are >concerned, and it's close enough to the Neanderthal engineer's use of >the dirac, that there is no measurable difference in effect....>r b-jHey, that's a nice touch. I guess during most of these discussions I'd always just kind of built that sort of thing into the assumptions so that it all works out, and consequently never really understood what all the hubbub was about. Thanks to this thread I'm finally seeing the distinction that has bundled a few folks panties over the years. My main reaction is that there are bigger things to get twisted over than this...yikes. On a somewhat related note (well, the 1 Planck Time bit reminds me of it)...somebody pointed out that the precision needed in Pi to be able to discern a single Angstrom at the radius of the known limit of the universe is only about twenty decimal places or so (I don't remember the exact number, but it was surprisingly low from what I thought it would be). So using some small finite pulse width like you've suggested may actually be quite practical in several ways. Eric Jacobsen Minister of Algorithms Abineau Communications http://www.ericjacobsen.org

Posted by ●August 18, 2007

On Aug 15, 5:55 am, "NewLine"wrote: > > 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). > > -- > > first of all, thanks to all the people that have responded, all your inputs > were very useful to me...also rbj I liked a lot the link to your > mathematical description of sampling & reconstruction. > > Below I tried to do continue with what Randy introduced above. Are there any > flaws in this? > > ---------------- > > my discrete signal, only k= 0 : N-1 values are non-zero (spaced T) > > s[k] > > discrete signal represented as series of scaled dirac's as if it were a > 'continuous signal' > > s(t) = SUM_{k=0}^_{k=N-1} s[k] * d(t-k*T) > > where d(t) is the dirac delta > > FT of s(t): > > S(f) = INT_{-inf}_{+inf} SUM_{k=0}^_{k=N-1} s[k] * d(t-k*T) > exp(-i*2*pi*f*t) dt > > S(f) = SUM_{k=0}^_{k=N-1} s[k] INT_{-inf}_{+inf} d(t-k*T) > exp(-i*2*pi*f*t) dt > > S(f) = SUM_{k=0}^_{k=N-1} s[k] exp(-i*2*pi*f*k*T) > > S(f) = SUM_{k=-inf}^_{k=+inf} s[k] exp(-i*2*pi*f*k*T) > > replacing 2*pi*f*T with the normalized radial frequency w (radians per > sample) this becomes: > > SUM_{k=-inf}^_{k=+inf} s[k] exp(-i*k*w) > > which is if I am not mistaken the DTFT.other than the naked delta functions which are objected to by most strict mathematicians (who insist on clothing the naked natives with integrals), it's fine. being a Neanderthal engineer, i generally approve of naked dirac delta functions. if any strict mathematician picks on you, tell them that d(t) is not strictly a dirac delta function (which ain't a real function, as far as they permit) but a simple rectangular function with 1 Planck Time in width and a (dimensionless) unit area (so the height is the reciprocal of the Planck Time). that's a true function, as far as the mathematicans are concerned, and it's close enough to the Neanderthal engineer's use of the dirac, that there is no measurable difference in effect. note, if you replace FT with LT, then, instead of a DTFT, you get the Z transform. they're are related. all descendents of Fourier. r b-j

Posted by ●August 17, 2007

Rune Allnorwrites: > On 15 Aug, 11:55, "NewLine"No. This integral exists in the sense of Lebesgue and is _zero_.> wrote: > ... > N 1 > X(w) = lim sum ----- exp(jw'n)exp(-jwn) [2.a] > N->inf n=-N 2N-1 > > N 1 > = lim sum ----- 1 [2.b] > N->inf n=-N 2N-1 > > / 2N-1 > | lim ------ = 1 w = w' > = | N->inf 2N-1 > | > \ 0 w =/= w' > > So X(w) == 0 "almost everywhere ," i.e. everywhere > except for one point, at w = w', where it equals 1. > It's a half-weird function if you are only used to > think in terms of continuous function, but for > somebody with a college maths course under the > belt, it is nothing extraordinary about X(w). > > .... > This was the problem Lebesgue solved. He developed > a way of integrating functions which are 0 "almost > everywhere" so that > > inf > integral_Lebesgue X(w) = 1. > -inf > > with X(w) above. > > The one problem with Lebesque's integral is that > it is far from easy to use (well, that's what > my maths teachers told me; I have no personal > experience with it)The theory of Lebesgue integration is not so simple as Riemann's, but the essential thing from the physicist's/engineer's point of view is: 1) For any function f:R --> R and any intervall (a,b), where any of the boundarys may be infinity, if the Riemann integral exists, then the Lebesgue integral exists too, and it's value is equal to the one of Riemann's. 2) If f is integrable over (a,b) in the sense of Lebesgue, and, if g is another function, that is equal to f on (a,b), except for a subset of measure zero, then g is also integrable on (a,b) and the value of the integral is the same as from f. Example (the standard): f(x) = 0 for all real numbers x. g(x) = 1 for all rational numbers and g(x) = 0 else. The integral over f always exists and is zero. The rational numbers are of measure zero (as is any countable set.) The Riemann integral over g does not exist over any non-trivial intervall. But for any intervall the integral over g in the sense of Lebesgue exists and is zero.> so Dirac set about to come > up with a way to integrate functions which are > 0 "almost everywhere" without having to resort > to the Lebesgue integral.Oh no. Distributions are not about integration. They are mathematical objects in their own right, a little resembling ``ordinary'' functions, and often much more usefull for modelling in theoretical physics or engineering than ordinary functions are.> > The result is Dirac's delta function. The function > > inf > X = integral d(t - t0) dt > -inf > > is basically nothing more than a tag, a note to > the analyst, to say that "add 1 to the integral > X when you pass t = t0."Well--- yes, something along these lines was suggested by Dirac. But how about differentiation? etc. etc. As long as you don't have a precise mathematical theory, what you can do with Dirac deltas is either trivial, or you are not sure wether some operation is valid. The mathematical understanding of Dirac's ``delta functions'' was chiefly developed by Laurent Schwartz, who published his textbook ``Th\'eorie des distributions'' not until 1950; about twenty years after Dirac's suggestion.> That's all there is to it. Once you get a clear > picture of why the Dirac delta is useful, you > also see why it is plain wrong to use it > outside the integral sign.This may be correct, when you restrict your understanding of the Dirac delta to that of a ``tag to the analyst''. But mathematicans have developed a much wider, more thorrow understanding, and there any distribution makes sense outside of an integral, as well as an ordinary function does. -- hw

Posted by ●August 16, 2007

> > What exactly are you starting out with? A set of samples, i.e., an > ordered set of numbers? > > If so, then you don't go from the those directly to a CTFT. The input > to a CTFT is a continuous-time signal, and that's not what you > have. So you have to first define how you're going to convert your > samples into a continuous-time signal before you can take the CTFT of > it. > > What is the point of your academic query? I mean, what the heck are you > trying > to figure out? > --I must admit that it probably was a very strange question. What I was trying to figure out is: Imagine a generator that can create continuous sine waves from -inf to +inf. And this signal goes into a block that will convolve the sine wave with another signal that is zero everywhere except at some discrete points, spaced T apart. So you could see it like sending a continuous signal in a filter that looks a lot like a digital filter, but that is actually able to process the continuous signal. I was wondering if the frequency response of such a block/filter would be periodic with 1/T. I am convinced it is because the filter is discrete, but want some check. Writing this down and thinking it over again, probably to make the model more in line with theory I have to place a sampler on the continuous signal, and then I have a complete discrete system (samples and filter)...which obviously is periodic with 1/T in the freq domain.

Posted by ●August 16, 2007

robert bristow-johnson wrote: (snip on DFT and DTFT)>>The other that always confuses me is, why isn't that a Fourier series?> why do you say that it isn't? the discrete samples in the time domain > are the fourier coefficients of the periodic spectrum in the freqency > domain. if you normalize out the 2*pi factor (by expressing frequency > in normalized cycles/sample rather than radians/sample), the period of > that periodic spectrum is 1, and the spacing between your discrete > coefs (which are the samples) is 1.My feeling is that it should be either a Fourier series, or a Fourier transform, and not both. One thing should not have two incompatible names. That is all. -- glen

Posted by ●August 15, 2007

"NewLine"writes: >> You've prematurely represented the signal as a sequence s[k]. Keep the >> original signal intact. By the way, I don't see the original signal >> here, so let's call it s_a(t) (for "analog"), and let s(t) be the >> "discretized" version, i.e., >> > > That's because my original signal (as in the original post) is a discrete > signal. > I guess your extension adds sampling a real nice analog to it...Hmm. I went back and re-read your original post and I might have misread you all along. What exactly are you starting out with? A set of samples, i.e., an ordered set of numbers? If so, then you don't go from the those directly to a CTFT. The input to a CTFT is a continuous-time signal, and that's not what you have. So you have to first define how you're going to convert your samples into a continuous-time signal before you can take the CTFT of it. What is the point of your academic query? I mean, what the heck are you trying to figure out? -- % Randy Yates % "She has an IQ of 1001, she has a jumpsuit %% Fuquay-Varina, NC % on, and she's also a telephone." %%% 919-577-9882 % %%%%% 'Yours Truly, 2095', *Time*, ELO http://home.earthlink.net/~yatescr

Posted by ●August 15, 2007

> You've prematurely represented the signal as a sequence s[k]. Keep the > original signal intact. By the way, I don't see the original signal > here, so let's call it s_a(t) (for "analog"), and let s(t) be the > "discretized" version, i.e., >That's because my original signal (as in the original post) is a discrete signal. I guess your extension adds sampling a real nice analog to it...

Posted by ●August 15, 2007

On Aug 15, 4:33 pm, glen herrmannsfeldtwrote: > > When I learned about delta functions, I was told that mathematicians > didn't like them, and often didn't believe in them.what they don't believe is that the dirac delta "function" is a true function as the mathematicians understand a function of a real variable. from the POV ofLebesgue integration (which is more general than the Riemann integration we learned in calculus), if two functions agree "almost everywhere" (which is "everywhere but for a countable number of discrete points"), then the integrals of the two functions over the same region (or limits of integration) must agree. well, we electrical engineers like to think of the dirac impulse as have zero value everywhere but at t=0, and its integral (from -eps to +eps) is 1. but the dirac impulse "function" agrees with the zero everywhere function at every point except t=0, and we know that the integral of 0 is 0. so this understanding of the dirac impulse violates that property of Lebesgue integration of two virtually identical functions.> There was a story > about some mathematician who rewrote a quantum mechanics book to remove > delta functions from all the explanations.yeah, some math guys really don't like it, and like our Neanderthal engineering treatment of the dirac impulse even less. i'm completely fine with the Neanderthal engineering usage of the dirac delta. r b-j