There are a number of authors and a number of contributors to this NG who claim that sampling of analogue signals is represented by multiplying the incoming waveform f(t) by a comb of Diracian Delta Functions of the form d(t - T) and who then go on to claim that each such sample gives rise to a contribution to the spectrum of the sampled signal of f(T).e^(-sT). As this claim, this apparently faulty meme, ought to be such a fundamental part of DSP, its foundation stone in fact, it should be possible to prove this assertion by appeal to Dirac's properties of his Delta Function, and to the Laplace transform. I think that such a proof should be a simple thing to be provided by that body of authors and contributors if the claim were to be true. In our training of the Laplacian method, we are presented with a whole range of such derivations, the spectra for d(t), u(t), t, sin(t), rect(t), sinc(t), etc etc etc. Why not a similar derivation for f(t).d(t - T)? It cannot be a difficult matter for those who make the claim that sampling is so represented! As that body of authors and contributors seem unable to provide such a proof and resort to side issues and rather silly and childish ad hominem attacks when challenged upon the matter the conclusion that I reach is that the claim is false, and that that body of authors and contributors hold a religious-like stance to the matter and respond with the emotional maladjustment which is the mark of all those who are the religious loonies of the world today. (11/9 and the resultant war brought on by the religionists Bush, Blair and Windsor being prime examples of catastrophes brought on by emotional maladjustment)

There are a number of authors and a number of contributors to this NG who claim that sampling of analogue signals is represented by multiplying the incoming waveform f(t) by a comb of Diracian Delta Functions of the form d(t - T) and who then go on to claim that each such sample gives rise to a contribution to the spectrum of the sampled signal of f(T).e^(-sT). As this claim, this apparently faulty meme, ought to be such a fundamental part of DSP, its foundation stone in fact, it should be possible to prove this assertion by appeal to Dirac's properties of his Delta Function, and to the Laplace transform. I think that such a proof should be a simple thing to be provided by that body of authors and contributors if the claim were to be true. In our training of the Laplacian method, we are presented with a whole range of such derivations, the spectra for d(t), u(t), t, sin(t), rect(t), sinc(t), etc etc etc. Why not a similar derivation for f(t).d(t - T)? It cannot be a...

Derivation Of The Spectrum Due To f(t).d(t

M

Marc W 21 years ago

Thanks for pointing that out, I have been doing more DSP lately (without sampling) and had somehow managed to confuse the real-time version of the delta with the discrete one ( delta(n) = { 1 if n = 0 { 0 otherwise )

After reading both of your explanations I realised that my working was incorrect as it was a weird hybrid.

Now that I have revised the correct continuous time version, I understand David's original working and also remember trying to do it the way Airy is doing it when I first learnt about laplace transforms. It wasn't fun lol.

Just to clarify to make sure im not off the rails again: Airy, I dont understand why you can't call f(t).delta(t-T) = g(t)? I personally cannot see anything wrong with it. g(t) itself will be undefined at T and zero at t = not T. But since it is being integrated, the sifting theorem should still work shouldn't it?

More importantly you should be able to go: g(t) = e^-(st).f(t) now you have: int(from 0 to inf) of ( g(t).delta(t-T) ) dt which is a basic sifting theorem problem which David illustrated previously.

Cheers Marc

Vote

R

robert bristow-johnson 21 years ago

in article 1VRqd.13100$ snipped-for-privacy@news02.tsnz.net, Richard at richman snipped-for-privacy@hotmail.com wrote on 11/29/2004 22:36:

but shucks! you gave it away! for free! i wanted to extract a promise (that the asshole would probably not keep) before telling him what he did wrong.

oh well, it's public knowledge anyway.

r b-j

Vote

A

Airy R. Bean 21 years ago

Perhaps you should pay attention to the discussion rather than resorting in the first instance to personal remarks?

Vote

A

Airy R. Bean 21 years ago

Is it necessarily true that if two integrals are equal, then the functions under those integrals are equal?

Vote

A

Airy R. Bean 21 years ago

Multiplying one or both sides of an identity by 1 does not value to your argument.

Vote

A

Airy R. Bean 21 years ago

Shame on you for making personal remarks.

The conclusion about your university-style email address is that you are a first-year man.

Vote

A

Airy R. Bean 21 years ago

Looking at the work done by Dirac, his Delta function is a function defined as a limit.

No problem there. All derivatives are so defined.

Vote

A

Airy R. Bean 21 years ago

You cannot prove something merely by defining it to be correct.

Vote

A

Airy R. Bean 21 years ago

To use your own argument, just saying it, does not prove it one way or the other.

Vote

A

Airy R. Bean 21 years ago

The Delta function was on a rigorous mathematical basis when Dirac defined it, although he spoke to the contrary.

The idea of a function being the limit of another function has always been entirely respectable and is found in the evaluation of every derivative.

The introduction of generalised functions and the Theory Of Distributions is entirely irrelevant when no mathematics resulting from that theory is applied.

Vote

A

Airy R. Bean 21 years ago

Therefore it is also true to say of you "But you're somehow convinced that you must be right, " and therefore it was an inappropriate thing for you to introduce.

Furthermore, if something that you read in one of your textbooks is your reply to a point of debate, then it is a simple matter for you to type it out. Far better than sneering at your correspondent!

Vote

A

Airy R. Bean 21 years ago

What other way is there to evaluate it that is not based upon the two methods I described?

Your reputation as a debater is not enhanced by your wont to sneer but without producing counter-argument.

I have never asserted that "X is the _only_ way to accomplish Y"

Vote

A

Airy R. Bean 21 years ago

The point at which I have stopped quoting you is the point at which you seem to have made a simple change of variable from "t" to "u" which does not seem to support your change of limits from a^b to 0^t (assuming that I have understood your notation for expressing limits)

How have you calculated your change of limits?

Vote

A

Airy R. Bean 21 years ago

It doesn't agree with integration by parts.

Vote

A

Airy R. Bean 21 years ago

In the evaluation of integral transforms, the sifting theory will work if it is present as one of the integrals.

Vote

A

Airy R. Bean 21 years ago

Vote

A

Airy R. Bean 21 years ago

Yes it's a constant.

Where did I c*ck up? Let's see.....

int(UV) = U.int(V) - int[dU.int(V)] giving..... int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... with "f(t).d(t-T)" as V and "e^(-sT)" as U ..... f(T).e^(-st) - int(e^(-st)/-s . f(T))..... f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... f(T).e^(-sT) - f(T).e^(-sT)....

Ah yes, typo, should read..... "with "f(t).d(t-T)" as V and "e^(-st)" as U ....."

also the next line is wrong, should read..... "f(T).e^(-st) - int(-s.e^(-st) . f(T))....."

However the effect of a consecutive differentiation followed by an integral is still to give -s/-s.

Giving.... int(UV) = U.int(V) - int[dU.int(V)] giving..... int(+/-inf)(f(t).d(t-T).e^(-st)) as ..... with "f(t).d(t-T)" as V and "e^(-st)" as U ..... f(T).e^(-st) - int(-s.e^(-st) . f(T))..... f(T).e^(-st) - -s.e^(-st)/-s . f(T)..... f(T).e^(-sT) - f(T).e^(-sT)....

> "Airy R. Bean" wrote:

Vote

A

Airy R. Bean 21 years ago

Confidential, sorry.

Vote

A

Airy R. Bean 21 years ago

Because of your childish behaviour quoted below, the rest of your article has been ignored. If you have something of value to contribute to the discussion, then please present it in a style more appropriate to an international forum.

Vote

A

Airy R. Bean 21 years ago

In your second equation, how come you've still got "t" in your expression of the Diracian, but you've replaced it by T in the other two terms?

It would seem to me that if a principle is to be applied, then it must be applied everywhere, so that you must end up with d(T -T) = d(0).

Vote

Derivation Of The Spectrum Due To f(t).d(t - T)?????

Join the Discussion

Didn't find your answer?