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

• posted

Top-posting makes the whole thing impractical.

• posted

Nothing wrong with top-posting.

It is the preferred method, because you don't have to page down through much repeated and already-seen history.

Bottom posted articles follow>

• posted

Of course not.

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David C. Ullrich

• posted

No problem - I assumed you were talking about someone else, wondered why I didn't see any posts from him... (the spelling with one l is much more common.)

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David C. Ullrich

• posted

Of course not. Here the question was about a _definition_. I didn't claim to have proved I was giving a correct definition.

The definition I gave _is_ correct, and perfectly standard, your ignorance of the matter notwithstanding.

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David C. Ullrich

• posted

are you afraid to see your mistake explicitly exposed?

"cluck, cluck"

r b-j

in article snipped-for-privacy@uni-berlin.de, Airy R. Bean at snipped-for-privacy@privacy.net wrote on 11/30/2004 04:17:

• posted

Yes, but the delta "function" is _not_ the limit of a sequence of ordinary functions, at least not in the sense that one speaks of limits in calculus. It's not a function at all.

But it is true that one can define the _action_ of a delta function as a limit - that is, the linear functional defined by the delta "function" is indeed a limit of the linear functionals defined by a sequence of ordinary functions. If you want to look at it that way it's very easy to show from that point of view that the Laplace transform is what it is:

Fix T, and say f_n(t) = n for T < t < T + 1/n, f_n(t) = 0 for other t. Then the f_n _do_ converge to delta(t-T) in a certain sense - in particular the Laplace transform of delta(t-T) _is_ the limit of the Laplace transform of f_n as n -> infinity. Let's say F_n is the LT of f_n. Then

F_n(s) = n int_T^{T+1/n} e^{-st} dt

= n (e^{-s(T+1/n)) - e^{-sT}) / (-s)

= e^{-sT} (1 - e^{-s/n}) / (s/n).

It's an easy calculus exercise to show that (1 - e^{-s/n}) / (s/n) -> 1 as n -> infinity. So the limit of F_n(s) is e^{-sT}.

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David C. Ullrich

• posted

The integration by parts that you're doing is _wrong_. The fact that integration by parts works is a theorem. Theorems have hypotheses - whatever delta(t-T) is, it's not a continuous function.

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David C. Ullrich

• posted

I explained exactly how to evaluate this Laplace transform correctly in my first post on the topic.

(I also just posted a more complicated proof, using the fact that the delta "function" is in some sense a limit of ordinary functions.)

Guffaw.

Uh, yes you did. In your reply to my first post you said in part

"But. to evaluate the (unilateral) Laplace transform, we either have to be able to present f(t) as a function of exponentials, in which case simple integration applies together with the adding of exponents, or we have to apply Integration by Parts. (or else for a time domain product we have to evaluate an s domain convolution)"

which does say that this and that are the only way to do something.

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David C. Ullrich

• posted

• posted

.....[Pantomime mode ON].....

Oh, yes! It _IS_ a function.

.....[Pantomime mode OFF].....

And I have never defined it as a limit of a sequence of functions any more than I would define a derivative as such a limit.

The limit employed is very much the same limit as used in calculus; t > 0.

• posted

Make your mind up - first you say that it is not a function, then you say that it is the delta function.

• posted

So you have argued a derivation, without using the concept of generalised functions which were in any case irrelevant, to arrive at a result that wasn't in dispute.

I wonder why?

• posted

I am not afraid of anything, be that anything you, or the truth. It is simply that I refuse to deal with you when you indulge yourself in infantile outbursts.

Shame on you.

You should know better, including not indulging in silly yah-boo-sucks tirades as in your latest rant below.

You do nothing for your reputation by so behaving.

• posted

Show where it is wrong, or else resort to infantile sneering.

• posted

.....[Pantomime mode ON]..... Oh, no! You didn't! .....[Pantomime mode OFF].....

• posted

So first of all you say that I said, "X is the _only_ way to accomplish Y"

Then you said that I said, "this and that are the only way to do something."

You're not consistent.

In any case, I suggested three ways of proceeding and certainly did not state that "X is the _only_ way to accomplish Y" nor did I state "this and that are the only way to do something.".

• posted

The result I arrived at was that the Laplace transform of delta(t-T) is e^{-sT}. That wasn't in dispute?

Because you're too stupid or stubborn to agree with the much simpler argument proving the same thing.

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David C. Ullrich

• posted

Ok. Took me a minute to find the place where you showed us the integration by parts:

" Certainly, if you integrate by parts, you would choose int(f(t).d(t-T)) as the integrated bit to yield f(T), but when I try this, I get 0!......

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)....

f(T).e^(-st) - int(e^(-st)/-s . f(T))

is already wrong. My guess is because of a confusion over definite integrals versus antiderivatives, ie indefinite integrals. The (definite) integral from

0 to infinity of f(t).d(t-T) is indeed f(T). But what we need here is an antiderivative.

An antiderivative of f(t).d(t-T) is given by F(t), where F(t) = 0 for t < T and f(T) for t > T. So that line should be

F(t).e^(-st) - int(e^(-st)/-s . F(t)).

Golly, I assumed that you had some idea how to do simple calculus - looking at this I realize you took the derivative wrong. What we really get is this:

F(t).e^(-st) - int(e^(-st)(-s) . F(t)). Now when we put in the limits t = 0 to t = infinity the first term vanishes since F(0) = 0 and e^(-st) tends to 0 as t -> infinty (at least if s > 0; if s < 0 this integration by parts is not going to work.) So out Laplace transform becomes

- int_0^infinity (e^(-st)/-s . F(t))

Since F(t) = 0 for t < T and f(T) for t > T this equals

f(T) int_T^infinity (e^(-st)s ).

Again, it's easy to evaluate the last integral; int_T^infinity (e^(-st)s ) = e^(-sT), so we finally get f(T)e^(-sT) for the Laplace transform.

So now we've found _two_ mistakes in your integration by parts, confusing antiderivatives and definite integrals and taking a very simple derivative incorrectly.

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David C. Ullrich

• posted

More properly, "It is preferred by some, because they don't have to page down through much repeated and already-seen history." It is not universally 'preferred' by all use-net users. In fact, there are several use-nettiquite articles on the subject.

... by those that don't want to scroll through the previous context. Others, that don't mind reading the postings in context don't mind at all. In fact, many prefer in-context posting as it is much clearer what the poster is replying to.

In the end, in-context posting, with snippage of unrelated, repeatedly quoted material is 'preferred' by many.

You don't speak for the entire use-net community, so don't pretend that you do.

daestrom

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