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

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

Reply to
Marc W
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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

Reply to
robert bristow-johnson

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

Reply to
Airy R. Bean

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

Reply to
Airy R. Bean

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

Reply to
Airy R. Bean

Shame on you for making personal remarks.

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

Reply to
Airy R. Bean

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.

Reply to
Airy R. Bean

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

Reply to
Airy R. Bean

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

Reply to
Airy R. Bean

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.

Reply to
Airy R. Bean

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!

Reply to
Airy R. Bean

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"

Reply to
Airy R. Bean

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?

Reply to
Airy R. Bean

It doesn't agree with integration by parts.

Reply to
Airy R. Bean

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

Reply to
Airy R. Bean

Reply to
Airy R. Bean

Yes it's a constant.

Where did I cock 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)....

  1. 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)....

  1. > "Airy R. Bean" wrote:
Reply to
Airy R. Bean

Confidential, sorry.

Reply to
Airy R. Bean

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.

Reply to
Airy R. Bean

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).

Reply to
Airy R. Bean

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