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

You poor folks in UKRA have our sincere condolences. How much would it cost us to have you keep him there? ;-)

LOL! The "ISP failure" again. Sure sign of a losing stand.

Reference: See your thread about a problem with an off-the-shelf amateur transceiver that you bought.

Amazing that you binned those posts in a group in which you have so many ongoing 'contributions' . Not.

Cheap. Very, very cheap.

Numerical Analysis is about bringing theoretical mathematics back to earth in the first place. Most practical problems are too complicated to enable a decent error analysis. Your statement is wishful thinking; it bears little resemblance to Numerics in practice.

Han de Bruijn

Have you noticed that it never seems to bother anyone when you say "shame on you" like this?

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David C. Ullrich

Ok, maybe it's not nonsense - I missed your point.

The idea that this has some relevance seems a little nonsensical, but never mind that.

Uh, suppose that x is a very small negative number. Then h(x) = 0, and Erf(x/eps) is very close to 1/2. So if Erf(x/eps) is close to h(x) then it follows that 0 is very close to 1/2. No need to worry about what happens when x = 0.

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David C. Ullrich

Ah, but the definition tells us that the Dirac function has a value of zero everywhere but at zero. Thus...

lim m-> 0 [int_-oo^(-m) delta dt] =0

If the Dirac function was *not* defined as zero everywhere else, then we could not deduce this. But because it is explictly defined to have a value of zero everywhere but the input value of zero, this is a simple deduction.

daestrom

No, you do *not*. It is elementary that if...

int_-oo^+oo delta(t) dt = 1

and... delta(t) = zero for all values of t 0 (a part of the definition

), then

lim m-> 0 [int_-oo^-m delta(t) dt = 0

And since for any definite integral... int_-oo^+oo g(t) dt = int_-oo^a g(t) dt + int_a^+oo g(t) dt

We can conclude that

int_a^+oo g(t) dt = int_-oo^+oo g(t) dt - int_-oo^a g(t) dt

Thus... int_+a^+oo f(t)delta(t) dt = int_-oo^+oo f(t) delta(t) dt - int_-oo^-a delta(t) dt

when lim ->a = 0

int_a^+oo (ft)delta(t) dt = f(a)

Once again you are working with an incomplete definition of the dirac function. There is more to the definition than just...

int_-oo^+oo delta(t) dt = 1

Look carefully at equation 3 of this page. It shows that in fact the delta function is...

int_(a-e)^(a+e) f(x)delta(x-a)dx = f(a) for all values of e > 0

And equation 4.... delta(x-a) = 0 for all values of x a.

These are *exactly* the two characteristics of the delta function that I exploited in my solution, yet you maintain they are not part of the definition. Quote a credible reference with your unusual 'definition' of the Dirac delta function please.

daestrom

It isn't that difficult to derive. But ARB seems to have made a couple of mistakes when he tried to integrate by parts and came to the wrong result. Yes, there seems to be a lot of personal attacks along the way :-(

His most recent objections seem to revolve around the particular definition he is using for the Dirac delta function. I replied with a couple of links showing that delta(t) = 0 for all values of t0. Hopefully he'll take the time to review the linked information carefully.

daestrom

Not true! Give me such a small number (x) and I'll give you a small number (eps) such that Erf(x/eps) is very close to 0 . How close do you want? Say | Erf(x/eps) - 0 | < delta. Right? An Erf function is monotonously increasing. Thus it can be inverted. Name of inverse function let be: Erf(-1). Then we find: | Erf(x/eps) | < delta ==>

| x/eps | > | Erf(-1)(delta) | ==> eps < | x / Erf(-1)(delta) | .

The radio amateur: 0 < h(0) < 1 and Erf(0) = 1/2 is in that range.

Han de Bruijn

It's not that I want to defend him, but this radio amateur has some point in finding a delta function to be the "same" as a bell shaped/ Gauss function with a very small value for its spread. As I pointed out to Ullrich, also the heaviside function cannot be distinguished, physically, from an error function with the same very small spread.

Han de Bruijn

Which was pretty much David's point. Just apply it to your "delta function" depending on eps.

If this has been Ullrich's point, then something must be very wrong with my understanding of English.

Han de Bruijn

That's true. Otoh given eps > 0 there exists x < 0 such that the two are far apart.

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David C. Ullrich

Stupid boy.

Stupid boy.

Stupid boy.

We are agreed then

, this is a simple deduction.

.....[Pantomime mode ON]..... Oh, yes! You do! .....[Pantomime mode OFF].....

Either you are dealing with the function as defined, or else you are creating your own function.

If you have something to present in defence of your argument, then present it yourself.