Re: How to define function (Delta(t))^2? Does this function has any meaning?

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 10/29/04 

Date: Fri, 29 Oct 2004 08:35:13 -0500

On 28 Oct 2004 13:43:32 -0700, carlos@colorado.edu (Carlos Felippa)
wrote:

>David C. Ullrich <ullrich@math.okstate.edu> wrote in message news:<1dp1o0to5bve6prbtbufv58q12p4bd6b1t@4ax.com>...
>> On 27 Oct 2004 16:26:13 -0700, carlos@colorado.edu (Carlos Felippa)
>> wrote:
>>
>> >"kiki" <lunaliu3@yahoo.com> wrote in message news:<cloo1n$8ne$1@news.Stanford.EDU>...
>> >> Any thoughts?
>> >>
>> >> thanks a lot
>> >
>> >Using the convolution theorem, the answer is +infinity.
>>
>> Actually there _is_ no answer. But if we're talking about
>> what the answer would be if we could just apply that
>> theorem in a situation where it really does not apply:
>>
>> It's worse than what you say. That theorem says (not) that
>> the Fourier transform of the answer would be +infinity.
>> Now what is the inverse transform of _that_? That's totally
>> meaningless.
>>
>>
>> ************************
>>
>> David C. Ullrich
>
>Actually I teach that theorem to aerospace sophomores in the Spring.

Wow. Can you tell us exactly what this theorem says, including
all the _hypotheses_?

(I'm not sure what your point in telling us this is, but if
your point is that yes the theorem does apply here the answer
is no, it doesn't. What are the hypotheses again?)

>Might be a good midterm quiz bonus question :-)
>
>Another way to visualize the +infinity result: draw the usual
>impulse-function triangle of unit area with base 2a and height 1/a.
>Square and integrate to get 2/(3a), then make a->0. Does this work
>for other impulse function shapes? I dont know.

************************

David C. Ullrich

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