Convolution with a constant value signal

On Aug 13, 12:46&#4294967295;pm, "Computer Man" <inva...@invalid.invalid> wrote:
> "Dave" <dspg...@netscape.net> wrote in message
>
> news:2ba5ecc3-9d8d-489a-89ba-d82fdbdd7b6c@a13g2000yqc.googlegroups.com...
>
> > In another one of his books he explains the
> > multiplication by a dirac delta train (i.e sampling) by saying that
> > the samples have an area (thus meaing integrating) equal to the value
> > of the continuous function at that time.
>
> Then it must be the only case in advanced mathematical analysis of
> an integration being claimed to happen but never appearing in any
> of the derivations!

Well it has been awhile since I read every advanced mathmatical
analysis, so I can't say for sure. Actually, he doesn't do the
integration he only offers it as an interpretation by saying the area
of the samples corresponds to the value of the function at that
particular point - which is something I haven't seen any of the other
authors do.


Rune, concerning the definition of integral - that is definitely
beyond my area of expertise. I can only offer a couple of ideas with
which I have a passing familiarity with and I don't even know if they
are valid.

1)  I thought that the square integrability (sp) was only a sufficient
condition, not a necessary one. Maybe I'm wrong - I definitely could
be.
2) Possibly switching to a Lebesgue integral makes a difference.

In the end - Don't shoot the messenger. I'm just telling you what I've
read - hopefully it was written by someone who knows more than I
do. :)

Cheers,
Dave

On Aug 13, 11:15&#4294967295;am, Dave <dspg...@netscape.net> wrote:
> On Aug 13, 9:12&#4294967295;am, Rune Allnor <all...@tele.ntnu.no> wrote:
>
>
>
> > On 13 Aug, 14:26, Dave <dspg...@netscape.net> wrote:
>
> > > On Aug 13, 4:12&#4294967295;am, Rune Allnor <all...@tele.ntnu.no> wrote:
>
> > > > On 12 Aug, 18:10, karl bezzoto <karl.bezz...@googlemail.com> wrote:
>
> > > > > Hello,
> > > > > In the following A is a real constant value signal and the symbol *
> > > > > denotes the convolution operation.
>
> > > > > We know that:
> > > > > A*x(n)----Fourier----> A.&(f).X(f), where &(f) denotes Dirac function.
>
> > > > Wrong. &#4294967295;FT{ax(t)} = aFT{x(t)}.
>
> > > Rune - he stated that the * meant convolution not multiplication, so
> > > his original statement is, I believe, correct.
>
> > I am sure you can come up with a reference for how
> > to convolve with a constant?
>
> Yes, See Papoulis "Signals, Systems and Transforms" 1977. The FT of a
> constant is the scaled (i.e. amplitude) dirac delta at zero frequency.
>
>
>
>
>
> > > > > however,
> > > > > A.X(0) &#4294967295; &#4294967295; <----Fourier---- A.&(f).X(f)
>
> > > > Wrong. Dirac's delta is a distribution, and should never
> > > > appear outside an integration sign.
>
> > > While you may not like the dirac function appearing by itself, it does
> > > have a FT since in taking the FT you are integrating over it.
>
> > I know quite a few (most?) of my opinions are generally
> > regarded as controversial. However, at least one other
> > person agrees with me, check out
>
> > Papoulis: "The Fourier Integral and its Applications",
> > (1961) page 269:
>
> > The problems about understanding the Dirac Delta are
> > down to "...the reluctance of the applied scientist to
> > accept the description of a physical quantity by a concept
> > that is not an ordinary function, but is specified by
> > certain properties of integral nature."
>
> > I think Papoulis nailed the cause of just about every
> > single comp.dsp semantics war over Dirac's Delta, decades
> > before USENET ever existed.
>
> Yes, I agree - I didn't want to see this thread dissolve into one of
> those. However, he still uses the dirac delta throughout his
> derivations without it appearing under an integral (see the book
> referenced about). In another one of his books he explains the
> multiplication by a dirac delta train (i.e sampling) by saying that
> the samples have an area (thus meaing integrating) equal to the value
> of the continuous function at that time.
>
> Cheers,
> Dave

Sorry I made a mistake the book I'm referring to is "Signal Analysis"
by Papoulis.

Cheers,
Dave