On Aug 13, 11:15�am, Dave <dspg...@netscape.net> wrote:
> On Aug 13, 9:12�am, Rune Allnor <all...@tele.ntnu.no> wrote:
>
>
>
> > On 13 Aug, 14:26, Dave <dspg...@netscape.net> wrote:
>
> > > On Aug 13, 4:12�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. �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) � � <----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
Reply by Dave●August 14, 20092009-08-14
On Aug 13, 12:46�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
Reply by Computer Man●August 13, 20092009-08-13
"Dave" <dspguy2@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!
Reply by Rune Allnor●August 13, 20092009-08-13
On 13 Aug, 17:15, Dave <dspg...@netscape.net> wrote:
> On Aug 13, 9:12�am, Rune Allnor <all...@tele.ntnu.no> wrote:
>
>
>
>
>
> > On 13 Aug, 14:26, Dave <dspg...@netscape.net> wrote:
>
> > > On Aug 13, 4:12�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. �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.
In that case you need to define what you mean both by
'FT' and 'convolution'. One of the requirements for a
signal is that
inf
integral |f(x)|^2 dx < inf.
-inf
This is clearly not satisfied if f is a non-zero constant.
Rune
Reply by karl bezzoto●August 13, 20092009-08-13
On 13 Aug, 16:15, Dave <dspg...@netscape.net> wrote:
> On Aug 13, 9:12�am, Rune Allnor <all...@tele.ntnu.no> wrote:
>
>
>
>
>
> > On 13 Aug, 14:26, Dave <dspg...@netscape.net> wrote:
>
> > > On Aug 13, 4:12�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. �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) � � <----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- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Thanks Dave. It is what i had in mind while using Dirac function
above. As you explained, the sampling of a signal is often formalised
first as a multiplication of the input signal with a dirac delta
train.
however as far as my question is concerned i can't still equal to the
2 equations given in my OP
Reply by Dave●August 13, 20092009-08-13
On Aug 13, 9:12�am, Rune Allnor <all...@tele.ntnu.no> wrote:
> On 13 Aug, 14:26, Dave <dspg...@netscape.net> wrote:
>
> > On Aug 13, 4:12�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. �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) � � <----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
Reply by Jerry Avins●August 13, 20092009-08-13
Computer Man wrote:
...
> (ISTR "That's not irrelevant, that's ir hippopotamus" from the Goon Show?)
Flanders and Swann's patter intro to "The Hippopotamus Song".
...
Glorious mud to you,
Jerry
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by Rune Allnor●August 13, 20092009-08-13
On 13 Aug, 14:26, Dave <dspg...@netscape.net> wrote:
> On Aug 13, 4:12�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. �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?
> > > however,
> > > A.X(0) � � <----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.
Rune
Reply by Dave●August 13, 20092009-08-13
On Aug 13, 4:12�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. �FT{ax(t)} = aFT{x(t)}.
Rune - he stated that the * meant convolution not multiplication, so
his original statement is, I believe, correct.
>
> > however,
> > A.X(0) � � <----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.
Cheers,
David
Reply by Computer Man●August 13, 20092009-08-13
"Rune Allnor" <allnor@tele.ntnu.no> wrote in message
news:d0f090ac-e82c-4fa9-8d81-e0e42d04b72b@w6g2000yqw.googlegroups.com...
> Wrong. Dirac's delta is a distribution, and should never
> appear outside an integration sign.
There are two discussion points here.
1. When the Delta Function is introduced (to electronic engineers, at least)
as Limit as T -> 0 of (U(t) - U(t-T)) / T then other waveforms (such as the
sampled pulse) can be discussed as being a scaled version of the Impulse
itself,
rather than as the effect of the impulse on another waveform, in which case
the
integral operation is irrelevant.
(ISTR "That's not irrelevant, that's ir hippopotamus" from the Goon Show?)
2. A discussion about distributions is unhelpful to the point of
irrelevance.
When the Delta Function is introduced (to electronic engineers, at least),
they are
invited to feel uncomfortable about it not being a proper function despite
that up
to that point they will have been handling steps, pulses, square and
triangular waves,
functions that are as equally unanalytic as the delta, without any
discomfort
whatsoever being felt or suggested.
To ease the suggested discomfort, a study of distributions and generalised
functions
is a further suggestion, after which, you come back to using the delta as an
ordinary function,
just as you were before the suggested disconfort, but without the
application of any new
mathematics that you might have encountered during your study.
Therefore the suggestion of distributions and / or generalised functions is
an irrelevant
sidetrack that adds nothing to the understanding or the application of
mathematics.