DSP Books

Thanks. Yep, there's a 2nd edition
of my book coming out. My publisher
predicts that it'll be available (copies
in the warehouse) during the third week
of March (next month.)

Rick, I suggest you hold off printing that book. There is a DSP 'expert' in
the UK (Airy R Bean) who claims that all current DSP theory is flawed as it
neglects the need for "Big K".

;-)

Regards

GW

Ryan H.

2004-02-24 02:05:52 UTC

What is the "Big K"?

Post by Graham W

Thanks. Yep, there's a 2nd edition
of my book coming out. My publisher
predicts that it'll be available (copies
in the warehouse) during the third week
of March (next month.)

Rick, I suggest you hold off printing that book. There is a DSP 'expert' in
the UK (Airy R Bean) who claims that all current DSP theory is flawed as it
neglects the need for "Big K".
;-)
Regards
GW

Brian Reay

2004-02-24 08:21:11 UTC

Post by Ryan H.
What is the "Big K"?

Don't worry about is Ryan, it is just a figment of a deluded mind (Airy R
Bean's). If you want a laugh I suggest a Google search of uk.radio.amateur

If you need laughter therapy, you can look up some of "Airy's" other pet
ideas- heat bands, nuclear emissions, complex numbers, negative frequency,
.........

--
73
Brian
G8OSN
www.g8osn.org.uk
www.amateurradiotraining.org.uk for FREE training material for all UK
amateur radio licences
www.phoenixradioclub.org.uk - a RADIO club specifically for those wishing
to learn more about amateur radio

RVMJ 99g

2004-02-24 17:04:45 UTC

Post by Ryan H.
What is the "Big K"?

It is - or appears to be - a fiddle-factor that Airy R. Bean uses to
link DSP theory to his perception of it.

Some have suggested his perception is flawed, but nevertheless he
keeps returning to the theme; or at least it wasn't too long ago
(perhaps a year?) since he last proposed it.

A Google search should turn up more info, but he was probably posting
under a different sock-puppet then.

--
from
RVMJ
(dot) 99g (at) BTinternet (dot) com

Airy R. Bean

2004-02-24 21:41:25 UTC

It was one of my posits for attempting to square up the
claims made in the literature for what is the mathematical
basis of sampling against the description that electrical
engineers will have been taught up to that point about the properties
of Dirac's Delta function.

For example, int(+/- inf) ( F(t) x d(t-T) ) yields f(T), which
is a constant function.

It does NOT yield f(T) x d(t-T) which is claimed as the basis
for sampling in many texts, (but not O & S)

(I am awaiting a copy of Dirac's 1930 work on Quantum Mechanics
in order to read up on the original development of the Delta Function,
but it seems to be a classic; the only copies located by my book-finder
have been priced at several hundred pounds)

My recent interest has been to attempt to lay the ghost as to
when and where the apparently erroneous explanations appeared.

As to those who repeatedly make rather silly childish sneers
about any attempts to improve one's knowledge and understanding,
I say this...."Empty vessels make most noise".

Post by Ryan H.
What is the "Big K"?

RVMJ 99g

2004-02-24 22:19:41 UTC

Post by Airy R. Bean
"Empty vessels make most noise".

I think we'd all agree with that, at least.

Since you managed to avail yourself of an expensive law book when a
libel case threatened, couldn't you do the same for the book you
mention? After all, what value do you place on knowledge?

--
from
RVMJ
(dot) 99g (at) BTinternet (dot) com

robert bristow-johnson

2004-02-25 03:42:55 UTC

Post by Ryan H.
What is the "Big K"?

(corrected capitalization typo)

Post by Airy R. Bean
It does NOT yield f(T) x d(t-T) which is claimed as the basis
for sampling in many texts, (but not O & S)

which textbooks say that

+inf
integral{ f(t)*d(t-T) dt} = f(T)*d(t-T) ???
-inf

saying that is not correct but, until you come up with a list of texts that
say that, i think that the complaint is a straw man.

now engineers *do* tend to be a sloppy bunch and we treat the Dirac impulse
like a normal "function" (i like to think of it as a rectangular function of
unit area and about one Planck Time in width) and we will sometimes say

f(t)*d(t-T) = f(T)*d(t-T) .

it's a little bit illegit since the Dirac impulse shouldn't really be laying
around naked anywhere without being surrounded by an integral. at least
that's what the math guys tell us. but i don't even have a problem with
that infraction of the rules because it is virtually true for a practical
impulse of non-zero but arbitrarily small width. both sides represent a
"function" (if the math guys will allow us to call it that, *they* call it a
"distribution") that is zero everywhere except at T and the area under the
"function" is f(T).

as if that would solve anything.

Post by Airy R. Bean
My recent interest has been to attempt to lay the ghost as to
when and where the apparently erroneous explanations appeared.

again, what legit engineering text says

+inf
integral{ f(t)*d(t-T) dt} = f(T)*d(t-T) ?
-inf

i haven't seen it.

besides, how does the concept of a "Big K" come out of that?

r b-j

RVMJ 99g

2004-02-25 09:13:57 UTC

Post by Airy R. Bean
My recent interest has been to attempt to lay the ghost as to
when and where the apparently erroneous explanations appeared.

again, what legit engineering text says
+inf
integral{ f(t)*d(t-T) dt} = f(T)*d(t-T) ?
-inf
i haven't seen it.
besides, how does the concept of a "Big K" come out of that?

As a measure of Bean's wider credibilty, you may not want to know (but
could find it informative) that in a new posting recently on
uk.radio.amateur, he described (hopefully only some of) his boyhood
experiences of 'wet farts'.

HTH

--
from
RVMJ
(dot) 99g (at) BTinternet (dot) com

Chimera

2004-02-25 22:43:21 UTC

Post by RVMJ 99g

Post by Airy R. Bean
My recent interest has been to attempt to lay the ghost as to
when and where the apparently erroneous explanations appeared.

again, what legit engineering text says
+inf
integral{ f(t)*d(t-T) dt} = f(T)*d(t-T) ?
-inf
i haven't seen it.
besides, how does the concept of a "Big K" come out of that?

Compared to some of his technical postings the wet fart ones seem quite
rational.

Chimera

Airy R. Bean

2004-02-25 11:04:00 UTC

Any text book that says that sampling at the instant T is
represented by f(t) x d(t-T) and then uses this to claim
to further claim that the transform of such sampling is f(T) x e^(-sT) after
taking the transform of d(t) as 1.

Post by robert bristow-johnson
which textbooks say that
+inf
integral{ f(t)*d(t-T) dt} = f(T)*d(t-T) ???
-inf

robert bristow-johnson

2004-02-26 08:08:40 UTC

Post by robert bristow-johnson
which textbooks say that
+inf
integral{ f(t)*d(t-T) dt} = f(T)*d(t-T) ???
-inf

Any text book

"Any" textbook? that gives me _wide_ latitude in response. alls i need to
do is find one that doesn't.

Post by Airy R. Bean
that says that sampling at the instant T is
represented by f(t) x d(t-T) and then uses this to claim
to further claim that the transform of such sampling is f(T) x e^(-sT) after
taking the transform of d(t) as 1.

no, they say that

Laplace{ f(t) * d(t-T) } = f(T) * e^(-sT)

which is, in fact, true. and it's a legit expression since the Dirac
impulse function is getting integrated. before applying the L.T., they may
be just a wee bit illegit if they let f(t)*d(t-T) hang around without
getting integrated w.r.t. "t", but that little infraction doesn't bother me
much because...

Post by robert bristow-johnson
now engineers *do* tend to be a sloppy bunch and we treat the Dirac impulse
like a normal "function" (i like to think of it as a rectangular function of
unit area and about one Planck Time in width)

Planck Time = 10^(-43) seconds.

in the ballpark. no need to be too precise.

Post by Airy R. Bean
If unit area, then height is OOO 10^(43)

not precisely. *dimensionless* unit area. then the height is 10^43 Hertz.
the height of a unit impulse always has dimension that is the reciprocal of
the independent variable (time, in this case).

Post by Airy R. Bean
Let us suppose that you are sampling a
waveform which has an instantaneous value
of 5,

let's say a constant value of 5 over 10^(-43) sec. is that close enough to
instantaneous for you? also, i'm taking you at your word for it and treat
"5" as 5 (dimensionless). not "5 volts" or something like that. if you're
tossing in another dimension and unit, you must be explicit.

Post by Airy R. Bean
then the resultant amplitude of the
claimed action of sampling of f(t) x d(t-T) is 5 x 10^(-43)

uh, did you mean 10^(43) or 10^(-43)? i presume the former. but, to be
precise, the amplitude of the sampled impulse is 5*10^43 (1/sec).

Post by Airy R. Bean
and not 5 as your subsequent analysis might claim.

you haven't seen my subsequent analysis yet. how do you know what it is?

the height of the rectangular impulse sampled waveform would be
5*10^43 (1/sec) and the width is still 10^(-43) sec, but when you describe
the *strength* of that impulse, you do so with its area (integrated w.r.t.
time) and that is simply 5. whether it is 5*10^43 (1/sec) with 10^(-43) sec
width or 2.5*10^43 (1/sec) with 2*10^(-43) sec width, the behavior is about
the same. (yet one is twice as tall as the other, hmmmm, how can both be
about the same Airy?)

Post by Airy R. Bean
In the mathematical analysis that intending engineers
will have come across up to the point where they
learn about sampling, where has the concept
of multiplying the instantaneous value of one
function against the _AREA_ of another come in?

fundamentally, you are multiplying the instantaneous value of one function
(the function getting sampled) against the instantaneous value of another
(the sampling impulse) which, except for the sampling "instant" (that
instant is one Planck Time in width as far as i'm concerned), throws away
all information about the function getting sampled (because it gets
multiplied by zero) except for around the sampling instance. at that point,
the height of the sampling impulse (10^43 1/sec) gets multiplied by the
instantaneous value of the function getting sampled, but the width is
unchanged. so with the width unchanged and height getting multiplied by
that instantaneous value, that is equivalent to the _AREA_ getting
multiplied by that very same instantaneous value.

Post by Airy R. Bean
If you, or anyone, wishes to introduce an "area" into
the field of integral transforms, then she or he would
be well-advised to support that area with a corresponding
integral to evaluate it.

hey, didn't i say that? (or i said the "math guys" say it.) big deal. the
sampling function (that infinite sum of unit impulses all spaced by the
sampling period) doesn't really mean anything real anyway. it is simply a
theoretical construction that shows how you can literally discard all
information between those sampling instances and still, mathematically, have
enough information left in those samples to reconstruct the original signal,
even in between the sampling instances. big deel.

okay, Airy. where is that Big K? i vanna see it now. (and try to respond
in a single post. i refuse to keep track of all of the separate little
chunks you expectorate at me. i can deal with one at a time.)

r b-j

Airy R. Bean

2004-02-26 10:36:45 UTC

Sorry, but you're talking absolute bollocks that fits in with no
part of the calculus.

Chimera

2004-02-26 11:07:49 UTC

Post by Airy R. Bean
Sorry, but you're talking absolute bollocks that fits in with no
part of the calculus.

Why don't you try THINKING before you dismiss a GOOD posting as bollocks.

Look at what I posted, what Robert has posted, at previous posting you also
dismissed and try THINKING.

Chimera.

Graham W

2004-02-26 11:45:25 UTC

Post by Airy R. Bean
Sorry, but you're talking absolute bollocks that fits in with no
part of the calculus.

Why don't you try THINKING before you dismiss a GOOD posting as bollocks.
Look at what I posted, what Robert has posted, at previous posting you also
dismissed and try THINKING.

Airy THINK????????????????????????????????

Some hope.

Rational thought rewires 2 undamaged brain cells. Who is going to lend them
to Airy and risk having them pickled?

73s de

GW

Chimera

2004-02-26 12:24:12 UTC

Post by Graham W

Post by Airy R. Bean
Sorry, but you're talking absolute bollocks that fits in with no
part of the calculus.

Why don't you try THINKING before you dismiss a GOOD posting as bollocks.
Look at what I posted, what Robert has posted, at previous posting you

also

Post by Chimera
dismissed and try THINKING.

Airy THINK????????????????????????????????
Some hope.
Rational thought rewires 2 undamaged brain cells. Who is going to lend them
to Airy and risk having them pickled?

He is a difficult student. My students seem to cope with normal alcohol
abuse and still handle this basic material.

Chimera.

Airy R. Bean

2004-02-26 11:41:22 UTC

At the accepted risk of stirring up the Twitterers Of Twaddle from their
slumber.....

-----OOOOO-----

The conversion from the continuous time world of analogue
to the discrete time world of digital causes some mathematical problems.

The action of sampling is to multiply an incoming analogue waveform
by a series of unity amplitude spikes (ideally of zero width) to yield,
still
in the analogue world, a set of samples of the incoming waveform, again
of zero width.

How do we represent these samples mathematically? In particular, how
do we analyse their frequency spectra?

The calculus available to us in the analogue world holds these samples to be
zero-integrable
and therefore their application into any integral transform (Fourier,
Laplace, etc)
will yield a zero spectrum.

However, we know that this is not true of the systems into which this
sampling takes
place, for after processing of these samples and re-application back to the
analogue
world through a DAC, the whole of the works of Mozart re-appear in all their
glory.

So, what is the nature of these samples, and how shall they be represented?
We have a mathematical model, integral calculus, that is failing us because
of a zero-integrable result in an area where the results (Mozart, above) are
far from zero!

The approach taken, (and we are free to adopt any approach that may appeal
to
us to model an aspect of engineering for which conventional mathematics has
failed us) is to model the sampled spikes as though they are a
multiple of the Analogue Unit Impulse (not to be confused with the Discrete
Unit Sample, more of that later). We need to be very cautious when we do
this,
because if we claim that the action of our sampling is to multiply the
incoming waveform by a Unit Impulse, there are many, many valid
mathematical objections that can be raised against such a claim, not the
least of which is the lack of the order of infinity in a sample that is only
a volt, or so, high.

So, the model (and it is only a model and not a REAL representation)
that we take is to say that the unit of unity that was the continuous
analogue world has become the unit of infinity in the discrete analogue
world. This
resolves our problem of zero-integrability, the properties of the Unit
Impulse, Dirac's Delta function, being well established.

It may help to think of some fiddle factor that has caused this huge
multiplication of size. I had previously referred to a non-specific,
"Big-K", although what I am implying now would, in fact, be the reciprocal
of what was previously posited as "Big-K". Such a fiddle factor would be
justified because even if though the scale of our analogue samples in unity
and not infinity, the shape of our sampled pulses is isomorphic with the
Unit
Impulse. We must, however, remember to remove this fiddle factor, this
Big-K, this reciprocal of Big-K; and we do this when applying the outputs
of our DSP to the DAC from which comes out our Mozart once again
in signals that are of the order of unity.....
The transfer function, or impulse response, of the combined effect of the
DAC and the sample-and-hold is taken to be a single pulse of unity
value(existing
till the time of the next conversion from the output of our DSP).
So, in response to a voltage spike of no width and of unity height, the
sample-and-hold
responds with its pulse lasting as long as the sample time. Now, in reality
we
have a spike of one volt high which we have been analysing as though it was
infinity
volts high. How do we mentally model this removal of Big-K?

Think of this......if a network responds with an output that is it impulse
response, then
what it was triggered with must have been an impulse!

-----OOOOO-----

IT IS IMPORTANT TO REMEMBER THAT THE ACTION OF SAMPLING
IS NOT TO MULTIPLY THE INCOMING ANALOGUE WAVEFORM BY A
COMB OF UNIT IMPULSES AND THEIR DELAYED SIBLINGS; IF THAT
WERE TO BE CLAIMED THEN IT COULD BE EASILY KNOCKED DOWN.

The action of sampling is only _REPRESENTED_ by such a multiplication and
only then to resolve the specific problem of zero-integrability.

-----OOOOO-----

In response to the OP.....

On the whole, it seems to me that despite your
undoubted standing in the DSP community, you
were one of those who swallowed the plausible story
about sampling, and because you understood well at that
point all that you had needed to know about the Delta function
up to that time, that you, as did so many others, accepted
statements made about the Delta function without objection and without
query.

It is only now, much later on, that you have allowed yourself to be
drawn into a challenge that was not aimed at you personally, and I
suspect that you cannot accept that there is a flaw
in your underlying knowledge; causing you to react with Freudian
Rationalisation by interjecting a rather silly adaptation
of the integral calculus, as quoted from you below.

-----OOOOO-----

This has, however, been an interesting series of threads over the
last few years. It is not an area in which I am yet professionally
involved, and with so many other things to occupy my time, the
ugly head got reared again whenever I drifted back round to the topic.

Based upon the explanation given above, I am satisfied. However, I
also remain satisfied that all the objections and protests that I raised
against the glib throw-away-lines of the text-books were also correct!

It is interesting that in response to my claim that the text-books were
dubious was that so many of you responded by indignantly requoting
the text-book position and thereby contributing nothing to
the discussion. More Freudian Rationalisation, perhaps?

-----OOOOO-----

I leave it to the Twitterers Of Twaddle to now exhibit their playground
habituation.
No doubt there will be an indignant mass fart of, "I told you so" when, in
fact, they told nothing of the kind.

Post by robert bristow-johnson
fundamentally, you are multiplying the instantaneous value of one function
(the function getting sampled) against the instantaneous value of another
(the sampling impulse) which, except for the sampling "instant" (that
instant is one Planck Time in width as far as i'm concerned), throws away
all information about the function getting sampled (because it gets
multiplied by zero) except for around the sampling instance. at that

point,

Post by robert bristow-johnson
the height of the sampling impulse (10^43 1/sec) gets multiplied by the
instantaneous value of the function getting sampled, but the width is
unchanged. so with the width unchanged and height getting multiplied by
that instantaneous value, that is equivalent to the _AREA_ getting
multiplied by that very same instantaneous value.

Chimera

2004-02-26 12:20:35 UTC

Post by Airy R. Bean
At the accepted risk of stirring up the Twitterers Of Twaddle from their
slumber.....
-----OOOOO-----
The conversion from the continuous time world of analogue
to the discrete time world of digital causes some mathematical problems.
The action of sampling is to multiply an incoming analogue waveform
by a series of unity amplitude

Let me stop you right THERE.

The spikes are unity AREA not AMPLITUDE

Post by Airy R. Bean
spikes (ideally of zero width) to yield,

Now ask yourself why the width is important. (Someone has told you this
before. Try using Google.)

<Big K nonsense snipped as even Airy should see it isn't required.>

Chimera

Brian Reay

2004-02-26 13:50:06 UTC

Let me stop you right THERE.
The spikes are unity AREA not AMPLITUDE

Post by Airy R. Bean
spikes (ideally of zero width) to yield,

Now ask yourself why the width is important. (Someone has told you this
before. Try using Google.)
<Big K nonsense snipped as even Airy should see it isn't required.>

Oh dear, you are going to upset Gareth now. We've been here before and he
can't seem to grasp the significance of the area and the width.

Have you noticed he was cross posting. I've limited this to uk.radio.amateur

Regards

Brian

--
73
Brian
G8OSN
www.g8osn.org.uk
www.amateurradiotraining.org.uk for FREE training material for all UK
amateur radio licences
www.phoenixradioclub.org.uk - a RADIO club specifically for those wishing
to learn more about amateur radio

Airy R. Bean

2004-02-26 15:57:48 UTC

Is your need for attention so great that you must reply
to the posts that you made under your own alter-ego?

Under your alter-ego you have mistakenly assumed that
the sampling spikes referred to are Unit Impulses, and your
motivation to be insulting has merely made you look rather silly.

If you are really that bothered that people in other NG who
are your technical superiors might criticise you, then perhaps
you'd be better off not posting at all?

Post by Airy R. Bean
At the accepted risk of stirring up the Twitterers Of Twaddle from their
slumber.....
The conversion from the continuous time world of analogue
to the discrete time world of digital causes some mathematical problems.
The action of sampling is to multiply an incoming analogue waveform
by a series of unity amplitude

Let me stop you right THERE.
The spikes are unity AREA not AMPLITUDE

Post by Airy R. Bean
spikes (ideally of zero width) to yield,

Now ask yourself why the width is important. (Someone has told you this
before. Try using Google.)

Oh dear, you are going to upset him now. We've been here before and he
can't seem to grasp the significance of the area and the width.
Have you noticed he was cross posting. I've limited this to

uk.radio.amateur

Chimera

2004-02-26 16:37:17 UTC

I didn't see Brian's post about the sampling spikes but you still haven't
answered my post about the same topic.

Sampling pulses ARE Unit Pulses. UNIT AREA. REPEAT UNIT AREA. If you don't
understand that I suggest take up another line of study.

Now, I could go on about the nature of the Unit Pulse but I suspect you
wouldn't understand it.

Chimera

Graham W

2004-02-26 18:24:56 UTC

Post by Airy R. Bean
Is your need for attention so great that you must reply
to the posts that you made under your own alter-ego?
Under your alter-ego you have mistakenly assumed that
the sampling spikes referred to are Unit Impulses, and your
motivation to be insulting has merely made you look rather silly.
If you are really that bothered that people in other NG who
are your technical superiors might criticise you, then perhaps
you'd be better off not posting at all?

Ad homin wots it now? Let us just remember who is the socket puppet here,
shall we, Gareth Alun Evans of Chippenham?

GW

repatch

2004-02-26 14:55:23 UTC

Plonk, again...

Post by Airy R. Bean
At the accepted risk of stirring up the Twitterers Of Twaddle from their

RVMJ 99g

2004-02-26 15:22:53 UTC

Post by Airy R. Bean
The conversion from the continuous time world of analogue
to the discrete time world of digital causes some mathematical problems.

This is to my recollection the third time over the years you have
raised the topic, and judging by the responses you have received, I am
beginning to suspect that the 'problems' lie between your ears.

--
from
RVMJ
(dot) 99g (at) BTinternet (dot) com

daestrom

2004-02-26 22:32:12 UTC

Post by Airy R. Bean
At the accepted risk of stirring up the Twitterers Of Twaddle from their
slumber.....
-----OOOOO-----

<snip>

Post by Airy R. Bean
However, we know that this is not true of the systems into which this
sampling takes
place, for after processing of these samples and re-application back to the
analogue
world through a DAC, the whole of the works of Mozart re-appear in all their
glory.

Does not the 'hold' section of a DAC add back in an integrable component,
explaining why the analog output has been restored? A 'hold' mechanism is
central to all DACs.

Post by Airy R. Bean
-----OOOOO-----
IT IS IMPORTANT TO REMEMBER THAT THE ACTION OF SAMPLING
IS NOT TO MULTIPLY THE INCOMING ANALOGUE WAVEFORM BY A
COMB OF UNIT IMPULSES AND THEIR DELAYED SIBLINGS; IF THAT
WERE TO BE CLAIMED THEN IT COULD BE EASILY KNOCKED DOWN.

Is not the digital sampling of the analog input actually a convolution of
the two functions? Convolution can often be represented by multiplication,
but it is distinctly different.
<snip>

daestrom

Airy R. Bean

2004-02-27 09:08:29 UTC

You'd be in one of those mathematical "trick" areas, like
when you prove that 1 = 2, if you started to integrate
something that was already zero.

Post by Airy R. Bean
However, we know that this is not true of the systems into which this
sampling takes
place, for after processing of these samples and re-application back to

the

Post by Airy R. Bean
analogue
world through a DAC, the whole of the works of Mozart re-appear in all

their

Post by Airy R. Bean
glory.

Does not the 'hold' section of a DAC add back in an integrable component,
explaining why the analog output has been restored? A 'hold' mechanism is
central to all DACs.

Chimera

2004-02-27 09:34:06 UTC

Post by Airy R. Bean
You'd be in one of those mathematical "trick" areas, like
when you prove that 1 = 2, if you started to integrate
something that was already zero.

Judging by your maths I would assume this is standard practice for you.

Chimera

daestrom

2004-02-27 20:59:23 UTC

Post by Airy R. Bean
You'd be in one of those mathematical "trick" areas, like
when you prove that 1 = 2, if you started to integrate
something that was already zero.

Not much of a 'trick'. A simple 'hold' section is nothing more than a
capacitor. And of course the voltage on a cap is well known to be
proportional to the integral of the net input current.

The net input current isn't *always* zero. There exists a pulse with a
width, however narrow, when the input is not zero. If you claim it is
*always* zero, then you haven't applied any pulse and have thrown away all
the information in the signal. No matter how narrow the pulse width, it is
never zero, and always has *some* area (definition of limit in this sense is
the width *approaches* zero but never reaches it).

daestrom

Airy R. Bean

2004-02-28 12:32:40 UTC

If you have a mathematical function that has no width
and exists only at discrete points then it is zero-integrable,
with the exception of the contrivance of Dirac's Delta function.

Post by Airy R. Bean
You'd be in one of those mathematical "trick" areas, like
when you prove that 1 = 2, if you started to integrate
something that was already zero.

Not much of a 'trick'. A simple 'hold' section is nothing more than a
capacitor. And of course the voltage on a cap is well known to be
proportional to the integral of the net input current.
The net input current isn't *always* zero. There exists a pulse with a
width, however narrow, when the input is not zero. If you claim it is
*always* zero, then you haven't applied any pulse and have thrown away all
the information in the signal. No matter how narrow the pulse width, it is
never zero, and always has *some* area (definition of limit in this sense is
the width *approaches* zero but never reaches it).

Chimera

2004-02-28 14:31:38 UTC

Post by Airy R. Bean
If you have a mathematical function that has no width
and exists only at discrete points then it is zero-integrable,
with the exception of the contrivance of Dirac's Delta function.

The "contrivance of Dirac's Delta function"? As we are talking about Dirac's
Delta function does that not make it rather relevant?

Your problem is that you don't understand even the basics of what you are
trying to study. Dirac's delta function has unit AREA and tends toward zero
width. The fact that it is defined as an AREA _repeat AREA _ then it is
intergrable. If not, then sampling would not work.

Do you have any technical qualifications in this field? Any at all?

Chimera

Airy R. Bean

2004-02-28 17:28:02 UTC

1. The person masquerading as "Chimera" has latched onto
the attributes of the Delta Function, which were never in dispute.
However, taken with her infantile stance of name-calling, she is
best ignored for the irrelevance that she undoubtedly is.
Examples of her infantile stance may be seen below.
2. Whereas what you say is true, there is a problem with
real sampling taken as points in that it yields samples that are
zero-integrable
and therefore impossible to determine the frequency spectrum for
by conventional mathematical analysis.
3. The spectrum of real sampling changes dependant upon the
width of the sampling pulse, and we are only interested in the
value of the sampled waveform at the rising edge, and a spectrum that
is independent of the sampling pulse width.
4. So, to represent that rising edge only, we treat our real samples
as existing only at that edge
5. As such, those samples are isomorphic to the Unit Impulse but
because of their small size are zero-integrable.
6. Therefore, in order to achieve an analysis, we replace the unity
of one that is the characteristic of the real world with the unity
of infinity which is the amplitude of the Unit Impulse.
7. This is a contrivance for modelling only, and it is important
to stress that real sampling is NOT multiplying by a Unit Impulse,
for this raises so many valid mathematical objections.

Facts are inconvenient? Only to those who have swallowed the
religious lie that real sampling _IS_ multiplying by a comb of impulses
rather then _IS MODELLED_ by such.

The "contrivance of Dirac's Delta function"? As we are talking about Dirac's
Delta function does that not make it rather relevant?
Your problem is that you don't understand even the basics of what you are
trying to study. Dirac's delta function has unit AREA and tends toward zero
width. The fact that it is defined as an AREA _repeat AREA _ then it is
intergrable. If not, then sampling would not work.
Do you have any technical qualifications in this field? Any at all?

Not to pile on here, but the fact is, if someone provides sampling data at
something a bit more than twice the bandwidth of the initial waveform, it

can

be exactly reconstructed.
Is this one of those discussions were facts are inconvenient?

Chimera

2004-02-28 18:51:29 UTC

Post by Airy R. Bean
1. The person masquerading as "Chimera" has latched onto
the attributes of the Delta Function, which were never in dispute.
However, taken with her infantile stance of name-calling, she is
best ignored for the irrelevance that she undoubtedly is.
Examples of her infantile stance may be seen below.
2. Whereas what you say is true, there is a problem with
real sampling taken as points in that it yields samples that are
zero-integrable
and therefore impossible to determine the frequency spectrum for
by conventional mathematical analysis.
3. The spectrum of real sampling changes dependant upon the
width of the sampling pulse, and we are only interested in the
value of the sampled waveform at the rising edge, and a spectrum that
is independent of the sampling pulse width.
4. So, to represent that rising edge only, we treat our real samples
as existing only at that edge
5. As such, those samples are isomorphic to the Unit Impulse but
because of their small size are zero-integrable.
6. Therefore, in order to achieve an analysis, we replace the unity
of one that is the characteristic of the real world with the unity
of infinity which is the amplitude of the Unit Impulse.
7. This is a contrivance for modelling only, and it is important
to stress that real sampling is NOT multiplying by a Unit Impulse,
for this raises so many valid mathematical objections.
Facts are inconvenient? Only to those who have swallowed the
religious lie that real sampling _IS_ multiplying by a comb of impulses
rather then _IS MODELLED_ by such.

1. If the attributes of the unit impulse were never in dispute,
why did you dispute them _AND_ insist that it had unit amplitude?

2. The unit impulse only approaches zero width, it never actually
becomes zero. This concept is as old as Newton,
surely you have come across it before?

3. Which part of the rising edge would that be? Even with your
warped version of the unit pulse (before you stopped disputing its
attributes)
you should see that if it is only its edge that is significant then its
amplitude
and width would be irrelevant. What we are really interested in is the POWER
in the waveform being sampled so the width can not be zero. Zero width =
zero power.

4. You might treat the samples as existing only at the edge but those
who understand sampling know that it is invalid. See 3, above.

5. New word "isomorphic". Not applicable in this case but a new word for
you.
Maybe you can use it in Scrabble. If you will excuse the pun, don't stretch
it to infinite amplitude......

6. Whoaaaa "the unity of infinity". A good title for a love song maybe but,
in this context, nonsense.

7. Whoaa, you have suddenly come to the real world. A new experience for
you, it would appear.
Please stay awhile and have a look around. You will find there are people
here that understand DSP.

You have never actually done any DSP, I assume?
Never implemented a A/D converter so never analysed why you need a sample
and hold?
While you are in the real world I suggest you try doing so. A bit of reality
helps
everyone. Don't overdo it, it seems to have been awhile for you.

Chimera.

jim <"N0sp"@

2004-02-29 14:00:11 UTC

The "contrivance of Dirac's Delta function"? As we are talking about Dirac's
Delta function does that not make it rather relevant?
Your problem is that you don't understand even the basics of what you are
trying to study. Dirac's delta function has unit AREA and tends toward zero
width. The fact that it is defined as an AREA _repeat AREA _ then it is
intergrable. If not, then sampling would not work.

No, this is completely incorrect. The dirac delta function is just a convenient
construct that simply encompasses the notion that the spectrum of the sampling
process is flat. Similarly the notion of a bandlimited function is a convenient
idealization. In reality there are lots of working sampling processes that don't
meet these ideal conditions. In fact its fairly safe to say that you will never
encounter a sampling process that meets these conditions perfectly.
In fact, a sampling process doesn't even have to come close to the idealized
dirac pulse to work. For instance digital cameras work just fine.

-jim

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Brian Reay

2004-02-29 14:36:59 UTC

The "contrivance of Dirac's Delta function"? As we are talking about Dirac's
Delta function does that not make it rather relevant?
Your problem is that you don't understand even the basics of what you are
trying to study. Dirac's delta function has unit AREA and tends toward zero
width. The fact that it is defined as an AREA _repeat AREA _ then it is
intergrable. If not, then sampling would not work.

I don't think there is a disconnect between Chimera's version and the point
you make. The feature that is common is the idea that sampling takes a
finite time which is, in itself, and integration process. Think of a sample
and hold- it integrates the signal over the time the sample gate is open, it
doesn't take a sample in zero time.

As Chimera and daestrom point out, the area of the unit pulse is
significant.

Airy R. Bean

2004-02-29 14:51:14 UTC

Not so. The accepted analyses of sampling use
only the value of f(T) and not the integrated
sum over the interval f(T) - f(T+dt). There is no integration of the
input function f(t) over the sampling period
in the standard analyses.

Whatever the width of the sampling pulse in your circuit, the
value of the input function at the rising edge only is used.

Post by Brian Reay
I don't think there is a disconnect between Chimera's version and the point
you make. The feature that is common is the idea that sampling takes a
finite time which is, in itself, and integration process. Think of a sample
and hold- it integrates the signal over the time the sample gate is open, it
doesn't take a sample in zero time.

Chimera

2004-02-29 17:02:25 UTC

Post by Airy R. Bean
Not so. The accepted analyses of sampling use
only the value of f(T) and not the integrated
sum over the interval f(T) - f(T+dt). There is no integration of the
input function f(t) over the sampling period
in the standard analyses.
Whatever the width of the sampling pulse in your circuit, the
value of the input function at the rising edge only is used.

Accepted by you only, to those of us with real experience your analysis is
laughable.

Your comment re the rising edge shows a lack of understanding that I have
never seen the like of.

If only the edge matters, why did Dirac specify the area as unity and not
some measure of the rise time of the unit impulse. It would then be the zero
rise time pulse not the unit pulse.

Chimera

jim <"N0sp"@

2004-02-29 17:49:02 UTC

Post by Airy R. Bean
Not so.

What's not so? Your post appears in response to mine, but you quote some one else.

Post by Airy R. Bean
The accepted analyses of sampling use
only the value of f(T) and not the integrated
sum over the interval f(T) - f(T+dt). There is no integration of the
input function f(t) over the sampling period
in the standard analyses.

Standard analysis of what? Is f(t) the notes Mozart scored? or is it the sound
waves coming out of the orchestra? or is it the current in the microphone? at what
point does f(t) become f(t)?

Post by Airy R. Bean
Whatever the width of the sampling pulse in your circuit, the
value of the input function at the rising edge only is used.

what circuit? who said anything about circuits.

-jim

Post by Brian Reay
I don't think there is a disconnect between Chimera's version and the

point

Post by Brian Reay
you make. The feature that is common is the idea that sampling takes a
finite time which is, in itself, and integration process. Think of a

sample

Post by Brian Reay
and hold- it integrates the signal over the time the sample gate is open,

Post by Brian Reay
doesn't take a sample in zero time.

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Airy R. Bean

2004-02-29 17:55:28 UTC

No - it's posted in response to Mr.Reay.

Post by Airy R. Bean
Not so.

What's not so? Your post appears in response to mine, but you quote some one else.

Post by Brian Reay
I don't think there is a disconnect between Chimera's version and the

point

Post by Brian Reay
you make. The feature that is common is the idea that sampling takes a
finite time which is, in itself, and integration process. Think of a

sample

Post by Brian Reay
and hold- it integrates the signal over the time the sample gate is open,

Post by Brian Reay
doesn't take a sample in zero time.

Chimera

2004-02-29 17:03:34 UTC

The "contrivance of Dirac's Delta function"? As we are talking about

Dirac's

Post by Chimera
Delta function does that not make it rather relevant?
Your problem is that you don't understand even the basics of what you

are

Post by Chimera
trying to study. Dirac's delta function has unit AREA and tends toward

zero

Post by Chimera
width. The fact that it is defined as an AREA _repeat AREA _ then it is
intergrable. If not, then sampling would not work.

No, this is completely incorrect. The dirac delta function is just a

convenient

Post by jim <"N0sp"@
construct that simply encompasses the notion that the spectrum of the

sampling

Post by jim <"N0sp"@
process is flat. Similarly the notion of a bandlimited function is a

convenient

Post by jim <"N0sp"@
idealization. In reality there are lots of working sampling processes

that

Post by Airy R. Bean
don't

Post by jim <"N0sp"@
meet these ideal conditions. In fact its fairly safe to say that you

will

Post by Airy R. Bean
never

Post by jim <"N0sp"@
encounter a sampling process that meets these conditions perfectly.
In fact, a sampling process doesn't even have to come close to the

idealized

Post by jim <"N0sp"@
dirac pulse to work. For instance digital cameras work just fine.

I don't think there is a disconnect between Chimera's version and the point
you make. The feature that is common is the idea that sampling takes a
finite time which is, in itself, and integration process. Think of a sample
and hold- it integrates the signal over the time the sample gate is open, it
doesn't take a sample in zero time.
As Chimera and daestrom point out, the area of the unit pulse is
significant.

Sanity at last.

Chimera

jim <"N0sp"@

2004-03-01 12:46:50 UTC

Post by Brian Reay
I don't think there is a disconnect between Chimera's version and the

point

Post by Brian Reay
you make. The feature that is common is the idea that sampling takes a
finite time which is, in itself, and integration process. Think of a

sample

Post by Brian Reay
and hold- it integrates the signal over the time the sample gate is open,

Post by Brian Reay
doesn't take a sample in zero time.
As Chimera and daestrom point out, the area of the unit pulse is
significant.

Sanity at last.

Sane maybe, but not a very good grasp of the sampling process. By the reasoning
you guys are using the Weather Service is going about it all wrong for sampling
rainfall. Instead of recording the rainfall in the gauge once a day and then
emptying it. what they should be doing is just sticking the gauge out the window
for a second at the same time each day. Mr Bean would go even further and not even
allow them to retract their arm before noting the measurement, but instead insist
that they record the measurement as soon as the arm is fully extended.
There's no integration involved - the contents of the rain are just recorded once
a day. The sampling theorem tells you that there exists a continuos function that
can be derived that interpolates these samples. It doesn't tell you whether the
rain gauge is in a desert or tropical island, or if the gauge is in inches or
millimeters. Those things will affect what the continuous function looks like. A
crack in the bottom of the rain gauge or placing the rain gauge where the water
runs off a roof will also give you a different function f(t). You can't possibly
imagine that the dirac function is a mapping that ecompasses all these variable
and yields f(t) ~ that pure function "rainfall".

-jim

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Chimera

2004-03-01 18:08:09 UTC

Post by Brian Reay
I don't think there is a disconnect between Chimera's version and the

point

Post by Brian Reay
you make. The feature that is common is the idea that sampling takes a
finite time which is, in itself, and integration process. Think of a

sample

Post by Brian Reay
and hold- it integrates the signal over the time the sample gate is open,

Post by Brian Reay
doesn't take a sample in zero time.
As Chimera and daestrom point out, the area of the unit pulse is
significant.

Sanity at last.

For the rain gauge, this equates to measuring the rain fall for on day, the
sampling period. From that you can determine no more than how much it rained
on one day (the sample time, which equates to the width of the sample
pulse).

For the unit pulse sampling a waveform, a single pulse sampling the waveform
once, all you determine is the value of wave form being sampled over the
sample pulse width. You cannot determine anything about the nature of the
waveform at other times.

In the case of the rain gauge, the water collect is the rain that fell in
the sample period (assuming there are no error sources). In the sample pulse
case, it is the power of the sampled waveform averaged over the pulse width.

Also, no single sample process is immune from the error due to things link
leaky gauges etc. It is possible to account for some error sources but can
we get Airy to understand the basics first.

Sampling theory (Nyquist) tells use that we must sample waveform at twice
its frequency of repetition, if we are to reconstructed or, as is more
appropriate in this analogy, be able to predict a value at some future time.

Airy's problem hasn't extended that far yet, he still can't grasp the way a
single sample works. The width of the Dirac pulse is finite AND its area 1.
If it is repeated, like a comb as he refers to, you get a series of samples,
each sample being the width of the sampling pulse and repeated at the
samplying frequency.

Chimera

daestrom

2004-03-01 22:41:39 UTC

Post by Brian Reay
I don't think there is a disconnect between Chimera's version and the

point

Post by Brian Reay
you make. The feature that is common is the idea that sampling takes a
finite time which is, in itself, and integration process. Think of a

sample

Post by Brian Reay
and hold- it integrates the signal over the time the sample gate is open,

Post by Brian Reay
doesn't take a sample in zero time.
As Chimera and daestrom point out, the area of the unit pulse is
significant.

Sanity at last.

Not so. The two methods you mention for rainfall are a *sampling*
technique, and a *measurement* technique. If you have the resources to
measure every item (in this case rate of rainfall over time), then 100%
measurement is the way to go. But if you don't have the resources to sample
continuously, you can sometimes get a very good approximation by sampling at
appropriate intervals and infering the continuous function based on the
sample points. To get an *accurate* reproduction you would need to sample
at least twice the frequency of interest. Considering that rain storms
typically come and go in the course of hours or faster, 'the Weather
Service' would have to sample more often than once a day.

And such a sample would be the *rate* of rainfall, not the total so far. To
find the total, one would have to *assume* the rate is changing between
samples by some known function f(t). Then one could integrate the total
rainfall with some degree of accuracy.

But since the weather is unpredictable, you may have to sample the rate of
rainfall quite often and then just *assume* the rate changes linearly
between samples.

Post by jim <"N0sp"@
Mr Bean would go even further and not even
allow them to retract their arm before noting the measurement, but instead insist
that they record the measurement as soon as the arm is fully extended.
There's no integration involved - the contents of the rain are just recorded once
a day. The sampling theorem tells you that there exists a continuos function that
can be derived that interpolates these samples. It doesn't tell you whether the
rain gauge is in a desert or tropical island, or if the gauge is in inches or
millimeters. Those things will affect what the continuous function looks like. A
crack in the bottom of the rain gauge or placing the rain gauge where the water
runs off a roof will also give you a different function f(t). You can't possibly
imagine that the dirac function is a mapping that ecompasses all these variable
and yields f(t) ~ that pure function "rainfall".

Because all those things imply that the rate of rainfall is a rather complex
f(t), to get a reasonable accuracy, one would have to estimate the highest
frequency component and sample at least twice this rate. No simple matter.
But the beauty of such a rain gauge is it never needs emptying :-)

If you empty a conventional gauge once a day, how do you calculate the
average rainfall for a month? Easy, you have ~30 samples of rainfall with
each sample measuring inches-of-rain per day. You just have a sample width
that is very close to the sample period (minus a second or two to empty it
each day).

Some weather observers empty the gauge more often than once a day. The more
often one empties it, the more the situation approaches a digital sample of
the volume/time and you have a rain-rate sample. The 'gauge' is just a
long-time-constant 'hold' device in a sample and hold system.

daestrom

Airy R. Bean

2004-02-29 14:40:13 UTC

Whoever is the lady masquerading as, "Chimera", she seems
to have become obsessed by the wrong end of the stick.

I have never disputed the facts below that she attributes to
the Delta function. I have held these to be true for years, but
it is their very truth that means that using a comb of Delta
Functions can only be an idealised model of sampling and never
the actuality of sampling.

Still, right from her very first appearance a couple of months ago,
"Chimera"'s motivation seems to have been to vent her spleen
at me rather than to introduce any technical enlightenment. The
google record speaks! Her aggressive jeering from the sidelines
makes her irrelevant and she is best ignored.

The "contrivance of Dirac's Delta function"? As we are talking about Dirac's
Delta function does that not make it rather relevant?
Your problem is that you don't understand even the basics of what you are
trying to study. Dirac's delta function has unit AREA and tends toward zero
width. The fact that it is defined as an AREA _repeat AREA _ then it is
intergrable.

.....[All of jim's remarks snipped!].....

Chimera

2004-02-29 16:57:14 UTC

Post by Airy R. Bean
Whoever is the lady masquerading as, "Chimera", she seems
to have become obsessed by the wrong end of the stick.
I have never disputed the facts below that she attributes to
the Delta function. I have held these to be true for years, but
it is their very truth that means that using a comb of Delta
Functions can only be an idealised model of sampling and never
the actuality of sampling.
Still, right from her very first appearance a couple of months ago,
"Chimera"'s motivation seems to have been to vent her spleen
at me rather than to introduce any technical enlightenment. The
google record speaks! Her aggressive jeering from the sidelines
makes her irrelevant and she is best ignored.

You have disputed the nature of the Delta function, you claimed the
amplitude was unity whereas it is the _AREA_ which is unity.

Have you ever actually had any formal education in maths or engineering? If
so, can I suggest you ask for a refund.

Chimera

Chimera

2004-02-29 16:53:10 UTC

The "contrivance of Dirac's Delta function"? As we are talking about Dirac's
Delta function does that not make it rather relevant?
Your problem is that you don't understand even the basics of what you are
trying to study. Dirac's delta function has unit AREA and tends toward zero
width. The fact that it is defined as an AREA _repeat AREA _ then it is
intergrable. If not, then sampling would not work.

Your link to the 'real world' is very apt and valid for the most part. But,
as Dr Reay has pointed out, sampling does not take place on zero time. If it
did, no energy would be transferred, which is clearly not tenable.

The _AREA_ of the unit pulse is relevant and, in the real world, equates to
the sample time.

Chimera

daestrom

2004-02-29 13:55:33 UTC

Ah... Your problem is you think the pulse has zero width. It is *not*
exactly zero. It can approach zero and get as close as you are physically
able to make it and/or want. But it *never* reaches zero. If it did, then
there would be no pulse at all.

The area under the pulse is what is important. If the pulse width is fixed
at any arbitrary size greater than zero, the pulse is definitely intergable.

daestrom

Airy R. Bean

2004-02-29 14:30:50 UTC

But it isn't my problem, nor anyone else's. It is a fact of the
action of sampling. We are interested in the amplitude of the
sampled function at the rising edge only.

If we consider the rising edge to be the point of sample, then
we become independent of the width of the sampling pulse in
whatever circuit you are using.

As such, we wish to analyse a single point. That we wish to analyse
through an integrable transform gives us a difficulty because such
points are zero-integrable. The solution is to model using the infinity
that is represented by the Unit Impulse's amplitude as our unity.

It is misleading to state that the area of the pulse is of interest, because
there is no part of the calculus that supports the instantaneous multiplying
of one
function by the area of another. This last ruse is one of the religious
icons of those who state that sampling _IS_ the effect of multiplying
the incoming waveform by a comb of Delta Functions rather than
stating that it is only _MODELLED_ by such. It is a religious icon
because it is a matter of blind faith and not a matter of the
underlying calculus. Like all religions, you need to contrive more
and more weird and ridiculous explanations to explain away your
initial assumption which is wrong.

Ah... Your problem is you think the pulse has zero width. It is *not*
exactly zero. It can approach zero and get as close as you are physically
able to make it and/or want. But it *never* reaches zero. If it did, then
there would be no pulse at all.
The area under the pulse is what is important. If the pulse width is fixed
at any arbitrary size greater than zero, the pulse is definitely intergable.

Frank Turner-Smith G3VKI

2004-02-29 19:16:20 UTC

Post by Airy R. Bean
But it isn't my problem, nor anyone else's. It is a fact of the
action of sampling. We are interested in the amplitude of the
sampled function at the rising edge only.
If we consider the rising edge to be the point of sample, then
we become independent of the width of the sampling pulse in
whatever circuit you are using.

OK, but if you sample the pulse during its rise time, how can you predict
its final amplitude?

--
;>)
73 de Frank Turner-Smith G3VKI - mine's a pint.
http://turner-smith.co.uk

Airy R. Bean

2004-02-27 09:10:42 UTC

It is only the convolution of the two functions once you
have reached the frequency domain. The objections that I
am raising to the plausible standard explanation prevent you
from reaching the frequency domain.

Post by Airy R. Bean
IT IS IMPORTANT TO REMEMBER THAT THE ACTION OF SAMPLING
IS NOT TO MULTIPLY THE INCOMING ANALOGUE WAVEFORM BY A
COMB OF UNIT IMPULSES AND THEIR DELAYED SIBLINGS; IF THAT
WERE TO BE CLAIMED THEN IT COULD BE EASILY KNOCKED DOWN.

Is not the digital sampling of the analog input actually a convolution of
the two functions? Convolution can often be represented by

multiplication,

Post by daestrom
but it is distinctly different.

Leigh

2004-02-26 16:56:05 UTC

Post by Airy R. Bean
Sorry, but you're talking absolute bollocks that fits in
with no part of the calculus.

Typical - "I'm no longer winning this argument" response from
Airy Arsehole!

--
#!/bin/sh {who;} {last;} {pause;} {grep;} {touch;} {unzip;}
mount /dev/girl -t {wet;} {fsck;} {fsck;} {fsck;} {fsck;} echo
yes yes yes {yes;} umount {/dev/girl;zip;} rm -rf {wet.spot;}
{sleep;} finger: permission denied

robert bristow-johnson

2004-02-27 06:39:23 UTC

Post by Frank Turner-Smith G3VKI

Post by Airy R. Bean
Did I go wrong somewhere above?

No, in your case the midwife went wrong and threw the wrong bit away.
...(_!_)...

that was just too easy, but gratifying nonetheless (like hitting that final
E major power chord with the Marshall cranked up to "arc-weld").

BTW, guys, from 1968 to about 1975, i used to be a ham, too. WB0CCA ("CCA"
was "Casselton's Commie Adolescent".)

r b-j

Post by Frank Turner-Smith G3VKI
Sorry, but you're talking absolute bollocks that fits in with no
part of the calculus.

no, i guess you're right and the calculus does not submit itself to the
axiom called "the associative property of multiplication".

(instantaneous_value*pulse_height)*pulse_width does not really equal
instantaneous_value*(pulse_height*pulse_width) , does it, Mr. Bean? and it
is absolute bollocks to say that

pulse_height * pulse_width = pulse_area

i'm learning so much.

Post by Frank Turner-Smith G3VKI

Walt Davidson

2004-02-27 08:55:57 UTC

On Fri, 27 Feb 2004 01:39:23 -0500, robert bristow-johnson

Post by robert bristow-johnson
BTW, guys, from 1968 to about 1975, i used to be a ham, too. WB0CCA ("CCA"
was "Casselton's Commie Adolescent".)

The "Good Old Boys" have locked people up for less than that in the U.
S. of A.

It used to be Reds under the bed. Now it's Ali Baba!

--
Walt Davidson Email: g3nyy @despammed.com

Airy R. Bean

2004-02-27 09:01:00 UTC

I think that the problem that you have, is that despite your
accepted position as an authority in the world of DSP, is that
you have accepted the explanation of sampling as a multiplication
by a comb of Delta Functions as a matter of blind faith and not as a
matter of technical rigour. Like all such matters of faith, in
particular the widespread mental illness that is religion of any
kind, you have to also accept an ever-greater series of
specious explanations to unexplain or talk around the
anomalies that are introduced because the matter of faith
is either plain wrong or just misleading.

You are uncomfortable that your bluff has been called (not as a
personal attack; you elected to join the discussion) but cannot
face up to the error in your religion, and so you resort to
rather silly outbursts such as you do below. The issue of
an area being a height times a width is not in dispute, nor has it
ever been. Neither has there been any previous discussion of associativity.

Your remarks about the instantaneous multiplication of the
amplitude of one function by the area of another is an example
of a specious explanation that is simply not supported by any
aspect of the calculus.

Your remarks about explaining things to yourself by means of
Planck time are also an example of speciousness - Planck Time
is a matter of practical reality whereas a comb of Delta Functions is
not. Why not resolve the difficulties that you yourself obviously have in
this
area by an appeal to the real truth about practical sampling?

Intellectual honesty is more respectable than the make-believe world
of religion.

A further example of the emotional state brought on by a religious basis
rather
than a technical basis is your resorting to ever-more silly emotional
outbursts.

Post by Airy R. Bean
Sorry, but you're talking absolute bollocks that fits in with no
part of the calculus.

no, i guess you're right and the calculus does not submit itself to the
axiom called "the associative property of multiplication".
(instantaneous_value*pulse_height)*pulse_width does not really equal
instantaneous_value*(pulse_height*pulse_width) , does it, Mr. Bean? and it
is absolute bollocks to say that
pulse_height * pulse_width = pulse_area
i'm learning so much.

Chimera

2004-02-27 09:30:45 UTC

Post by Airy R. Bean
I think that the problem that you have, is that despite your
accepted position as an authority in the world of DSP, is that
you have accepted the explanation of sampling as a multiplication
by a comb of Delta Functions as a matter of blind faith and not as a
matter of technical rigour. Like all such matters of faith, in
particular the widespread mental illness that is religion of any
kind, you have to also accept an ever-greater series of
specious explanations to unexplain or talk around the
anomalies that are introduced because the matter of faith
is either plain wrong or just misleading.

So you haven't figured out why the unit impulse AREA is unity and not its
amplitude yet, I take it.

Nor, I assume, have you figured out why the width must tend to 0.

You really don't understand this do you?

Let me help you some more....... What does the area under a waveform in the
time domain represent?

Chimera

James Calivar

2004-02-27 17:11:21 UTC

So you haven't figured out why the unit impulse AREA is unity and not its
amplitude yet, I take it.
Nor, I assume, have you figured out why the width must tend to 0.

Ooh let me try - is it because that's by definition?

Post by Chimera
You really don't understand this do you?
Let me help you some more....... What does the area under a waveform in the
time domain represent?
Chimera

Going out on a limb here - is it total energy?

Airy R. Bean

2004-02-27 17:29:58 UTC

The silly person masquerading as, "Chimera" seems to
have become obsessed with a side issue.

She was kill-filed following her rather immature campaign
of name-calling. There's no place in public forum for infantile
behaviour such as she exhibits.

Post by James Calivar

So you haven't figured out why the unit impulse AREA is unity and not its
amplitude yet, I take it.
Nor, I assume, have you figured out why the width must tend to 0.

Ooh let me try - is it because that's by definition?

Post by Chimera
You really don't understand this do you?
Let me help you some more....... What does the area under a waveform in

the

Post by Chimera
time domain represent?

Going out on a limb here - is it total energy?

huLLy

2004-02-27 18:16:35 UTC

Post by Airy R. Bean
The silly person masquerading as, "Chimera" seems to
have become obsessed with a side issue.
She was kill-filed following her rather immature campaign
of name-calling. There's no place in public forum for infantile
behaviour such as she exhibits.

How did you manage to kill file yourself then? It hasn't worked, btw..

Chimera

2004-02-27 19:00:08 UTC

Now, if there was any doubt that you really don't have a clue then you have
confirmed it.

The unit pulse being an AREA is not a "side issue".

Chimera

g***@nospamblueyonder.co.uk

2004-02-27 23:46:05 UTC

The silly person masquerading as, "Airy R Bean" seems to
have become obsessed with a side issue.

She was kill-filed following her rather immature campaign
of name-calling. There's no place in public forum for infantile
behaviour such as she exhibits.

Mr T

2004-02-28 07:36:28 UTC

You should know best after all the crap you come out with.

Post by g***@nospamblueyonder.co.uk
The silly person masquerading as, "Airy R Bean" seems to
have become obsessed with a side issue.
She was kill-filed following her rather immature campaign
of name-calling. There's no place in public forum for infantile
behaviour such as she exhibits.

Airy R. Bean

2004-02-28 12:35:35 UTC

You behave like a 5 year old.....

Leigh

2004-02-28 18:49:00 UTC

Post by Airy R. Bean
You behave like a 5 year old.....

That's two years older than the characteristics you display.

Chimera

2004-02-27 19:02:18 UTC

Post by James Calivar

Post by Chimera
You really don't understand this do you?
Let me help you some more....... What does the area under a waveform in

the

Post by Chimera
time domain represent?
Chimera

Going out on a limb here - is it total energy?

Well done, normally referred to as power.

Would you like to explain that to Airy.

Chimera

Chimera

2004-02-27 12:01:33 UTC

Turner-Smith

Post by Frank Turner-Smith G3VKI

Post by Airy R. Bean
Did I go wrong somewhere above?

No, in your case the midwife went wrong and threw the wrong bit away.
...(_!_)...

that was just too easy, but gratifying nonetheless (like hitting that final
E major power chord with the Marshall cranked up to "arc-weld").
BTW, guys, from 1968 to about 1975, i used to be a ham, too. WB0CCA ("CCA"
was "Casselton's Commie Adolescent".)

At one time many of our students were radio hams but not so much these days.
Sometimes they were an odd lot. Enough to put you off the hobby.

Chimera

Airy R. Bean

2004-02-26 10:44:41 UTC

.....[PANTOMIME MODE ON].....
Oh no! They don't!
.....[PANTOMIME MODE OFF].....

They present the doubtful claim of sampling being
given by simple multiplication against a delayed Delta
Function, and then directly state that therefore
the transform is f(T) x e^(-sT)

Post by Airy R. Bean
Any text book that says that sampling at the instant T is
represented by f(t) x d(t-T) and then uses this to claim
to further claim that the transform of such sampling is f(T) x e^(-sT) after
taking the transform of d(t) as 1.

no, they say that
Laplace{ f(t) * d(t-T) } = f(T) * e^(-sT)

Airy R. Bean

2004-02-26 10:38:06 UTC

Either the accounts are legit, or they are not.

And if you're not bothered about the legitimacy of the
mathematics......

Post by robert bristow-johnson
they may
be just a wee bit illegit if they let f(t)*d(t-T) hang around without
getting integrated w.r.t. "t", but that little infraction doesn't bother me
much because...

Airy R. Bean

2004-02-26 10:41:33 UTC

1. Yes. My typo. No doubt the Twitterers Of Twaddle will
dine out on it for years to come.
As always, in their case, "Empty vessels make most noise".

2.Yes, because with the normalised current of 1 Amp being
1 Coulomb/sec, the volts can be regarded as Joules/sec

Post by Airy R. Bean
then the resultant amplitude of the
claimed action of sampling of f(t) x d(t-T) is 5 x 10^(-43)

uh, did you mean 10^(43) or 10^(-43)? i presume the former. but, to be
precise, the amplitude of the sampled impulse is 5*10^43 (1/sec).

RVMJ 99g

2004-02-26 15:32:00 UTC

Post by Airy R. Bean
2.Yes, because with the normalised current of 1 Amp being
1 Coulomb/sec, the volts can be regarded as Joules/sec

You need to quote something else to make 1 watt = 1 volt.

How many dB would that be?

--
from
RVMJ
(dot) 99g (at) BTinternet (dot) com

Frank Turner-Smith G3VKI

2004-02-26 18:12:10 UTC

Post by Airy R. Bean
1. Yes. My typo. No doubt the Twitterers Of Twaddle will
dine out on it for years to come.
As always, in their case, "Empty vessels make most noise".

Judging by the quantity of crap you have posted over the years I consider
that a true statement.
...(_!_)...

Airy R. Bean

2004-02-26 17:41:24 UTC

Evaluating Laplace{ f(t) * d(t) }

int (+inf/0)( f(t) x d(t) x e^(-st))

integrating by parts.... int(uv) = u x int(v) - int( du x int(v))....

let f(t) = u and d(t) x e^(-st) = v......

We get.....
f(t) x int(d(t) x e^(-st)) - int(f'(t) x int(d(t) x e^(-st)))

However, by definition, int(d(t) x e^(-st)) is 1, so
the above simplifies to

f(t) - int(f'(t))

= f(t) - f(t)

= 0
(and not even f(0) and not even needing to apply the limits)

Did I go wrong somewhere above?

Post by robert bristow-johnson
no, they say that
Laplace{ f(t) * d(t-T) } = f(T) * e^(-sT)

Frank Turner-Smith G3VKI

2004-02-26 18:10:32 UTC

Post by Airy R. Bean
Did I go wrong somewhere above?

No, in your case the midwife went wrong and threw the wrong bit away.
...(_!_)...

Airy R. Bean

2004-02-25 11:19:49 UTC

Planck Time = 10^(-43) seconds.

If unit area, then height is OOO 10^(43)

Let us suppose that you are sampling a
waveform which has an instantaneous value
of 5, then the resultant amplitude of the
claimed action of sampling of f(t) x d(t-T) is 5 x 10^(-43) and not
5 as your subsequent analysis might claim.

l***@eternal-flames.gov

2004-02-25 20:01:56 UTC

On Wed, 25 Feb 2004 11:19:49 -0000, as the pitiful wreck that had once

Post by Airy R. Bean
Planck Time = 10^(-43) seconds.

Not to be confused with 'Thick as a plank time' which refers to the
content of most of your "technical" postings.

Old Nick.

Chimera

2004-02-25 22:41:40 UTC

Post by Airy R. Bean
Planck Time = 10^(-43) seconds.
If unit area, then height is OOO 10^(43)
Let us suppose that you are sampling a
waveform which has an instantaneous value
of 5, then the resultant amplitude of the
claimed action of sampling of f(t) x d(t-T) is 5 x 10^(-43) and not
5 as your subsequent analysis might claim.

Having looked at responses to your utterings via Google, I know you have had
this correctly explained to you before.

However, you seem to ignore valid responses that point out things you
haven't worked out for yourself.

I suggest you ask yourself why the unit impulse must have unit area and is
considered to be infinitesimally narrow.

Grasp those key points and it all becomes clear.

Chimera

Airy R. Bean

2004-02-25 11:24:22 UTC

You "like to think"?????

That sounds like a Freudian rationalisation to
deceive yourself over something which you don't understand
either.

Post by robert bristow-johnson
(i like to think of it as a rectangular function of
unit area and about one Planck Time in width)

Airy R. Bean

2004-02-25 12:02:12 UTC

All of which seem to suggest that your model of sampling
is mathematically unsound, and that you know it to be so.

OK, the model that you use seems to produce the right
answers for you and pays for your daily bread, but as an
answer to a protest about mathematical rigour?

(i like to think .... one Planck Time in width)
.....the Dirac impulse shouldn't really be laying
around naked anywhere without being surrounded by an integral
.....virtually true.....
.....the area under the "function" is f(T).

Chimera

2004-02-25 22:46:44 UTC

Post by Airy R. Bean
All of which seem to suggest that your model of sampling
is mathematically unsound, and that you know it to be so.
OK, the model that you use seems to produce the right
answers for you and pays for your daily bread, but as an
answer to a protest about mathematical rigour?

There is nothing wrong with the model, just your understanding of it.

Figure out why must the area of the unit impulase by unity and its width
tend to 0 and you will understand much more.

Chimera

Airy R. Bean

2004-02-25 11:57:20 UTC

In the mathematical analysis that intending engineers
will have come across up to the point where they
learn about sampling, where has the concept
of multiplying the instantaneous value of one
function against the _AREA_ of another come in?

If you, or anyone, wishes to introduce an "area" into
the field of integral transforms, then she or he would
be well-advised to support that area with a corresponding
integral to evaluate it.

...... and the area under the "function" is f(T).

Airy R. Bean

2004-02-25 12:37:00 UTC

What's the legit version?

.....we will sometimes say
f(t)*d(t-T) = f(T)*d(t-T) .
it's a little bit illegit.......

I. R. Khan

2004-02-26 01:06:29 UTC

Robert, this is a nice post. Why don't other also challenge/reply his
claims/questions technically, instead of making personal attacks?

Regards,
Ishtiaq.

Post by robert bristow-johnson
which textbooks say that
+inf
integral{ f(t)*d(t-T) dt} = f(T)*d(t-T) ???
-inf
saying that is not correct but, until you come up with a list of texts that
say that, i think that the complaint is a straw man.
now engineers *do* tend to be a sloppy bunch and we treat the Dirac impulse
like a normal "function" (i like to think of it as a rectangular function of
unit area and about one Planck Time in width) and we will sometimes say
f(t)*d(t-T) = f(T)*d(t-T) .
it's a little bit illegit since the Dirac impulse shouldn't really be laying
around naked anywhere without being surrounded by an integral. at least
that's what the math guys tell us. but i don't even have a problem with
that infraction of the rules because it is virtually true for a practical
impulse of non-zero but arbitrarily small width. both sides represent a
"function" (if the math guys will allow us to call it that, *they* call it a
"distribution") that is zero everywhere except at T and the area under the
"function" is f(T).

as if that would solve anything.

Post by Airy R. Bean
My recent interest has been to attempt to lay the ghost as to
when and where the apparently erroneous explanations appeared.

again, what legit engineering text says
+inf
integral{ f(t)*d(t-T) dt} = f(T)*d(t-T) ?
-inf
i haven't seen it.
besides, how does the concept of a "Big K" come out of that?
r b-j

MattD..

2004-02-26 09:27:04 UTC

When the packet hit the pocket on the socket on the port, My server said to
I. R. Khan, "What message have you brought?" The client it responded in the
only way it knew, Sort out these ones and zeros, 'cause I haven't got a

Post by I. R. Khan
Robert, this is a nice post. Why don't other also challenge/reply his
claims/questions technically, instead of making personal attacks?

They have, many times. Watch closely and all will become clear...

--
MattD..

Steve Underwood

2004-02-26 10:42:23 UTC

Post by I. R. Khan
Robert, this is a nice post. Why don't other also challenge/reply his
claims/questions technically, instead of making personal attacks?

No, it isn't a nice post. Its termed "feeding the troll". It is not
considered good practice on the Internet. I'm guilty of it too at
times. However, making a meaningful technical response to people like
Mr Bean is actually the least meaningful thing you can do. Sometimes
each of us forgets that. Robert is a smart guy, though. He'll come to
his senses soon and shut up. :-)

Remember: Don't argue with an idiot. He'll drag you down to his level,
then beat you with experience.

Regards,
Steve

Graham W

2004-02-26 11:47:32 UTC

Post by I. R. Khan
Robert, this is a nice post. Why don't other also challenge/reply his
claims/questions technically, instead of making personal attacks?

No, it isn't a nice post. Its termed "feeding the troll". It is not
considered good practice on the Internet. I'm guilty of it too at
times. However, making a meaningful technical response to people like
Mr Bean is actually the least meaningful thing you can do. Sometimes
each of us forgets that. Robert is a smart guy, though. He'll come to
his senses soon and shut up. :-)
Remember: Don't argue with an idiot. He'll drag you down to his level,
then beat you with experience.

Airy doesn't qualify as an idiot but he seems to be trying to work up to
that level.

73s de

GW

jim <"N0sp"@

2004-02-26 16:29:11 UTC

Post by I. R. Khan
Robert, this is a nice post. Why don't other also challenge/reply his
claims/questions technically, instead of making personal attacks?

Well, I for one, found Robert's post to be quite worthwhile reading and your
postings to this thread to be useless stupidity. I really don't need you or the
other sniper's in this thread to tell me how I should be interpreting someone
elses post. If you believe shutting up is the thing to do, then do it. Endless
jabbering about this being the appropiate time to shut-up is taking stupidity to
the extreme.

-jim

-----= Posted via Newsfeeds.Com, Uncensored Usenet News =-----
http://www.newsfeeds.com - The #1 Newsgroup Service in the World!
-----== Over 100,000 Newsgroups - 19 Different Servers! =-----

robert bristow-johnson

2004-02-27 06:39:19 UTC

Post by I. R. Khan
Robert, this is a nice post. Why don't other also challenge/reply his
claims/questions technically, instead of making personal attacks?

actually, thanks guys (I.R., Steve, and jim). i think Steve is right and
i'm leaving the Beanie alone. i'm usually not one to back away from a fight
on comp.dsp when i think i'm likely to be correct (and this Dirac thing has
been a topic of heated conversation in the past - Google "Jay Rabkin" and
myself to see). but this time it's different. (did you see his attempt to
do an integral with a Dirac impulse in it? "integrating by parts...."
whooo, boy! it's embarrassing.)

anyway, it seems that Beanie has a history with the U.K. ham radio guys, and
on occasion comp.dsp attracts someone crossing over from another group where
that person's rep suggests to us comp.dspers that maybe we should learn from
their experience.

anyway, i thought i'd give him a chance. ya know, first give the benefit of
doubt until you learn differently. (now i know differently.)

Post by Steve Underwood
Sorry, but you're talking absolute bollocks that fits in
with no part of the calculus.

Typical - "I'm no longer winning this argument" response from
Airy Arsehole!

it looks like Leigh got it pretty much correct (except i didn't see Beanie
*ever* winning the argument, but maybe he was in his own mind).

r b-j

Airy R. Bean

2004-02-27 09:06:42 UTC

A further illustration in my critique of you as a
religious man and not as a rigorous techie is given
by your rather silly outburst below.

If the matters which I raised are erroneous, then they
will be dismissed by simple answers, and not by religious
nonsense (instantaneous multiplication of the amplitude of
one function by the area of another!!!!!) nor by silly emotional
outbursts such as you adopt below.

You do yourself no favours by such outbursts.

If you could not follow my integration-by-parts and can only
reply by sneering, that says so much about you.

Post by robert bristow-johnson
actually, thanks guys (I.R., Steve, and jim). i think Steve is right and
i'm leaving the Beanie alone. i'm usually not one to back away from a fight
on comp.dsp when i think i'm likely to be correct (and this Dirac thing has
been a topic of heated conversation in the past - Google "Jay Rabkin" and
myself to see). but this time it's different. (did you see his attempt to
do an integral with a Dirac impulse in it? "integrating by parts...."
whooo, boy! it's embarrassing.)
anyway, it seems that Beanie has a history with the U.K. ham radio guys, and
on occasion comp.dsp attracts someone crossing over from another group where
that person's rep suggests to us comp.dspers that maybe we should learn from
their experience.
anyway, i thought i'd give him a chance. ya know, first give the benefit of
doubt until you learn differently. (now i know differently.)
it looks like Leigh got it pretty much correct (except i didn't see Beanie
*ever* winning the argument, but maybe he was in his own mind).

Chimera

2004-02-27 09:32:32 UTC

Post by Airy R. Bean
A further illustration in my critique of you as a
religious man and not as a rigorous techie is given
by your rather silly outburst below.
If the matters which I raised are erroneous, then they
will be dismissed by simple answers, and not by religious
nonsense (instantaneous multiplication of the amplitude of
one function by the area of another!!!!!) nor by silly emotional
outbursts such as you adopt below.

It isn't amplitude that is multiplied it is AREA. The unit impulse has
UNIT AREA.

Do a search on the web if you don't believe me.

Think POWER

Chimera

g***@nospamblueyonder.co.uk

2004-02-27 23:46:04 UTC

On Fri, 27 Feb 2004 09:06:42 -0000, "Airy R. Bean"

Post by Airy R. Bean
A further illustration in my critique of you as a
religious man and not as a rigorous techie is given
by your rather silly outburst below.

You seem to have a real problem making friends of any sort!

Frank Turner-Smith G3VKI

2004-02-28 00:06:10 UTC

Post by g***@nospamblueyonder.co.uk
On Fri, 27 Feb 2004 09:06:42 -0000, "Airy R. Bean"

Post by Airy R. Bean
A further illustration in my critique of you as a
religious man and not as a rigorous techie is given
by your rather silly outburst below.

You seem to have a real problem making friends of any sort!

and why does that NOT surprise me at all?

--
;>)
73 de Frank Turner-Smith G3VKI - mine's a pint.
http://turner-smith.co.uk

Mr T

2004-02-28 07:36:29 UTC

So do you.

Post by g***@nospamblueyonder.co.uk
On Fri, 27 Feb 2004 09:06:42 -0000, "Airy R. Bean"

Post by Airy R. Bean
A further illustration in my critique of you as a
religious man and not as a rigorous techie is given
by your rather silly outburst below.

You seem to have a real problem making friends of any sort!

Airy R. Bean

2004-02-28 12:34:08 UTC

If they behave like 5 year olds, as you do, who'd want 'em?

Post by g***@nospamblueyonder.co.uk
You seem to have a real problem making friends of any sort!

Steve Underwood

2004-02-27 17:36:16 UTC

Post by I. R. Khan
Robert, this is a nice post. Why don't other also challenge/reply his
claims/questions technically, instead of making personal attacks?

As usual, Robert's post was thoughtful and written with style. The
snag is, it was in the wrong place. Would you try to extinguish a fire
with gasolene, or starve it of its fuel?

Regards,
Steve

Chimera

2004-02-27 19:06:20 UTC

Post by Steve Underwood
As usual, Robert's post was thoughtful and written with style. The
snag is, it was in the wrong place. Would you try to extinguish a fire
with gasolene, or starve it of its fuel?

Yep, Robert knows his stuff. I can't agree that we should leave "Airy" to
post his nonsense. REAL students use the newsgroups and may be misled by his
nonsense.

I love the area of the unit pulse being a side issue.

Chimera

Steve Underwood

2004-02-28 05:08:09 UTC

Yep, Robert knows his stuff. I can't agree that we should leave "Airy" to
post his nonsense. REAL students use the newsgroups and may be misled by his
nonsense.
I love the area of the unit pulse being a side issue.
Chimera

You have a good point. However, news groups have largely become an
abandoned wasteland. Only a few remain fertile - generally the more
technical ones. comp.dsp is largely clear of rubbish. The last time I
tried comp.fpga it was too. A key reason most groups have decayed is
the noise level, and a large part of that comes from feeding the
trolls. If you don't feed them, they tend to wither. I think nobody
here wants censored newsgroups, so a perfect solution isn't viable. On
balance simply ignoring the trolls seems to work best.

You seem to think Mr Bean is beneatth the idiot level. A few things he
says make we wonder. Its possible he is an intelligent guy, with a
really warped sense of fun. Who knows?

Regards,
Steve

Frank Turner-Smith G3VKI

2004-02-28 08:54:54 UTC

Post by Steve Underwood
You seem to think Mr Bean is beneatth the idiot level. A few things he
says make we wonder. Its possible he is an intelligent guy, with a
really warped sense of fun. Who knows?
Regards,
Steve

The words 'Airy' and 'intelligent' are mutually exclusive.

--
;>)
73 de Frank Turner-Smith G3VKI - mine's a pint.
http://turner-smith.co.uk

Airy R. Bean

2004-02-26 10:47:13 UTC

I'm afraid that there is a lot of history with those Twitterers Of Twaddle.

They are best ignored for being the escapees from the school playground
that they undoubtedly are.

Post by I. R. Khan
Robert, this is a nice post. Why don't other also challenge/reply his
claims/questions technically, instead of making personal attacks?

l***@eternal-flames.gov

2004-02-25 07:46:22 UTC

On Tue, 24 Feb 2004 21:41:25 -0000, as the pitiful wreck that had once

I think you should master some of the simpler things in life...
English, Latin, potty-training etc., before you worry about what
electrical engineers have, or have not, been taught.

Old Nick.

Randy Yates

2004-02-25 12:13:45 UTC

Post by Airy R. Bean
It was one of my posits for attempting to square up the
claims made in the literature for what is the mathematical
basis of sampling against the description that electrical
engineers will have been taught up to that point about the properties
of Dirac's Delta function.
For example, int(+/- inf) ( F(t) x d(t-T) ) yields f(T), which
is a constant function.
It does NOT yield f(T) x d(t-T) which is claimed as the basis
for sampling in many texts, (but not O & S)

I have seen this model of sampling presented in several courses
and in no case was it claimed that \int_{-infty}^{+\infty} f(t) d(t-T) dt
is equal to f(T)*d(t-T). Rather, it is always evaluated as f(T). There
is even a common name for this property of the Dirac delta function:
the "sifting" property.

I'd toss any such text that teaches this.

--RY

Post by Airy R. Bean
(I am awaiting a copy of Dirac's 1930 work on Quantum Mechanics
in order to read up on the original development of the Delta Function,
but it seems to be a classic; the only copies located by my book-finder
have been priced at several hundred pounds)
My recent interest has been to attempt to lay the ghost as to
when and where the apparently erroneous explanations appeared.
As to those who repeatedly make rather silly childish sneers
about any attempts to improve one's knowledge and understanding,
I say this...."Empty vessels make most noise".

Post by Ryan H.
What is the "Big K"?

--
% Randy Yates % "Watching all the days go by...
%% Fuquay-Varina, NC % Who are you and who am I?"
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Airy R. Bean

2004-02-25 12:26:59 UTC

f(T) is a constant function and contains no impulse,
and therefore its transformation is f(T)/s, giving
a delayed transformation of (f(T) x e^(-sT))/s
and not just (f(T) x e^(-sT)) as claimed by those
who suggest that sampling is represented by
f(t) x d(t-T).

Post by Randy Yates
I have seen this model of sampling presented in several courses
and in no case was it claimed that \int_{-infty}^{+\infty} f(t) d(t-T) dt
is equal to f(T)*d(t-T). Rather, it is always evaluated as f(T). There
the "sifting" property.

robert bristow-johnson

2004-02-24 01:27:57 UTC