On Aug 18, 2:28 pm, Eric Jacobsen <eric.jacob...@ieee.org> wrote:
> On Sat, 18 Aug 2007 01:21:48 -0700, robert bristow-johnson
>
> <r...@audioimagination.com> wrote:
> > if any strict mathematician
> >picks on you, tell them that d(t) is not strictly a dirac delta
> >function (which ain't a real function, as far as they permit) but a
> >simple rectangular function with 1 Planck Time in width and a
> >(dimensionless) unit area (so the height is the reciprocal of the
> >Planck Time).  that's a true function, as far as the mathematicans are
> >concerned, and it's close enough to the Neanderthal engineer's use of
> >the dirac, that there is no measurable difference in effect.
>
>    I guess during most of these discussions
> I'd always just kind of built that sort of thing into the assumptions
> so that it all works out, and consequently never really understood
> what all the hubbub was about.

you see, strictly speaking, a function f(x) is just that: a mapping of
x to f(x).  it is not allowed to worry about meta-data like *how* f(x)
came to be (regarding the limit of nascent delta functions).   all we
are allowed to think about f(x) = delta(x) is that for every x != 0,
then what comes outa the mapping is f(x)=0.  then it doesn't matter
what definable number f(0) is, the integral is zero.  if f(0) is not
defined, then f(x) is not integrable for any region containing x=0.
that's the problem.

>   Thanks to this thread I'm finally
> seeing the distinction that has bundled a few folks panties over the
> years.   My main reaction is that there are bigger things to get
> twisted over than this...yikes.

it's a problem when they say if f(x)=delta(x) and g(x)=0, we cannot
say that the integral of f(x) is 1 while the integral of g(x) is
zero.  i want to be able to say it is so, even if my Real Analysis
prof would give me an F.

> On a somewhat related note (well, the 1 Planck Time bit reminds me of
> it)...somebody pointed out that the precision needed in Pi to be able
> to discern a single Angstrom at the radius of the known limit of the
> universe is only about twenty decimal places or so (I don't remember
> the exact number, but it was surprisingly low from what I thought it
> would be).

geez,

10^10 Ang/m * 299792458 m/s * 31557600 s/yr * 13.7x10^9 yr

what does that come out to be?  i think lot more than 20 digits.  more
like 40.

>   So using some small finite pulse width like you've
> suggested may actually be quite practical in several ways.

it's just to get the mathematicians offa my back.

i remember taking "Real Analysis" and learning about the Lebesgue
Integral for the first time.  i knew right away i had a conceptual
problem when we we told that if f(x) = g(x) for all but a countable
number of discrete x points (that's what "almost everywhere" means)
that the integrals of f() and g() must be the same.  it's been an icky
bugger ever since.  so i decided to adopt the bug as a house pet.

r b-j

On Aug 18, 1:21 am, robert bristow-johnson <r...@audioimagination.com>
wrote:
> On Aug 15, 5:55 am, "NewLine"
>
>
>
> <qw897jhk_remove_this_and_t...@skynet.be> wrote:
> > > You're on the right track. When you model the discrete signal x(nT) as
> > > a series of scaled diracs, then plug that into the CTFT, the integral
> > > from the CTFT turns them into a sum of numbers ala DTFT. Remember the
> > > sifting property of the Dirac? \int_{-infty}^{+\infty} x(t)\delta(t - tau)
> > > dt = x(tau).
> > > --
>
> > first of all, thanks to all the people that have responded, all your inputs
> > were  very useful to me...also rbj I liked a lot the link to your
> > mathematical description of sampling & reconstruction.
>
> > Below I tried to do continue with what Randy introduced above. Are there any
> > flaws in this?
>
> > ----------------
>
> > my discrete signal, only k= 0 : N-1 values are non-zero (spaced T)
>
> >      s[k]
>
> > discrete signal represented as series of scaled dirac's as if it were a
> > 'continuous signal'
>
> >      s(t) = SUM_{k=0}^_{k=N-1} s[k] * d(t-k*T)
>
> >      where d(t) is the dirac delta
>
> > FT of s(t):
>
> >      S(f) = INT_{-inf}_{+inf} SUM_{k=0}^_{k=N-1} s[k] * d(t-k*T)
> > exp(-i*2*pi*f*t) dt
>
> >      S(f) = SUM_{k=0}^_{k=N-1} s[k] INT_{-inf}_{+inf} d(t-k*T)
> > exp(-i*2*pi*f*t) dt
>
> >      S(f) = SUM_{k=0}^_{k=N-1} s[k] exp(-i*2*pi*f*k*T)
>
> >      S(f) = SUM_{k=-inf}^_{k=+inf} s[k] exp(-i*2*pi*f*k*T)
>
> > replacing 2*pi*f*T with the normalized radial frequency w (radians per
> > sample) this becomes:
>
> >      SUM_{k=-inf}^_{k=+inf} s[k] exp(-i*k*w)
>
> > which is if I am not mistaken the DTFT.
>
> other than the naked delta functions which are objected to by most
> strict mathematicians (who insist on clothing the naked natives with
> integrals), it's fine.  being a Neanderthal engineer, i generally
> approve of naked dirac delta functions.  if any strict mathematician
> picks on you, tell them that d(t) is not strictly a dirac delta
> function (which ain't a real function, as far as they permit) but a
> simple rectangular function with 1 Planck Time in width and a
> (dimensionless) unit area (so the height is the reciprocal of the
> Planck Time).  that's a true function, as far as the mathematicans are
> concerned, and it's close enough to the Neanderthal engineer's use of
> the dirac, that there is no measurable difference in effect.

Given the uncertainty of measuring such small units of
time mapped to the actual universe, your rectangular
Planck-width function may actually become a probability
distribution of some sort, thus still needing integration
to show equivalence with the use of the ideal Platonic
rectangle.


IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

On Sat, 18 Aug 2007 01:21:48 -0700, robert bristow-johnson
<rbj@audioimagination.com> wrote:


> if any strict mathematician
>picks on you, tell them that d(t) is not strictly a dirac delta
>function (which ain't a real function, as far as they permit) but a
>simple rectangular function with 1 Planck Time in width and a
>(dimensionless) unit area (so the height is the reciprocal of the
>Planck Time).  that's a true function, as far as the mathematicans are
>concerned, and it's close enough to the Neanderthal engineer's use of
>the dirac, that there is no measurable difference in effect.

...

>r b-j

Hey, that's a nice touch.    I guess during most of these discussions
I'd always just kind of built that sort of thing into the assumptions
so that it all works out, and consequently never really understood
what all the hubbub was about.   Thanks to this thread I'm finally
seeing the distinction that has bundled a few folks panties over the
years.   My main reaction is that there are bigger things to get
twisted over than this...yikes.

On a somewhat related note (well, the 1 Planck Time bit reminds me of
it)...somebody pointed out that the precision needed in Pi to be able
to discern a single Angstrom at the radius of the known limit of the
universe is only about twenty decimal places or so (I don't remember
the exact number, but it was surprisingly low from what I thought it
would be).   So using some small finite pulse width like you've
suggested may actually be quite practical in several ways.

Eric Jacobsen
Minister of Algorithms
Abineau Communications
http://www.ericjacobsen.org

On Aug 15, 5:55 am, "NewLine"
<qw897jhk_remove_this_and_t...@skynet.be> wrote:
> > You're on the right track. When you model the discrete signal x(nT) as
> > a series of scaled diracs, then plug that into the CTFT, the integral
> > from the CTFT turns them into a sum of numbers ala DTFT. Remember the
> > sifting property of the Dirac? \int_{-infty}^{+\infty} x(t)\delta(t - tau)
> > dt = x(tau).
> > --
>
> first of all, thanks to all the people that have responded, all your inputs
> were  very useful to me...also rbj I liked a lot the link to your
> mathematical description of sampling & reconstruction.
>
> Below I tried to do continue with what Randy introduced above. Are there any
> flaws in this?
>
> ----------------
>
> my discrete signal, only k= 0 : N-1 values are non-zero (spaced T)
>
>      s[k]
>
> discrete signal represented as series of scaled dirac's as if it were a
> 'continuous signal'
>
>      s(t) = SUM_{k=0}^_{k=N-1} s[k] * d(t-k*T)
>
>      where d(t) is the dirac delta
>
> FT of s(t):
>
>      S(f) = INT_{-inf}_{+inf} SUM_{k=0}^_{k=N-1} s[k] * d(t-k*T)
> exp(-i*2*pi*f*t) dt
>
>      S(f) = SUM_{k=0}^_{k=N-1} s[k] INT_{-inf}_{+inf} d(t-k*T)
> exp(-i*2*pi*f*t) dt
>
>      S(f) = SUM_{k=0}^_{k=N-1} s[k] exp(-i*2*pi*f*k*T)
>
>      S(f) = SUM_{k=-inf}^_{k=+inf} s[k] exp(-i*2*pi*f*k*T)
>
> replacing 2*pi*f*T with the normalized radial frequency w (radians per
> sample) this becomes:
>
>      SUM_{k=-inf}^_{k=+inf} s[k] exp(-i*k*w)
>
> which is if I am not mistaken the DTFT.

other than the naked delta functions which are objected to by most
strict mathematicians (who insist on clothing the naked natives with
integrals), it's fine.  being a Neanderthal engineer, i generally
approve of naked dirac delta functions.  if any strict mathematician
picks on you, tell them that d(t) is not strictly a dirac delta
function (which ain't a real function, as far as they permit) but a
simple rectangular function with 1 Planck Time in width and a
(dimensionless) unit area (so the height is the reciprocal of the
Planck Time).  that's a true function, as far as the mathematicans are
concerned, and it's close enough to the Neanderthal engineer's use of
the dirac, that there is no measurable difference in effect.

note, if you replace FT with LT, then, instead of a DTFT, you get the
Z transform.  they're are related.  all descendents of Fourier.

r b-j

Rune Allnor <allnor@tele.ntnu.no> writes:

> On 15 Aug, 11:55, "NewLine" <qw897jhk_remove_this_and_t...@skynet.be>
> wrote:
> ...
>                N     1
> X(w) = lim    sum  -----  exp(jw'n)exp(-jwn)   [2.a]
>       N->inf  n=-N 2N-1
>
>               N     1
>      = lim   sum  -----  1                    [2.b]
>       N->inf n=-N  2N-1
>
>        /          2N-1
>        |   lim   ------ = 1      w = w'
>      = |   N->inf 2N-1
>        |
>        \                  0      w =/= w'
>
> So X(w) == 0 "almost everywhere ," i.e. everywhere
> except for one point, at w = w', where it equals 1.
> It's a half-weird function if you are only used to
> think in terms of continuous function, but for
> somebody with a college maths course under the
> belt, it is nothing extraordinary about X(w).
>
> ....
> This was the problem Lebesgue solved. He developed
> a way of integrating functions which are 0 "almost
> everywhere" so that
>
>   inf
> integral_Lebesgue X(w) = 1.
>  -inf

No. This integral exists in the sense of Lebesgue and is _zero_.

>
> with X(w) above.
>
> The one problem with Lebesque's integral is that
> it is far from easy to use (well, that's what
> my maths teachers told me; I have no personal
> experience with it) 

The theory of Lebesgue integration is not so simple as Riemann's, but
the essential thing from the physicist's/engineer's point of view is:

1) For any function f:R --> R and any intervall (a,b), where any of
the boundarys may be infinity, if the Riemann integral exists, then
the Lebesgue integral exists too, and it's value is equal to the one
of Riemann's.

2) If f is integrable over (a,b) in the sense of Lebesgue, and, if
g is another function, that is equal to f on (a,b), except for a
subset of measure zero, then g is also integrable on (a,b) and the
value of the integral is the same as from f.

Example (the standard): 

f(x) = 0 for all real numbers x.
g(x) = 1 for all rational numbers and g(x) = 0 else.

The integral over f always exists and is zero.  The rational numbers
are of measure zero (as is any countable set.)  The Riemann integral
over g does not exist over any non-trivial intervall.  But for any
intervall the integral over g in the sense of Lebesgue exists and is
zero.

>                     so Dirac set about to come
> up with a way to integrate functions which are
> 0 "almost everywhere" without having to resort
> to the Lebesgue integral.

Oh no.  Distributions are not about integration.  They are
mathematical objects in their own right, a little resembling
``ordinary'' functions, and often much more usefull for modelling in
theoretical physics or engineering than ordinary functions are.

>
> The result is Dirac's delta function. The function
>
>    inf
> X = integral d(t - t0) dt
>    -inf
>
> is basically nothing more than a tag, a note to
> the analyst, to say that "add 1 to the integral
> X when you  pass t = t0."

Well--- yes, something along these lines was suggested by Dirac.  But
how about differentiation? etc. etc.  As long as you don't have a
precise mathematical theory, what you can do with Dirac deltas is
either trivial, or you are not sure wether some operation is valid.

The mathematical understanding of Dirac's ``delta functions'' was
chiefly developed by Laurent Schwartz, who published his textbook
``Th\'eorie des distributions'' not until 1950; about twenty years
after Dirac's suggestion.  

> That's all there is to it. Once you get a clear
> picture of why the Dirac delta is useful, you
> also see why it is plain wrong to use it
> outside the integral sign.

This may be correct, when you restrict your understanding of the Dirac
delta to that of a ``tag to the analyst''.  But mathematicans have
developed a much wider, more thorrow understanding, and there any
distribution makes sense outside of an integral, as well as an
ordinary function does.

-- 
hw

>
> What exactly are you starting out with? A set of samples, i.e., an
> ordered set of numbers?
>
> If so, then you don't go from the those directly to a CTFT. The input
> to a CTFT is a continuous-time signal, and that's not what you
> have. So you have to first define how you're going to convert your
> samples into a continuous-time signal before you can take the CTFT of
> it.
>
> What is the point of your academic query? I mean, what the heck are you 
> trying
> to figure out?
> -- 

I must admit that it probably was a very strange question.

What I was trying to figure out is:

Imagine a generator that can create continuous sine waves from -inf to +inf.
And this signal goes into a block that will convolve the sine wave with 
another signal that is zero everywhere except at some discrete points, 
spaced T apart.

So you could see it like sending a continuous signal in a filter that looks 
a lot like a digital filter, but that is actually able to process the 
continuous signal.

I was wondering if the frequency response of such a block/filter would be 
periodic with 1/T. I am convinced it is because the filter is discrete, but 
want some check.

Writing this down and thinking it over again, probably to make the model 
more in line with theory I have to place a sampler on the continuous signal, 
and then I have a complete discrete system (samples and filter)...which 
obviously is periodic with 1/T in the freq domain.

robert bristow-johnson wrote:

(snip on DFT and DTFT)

>>The other that always confuses me is, why isn't that a Fourier series?

> why do you say that it isn't?  the discrete samples in the time domain
> are the fourier coefficients of the periodic spectrum in the freqency
> domain.  if you normalize out the 2*pi factor (by expressing frequency
> in normalized cycles/sample rather than radians/sample), the period of
> that periodic spectrum is 1, and the spacing between your discrete
> coefs (which are the samples) is 1.

My feeling is that it should be either a Fourier series, or a
Fourier transform, and not both.

One thing should not have two incompatible names.  That is all.

-- glen

"NewLine" <qw897jhk_remove_this_and_this@skynet.be> writes:

>> You've prematurely represented the signal as a sequence s[k]. Keep the
>> original signal intact. By the way, I don't see the original signal
>> here, so let's call it s_a(t) (for "analog"), and let s(t) be the
>> "discretized" version, i.e.,
>>
>
> That's because my original signal (as in the original post) is a discrete 
> signal.
> I guess your extension adds sampling a real nice analog to it... 

Hmm. I went back and re-read your original post and I might have
misread you all along.

What exactly are you starting out with? A set of samples, i.e., an
ordered set of numbers?

If so, then you don't go from the those directly to a CTFT. The input
to a CTFT is a continuous-time signal, and that's not what you
have. So you have to first define how you're going to convert your
samples into a continuous-time signal before you can take the CTFT of
it.

What is the point of your academic query? I mean, what the heck are you trying
to figure out?
-- 
%  Randy Yates                  % "She has an IQ of 1001, she has a jumpsuit
%% Fuquay-Varina, NC            %            on, and she's also a telephone."
%%% 919-577-9882                % 
%%%% <yates@ieee.org>           %        'Yours Truly, 2095', *Time*, ELO   
http://home.earthlink.net/~yatescr

> You've prematurely represented the signal as a sequence s[k]. Keep the
> original signal intact. By the way, I don't see the original signal
> here, so let's call it s_a(t) (for "analog"), and let s(t) be the
> "discretized" version, i.e.,
>

That's because my original signal (as in the original post) is a discrete 
signal.
I guess your extension adds sampling a real nice analog to it... 

On Aug 15, 4:33 pm, glen herrmannsfeldt <g...@ugcs.caltech.edu> wrote:
>
> When I learned about delta functions, I was told that mathematicians
> didn't like them, and often didn't believe in them.

what they don't believe is that the dirac delta "function" is a true
function as the mathematicians understand a function of a real
variable.  from the POV ofLebesgue integration (which is more general
than the Riemann integration we learned in calculus), if two functions
agree "almost everywhere" (which is "everywhere but for a countable
number of discrete points"), then the integrals of the two functions
over the same region (or limits of integration) must agree.  well, we
electrical engineers like to think of the dirac impulse as have zero
value everywhere but at t=0, and its integral (from -eps to +eps) is
1. but the dirac impulse "function" agrees with the zero everywhere
function at every point except t=0, and we know that the integral of 0
is 0.  so this understanding of the dirac impulse violates that
property of Lebesgue integration of two virtually identical functions.

>  There was a story
> about some mathematician who rewrote a quantum mechanics book to remove
> delta functions from all the explanations.

yeah, some math guys really don't like it, and like our Neanderthal
engineering treatment of the dirac impulse even less.  i'm completely
fine with the Neanderthal engineering usage of the dirac delta.


r b-j