Stan Pawlukiewicz
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 Posted:
Wed Dec 21, 2005 12:51 am Post subject:
Re: A FAQ for Xmas |
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robert bristow-johnson wrote:
| Quote: | Beanie regurjitates:
In our sampling, we take an infinitesimally small width but
only an amplitude which is of the order of the analogue signal,
and therefore the volt/time product of this is zero and it is zero
integrable using the calculus. The volt/time product is an
infinite number of orders of magnitude less than unity, so would you
justify claiming that it is a scaled Unit Impulse?
Howard Long wrote:
Folks
Maybe I'm being way to simplistic here, but isn't this just a basic calculus
issue? Although t tends to zero, v tends to infinity. With only these two
terms you are left without a full definition: you need to state that v*t
tends to 1 as t tends to zero. This limit is part of the definition of the
Dirac Delta function. They teach the introductory calculus concept of limits
at school at about the age of 16.
Although it is an abstract concept in that we can never physically attain
infinite amplitude with zero width, as we approach closer and closer to that
experimentally, we can see that the unit impulse model fits closer and
closer.
there is not just the physical impossibility of a real infinite height
impulse, but the different mathematical understanding of the dirac
delta function that engineers have vs. what the pure mathematicians
say. there is a theorem in real analysis (where we learn about
Lebesgue integration) that says that for a function that is zero
"almost everywhere" (that is everywhere but a countable number of
points) that the integral of such a function is zero. but we engineers
look at the dirac delta as a function that is zero everywhere but at
t=0, yet its integral (over all t) is one. so something is not
connecting there. the math guyz like to call the dirac delta a
"distribution" (not a "function") and really just define it as a
notation that when you put it alongside ("multiply") a legit function
and integral, that construction is "functional" that maps that function
f(t) to a number which is f(0) (or f(a) for delta(t-a)). for the math
guys, the dirac delta has little meaning outside of an integral.
Beanie is just trolling, but he wraps himself in this issue to draw
attention. i was hoping that ignoring the troll would make him go
away, but it's not happening. i'm still not removing him from my
killfile, though.
i don't particularly like the pure mathematicians view of the dirac
delta, but i do not dispute their correctness about it (when we define
a function, it is a mapping of a number to another number and that
mapping doesn't really get to "remember" the limit of how it got
defined, and without the nascent dirac impulse - those in the limit -
there is no way to represent that finite area in an infinitely thin
pulse). they way i deal with in conceptually is, for the purpose of
*any* real or practical or engineering context, i just define the dirac
impulse as one of those nascent approximations to it (like one
femto-second wide and the reciprocal of that high). it will make
absolutely no numerical difference for any practical situation and then
i am allowed to have a bunch of dirac impulses outside of the integral
that i can manipulate as if they were some "normal" functions.
r b-j
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One of my books, Modern Electrical Communication Theory, by Stark and
Tutuer, goes though a development of sampling theory with (somewhat)
arbitrary sampling functions. If one accepts that the impulse has a
Fourier transform of 1 everywhere, it is really easy to see as a special
case that the delta provides the lowest bound on sample rate for perfect
reconstruction of a perfectly band limited waveform. The point being
made is that the bound known as the Nyquist rate is what all our text
books are trying to impart. |
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