On 3/27/14 6:53 AM, Sampo Syreeni wrote:
On 2014-03-27, Stefan Sullivan wrote:
Actually, yes there IS a requirement that it be periodic.
No, there is not. And the Shannon-Nyquist theorem isn't typically
proven under under any such assumption.
the *sampling* function is periodic (that's why we call it "uniform
sampling"), but the function being sampled, x(t), is just any reasonably
"well-behaved" function of t.
Furthermore, it generalizes to settings where periodicity isn't even
an option.
oh, it outa be an *option*. we know how to take the Fourier transform
of a sinusoid. (it's not square integrable, but we hand-wave our way
through it anyway.)
Granted, it *is* possible to prove the sampling theorem starting with
a square integrable, continuous, periodic function. That would in fact
be the most classical treatment of the subject, starting with Fourier
himself, and the way he utilized his series in the context of the heat
equation. But if you try to treat any general, real signals that way,
you'll have to pass the period to the limit in infinity, and that then
makes some of the attendant calculus unwieldy.
Nowadays pretty much nobody bothers with those details, except perhaps
to show the connection between the periodic in time, original series,
and the continuous time, far easier and more generalizable transform.
back, before i was banned from editing wikipedia (now i just edit it
anonymously), i spelled out the mathematical basis for the sampling and
reconstruction theorem in this version of the page:
https://en.wikipedia.org/w/index.php?title=Nyquist%E2%80%93Shannon_sampling_theorem&oldid=70176128
since then two particular editors (Dicklyon and BobK) have really made
the mathematical part less informative and useful. they just refer you
to the Poisson summation formula as the mathematical basis.
the only lacking in this proof is the same lacking that most electrical
engineering texts make with the Dirac delta function (or the Dirac comb,
which is the sampling function). to be strictly legitimate, i'm not
s'pose to have "naked" Dirac impulses in a mathematical expression. i
am simply treating Dirac impulses just like we do for the nascent delta
functions of very tiny, but non-zero width.
the Dirac delta is, strictly speaking, not really a "function", as the
mathematicians would put it. strictly speaking, if you integrate a
function that is zero "almost everywhere", the integral is zero, but we
lazy-ass electrical engineers say that the Dirac delta is a "function"
that is zero everywhere except the single point when t=0 and we say the
integral is 1.
so, if you can get past that, the "proof" in the past version of the
article above (and similar to that in many EE texts) is good enough.
--
r b-j [email protected]
"Imagination is more important than knowledge."
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