On 27/03/2014 3:23 PM, Doug Houghton wrote:
Is that making any sense? I'm struggling with the fine points. I bet
this is obvious if you understand the math in the proof.
I'm following along, vaguely.
My take is that this conversation is not making enough sense to give you
the certainty you seem to be looking for.
Your question seems to be very particular regarding specifics of the
definitions used in a theorem, but you have not quoted the theorem or
the definitions.
Most of the answers so far seem to be talking about interpretations and
consequences of the theorem.
May I suggest: quote the version of the theorem that you're working from
and the definitions of terms used that you're assuming, and we can go
from there. It may also be helpful to see the proof that you are working
from, perhaps someone can help unpack that.
Here's a version of the theorem that you may or may not be happy with:
SHANNON-NYQUIST SAMPLING THEOREM [1]
If a function x(t) contains no frequencies higher than B hertz,
it is completely determined by giving its ordinates at a series
of points spaced 1/(2B) seconds apart.
---
I'm loath to contribute my limited interpretation, but let me try (feel
free to ignore or ideally, correct me):
x(t) is an infinite duration continuous-time function.
a "frequency" is defined to be an infinite duration complex
sinusoid with a particular period.
The theorem is saying something about an infinite duration continuous
time signal x(t), and expressing a constraint on that signal in terms of
the signal's "frequency components".
To be able to talk about the frequency components of x(t) we can use a
continuous Fourier representation of the signal, i.e. the Fourier
transform [2], say x'(w), a complex valued function, w is a (continuous)
real-valued frequency parameter:
+inf
x'(w) = integrate x(t)*e^(-2*pi*i*t*w) dt
-inf
The Fourier transform can represent any continuous signal that is
integrable and continuous (I deduce this from the invertability of the
Fourier transform [3]). One consequence of this is that any practical
analog signal x(t) may be represented by its Fourier transform.
The theorem expresses a constraint the frequencies for which the Fourier
transform may be non-zero. Specifically, it requires that x'(w) = 0 for
all w < -N and all w > N, where N is the Nyquist frequency.
Note specifically that we are dealing with the continuous Fourier
transform, therefore there is no requirement for x(t) to be periodic or
of finite temporal extent.
The theorem also does not say anything about the time extent of the
discrete time signal (it is assumed to be infinite too).
That's my take on it anyway.
Ross.
[1]
http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Nyquist%E2%80%93Shannon_sampling_theorem.html
[2] http://en.wikipedia.org/wiki/Fourier_transform
[3] http://en.wikipedia.org/wiki/Fourier_inversion_formula
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