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Peter K.
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 Posted:
Tue Nov 29, 2005 7:29 am Post subject:
Re: question about non-uniform sampling? |
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David Tweed <dtweed@acm.org> writes:
| Quote: | I think we're all violently in agreement.
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Don't ya hate that? :-)
Ciao,
Peter K. |
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Rob Johnson
Guest
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| Posted:
Tue Nov 29, 2005 9:12 am Post subject:
Re: question about non-uniform sampling? |
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In article <438B0344.815A671@acm.org>,
David Tweed <dtweed@acm.org> wrote:
| Quote: | robert bristow-johnson wrote:
David Tweed wrote:
robert bristow-johnson wrote:
it seems to me that you need to solve a system of an infinite
number of equations that have an infinite number of unknowns,
No, you only have one equation for each nonuniform sample you took.
yeah, you're right, but i keep imagining the premise of the OP was that
many, or all of the samples are, in general, not lying at uniformly
spaced times.
The solution is the corresponding set (same number) of uniform samples.
You have to evaluate the sinc function N^2 times to get the coefficients
for the equations.
whereas you don't have more than one equation for only one out-of-step
sample, you still have to (if you're math is gonna be perfect) evaluate
an infinite number of sinc(.) functions. of course, we would say only
a certain finite number of neighboring samples can affect the sole
out-of-step sample, but if one is hoping to extrapolate the first
movement of Beethoven's Ninth into even the second movement (my fav
since the Huntley-Brinkley Report on NBC), i think that person would
need all infinity sinc(.) functions. and would need infinite
precision.
In order to make this tractable, you have to assume that the all of
the sinc coefficients outside of some interval (before the beginning
and after the end of the symphony) are all zero. I thought that was
implicit.
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However, if the signal is band-limited, it cannot be time-limited, and
so even though the coefficients of the sinc functions can be assumed 0
outside the time of the symphony, the actual signal of the symphony
does not vanish. It may be that to minimize the energy outside the
time of the symphony, the coefficients of the sinc functions outside
the time of the symphony need to be something other than 0.
Rob Johnson <rob@trash.whim.org>
take out the trash before replying |
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David Tweed
Guest
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| Posted:
Tue Nov 29, 2005 5:16 pm Post subject:
Re: question about non-uniform sampling? |
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Rob Johnson wrote:
| Quote: | However, if the signal is band-limited, it cannot be time-limited,
and so even though the coefficients of the sinc functions can be
assumed 0 outside the time of the symphony, the actual signal of
the symphony does not vanish. It may be that to minimize the
energy outside the time of the symphony, the coefficients of the
sinc functions outside the time of the symphony need to be something
other than 0.
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I'm not interested in minimizing the energy, only the information
content (the number of unknowns). The simplest way to do that is
to make them zero. If you want to assign nonzero values to meet
some additional criterion, as long as they're derived from what
you know about the interval in question, I guess that doesn't
change the overall information content.
-- Dave Tweed |
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glen herrmannsfeldt
Guest
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| Posted:
Sat Dec 10, 2005 9:16 am Post subject:
Re: question about non-uniform sampling? |
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Rob Johnson wrote:
(snip)
| Quote: | As has been discussed in the thread titled "how do you prove a signal
that is time-limited cannot be bandlimited?", a band-limited function
is entire, and an entire function is totally determined by its value
on any open interval. Therefore, if you can reconstruct a band-limited
function for even a small interval of time, you can reconstruct it for
all time.
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The physics solution is a periodic function that goes to zero at the
ends. If the requirement is that it have the specified values over a
finite range, with no requirement outside that range then a periodic
solution works. That is what the Fourier series does. Some people
don't like that solution, though it seems to be the favorite for car
CD players.
| Quote: | However, unlike the uniformly-sampled case, in which the signal
outside the interval does not affect the samples, we must assume
that the signal is zero* outside our interval of interest.
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Why must we assume that? If I want to record a CD I only care that
the signal is right for the length of the CD. What happens before or
after doesn't matter to my CD player.
| Quote: | In the uniformly-sampled case, the same band-limited assumption is
made. It is the limit on the band of the signal that allows us to
reconstruct it exactly from the samples. How can we assume that the
signal is zero outside our interval of interest? This would mean we
are assuming a signal that is band limited can also be time limited.
Consider the case of a signal of finite bandwidth constructed
by applying random scaling coefficients to a regular sequence
of sinc pulses. In the uniformly-sampled case, the pulses
outside our interval won't affect our samples. But in the
nonuniform case, we can't make that same argument.
Why won't the sinc pulses outside the interval affect the samples?
sinc functions, being entire, effect the whole real line, except for
isolated zeroes. If sinc functions cancel, even on a small open set,
they cancel everywhere.
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Sinc are the solution when it is not time limited. The solution will
be (slightly) different for the time limited case, closer to sinc as the
time gets longer.
| Quote: | Certainly it is possible to construct a band-limited function
that agrees with the sampled function at finitely many sampled
points. There are infinitely many functions with the same band
limit that agree with the sampled function at the same sample
points. However, this is different than reconstructing the
sampled signal.
Not quite. There's only one function with the exact same band
limits. There are infinitely many functions with the same
bandwidth. As in the uniformly-sampled case, they are called
images. As long as you know which image you're interested in,
you can select the correct interpolation function to reconstruct
the original signal exactly.
This would imply that there is a finite dimensional set of functions
whose Fourier Transform are supported in the given band. This is
definitely not true.
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For time limited signals, use the Fourier series instead.
| Quote: | Suppose we add one more sample point (x,y). Since we are dealing with
non-uniform samples, this should be no problem. Using the same method
to reconstruct the original signal, we can find a signal that matches
the original signal at the other sample points and at (x,y) which has
the same band limits. Since we can do this for any y, we can find
infinitely many functions with the same band limits which match the
original signal at the other sample points.
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Remember that infinity+1 and infinity-1 are still equal to infinity.
-- glen |
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