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lucy
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 Posted:
Sat Nov 12, 2005 9:15 am Post subject:
question about non-uniform sampling? |
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Hi all,
Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
What is the Nyquist sampling rate in the non-uniform case?
Thanks a lot!
-L |
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Michel Rouzic
Guest
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| Posted:
Sat Nov 12, 2005 9:15 am Post subject:
Re: question about non-uniform sampling? |
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lucy wrote:
| Quote: | What is the Nyquist sampling rate in the non-uniform case?
|
um.. i'll take the risk of trying to answer and say that it must be the
same one as if you had an uniform sampling rate matching to the
shortest distance between two samples in your non-uniformly sampled
signal. but i wouldnt be surprised if my answer was wrong or off-topic
(i'm a newbie kinda) |
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Joseph Fagan
Guest
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| Posted:
Sat Nov 12, 2005 10:36 am Post subject:
Re: question about non-uniform sampling? |
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"Michel Rouzic" <Michel0528@yahoo.fr> wrote in message
news:1131784415.779792.104570@g14g2000cwa.googlegroups.com...
| Quote: |
lucy wrote:
What is the Nyquist sampling rate in the non-uniform case?
um.. i'll take the risk of trying to answer and say that it must be the
same one as if you had an uniform sampling rate matching to the
shortest distance between two samples in your non-uniformly sampled
signal. but i wouldnt be surprised if my answer was wrong or off-topic
(i'm a newbie kinda)
"shortest distance" - do you mean "longest"? |
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Jerry Avins
Guest
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| Posted:
Sat Nov 12, 2005 5:16 pm Post subject:
Re: question about non-uniform sampling? |
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lucy wrote:
| Quote: | Hi all,
Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
What is the Nyquist sampling rate in the non-uniform case?
Thanks a lot!
|
This has been the subject of a few threads in the past. You might try
Google Groups for some insight. Theory says yes if the signal is
stationary, but practice is difficult. Given that the signal is
bandlimited, a system of samples can be solved as n simultaneous
equations. Spacing the samples uniformly in time simplifies the math but
other solutions are possible. If the signal isn't stationary, then the
samples must be close enough -- whatever that means -- to track the
changes. When the samples bunch too closely, the disturbing effects of
noise and truncation become overwhelmingly prominent.
So yes; in theory, there are many circumstances in which it can be done.
In practice, it is difficult. Fortunately, I have not so far had a need
to try.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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David Tweed
Guest
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| Posted:
Sat Nov 12, 2005 5:16 pm Post subject:
Re: question about non-uniform sampling? |
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lucy wrote:
| Quote: | Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
|
Yes, but it isn't easy.
| Quote: | What is the Nyquist sampling rate in the non-uniform case?
|
Believe it or not, it's the same as the uniform case ... the number
of samples over the time interval must exceed twice the bandwidth
of the signal.
See here (questions 2 and 3) for a little more detail:
http://www.circuitcellar.com/library/eq/136/index.asp
-- Dave Tweed |
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robert bristow-johnson
Guest
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| Posted:
Sun Nov 13, 2005 1:16 am Post subject:
Re: question about non-uniform sampling? |
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in article 4375FA3F.921CBB9A@acm.org, David Tweed at dtweed@acm.org wrote on
11/12/2005 09:32:
| Quote: | lucy wrote:
Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
Yes, but it isn't easy.
What is the Nyquist sampling rate in the non-uniform case?
Believe it or not, it's the same as the uniform case ... the number
of samples over the time interval must exceed twice the bandwidth
of the signal.
See here (questions 2 and 3) for a little more detail:
http://www.circuitcellar.com/library/eq/136/index.asp
|
hi Dave,
could you take a look at the paper that Bob Adams did in 1992 that i
reference here:
http://groups.google.com/group/comp.dsp/msg/ae7fe00eb3c8622b
i haven't cracked your brief analysis, but does that accomplish what i was
hoping would be shown that if your average sample rate is more than twice
the bandwidth, then random sampling will also be sufficient for
reconstruction?
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge." |
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glen herrmannsfeldt
Guest
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| Posted:
Sun Nov 13, 2005 5:16 pm Post subject:
Re: question about non-uniform sampling? |
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lucy wrote:
| Quote: | Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
What is the Nyquist sampling rate in the non-uniform case?
|
As others have said, if the average rate is fast enough it works.
It then gets back to your other question, which is end effects for a
time limited signal.
I had an example in the past where someone goes into a concert hall and
samples the signal at twice the required rate. This, then, is
theoretically enough to include the following concert, which is not
sampled at all.
As you previously mentioned, time limited signals can't be band limited,
so this only works for non-time limited signals. That means in infinite
number of sample points. It also requires that the signal not be
quantized, or at least not too coarsely.
(Depending on how non-uniform it is.)
-- glen |
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Guest
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| Posted:
Mon Nov 14, 2005 5:12 pm Post subject:
Re: question about non-uniform sampling? |
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A specific, and limited, example is when you do have a reference that
indicates where in its cycle the signal is, regardless of time.
For instance with rotating machinery, you might have a reference signal
that tells you each time one of the shafts rotates to a given position.
Then, you can use that to resample the (time) signals you measured, so
that they are evenly spaced with respect to the rotation (usually
within one cycle of some part of the machinery). This then lets you
enforce that your samples always happen to sample complete cycles of
the rotation, and that gives you the happy effect that the signal is
then ideally sampled. (This corresponds to a case that I describe on
our web site: http://www.bores.com/courses/intro/freq/3_exact.htm )
This technique was called 'order processing' and was developed and
publicised by Hewlett Packard some years ago - I don't know if
references are still available or if the method is widely used still.
Chris
==============================
Chris Bore
BORES Signal Processing
www.bores.com |
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Jerry Avins
Guest
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| Posted:
Mon Nov 14, 2005 5:15 pm Post subject:
Re: question about non-uniform sampling? |
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chris_bore@yahoo.co.uk wrote:
| Quote: | A specific, and limited, example is when you do have a reference that
indicates where in its cycle the signal is, regardless of time.
For instance with rotating machinery, you might have a reference signal
that tells you each time one of the shafts rotates to a given position.
Then, you can use that to resample the (time) signals you measured, so
that they are evenly spaced with respect to the rotation (usually
within one cycle of some part of the machinery). This then lets you
enforce that your samples always happen to sample complete cycles of
the rotation, and that gives you the happy effect that the signal is
then ideally sampled. (This corresponds to a case that I describe on
our web site: http://www.bores.com/courses/intro/freq/3_exact.htm )
This technique was called 'order processing' and was developed and
publicised by Hewlett Packard some years ago - I don't know if
references are still available or if the method is widely used still.
|
That amounts to sampling that's uniform in space, rather than in time.
The results have spatial significance, rather than temporal. It's so
standard in image processing that we don't even think about it.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Ron N.
Guest
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| Posted:
Tue Nov 15, 2005 12:48 am Post subject:
Re: question about non-uniform sampling? |
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lucy wrote:
| Quote: | Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
|
Interesting homework question old man.
The answer is probably no.
All the sample points could end up at the same point in time without
further specification on the non-uniformity.
| Quote: | What is the Nyquist sampling rate in the non-uniform case?
|
DC or infinity?
YMMV.
--
rhn |
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cacak5
Guest
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| Posted:
Tue Nov 15, 2005 1:15 am Post subject:
Re: question about non-uniform sampling? |
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lucy wrote:
| Quote: | Hi all,
Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
What is the Nyquist sampling rate in the non-uniform case?
Thanks a lot!
-L
|
This was a popular problem in the 1960's. A fairly thorough
presentation is made in the book "Discrete Time Systems - An
Introduction to the Theory" by Herbert Freeman, Wiley and Sons 1965;
section 4.8 - Nonuniform Sampling.
The main result for this issue is :
" a function f(t), bandlimited to -B/2<=w<=B/2 (where B=2pi/T) can be
uniquely reconstructed from a set of samples which are nonuniformly
spaced but satisfy the condition that there be precisely N distinct
samples to every interval of length NT, where N is some finite integer
.."
The actual reconstruction is shown in the book and is straight
forward; but really ugly.
This may be a more severe condition that it seems at first;
especially the "precisely N distinct ..." . There are probably better
results since this book was published forty years ago.
MzF |
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bergers@aol.com
Guest
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| Posted:
Tue Nov 15, 2005 8:03 am Post subject:
Re: question about non-uniform sampling? |
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If you a theoretical discuss, then read the following article:
H.J. Landau, "Sampling, Data Transmission, and the Nyquist Rate",
Proceedings of the IEEE, Vol 55, No 10, October 1967.
I quote from the abstract (note in the brackets is my comments):
"...In this paper we draw a distinction between reconstructioning a
signal from its samples, and doing so in a stable way [small errors in
sampling don't led to large reconstruction errors], we argue that only
stable sampling is meaningful in practice. We then prove that:
1) stable sampling cannot be performed at a rate lower than the
Nyquist,
2) data cannot be transmitted as samples as a rate higher that the
Nyquist,
regardless of the location of sampling instants, the nature of the set
of frequencies which the signals occupy, or the method of construction.
These conclusions apply not merely to finite-energy, but also to
bounded signals."
Scott |
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Carlos Moreno
Guest
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| Posted:
Tue Nov 15, 2005 9:15 am Post subject:
Re: question about non-uniform sampling? |
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lucy wrote:
| Quote: | Hi all,
Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
What is the Nyquist sampling rate in the non-uniform case?
|
There's an easy way to look at this.
Imagine that you have a sampled signal -- standard, uniform
sampling at a rate within Nyquist.
That means that the value of the signal at *every* point is
fully and exactly determined by all the samples (the infinity
of samples), by a linear relationship:
+oo
x(t) = sum { a_k x(kTs) }
k = -oo
Where the value of each a_k depends on the value of t (your
sinc function, etc. etc.)
So, if you can find x(t) as a linear function of all the x_k,
then you could solve for one of the x_k's as a function of all
the other x_k's and x(t):
x(nTs) = ( x(t) - sum {a_k x(kTs)} ) / a_n
k != i
So, that means that you can take *one* sample at *any* position
different from its "corresponding" position if you were always
sampling at uniform spacing, and you can then determine the
value of the signal at the point of the missing sample (one of
the points of uniform sampling). But if you can determine the
missing sample, then you now have all the samples of a uniformly-
sampled signal, and hence you can fully and exactly reconstruct
the analog signal.
Now, if you take two samples (instead of one) at the non-
corresponding positions, the above reasoning can be used to
show that now you have a set of two linear equations with two
unknowns. Provided that the two signals are sampled at
different points, and different from any other sample in the
uniform-sampling points, the equations are independent, and
the system has a unique solution.
The reasoning can be extended to any number N, no matter how
large.
I know this is not rigurous -- in particular, this shows that
the trick works for N samples taken at positions other than
the corresponding positions, no matter how large; but this
proves nothing about an "infinity" of samples taken non-
uniformly... Still, the result does suggest that you still
need the amount of samples that totals the same amount of
samples required in uniform sampling (suggesting that your
Nyquist condition is given by the average sampling rate).
HTH,
Carlos
-- |
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Jerry Avins
Guest
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| Posted:
Tue Nov 15, 2005 5:16 pm Post subject:
Re: question about non-uniform sampling? |
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Carlos Moreno wrote:
| Quote: | lucy wrote:
Hi all,
Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
What is the Nyquist sampling rate in the non-uniform case?
There's an easy way to look at this.
Imagine that you have a sampled signal -- standard, uniform
sampling at a rate within Nyquist.
That means that the value of the signal at *every* point is
fully and exactly determined by all the samples (the infinity
of samples), by a linear relationship:
+oo
x(t) = sum { a_k x(kTs) }
k = -oo
Where the value of each a_k depends on the value of t (your
sinc function, etc. etc.)
So, if you can find x(t) as a linear function of all the x_k,
then you could solve for one of the x_k's as a function of all
the other x_k's and x(t):
x(nTs) = ( x(t) - sum {a_k x(kTs)} ) / a_n
k != i
So, that means that you can take *one* sample at *any* position
different from its "corresponding" position if you were always
sampling at uniform spacing, and you can then determine the
value of the signal at the point of the missing sample (one of
the points of uniform sampling). But if you can determine the
missing sample, then you now have all the samples of a uniformly-
sampled signal, and hence you can fully and exactly reconstruct
the analog signal.
Now, if you take two samples (instead of one) at the non-
corresponding positions, the above reasoning can be used to
show that now you have a set of two linear equations with two
unknowns. Provided that the two signals are sampled at
different points, and different from any other sample in the
uniform-sampling points, the equations are independent, and
the system has a unique solution.
The reasoning can be extended to any number N, no matter how
large.
I know this is not rigurous -- in particular, this shows that
the trick works for N samples taken at positions other than
the corresponding positions, no matter how large; but this
proves nothing about an "infinity" of samples taken non-
uniformly... Still, the result does suggest that you still
need the amount of samples that totals the same amount of
samples required in uniform sampling (suggesting that your
Nyquist condition is given by the average sampling rate).
HTH,
|
It doesn't. There's a limit to how non-uniform the sampling can be
allowed to be. The example given above, of an hour's worth of music
sampled for half an hour at twice the minimum rate for the bandwidth, is
an adequate counterexample to what you claim is a general case.
Jerry
--
Engineering is the art of making what you want from things you can get.
ŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻŻ |
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Steve Underwood
Guest
|
| Posted:
Tue Nov 15, 2005 5:16 pm Post subject:
Re: question about non-uniform sampling? |
|
|
Jerry Avins wrote:
| Quote: | Carlos Moreno wrote:
lucy wrote:
Hi all,
Can non-uniform sampled signal be used to perfectly reconstruct the
original continuous time signal?
What is the Nyquist sampling rate in the non-uniform case?
There's an easy way to look at this.
Imagine that you have a sampled signal -- standard, uniform
sampling at a rate within Nyquist.
That means that the value of the signal at *every* point is
fully and exactly determined by all the samples (the infinity
of samples), by a linear relationship:
+oo
x(t) = sum { a_k x(kTs) }
k = -oo
Where the value of each a_k depends on the value of t (your
sinc function, etc. etc.)
So, if you can find x(t) as a linear function of all the x_k,
then you could solve for one of the x_k's as a function of all
the other x_k's and x(t):
x(nTs) = ( x(t) - sum {a_k x(kTs)} ) / a_n
k != i
So, that means that you can take *one* sample at *any* position
different from its "corresponding" position if you were always
sampling at uniform spacing, and you can then determine the
value of the signal at the point of the missing sample (one of
the points of uniform sampling). But if you can determine the
missing sample, then you now have all the samples of a uniformly-
sampled signal, and hence you can fully and exactly reconstruct
the analog signal.
Now, if you take two samples (instead of one) at the non-
corresponding positions, the above reasoning can be used to
show that now you have a set of two linear equations with two
unknowns. Provided that the two signals are sampled at
different points, and different from any other sample in the
uniform-sampling points, the equations are independent, and
the system has a unique solution.
The reasoning can be extended to any number N, no matter how
large.
I know this is not rigurous -- in particular, this shows that
the trick works for N samples taken at positions other than
the corresponding positions, no matter how large; but this
proves nothing about an "infinity" of samples taken non-
uniformly... Still, the result does suggest that you still
need the amount of samples that totals the same amount of
samples required in uniform sampling (suggesting that your
Nyquist condition is given by the average sampling rate).
HTH,
It doesn't. There's a limit to how non-uniform the sampling can be
allowed to be. The example given above, of an hour's worth of music
sampled for half an hour at twice the minimum rate for the bandwidth, is
an adequate counterexample to what you claim is a general case.
Jerry
|
The practicality of non-uniform sample isn't a whole lot different
whether we are talking about minor non-uniformity or some extreme. As
soon as sampling is even a little non-uniform it is highly sensitive to
sampling error and converter noise. As it becomes more non-uniform it
quickly becomes totally impractical to make sense of the kind of samples
you can get in the real world. There is nothing wrong with any extreme
of non-uniformity in a purely mathematical sense. That is in a world
with infinite sampling precision and no noise due to the converter itself.
Regards,
Steve |
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