Strange Discrete Fourier Transform question

Suppose that I have a finite-length discrete-time signal that is not
necessarily sampled on a uniform time grid. I will characterize this
signal as a series of time-amplitude pairs with time = t_sub_k and
amplitude = a_sub_k, where k is an integer from 1 to N that selects a
particular time-amplitude pair.

It is easy to take the DFT of this sequence even though the time grid
may not be uniform.

FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))

Now suppose that I transpose the role of amplitude and time; that is, I
form a new time-amplitude series that is related to the old series by;

a2_sub_k = t_sub_k
t2_sub_k = a_sub_k

Then I proceed to take the DFT of this new series, producing a new
function FT2(w).

Given FT(w), is there ANYTHING I can say about FT2(w) that is "general"
in nature and not tied to a specific time-amplitude series?

In particular, I am investigating series that are sampled on a log time
grid, where t_sub_k = log(k).

Obviously if we start with a_sub_k = t_sub_k (for example, a uniformly
sampled ramp signal) then the two DFT results are the same. But I am
wondering if there are some classes of non-uniform time grids in
conjunction with particular amplitude values that cause FT(w) and
FT2(w)  to have some relationship in terms of mapping poles and zeroes
from one domain into the other.

Any thoughts welcome!


Bob Adams

robert.w.adams@verizon.net skrev:
> Suppose that I have a finite-length discrete-time signal that is not
> necessarily sampled on a uniform time grid. I will characterize this
> signal as a series of time-amplitude pairs with time = t_sub_k and
> amplitude = a_sub_k, where k is an integer from 1 to N that selects a
> particular time-amplitude pair.
>
> It is easy to take the DFT of this sequence even though the time grid
> may not be uniform.
>
> FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))
>
> Now suppose that I transpose the role of amplitude and time; that is, I
> form a new time-amplitude series that is related to the old series by;
>
> a2_sub_k = t_sub_k
> t2_sub_k = a_sub_k
>
> Then I proceed to take the DFT of this new series, producing a new
> function FT2(w).
>
> Given FT(w), is there ANYTHING I can say about FT2(w) that is "general"
> in nature and not tied to a specific time-amplitude series?

Interchanging the time and amplitude destroys any hope of
saying anything useful about the time series. For an example,
choose

x[n] = sin(2*pi*k*n/N)

for some N, n = 0,...,N-1 and k < N/2. Try to do this "trick"
of yours, and deduce any useful information about the original
time series.

Rune

<robert.w.adams@verizon.net> wrote in message 
news:1154260055.670375.100450@h48g2000cwc.googlegroups.com...
> Suppose that I have a finite-length discrete-time signal that is not
> necessarily sampled on a uniform time grid. I will characterize this
> signal as a series of time-amplitude pairs with time = t_sub_k and
> amplitude = a_sub_k, where k is an integer from 1 to N that selects a
> particular time-amplitude pair.
>
> It is easy to take the DFT of this sequence even though the time grid
> may not be uniform.
>
> FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))
>
> Now suppose that I transpose the role of amplitude and time; that is, I
> form a new time-amplitude series that is related to the old series by;
>
> a2_sub_k = t_sub_k
> t2_sub_k = a_sub_k
>
> Then I proceed to take the DFT of this new series, producing a new
> function FT2(w).
>
> Given FT(w), is there ANYTHING I can say about FT2(w) that is "general"
> in nature and not tied to a specific time-amplitude series?
>
> In particular, I am investigating series that are sampled on a log time
> grid, where t_sub_k = log(k).
>
> Obviously if we start with a_sub_k = t_sub_k (for example, a uniformly
> sampled ramp signal) then the two DFT results are the same. But I am
> wondering if there are some classes of non-uniform time grids in
> conjunction with particular amplitude values that cause FT(w) and
> FT2(w)  to have some relationship in terms of mapping poles and zeroes
> from one domain into the other.

Bob,

You said a DFT was easy as:

FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))

where the values of t_sub_k are apparently ordered but arbitrary.
I note that "w" isn't in any way discrete - ????

So, I think this isn't a DFT as we know it.

Start with a DFT definition with
N the number of samples
T the temporal epoch,
fs the sampling frequency
T/N = 1/fs the temporal sample interval
1/T = fs/N the frequency sample interval

And the DFT is:

F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*(kT/N)]  ;n=0,N-1

If you want to have arbitrary values of "w" then you have a continuous 
Fourier Transform of a discrete sequence:

F(w) = sum k=1 to N a_sub_k*exp(-j*w*kT/N)   which almost looks like what 
you have above.

It's not discrete in frequency.  Just to be clear.  I don't see how that's 
"easy" to do using arrays of numbers in a computer - which seems to be where 
you're heading with this.

Then, if you want to make the sample times arbitrary then you would have 
what we started with:

F(w) = sum k=1 to N a_sub_k*exp(-j*w*t_sub_k)

It's still not discrete in frequency - which is what you need to get a 
finite sequence.

So, it's common to assume that the temporal sequence is periodic on T and 
the sum generating the transform is finite:  If this is done you get 
something like:

F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*t_sub_k]  ;n=0,N-1  !!!

The problem with this formulation is that the temporal resolution hasn't 
been defined on a regular grid and that's necessary for the Fourier 
Transform to be periodic instead of just infinite.  If it's infinite then 
this formulation can't be inverse transformed because it's incomplete - not 
all the information is there.

Another "problem" to ponder here is that the times aren't (i.e. need not be) 
ordered - although in a Fourier Transform they are ordered.  That must have 
some bearing, eh?

You might arbitrarily assign a time grid for the samples that sufficiently 
captures the sample times on that grid.  You might use 100 times the 
narrowest separation between samples with the idea that the other 
separations won't be too far off if forced onto the grid.
This results in maybe 1000*N sample points in time (?) and you put the 
sample amplitudes on the nearest grid point in time.

Now you can compute a DFT that's suitable (because you decided on the 
parameters) - albeit a fairly long one.

OK.

Now to switching between time and amplitude as you suggested:

Note that f(t) in the Fourier Transform is ordered in time.
Note that the values of t are unique.

Note that amplitudes of signals or data generally aren't unique.
That suggests flipping them results in a situation that doesn't fit the 
formulation.
One is a set of values and the other is an index or scale.  Really very 
different things.

Fred

Bob, you always bring such hard questions here.

robert.w.adams@verizon.net wrote:
> Suppose that I have a finite-length discrete-time signal that is not
> necessarily sampled on a uniform time grid. I will characterize this
> signal as a series of time-amplitude pairs with time = t_sub_k and
> amplitude = a_sub_k, where k is an integer from 1 to N that selects a
> particular time-amplitude pair.
>
> It is easy to take the DFT of this sequence even though the time grid
> may not be uniform.
>
> FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))
>
> Now suppose that I transpose the role of amplitude and time; that is, I
> form a new time-amplitude series that is related to the old series by;
>
> a2_sub_k = t_sub_k
> t2_sub_k = a_sub_k
>

okay, but now there is no hope of the  t2_sub_k being increasing with
k.  before i hwas thinking that  t_sub_k was a strictly increasing
sequence.

> Then I proceed to take the DFT of this new series, producing a new
> function FT2(w).
>
> Given FT(w), is there ANYTHING I can say about FT2(w) that is "general"
> in nature and not tied to a specific time-amplitude series?
>
> In particular, I am investigating series that are sampled on a log time
> grid, where t_sub_k = log(k).
>
> Obviously if we start with a_sub_k = t_sub_k (for example, a uniformly
> sampled ramp signal) then the two DFT results are the same. But I am
> wondering if there are some classes of non-uniform time grids in
> conjunction with particular amplitude values that cause FT(w) and
> FT2(w)  to have some relationship in terms of mapping poles and zeroes
> from one domain into the other.

(pain)

> Any thoughts welcome!

errrrg....  i;m an engunir.  i kin spel.

r b-j







i s'pose if it was an easy question, you would know the answer.

Fred Marshall wrote:
> <robert.w.adams@verizon.net> wrote in message
> news:1154260055.670375.100450@h48g2000cwc.googlegroups.com...
> > Suppose that I have a finite-length discrete-time signal that is not
> > necessarily sampled on a uniform time grid. I will characterize this
> > signal as a series of time-amplitude pairs with time = t_sub_k and
> > amplitude = a_sub_k, where k is an integer from 1 to N that selects a
> > particular time-amplitude pair.
> >
> > It is easy to take the DFT of this sequence even though the time grid
> > may not be uniform.
> >
> > FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))
> >
> > Now suppose that I transpose the role of amplitude and time; that is, I
> > form a new time-amplitude series that is related to the old series by;
> >
> > a2_sub_k = t_sub_k
> > t2_sub_k = a_sub_k
> >
> > Then I proceed to take the DFT of this new series, producing a new
> > function FT2(w).
> >
> > Given FT(w), is there ANYTHING I can say about FT2(w) that is "general"
> > in nature and not tied to a specific time-amplitude series?
> >
> > In particular, I am investigating series that are sampled on a log time
> > grid, where t_sub_k = log(k).
> >
> > Obviously if we start with a_sub_k = t_sub_k (for example, a uniformly
> > sampled ramp signal) then the two DFT results are the same. But I am
> > wondering if there are some classes of non-uniform time grids in
> > conjunction with particular amplitude values that cause FT(w) and
> > FT2(w)  to have some relationship in terms of mapping poles and zeroes
> > from one domain into the other.
>
> Bob,
>
> You said a DFT was easy as:
>
> FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))
>
> where the values of t_sub_k are apparently ordered but arbitrary.
> I note that "w" isn't in any way discrete - ????
>
> So, I think this isn't a DFT as we know it.
>
> Start with a DFT definition with
> N the number of samples
> T the temporal epoch,
> fs the sampling frequency
> T/N = 1/fs the temporal sample interval
> 1/T = fs/N the frequency sample interval
>
> And the DFT is:
>
> F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*(kT/N)]  ;n=0,N-1
>
> If you want to have arbitrary values of "w" then you have a continuous
> Fourier Transform of a discrete sequence:
>
> F(w) = sum k=1 to N a_sub_k*exp(-j*w*kT/N)   which almost looks like what
> you have above.
>
> It's not discrete in frequency.  Just to be clear.  I don't see how that's
> "easy" to do using arrays of numbers in a computer - which seems to be where
> you're heading with this.
>
> Then, if you want to make the sample times arbitrary then you would have
> what we started with:
>
> F(w) = sum k=1 to N a_sub_k*exp(-j*w*t_sub_k)
>
> It's still not discrete in frequency - which is what you need to get a
> finite sequence.
>
> So, it's common to assume that the temporal sequence is periodic on T and
> the sum generating the transform is finite:  If this is done you get
> something like:
>
> F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*t_sub_k]  ;n=0,N-1  !!!
>
> The problem with this formulation is that the temporal resolution hasn't
> been defined on a regular grid and that's necessary for the Fourier
> Transform to be periodic instead of just infinite.  If it's infinite then
> this formulation can't be inverse transformed because it's incomplete - not
> all the information is there.
>
> Another "problem" to ponder here is that the times aren't (i.e. need not be)
> ordered - although in a Fourier Transform they are ordered.  That must have
> some bearing, eh?
>
> You might arbitrarily assign a time grid for the samples that sufficiently
> captures the sample times on that grid.  You might use 100 times the
> narrowest separation between samples with the idea that the other
> separations won't be too far off if forced onto the grid.
> This results in maybe 1000*N sample points in time (?) and you put the
> sample amplitudes on the nearest grid point in time.
>
> Now you can compute a DFT that's suitable (because you decided on the
> parameters) - albeit a fairly long one.
>
> OK.
>
> Now to switching between time and amplitude as you suggested:
>
> Note that f(t) in the Fourier Transform is ordered in time.
> Note that the values of t are unique.
>
> Note that amplitudes of signals or data generally aren't unique.
> That suggests flipping them results in a situation that doesn't fit the
> formulation.
> One is a set of values and the other is an index or scale.  Really very
> different things.
>
> Fred



Fred


Yes, you are correct that I should not have used the term DFT; I was
thinking in terms of the Fourier Transform of impulsive time-domain
finite-length sequences where the times are not uniformly distributed.
Such signals have a continuous spectrum that does not repeat.

I think I made the problem too general; without any restriction on the
amplitude sequence, it's true that after you flip time and amplitude
you would likely get nonsense. I am more interested in special cases. A
trivial one is where you sample a linear ramp signal, in which case
time and amplitude are the same at every point and therefore reversing
them causes no change. But I think there are other cases where there
are some interesting relationships before and after the time-amplitude
switch.

It would take a long time to explain why I am interested in this; I
have developed an unhealthy obsession with solving the Reimann
Hypothesis, and have transformed it into a signal-processing problem
with log-time sampling (see my 2005 ISCAS paper for more detail).


Regards

Bob Adams

robert.w.adams@verizon.net wrote:
...
> It would take a long time to explain why I am interested in this; I
> have developed an unhealthy obsession with solving the Reimann
> Hypothesis, and have transformed it into a signal-processing problem
> with log-time sampling (see my 2005 ISCAS paper for more detail).

Bob, is there a free place to get that ISCAS paper?

might we bug you for a pdf if there is none?

r b-j

Fred Marshall wrote:
> <robert.w.adams@verizon.net> wrote in message 
> news:1154260055.670375.100450@h48g2000cwc.googlegroups.com...

>>Suppose that I have a finite-length discrete-time signal that is not
>>necessarily sampled on a uniform time grid. I will characterize this
>>signal as a series of time-amplitude pairs with time = t_sub_k and
>>amplitude = a_sub_k, where k is an integer from 1 to N that selects a
>>particular time-amplitude pair.

>>It is easy to take the DFT of this sequence even though the time grid
>>may not be uniform.

>>FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))


(snip)

> You said a DFT was easy as:

> FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))

> where the values of t_sub_k are apparently ordered but arbitrary.
> I note that "w" isn't in any way discrete - ????

> So, I think this isn't a DFT as we know it.

> Start with a DFT definition with
> N the number of samples
> T the temporal epoch,
> fs the sampling frequency
> T/N = 1/fs the temporal sample interval
> 1/T = fs/N the frequency sample interval

> And the DFT is:

> F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*(kT/N)]  ;n=0,N-1

Well, it would have been easier if it was stated that the samples
were of a periodic function.  My claim for a finite duration
series is that it can be assumed periodic with a period greater
than or equal to the duration without loss of information.

If one considers a Fourier integral with delta functions at
the appropriate points, it seems to make some sense to me.

It transforms N sample values into N frequency values.
The question, then, is does the inverse transform, the sum
of the frequency components evaluated at the original positions,
give the original signal? (see below)

> If you want to have arbitrary values of "w" then you have a continuous 
> Fourier Transform of a discrete sequence:

> F(w) = sum k=1 to N a_sub_k*exp(-j*w*kT/N)   which almost looks like what 
> you have above.

> It's not discrete in frequency.  Just to be clear.  I don't see how that's 
> "easy" to do using arrays of numbers in a computer - which seems to be where 
> you're heading with this.

> Then, if you want to make the sample times arbitrary then you would have 
> what we started with:

> F(w) = sum k=1 to N a_sub_k*exp(-j*w*t_sub_k)

> It's still not discrete in frequency - which is what you need to get a 
> finite sequence.

> So, it's common to assume that the temporal sequence is periodic on T and 
> the sum generating the transform is finite:  If this is done you get 
> something like:

> F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*t_sub_k]  ;n=0,N-1  !!!

> The problem with this formulation is that the temporal resolution hasn't 
> been defined on a regular grid and that's necessary for the Fourier 
> Transform to be periodic instead of just infinite.  If it's infinite then 
> this formulation can't be inverse transformed because it's incomplete - not 
> all the information is there.

I am not convinced.

If I have N samples amplitudes taken at N times, I want the N frequency
components such that when evaluated at the original N times I get the
original N amplitudes.  I am not sure yet that the OP's transform does
that, and so far I think it doesn't.

It is possible, though unlikely, that the transform is singular, such
that it can't be inverted.  It is more likely to be ill-conditioned,
which will happen if too many data points are too close together.
If the transform isn't singular then there is enough data.
The aliasing rules aren't simple, though.  You can have any N 
frequencies, not necessarily the first N harmonics (starting
from zero).

The nice thing about the Fourier series is that orthogonality makes
the inverse easy.  Without orthogonality one must compute the
inverse of the transform matrix.

> Another "problem" to ponder here is that the times aren't (i.e. need not be) 
> ordered - although in a Fourier Transform they are ordered.  That must have 
> some bearing, eh?

I don't think that matters much.  It is just like exchanging rows or 
columns of a matrix which, if done right, doesn't affect the answer.
Consider and N equations and N unknowns problem.  The answer doesn't
depend on the order I write the equations down.

-- glen

"glen herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message 
news:GbOdnc4e684Ae0rZnZ2dnUVZ_u-dnZ2d@comcast.com...
> Fred Marshall wrote:
>> <robert.w.adams@verizon.net> wrote in message 
>> news:1154260055.670375.100450@h48g2000cwc.googlegroups.com...
>
>>>Suppose that I have a finite-length discrete-time signal that is not
>>>necessarily sampled on a uniform time grid. I will characterize this
>>>signal as a series of time-amplitude pairs with time = t_sub_k and
>>>amplitude = a_sub_k, where k is an integer from 1 to N that selects a
>>>particular time-amplitude pair.
>
>>>It is easy to take the DFT of this sequence even though the time grid
>>>may not be uniform.
>
>>>FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))
>
>
> (snip)
>
>> You said a DFT was easy as:
>
>> FT(w) = SUM k=1 to N (a_sub_k*exp(-j*w*t_sub_k))
>
>> where the values of t_sub_k are apparently ordered but arbitrary.
>> I note that "w" isn't in any way discrete - ????
>
>> So, I think this isn't a DFT as we know it.
>
>> Start with a DFT definition with
>> N the number of samples
>> T the temporal epoch,
>> fs the sampling frequency
>> T/N = 1/fs the temporal sample interval
>> 1/T = fs/N the frequency sample interval
>
>> And the DFT is:
>
>> F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*(kT/N)]  ;n=0,N-1
>
> Well, it would have been easier if it was stated that the samples
> were of a periodic function.  My claim for a finite duration
> series is that it can be assumed periodic with a period greater
> than or equal to the duration without loss of information.
>
> If one considers a Fourier integral with delta functions at
> the appropriate points, it seems to make some sense to me.
>
> It transforms N sample values into N frequency values.
> The question, then, is does the inverse transform, the sum
> of the frequency components evaluated at the original positions,
> give the original signal? (see below)
>
>> If you want to have arbitrary values of "w" then you have a continuous 
>> Fourier Transform of a discrete sequence:
>
>> F(w) = sum k=1 to N a_sub_k*exp(-j*w*kT/N)   which almost looks like what 
>> you have above.
>
>> It's not discrete in frequency.  Just to be clear.  I don't see how 
>> that's "easy" to do using arrays of numbers in a computer - which seems 
>> to be where you're heading with this.
>
>> Then, if you want to make the sample times arbitrary then you would have 
>> what we started with:
>
>> F(w) = sum k=1 to N a_sub_k*exp(-j*w*t_sub_k)
>
>> It's still not discrete in frequency - which is what you need to get a 
>> finite sequence.
>
>> So, it's common to assume that the temporal sequence is periodic on T and 
>> the sum generating the transform is finite:  If this is done you get 
>> something like:
>
>> F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*t_sub_k]  ;n=0,N-1  !!!
>
>> The problem with this formulation is that the temporal resolution hasn't 
>> been defined on a regular grid and that's necessary for the Fourier 
>> Transform to be periodic instead of just infinite.  If it's infinite then 
>> this formulation can't be inverse transformed because it's incomplete - 
>> not all the information is there.
>
> I am not convinced.
>
> If I have N samples amplitudes taken at N times, I want the N frequency
> components such that when evaluated at the original N times I get the
> original N amplitudes.  I am not sure yet that the OP's transform does
> that, and so far I think it doesn't.
>
> It is possible, though unlikely, that the transform is singular, such
> that it can't be inverted.  It is more likely to be ill-conditioned,
> which will happen if too many data points are too close together.
> If the transform isn't singular then there is enough data.
> The aliasing rules aren't simple, though.  You can have any N frequencies, 
> not necessarily the first N harmonics (starting
> from zero).
>
> The nice thing about the Fourier series is that orthogonality makes
> the inverse easy.  Without orthogonality one must compute the
> inverse of the transform matrix.
>
>> Another "problem" to ponder here is that the times aren't (i.e. need not 
>> be) ordered - although in a Fourier Transform they are ordered.  That 
>> must have some bearing, eh?
>
> I don't think that matters much.  It is just like exchanging rows or 
> columns of a matrix which, if done right, doesn't affect the answer.
> Consider and N equations and N unknowns problem.  The answer doesn't
> depend on the order I write the equations down.
>
> -- glen

Glen,

First:

>"My claim for a finite duration
> series is that it can be assumed periodic with a period greater
> than or equal to the duration without loss of information."

This was dealt with later..... at a point where it seemed to fit into the 
discussion.

Second:

>> F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*t_sub_k]  ;n=0,N-1  !!!
>
>> The problem with this formulation is that the temporal resolution hasn't 
>> been defined on a regular grid and that's necessary for the Fourier 
>> Transform to be periodic instead of just infinite.  If it's infinite then 
>> this formulation can't be inverse transformed because it's incomplete - 
>> not all the information is there.
>
> "I am not convinced."

You are not convinced about what aspect of the assertion?

- Is it not correct that the temporal resolution hasn't been defined (by the 
OP) on a regular grid.
It would be even more correct to say that there is *no* temporal resolution 
defined in this formulation.

- Is it not correct that a regular grid of samples in time is necessary for 
the Fourier Transform to be periodic?  You can pick whatever irreducible 
temporal resolution you like and whatever number of nonzero temporal samples 
you like - it's the temporal resolution that is tied to the frequency 
periodicity.

- Is it not correct that the Fourier Transform is of infinite extent if 
there is *no* temporal period?  For example, what if two samples are 
separated by 1 time unit and two samples are separated by pi time units. 
Then there is no common temporal division possible (in theory).  In that 
case, the Fourier Transform is of infinite extent, no?

- Is it not correct that a number of samples selected out of an infinite 
Fourier Transform (or computed as distinct points from the time domain) do 
not, in general, sufficiently define the frequency domain to support 
computing an inverse transform?

Fred

Fred Marshall wrote:

(snip of discussion of Fourier transform with non-uniform
spaced sample points)

> First:

>>"My claim for a finite duration
>>series is that it can be assumed periodic with a period greater
>>than or equal to the duration without loss of information."

> This was dealt with later..... at a point where it seemed to fit into the 
> discussion.

> Second:

>>>F(n*fs/N) = sum k=1 to N a_sub_k*exp[-j*(nfs/N)*t_sub_k]  ;n=0,N-1  !!!

>>>The problem with this formulation is that the temporal resolution hasn't 
>>>been defined on a regular grid and that's necessary for the Fourier 
>>>Transform to be periodic instead of just infinite.  If it's infinite then 
>>>this formulation can't be inverse transformed because it's incomplete - 
>>>not all the information is there.

>>"I am not convinced."

> You are not convinced about what aspect of the assertion?

> - Is it not correct that the temporal resolution hasn't been defined
>  (by the OP) on a regular grid.
> It would be even more correct to say that there is *no* temporal resolution 
> defined in this formulation.

More often discussed here is sampled data with non-uniform sampling,
and the effect on the Nyquist frequency.

In the case of the Fourier transform the boundary conditions
are periodic by definition.  For the given problem, then, the
question should be what are the amplitudes of the appropriate
frequency components.

If you agree that the boundary conditions are periodic, then the
temporal resolution pretty much depends on the period and the
number of sample points, assuming the transform isn't singular.
(You can't have two sample points at the same time, for example.)

> - Is it not correct that a regular grid of samples in time is necessary for 
> the Fourier Transform to be periodic?  You can pick whatever irreducible 
> temporal resolution you like and whatever number of nonzero temporal samples 
> you like - it's the temporal resolution that is tied to the frequency 
> periodicity.

Resolution comes from the period and number of samples.  If there
is noise in the signal, it is better to have equally spaced points.

> - Is it not correct that the Fourier Transform is of infinite extent if 
> there is *no* temporal period?  For example, what if two samples are 
> separated by 1 time unit and two samples are separated by pi time units. 
> Then there is no common temporal division possible (in theory).  In that 
> case, the Fourier Transform is of infinite extent, no?

Maybe it shouldn't be called a Fourier transform anymore, but consider
sample points spaced [ 1, pi, 1, pi, 1, pi , etc.]  A signal could
have a period of pi+1.   As for the OP's question, any period
longer than the extent of the data should work.

> - Is it not correct that a number of samples selected out of an infinite 
> Fourier Transform (or computed as distinct points from the time domain) do 
> not, in general, sufficiently define the frequency domain to support 
> computing an inverse transform?

Well, this gets back to sampling theory itself.  If it can be 
represented with a finite number of uniformly space points, it should
also work with a finite number of non-uniformly spaced points.

I don't believe that the OP's transform is correct, though.

Consider the N point DFT as a NxN matrix that when multiplied by
an N point vector generates an N point transform.  The inverse
of the transform is the inverse of the matrix.  For the usual
DFT, the matrix and its inverse are simple to compute.

For non-uniform points, one desires the transform matrix that
is the inverse of the transform that, given the sample points and 
component amplitudes gives the sample data.  The two are not
simply related the way the DFT and IDFT are.

-- glen

glen herrmannsfeldt wrote:
> Fred Marshall wrote:

   ...

>> - Is it not correct that a number of samples selected out of an 
>> infinite Fourier Transform (or computed as distinct points from the 
>> time domain) do not, in general, sufficiently define the frequency 
>> domain to support computing an inverse transform?
> 
> Well, this gets back to sampling theory itself.  If it can be 
> represented with a finite number of uniformly space points, it should
> also work with a finite number of non-uniformly spaced points.

Surely not *any* amount of non-uniformity. This gets back to a previous 
discussion of this topic. Speech uniformly sampled for a minute at 8 KHz 
amounts to 480,000 samples; that many are generated in less than a 
second by a .5 MHz sampler. Does anyone suppose that a whole minute of 
speech can be reproduced from second's worth of samples? With enough 
processing power and ram, one could then reproduce an hour's worth of 
information from samples taken in the first minute. AMEX, here I come!

Jerry
-- 
Engineering is the art of making what you want from things you can get.
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