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Re: [Phys-l] some questions related to sampling

On 01/16/2009 01:10 PM, Stefan Jeglinski wrote:

My take is this. First, a "conventional approach" (N_f = N_t = N) is
straightforward. N samples "maps" into N frequencies and vice versa
with the inverse transform. I like your term "barely invertible."


If we then increase the resolution in frequency-space (f-space) by
increasing N_f as discussed, it might be assumed that we can achieve
a higher resolution result in the time domain when inverse
transforming back to it.

Not greater resolution, but rather greater _span_ in the
time domain.

But it doesn't work like that. We can't get
back the original waveform. Rather, we might I suppose (?), but there
is additional content created in the time domain, in particular
imaginary content, whereas the original time domain contained only
real data. Not good.

Maybe not good, but not terrible, either. The "additional"
content is all zeros.

And the way I look at it, the DFT was already guilty of
creating "additional" content, in the sense that it always
assumes that the input signal is periodic, i.e. a periodic
continuation of the time-domain signal is assumed to exist.

The increased-resolution transform is no worse; it just
assumes a _different_ continuation of the time domain
data. It continues it with zeros for a while, and then
continues periodically (with a longer period).

Two points:

1. If we abandon trying to write the transforms as summations, and
instead move to a matrix formulation, it is easy to move back and
forth in the following sense:

F = E . T


Tangential remark: If we are writing T as a column vector
and F as a row vector, note that E is _not_ the usual sort
of matrix that takes in a vector and puts out a vector of
the same kind. If we use what Misner, Thorn, & Wheeler
call "index gymnastics", T has an upstairs index, F, has
a downstairs index, and E has two downstairs indices, unlike
the usual sort of matrices that have one of each. Naturally
the inverse transform has two upstairs indices.

Here, T is the vector of samples in the time domain, and F is the
vector of frequencies (meaning the vector of transformed values in
f-space, not the frequencies themselves - words fail).

F is the vector of ordinates in the frequency domain.

E is the
square matrix, constructed of the exponential terms that appear in
the summations. For N_f = N_t = N, this is just the equation for the
discrete fourier transform, and its inverse is T = E^-1 . F, where
E^-1 is just the conventional inverse of a square matrix.

Now, increasing the f-space resolution amounts to enlarging the
number of terms in vector F, and the number of rows in E,

This is a question of taste, of the sort that ought not be
argued, but just to avoid confusion let me say that I have
been thinking of F as a _row_ vector and therefore here we
are increasing the number of _columns_ in F and by the same
token increasing the number of _columns_ in E (not "rows
in E"). This is how my demo spreadsheet lays things out.

such that T
remains intact. We can then still write T = E^-1 . F, where E^-1 is
now the pseudoinverse of the non-square matrix E. By doing so, we
recover the original time-domain samples without any "additional
content," no matter how much we increase the resolution in f-space. I
am a bit concerned about whether the pseudoinverse might introduce
issues, but AFAICT in the practical examples I've tried, the
calculation seems to be above suspicion.

That's true.

There's nothing magical about it.

What we call "the" pseudoinverse is just one way among many of
inverting a non-square matrix.

Suppose the forward transform increases the resolution by a
factor of 4, i.e. Nf/Nt = 4. Then the non-square forward
transform E consists of 4 fairly ordinary square transforms
(with a little bit of heterodyning). They are not stacked
side-by-side but rather collated i.e. intercalated. Any
one of the four is easily invertible. You have to take into
account the fact that three of the four have nonstandard
abscissas, but we know how to do that.

The key requirement is that the inverse transform not
produce too many rows. The pseudoinverse doesn't magically
accomplish that; it only does it because you instructed
it to do so, as part of the definition of pseudoinverse.
Any other row-reduction strategy would have worked equally
well, including simple subsampling of the frequency-domain

2. There remains the question of increasing the resolution in the
time domain (t-space). Well, if you have sampled at higher than the
Nyquist rate, and confirmed this by looking at the transform and
seeing no overlapped aliases, the sampling theorem teaches that you
can recover the time domain at any resolution desired, via the
interpolation formula involving the sinc function. No information
from the frequency domain (other than to confirm a lack of alias
overlap) is needed.

With respect to these 2 points, I do not see a way to perform a DFT,
increase the resolution in f-space, and then do an IDFT that achieves
a resolution in t-space that is higher than that given by the
original samples. But I would be interested in such an approach. We
are in fact inching toward the real question I want to ask, but I
want to first see what comments there might be about my above

If the time-domain data is the original data, no amount
of Fourier transforms or other math will ever create more
information. There will be no "loaves and fishes" miracle
where you create more just by rearranging things. The
second law of thermodynamics forbids it.

So all we are talking about here are various heuristics for
_interpolating_ between points in the time domain.

Interpolation is easy if you know the original signal was
band-limited before it was sampled ... i.e. no aliasing.

You cannot safely decide whether a signal is band-limited
or not by looking at the sampled data! That would be like
asking the drunkard whether he is drunk. For example, if
I have a 100.01 Hz wave and sample it at 10.00 Hz, it will
look like a beautiful 0.01 Hz signal. Ooops.

If you want to be sure that the original data is band-limited,
rely on the physics, not on the math. That is, put a filter
on it! A real, physical, analog filter. Then you can be

And that BTW answers the question about interpolation. The
analog filter is in control. The details of the design of
the analog filter dictate the details of the interpolation.