On Jun 28, 4:48 am, "Vista" <a...@gmai.com> wrote:
> But I can take more samples on the F(v) and refine FFT in
> order to zoom in ...

If you sample a wider bandwidth of F(w), still sampling
at above the Nyquist rate for F(w), and that new bandwidth
portion is non-zero, then more high frequency detail will
appear in your time domain signal f(t).

It's like turning the Treble control on your stereo back up.


IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

<snip, terminolgy?...

Maybe the OP is trying to describe Zoom-FFT?  Maybe not, but worth a
punt.  This sort of well-known stuff: http://www.numerix-dsp.com/zoomfft.html

But I can take more samples on the F(v) and refine FFT in order to zoom in , 
am I right?

"Srikanth" <skt@xdtech.com> wrote in message 
news:1183005029.716692.222720@i13g2000prf.googlegroups.com...
> As mentioned above, you are doing a band pass. The f_hat as a result
> of a band pass need not 'look' much like f(t), it depends on how wide
> your band is - make it tooo narrow, and all you will have is one
> sinusoidal signal. Make it very wide and you get something that looks
> more like f(t). This does not give you a higher 'resolution' in time.
> Assuming you have the same number of samples (since in FFT, IFFT, time
> doesn't really make sense. It is always in terms of samples), your
> f_hat will have the same length as f i.e. a=c, b=d.
> You can't better the resolution actually, since you already got your
> samples, as mentioned above. you only have information of f(t) at the
> samples. information about f(t) at every other instant of time is
> lost. If you want to interpolate, there is no need to go to the
> frequency domain
>
> HTH
> Srikanth
> 

On Jun 27, 8:44 pm, "Vista" <a...@gmai.com> wrote:
> When I truncate/extract out F(w) for w in [-B, B] and use step
> size deltaB to sample it and then do IFFT, what is the portion
> of f(t) I see?  Say f(t) for t in [a, b]. What are a and b?

-inf to +inf, folded up into the width of your IFFT.

If you multiply by a rectangular [-B, B] filter response in the
frequency domain, that is the same as convolution with a Sinc
function in the time domain.  The narrower the rectangle [-B, B],
the wider the main lobe of your Sinc.  The wider the main lobe
of your Sinc, the "blurrier" and more spread out the result.
A finite width IFFT will also add the wrapped tails of the Sinc
convolution to your result, since a Sinc has infinite extent,
and the tails of the periodic images of the Sinc will also be
convolved into your IFFT aperture as if they were wrapped
around or folded.

As mentioned by a previous poster, Oli, if f(t) is sufficiently
time-limited, then F(w) might be smooth enough to be sampled
without aliasing given a small enough deltaB.  If not...

> Now suppose I find there is some fine structure in [c, d],
> which is shown from the visual display of f_hat(t), t in
> [a, b]. And a<c<d<b.

Given that [a, b] is -inf to +inf, then this will be true
for any [c, d]

> How to do IFFT targeting at f(t) on [c, d] with higher resolution?

You can do a sparse matrix DFT using a wider [-B, B] rectangular
window in the frequency domain.  But you can't increase deltaB
past some limit without aliasing.

Posted to comp.dsp only.

IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

As mentioned above, you are doing a band pass. The f_hat as a result
of a band pass need not 'look' much like f(t), it depends on how wide
your band is - make it tooo narrow, and all you will have is one
sinusoidal signal. Make it very wide and you get something that looks
more like f(t). This does not give you a higher 'resolution' in time.
Assuming you have the same number of samples (since in FFT, IFFT, time
doesn't really make sense. It is always in terms of samples), your
f_hat will have the same length as f i.e. a=c, b=d.
You can't better the resolution actually, since you already got your
samples, as mentioned above. you only have information of f(t) at the
samples. information about f(t) at every other instant of time is
lost. If you want to interpolate, there is no need to go to the
frequency domain

HTH
Srikanth

Hi folks,

This is not a joke post. I really have this question in my mind for long 
time.

When I truncate/extract out F(w) for w in [-B, B] and use step size deltaB 
to sample it and then do IFFT, what is the portion of f(t) I see? Say f(t) 
for t in [a, b]. What are a and b?

Now suppose I find there is some fine structure in [c, d], which is shown 
from the visual display of f_hat(t), t in [a, b]. And a<c<d<b.

How to do IFFT targeting at f(t) on [c, d] with higher resolution?

And so on and so forth?

Thanks!


"Vista" <abc@gmai.com> wrote in message 
news:f5sk5g$qmm$1@news.Stanford.EDU...
> Hi all,
>
> Suppose I have a signal f(t), t is in [0, +infinity).
>
> And I have its spectrum F(w).
>
> Let's say I found out that its main spectrum has 99.9% in [-B, B].
>
> So I truncate/extract out the portion of F(w), for w in [-B, B], and 
> discretized the interval into small grids with step size deltaB.
>
> And I then do the inverse FFT on the above samples of F(w), let's call the 
> inverse FFT reconstruction f_hat.
>
> Which part of f(t) does this inverse FFT f_hat represent?
>
> If I want to zoom into a certain part of f(t), how can I use inverse FFT 
> to do that?
>
> Let's say I only need [a, b] where 0<a<b< infinity, in the time domain,
>
> I only need to visualize the f(t), for t in [a, b],
>
> how do I sample F(w) and do the inverse FFT?
>
> -------------------------
>
> More generally, I am actually thinking of desing a "spectrum analyzer" 
> with zoom-in feactures:
>
> as you can see, first do a coarse level inverse FFT to gain a big picture 
> of the f(t) curve, and then allow user to zoom into a particular part of 
> the f(t) curve and display it with higher resolution of IFFT.
>
> How to do all of these?
>
> Thanks a lot!
> 

Hi folks,

This is not a joke post. I really have this question in my mind for long 
time.

When I truncate/extract out F(w) for w in [-B, B] and use step size deltaB 
to sample it and then do IFFT, what is the portion of f(t) I see? Say f(t) 
for t in [a, b]. What are a and b?

Now suppose I find there is some fine structure in [c, d], which is shown 
from the visual display of f_hat(t), t in [a, b]. And a<c<d<b.

How to do IFFT targeting at f(t) on [c, d] with higher resolution?

And so on and so forth?

Thanks!


"Vista" <abc@gmai.com> wrote in message 
news:f5sk5g$qmm$1@news.Stanford.EDU...
> Hi all,
>
> Suppose I have a signal f(t), t is in [0, +infinity).
>
> And I have its spectrum F(w).
>
> Let's say I found out that its main spectrum has 99.9% in [-B, B].
>
> So I truncate/extract out the portion of F(w), for w in [-B, B], and 
> discretized the interval into small grids with step size deltaB.
>
> And I then do the inverse FFT on the above samples of F(w), let's call the 
> inverse FFT reconstruction f_hat.
>
> Which part of f(t) does this inverse FFT f_hat represent?
>
> If I want to zoom into a certain part of f(t), how can I use inverse FFT 
> to do that?
>
> Let's say I only need [a, b] where 0<a<b< infinity, in the time domain,
>
> I only need to visualize the f(t), for t in [a, b],
>
> how do I sample F(w) and do the inverse FFT?
>
> -------------------------
>
> More generally, I am actually thinking of desing a "spectrum analyzer" 
> with zoom-in feactures:
>
> as you can see, first do a coarse level inverse FFT to gain a big picture 
> of the f(t) curve, and then allow user to zoom into a particular part of 
> the f(t) curve and display it with higher resolution of IFFT.
>
> How to do all of these?
>
> Thanks a lot!
> 

On Jun 26, 11:06 pm, "Vista" <a...@gmai.com> wrote:
>
> Suppose I have a signal f(t), t is in [0, +infinity).
>
> And I have its spectrum F(w).
>
> Let's say I found out that its main spectrum has 99.9% in [-B, B].
>
> So I truncate/extract out the portion of F(w), for w in [-B, B], and
> discretized the interval into small grids with step size deltaB.
>
> And I then do the inverse FFT on the above samples of F(w), let's call the
> inverse FFT reconstruction f_hat.
>
> Which part of f(t) does this inverse FFT f_hat represent?
>

It depends on what you mean by "focus".  If you mean "set all the
unwanted spectral components to zero" and then you IFFT the result,
f_hat(nT) represents a filtered version of the original f(nT) over the
entire interval represented by the samples of f(nT).

If, by "focus", you mean "discard the unwanted spectral components"
and then you IFFT the result (containing a smaller number of spectral
components), f_hat(nT_prime) represents a decimated and filtered
version of the original f(nT), still over the entire interval
represented by the samples of f(nT).

On Jun 27, 7:14 pm, Oli Charlesworth <c...@olifilth.co.uk> wrote:
> On Jun 27, 4:06 am, "Vista" <a...@gmai.com> wrote:
>
> > Hi all,
>
> > Suppose I have a signal f(t), t is in [0, +infinity).
>
> > And I have its spectrum F(w).
>
> > Let's say I found out that its main spectrum has 99.9% in [-B, B].
>
> > So I truncate/extract out the portion of F(w), for w in [-B, B], and
> > discretized the interval into small grids with step size deltaB.
>
> > And I then do the inverse FFT on the above samples of F(w), let's call the
> > inverse FFT reconstruction f_hat.
>
> > Which part of f(t) does this inverse FFT f_hat represent?
>
> Just as sampling a non-bandlimited function in the time domain causes
> time-domain aliasing, the dual occurs in your scenario.  You are
> sampling a non-time-limited function in the frequency domain, which
> will cause frequency-domain aliasing.
>
> --
> Oli

It's very interesting.
We've all got different ideas about what Vista is trying to do.
The only thing we all agree on is that he/she is going about it the
wrong way.
Perhaps Vista is a troll, deliberately posing an inane question, then
sitting back to watch the fun.

On Jun 27, 4:06 am, "Vista" <a...@gmai.com> wrote:
> Hi all,
>
> Suppose I have a signal f(t), t is in [0, +infinity).
>
> And I have its spectrum F(w).
>
> Let's say I found out that its main spectrum has 99.9% in [-B, B].
>
> So I truncate/extract out the portion of F(w), for w in [-B, B], and
> discretized the interval into small grids with step size deltaB.
>
> And I then do the inverse FFT on the above samples of F(w), let's call the
> inverse FFT reconstruction f_hat.
>
> Which part of f(t) does this inverse FFT f_hat represent?

Just as sampling a non-bandlimited function in the time domain causes
time-domain aliasing, the dual occurs in your scenario.  You are
sampling a non-time-limited function in the frequency domain, which
will cause frequency-domain aliasing.


--
Oli