Comparing FFT's of different resolutions

Assume a signal such that an FFT done today gives identical result to 
one done last year [sample rate and number of samples held constant].

Now do 2 FFTs on this signal.
One with   1 second worth of data ( ie resolution of   1 Hz )
One with .01 second worth of data ( ie resolution of 100 Hz )

How would they be related?
[I have a conjecture, but my description would muddy the issue.]

This group is teaching me that a carefully phrased question is half the 
battle ;)

Direction to suitable resource is acceptable.
  [u can tell me where to go ;]
Thank you.

Richard Owlett <rowlett@atlascomm.net> wrote in news:12i5akrmn1h2t82
@corp.supernews.com:

> Assume a signal such that an FFT done today gives identical result to 
> one done last year [sample rate and number of samples held constant].
> 
> Now do 2 FFTs on this signal.
> One with   1 second worth of data ( ie resolution of   1 Hz )
> One with .01 second worth of data ( ie resolution of 100 Hz )
> 
> How would they be related?
> [I have a conjecture, but my description would muddy the issue.]
> 
> This group is teaching me that a carefully phrased question is half the 
> battle ;)
> 
> Direction to suitable resource is acceptable.
>   [u can tell me where to go ;]
> Thank you.

By all means, throw in your thought process.  It shows that you've put 
some thought into it, and you're not merely trying for a homework answer.

You're missing some data-- "resolution" is being used incorrectly here.  
We need to know sample rate and the number of points collected, and your 
use of "resolution" is muddying things up.

-- 
Scott
Reverse name to reply

Scott Seidman wrote:

> Richard Owlett <rowlett@atlascomm.net> wrote in news:12i5akrmn1h2t82
> @corp.supernews.com:
> 
> 
>>Assume a signal such that an FFT done today gives identical result to 
>>one done last year [sample rate and number of samples held constant].
>>
>>Now do 2 FFTs on this signal.
>>One with   1 second worth of data ( ie resolution of   1 Hz )
>>One with .01 second worth of data ( ie resolution of 100 Hz )
>>
>>How would they be related?
>>[I have a conjecture, but my description would muddy the issue.]
>>
>>This group is teaching me that a carefully phrased question is half the 
>>battle ;)
>>
>>Direction to suitable resource is acceptable.
>>  [u can tell me where to go ;]
>>Thank you.
> 
> 
> By all means, throw in your thought process.  It shows that you've put 
> some thought into it, and you're not merely trying for a homework answer.


It ain't homework as I'm approaching retirement ;)
That you asked that, tells me I'm more confused than I thought.


> 
> You're missing some data-- "resolution" is being used incorrectly here.  
> We need to know sample rate and the number of points collected, and your 
> use of "resolution" is muddying things up.
> 

I'll conform to adage:

It is better to be silent & thought a fool, than speak and remove all doubt.

Richard Owlett <rowlett@atlascomm.net> wrote in
news:4522D645.9010304@atlascomm.net: 

> It is better to be silent & thought a fool, than speak and remove all
> doubt. 
> 

Not at all.  Think of the problem in terms of the sampling rate and the 
number of points collected.  Then, you can transform that into the 
resolution of the fft, what frequency range the fft can represent, and any 
aliasing considerations that might be important, and then you're there.

The fact that it sounds like homework merely means that you're asking 
questions of the type that might come up in a class, which would mean that 
they're central issues worthy of consideration.

-- 
Scott
Reverse name to reply

Richard wrote :

> Assume a signal such that an FFT done today gives identical result to 
> one done last year [sample rate and number of samples held constant].
> 
> Now do 2 FFTs on this signal.
> One with 1 second worth of data ( ie resolution of 1 Hz )
> One with .01 second worth of data ( ie resolution of 100 Hz )
> 
> How would they be related?

Firstly, let me say that it is unusual to do two FFTs with a ratio of 100
between them. For this it would be more customary to do a DFT. That aside,
let me try and answer your question based purely on the facts you present.

1. The data source is the same for the two transforms (i.e. the sample
rate is unchanged)
2. The input frames for the two transforms differ by factor 100 in
duration (and consequently the number of samples)

From this it can be deduced that the transform point-sizes will differ by
factor 100, therefore the spectral bin-bandwidths (resolution if you like)
will also vary by factor 100 (as you correctly state)

What will also change will be the processing gain afforded by the
transform(s) - the ability to distinguish signal from the noise floor of
the spectrum.

This will change by the factor 10*Log(Bandwidth ratio)dB = 20dB

Therefore, your 0.01 second data frame will produce a spectrum whose S/N
ratio has worsened by 20dB.

Jeff

Richard Owlett skrev:
> Assume a signal such that an FFT done today gives identical result to
> one done last year [sample rate and number of samples held constant].
>
> Now do 2 FFTs on this signal.
> One with   1 second worth of data ( ie resolution of   1 Hz )
> One with .01 second worth of data ( ie resolution of 100 Hz )
>
> How would they be related?
> [I have a conjecture, but my description would muddy the issue.]

Well... the finer spectrum would have all the sample locations as the
coarser one, and another 9 samples in between... but that might not
have been what you asked?

> This group is teaching me that a carefully phrased question is half the
> battle ;)

Yes. Maybe we have taught too well -- there isn't really much to go
on, here. 

Rune

Sparks wrote:

> Richard wrote :

>>Assume a signal such that an FFT done today gives identical result to 
>>one done last year [sample rate and number of samples held constant].

>>Now do 2 FFTs on this signal.
>>One with 1 second worth of data ( ie resolution of 1 Hz )
>>One with .01 second worth of data ( ie resolution of 100 Hz )

>>How would they be related?

> Firstly, let me say that it is unusual to do two FFTs with a ratio of 100
> between them. For this it would be more customary to do a DFT. 

FFT works by factoring the number of points, and doing sub-transforms
on those factors.  It is best if those factors are all 2, but they
don't have to be.  There are routines around that can do other factors.

 > That aside,
> let me try and answer your question based purely on the facts you present.

> 1. The data source is the same for the two transforms (i.e. the sample
> rate is unchanged)
> 2. The input frames for the two transforms differ by factor 100 in
> duration (and consequently the number of samples)

> From this it can be deduced that the transform point-sizes will differ by
> factor 100, therefore the spectral bin-bandwidths (resolution if you like)
> will also vary by factor 100 (as you correctly state)

Yes, I disagree with the previous post that said it wasn't resolution.
While it does make sense to give the sampling rate (or ratio of rates),
the resolution does come from the total length.

If the sampling rates are the same, I believe that every 100th sample
of the high resolution transform will equal the points of the low 
resolution transform.   If the sample rates are different, then that
changes by the sampling rate ratio.

-- glen

glen herrmannsfeldt wrote:
> If the sampling rates are the same, I believe that every 100th sample
> of the high resolution transform will equal the points of the low
> resolution transform.

False. It is trivial to come up with counter-examples to this. See my
post.

Steven

stevenj@alum.mit.edu skrev:
> glen herrmannsfeldt wrote:
> > If the sampling rates are the same, I believe that every 100th sample
> > of the high resolution transform will equal the points of the low
> > resolution transform.
>
> False. It is trivial to come up with counter-examples to this. See my
> post.
>
> Steven

Can't see any other posts of yours in this thread? At least not through

groups.google.com.

Rune

Richard Owlett wrote:
> Assume a signal such that an FFT done today gives identical result to
> one done last year [sample rate and number of samples held constant].
>
> Now do 2 FFTs on this signal.
> One with   1 second worth of data ( ie resolution of   1 Hz )
> One with .01 second worth of data ( ie resolution of 100 Hz )
>
> How would they be related?

Assuming the data overlap, the shorter spectrum is exactly given by
convolving the longer spectrum with a sinc function (corresponding to a
0.01s rectangular window) and then taking every 100th bin.  This is
*not* the same as every 100th bin of the longer spectrum, contrary to
what other posters have implied (except for very special data).

However, this convolution and decimation of the spectrum can lead to
very different results, depending on the data.

For example, if the first 0.01 seconds of the data are zero, and the
remaining 0.99 seconds are nonzero, then obviously the spectra in the
two cases will look completely different.

On the other hand, suppose that the data are random with some
time-invariant distribution and temporal correlation function.  In this
case, *on average* the longer transform will have similar mean shape
(albeit with 100 times the spectral resolution of the shorter
transform) and nearly the same variance ... but if you look at the
spectrum from a *single* longer transform then it should look 100 times
more oscillatory because the spectral bins are closer together and are
uncorrelated random variables, and the variance has not decreased.

With other data models you will get other answers.

Cordially,
Steven G. Johnson