Calculation FFT...

Let's assume at us is available 1024 discrete samples. Further it is
carried out FFT. Whether probably from received 1024 spectral samples to
receive 256 spectral samples which are equivalent as if we have executed
FFT only on 256 input discrete samples?

On 5/19/2010 2:39 PM, alex65111 wrote:
> Let's assume at us is available 1024 discrete samples. Further it is
> carried out FFT. Whether probably from received 1024 spectral samples to
> receive 256 spectral samples which are equivalent as if we have executed
> FFT only on 256 input discrete samples?

I'm sorry. I can decode the first two sentences, but not the last one. I 
simply can't imagine what you are trying to ask.

Jerry
-- 
"I view the progress of science as ... the slow erosion of the tendency
  to dichotomize." --Barbara Smuts, U. Mich.
�����������������������������������������������������������������������

>On 5/19/2010 2:39 PM, alex65111 wrote:
>> Let's assume at us is available 1024 discrete samples. Further it is
>> carried out FFT. Whether probably from received 1024 spectral samples
to
>> receive 256 spectral samples which are equivalent as if we have
executed
>> FFT only on 256 input discrete samples?
>
>I'm sorry. I can decode the first two sentences, but not the last one. I 
>simply can't imagine what you are trying to ask.
>
>Jerry
>-- 
>"I view the progress of science as ... the slow erosion of the tendency
>  to dichotomize." --Barbara Smuts, U. Mich.
>???????????????????????????????????????????????????????????????????????
>


Let's assume there are 1024 samples X = [X1...X1024].
Y=FFT(X), Y =[Y1...Y1024].
It is required from Y to receive Z, where Z=FFT(X1...X256) 
or 
Z=FFT((X1...X256 + X257...X512 + X513...X768 + X769...X1024)/4).

On 5/19/2010 3:42 PM, alex65111 wrote:
>> On 5/19/2010 2:39 PM, alex65111 wrote:
>>> Let's assume at us is available 1024 discrete samples. Further it is
>>> carried out FFT. Whether probably from received 1024 spectral samples
> to
>>> receive 256 spectral samples which are equivalent as if we have
> executed
>>> FFT only on 256 input discrete samples?
>>
>> I'm sorry. I can decode the first two sentences, but not the last one. I
>> simply can't imagine what you are trying to ask.
>>
>> Jerry
>> --
>> "I view the progress of science as ... the slow erosion of the tendency
>>   to dichotomize." --Barbara Smuts, U. Mich.
>> ???????????????????????????????????????????????????????????????????????
>>
>
>
> Let's assume there are 1024 samples X = [X1...X1024].
> Y=FFT(X), Y =[Y1...Y1024].
> It is required from Y to receive Z, where Z=FFT(X1...X256)
> or
> Z=FFT((X1...X256 + X257...X512 + X513...X768 + X769...X1024)/4).

Reconstitute the original samples with IFFT(Y) and go from there.

Jerry
-- 
"I view the progress of science as ... the slow erosion of the tendency
  to dichotomize." --Barbara Smuts, U. Mich.
�����������������������������������������������������������������������

>Reconstitute the original samples with IFFT(Y) and go from there.
>
>Jerry
>-- 
>"I view the progress of science as ... the slow erosion of the tendency
>  to dichotomize." --Barbara Smuts, U. Mich.
>???????????????????????????????????????????????????????????????????????
>

The problem to receive result without reconstruction, operating only with
spectral samples.

>Reconstitute the original samples with IFFT(Y) and go from there.
>
>Jerry
>-- 
>"I view the progress of science as ... the slow erosion of the tendency
>  to dichotomize." --Barbara Smuts, U. Mich.
>???????????????????????????????????????????????????????????????????????
>

The problem to receive result without reconstruction, operating only with
spectral samples.

It turns out so that at the decision of one stage of a problem to have to
count FFT the large size, and at the second stage it is required on the
same data FFT of the smaller size. Accordingly it would be desirable to
count not anew FFT, and to use result from the first stage.

On 5/19/2010 4:16 PM, alex65111 wrote:
> It turns out so that at the decision of one stage of a problem to have to
> count FFT the large size, and at the second stage it is required on the
> same data FFT of the smaller size. Accordingly it would be desirable to
> count not anew FFT, and to use result from the first stage.

As far as I know, it doesn't work that way.

Jerry
-- 
"I view the progress of science as ... the slow erosion of the tendency
  to dichotomize." --Barbara Smuts, U. Mich.
�����������������������������������������������������������������������

On May 19, 12:42&#4294967295;pm, "alex65111" <alex65111@n_o_s_p_a_m.list.ru>
wrote:
>...
>
> Let's assume there are 1024 samples X = [X1...X1024].
> Y=FFT(X), Y =[Y1...Y1024].
> It is required from Y to receive Z, where Z=FFT(X1...X256)
> or
> Z=FFT((X1...X256 + X257...X512 + X513...X768 + X769...X1024)/4).

It is not clear to me what you mean by the last line. If the meaning
is:

Z = (FFT(x1 ... X256) + FFT(X257 ... X512) + FFT(X513 ... X768) +
FFT(X769 ... X1024)) / 4

then you can get that result from (Y1, Y5, Y9, ... Y1021) / 4. This is
the result you get from coherently averaging the FFTs of contiguous
blocks of rectangularly windowed FFTs.

If you are looking for something with "wider" bins you need to do
something else.

If you are only interested in power, you can form sums of the powers
of adjacent bins.

The more you can narrow down the application of the 256 values, the
more likely there is to be a simple solution or approximation. So, if
you want something else, what are you going to do with it?

Dale B. Dalrymple

>If you are looking for something with "wider" bins you need to do
>something else.
>
>If you are only interested in power, you can form sums of the powers
>of adjacent bins.
>
>The more you can narrow down the application of the 256 values, the
>more likely there is to be a simple solution or approximation. So, if
>you want something else, what are you going to do with it?
>
>Dale B. Dalrymple

Yes, it is necessary to look with "wider" bins and only power.  
It seems that "sums of the powers of adjacent bins" this that that is
necessary.