SNR calculation using FFT

chengnoon <chengnoon@gmail.com> wrote:
(snip)
 
> But the signal considered here is the ideal single tone frequency, for
> example single frequency signal A*cos(2*pi*fc), and it only have one single
> bin regardless the fft length.

Only if the frequency is an integer multiple of the reciprocal of
the transform length.  Otherwise you have to approximate it as
filling up some neighbor bins.

-- glen

chengnoon wrote:
>> chengnoon wrote:
>>>> No.
>>>>
>>>> You have to use the ratio of the average signal power to the average
>>>> noise power. The average noise power is relatively independent of N.
>>>>
>>>> Hope this helps.
>>>>
>>>> Greg
>>>>
>>> I donot think so. SNR is defined as the ratio of a signal power to the
>>> noise power corrupting the signal. And, the noise power is refer to
> the
>>> total noise power, not the average noise power, in desired bandwidth.
>> I think you misunderstand, probably because the explanation wasn't 
>> complete enough. The bins have dimension. The area is amplitude times 
>> bandwidth. As more FFT points are added, the bandwidth of a bin 
>> decreases. The bin power is bin amplitude times bin bandwidth. The total
> 
>> noise power is noise amplitude times total bandwidth. The total 
>> bandwidth is independent of number of bins. Can you take it from here?
>>
>> Jerry
> 
> 
> But the signal considered here is the ideal single tone frequency, for
> example single frequency signal A*cos(2*pi*fc), and it only have one single
> bin regardless the fft length.

Right. That's why you accept it as it comes. The noise, on the other 
hand is distributed on all the bins, that's why, in order to get its 
power across the band, you average the bin magnitudes.

I think that strictly, you should average the squares. If the variation 
from bin to bin isn't large, it doesn't matter much.

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