>> 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.
�����������������������������������������������������������������������
Reply by glen herrmannsfeldt●January 27, 20102010-01-27
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
Reply by glen herrmannsfeldt●January 27, 20102010-01-27
Jerry Avins <jya@ieee.org> wrote:
(snip)
> 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?
Well, for a truly periodic signal (unlikely in this case) the
bins are delta functions. They still have area, though.
(Zero width and finite area.) For non-periodic signals you
approximate the bins as a histogram would come out.
-- glen
Reply by chengnoon●January 27, 20102010-01-27
>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.
Thanks.
Reply by Jerry Avins●January 27, 20102010-01-27
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
--
Engineering is the art of making what you want from things you can get.
�����������������������������������������������������������������������
Reply by chengnoon●January 27, 20102010-01-27
>
>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.
Reply by Greg Heath●January 27, 20102010-01-27
On Jan 26, 8:50�pm, "chengnoon" <chengn...@gmail.com> wrote:
> >The correct expression for SNR would normalize the denominator by the
> >number of bins.
>
> >So,
> >SNR =3D 10*log10((N_FFT-1)*|disired single bin magnitude|^2/sum(|all
> >bins magnitude other than desired|^2))
>
> >Regards,
> >Dilip.
>
> May be I am not present the question clearly.
> Regardless of the fft lenght difference, the ratio
> |desired signal single bin magnitude|/|any other signal single bin
> magnitude|
> is a constant.
>
> So, the SNR calculation should be like this:
> SNR = 10*log10(A)
>
> where, A = M_sig/(M_1+M_2+M_3+...+M_N)
> and N is the fft length.
>
> As the fact that M_sig/M_n is a constant regardless of the fft length N.
>
> So, we get the SNR which depends on the fft length N, i.e., the longer fft
> length, the worse SNR.
>
> This is question.
>
> Thanks a lot.
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
Reply by bharat pathak●January 26, 20102010-01-26
Cheng,
The SNR calculation using FFT method does not depend on the
number of points (assuming it to be large to begin with).
Couple of things that needs to be considered while computing
FSNR (I call it Frequency domain SNR Calc, time domain would
be called TSNR).
1. Windowing of data prior to feeding it to FFT. This helps
in reducing spectral leakage to a large extent. I have
used blackman-harris window in my routine.
2. Due to windowing effect, the peak of sinetone will not be
seen at -6dB, hence a small correction is needed if you
wish to see the peak of sine-tone at -6db. This is just
like applying a scaling factor. This is just for visual
purpose.
3. Due to windowing and the input sine wave not being integral
multiple, there will be spread in peak, (it will not be
available at single point in FFT spectrum). Hence to compute
signal power we need to use couple of bins adjacent to the
peak point to consider signal power. This spread is completely
dependent on the type of window used and to be more precise
on the main lobe width of the window.
4. I have tested my routine for quite few cases and it seems
to give consistent results.
Regards
Bharat Pathak
Reply by chengnoon●January 26, 20102010-01-26
>The correct expression for SNR would normalize the denominator by the
>number of bins.
>
>So,
>SNR =3D 10*log10((N_FFT-1)*|disired single bin magnitude|^2/sum(|all
>bins magnitude other than desired|^2))
>
>Regards,
>Dilip.
>
May be I am not present the question clearly.
Regardless of the fft lenght difference, the ratio
|desired signal single bin magnitude|/|any other signal single bin
magnitude|
is a constant.
So, the SNR calculation should be like this:
SNR = 10*log10(A)
where, A = M_sig/(M_1+M_2+M_3+...+M_N)
and N is the fft length.
As the fact that M_sig/M_n is a constant regardless of the fft length N.
So, we get the SNR which depends on the fft length N, i.e., the longer fft
length, the worse SNR.
This is question.
Thanks a lot.
Reply by Dilip Warrier●January 26, 20102010-01-26
On Jan 26, 1:31�am, "chengnoon" <chengn...@gmail.com> wrote:
> Hi all,
>
> Currently, I use FFT to calculate the input single tone signal SNR and
> have a question about it posting below:
>
> SNR = 10*log10(|disired single bin magnitude|^2/sum(|all bins magnitude
> other than desired|^2))
>
> 1. if I set N1 as the fft number at 1st time, so the number of elements in
> sum() @ denominator is N1;
> 2. if I set N2 as the fft number at 2nd time, and the number of elementsin
> sum() @ denominator is N2;
>
> and in every calculation, the ratio (|disired single bin magnitude|^2/|any
> bin magnitude other than desired| is roughtly a constant. So, through this
> dedection, we got the conclusion that: the larger FFT number, the worse SNR
> we get. And, this is contradict with the theory!!!
>
> What's wrong with my SNR calculation?
>
> Thanks,
> chengnoon
The correct expression for SNR would normalize the denominator by the
number of bins.
So,
SNR = 10*log10((N_FFT-1)*|disired single bin magnitude|^2/sum(|all
bins magnitude other than desired|^2))
Regards,
Dilip.