SNR calculation using FFT

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

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?

1. You don't specify if you know the tone frequency or
    have to estimate it
2. The tone frequency has both positive and negative
   freqency fft components
3. What if the tone frequency is between fft bin frequencies?
4. SNR corresponds to the ratio of AVERAGE powers, not total power.

Hope this helps.

Greg

On Jan 26, 1:31&#4294967295;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.

>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.

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



    

On Jan 26, 8:50&#4294967295;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

>
>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.

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.
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>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.

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