Re: How to smooth FFT data on logarithmic frequency bins (octave bands)?

[Matthias Brennwald]

Am 0077.2015 00:14 schrieb "Juan Pablo Carbajal" <address@hidden>:
> >
> > On Fri, Jul 3, 2015 at 11:20 PM, Matthias Brennwald
> > <address@hidden> wrote:
> > > Hi all
> > >
> > > I calculated the frequency response of a loudspeaker from the impulse response. Now I'd like to smooth the response, e.g. in a 1/24 octave band (octave, not Octave). I do this by defining the frequency bins (e.g., 1/24 octave bins) and then loop through these bins, find the data in the frequency range of each bin, and calculate the means of each bin (see code example below). This (i) slow and (ii) not elegant.
> > >
> > > Does someone have an idea on how to improve this? Is there a way to transform the data such that the loop can be avoided?
> > >
> > > In addition, it would be nice to use a more elaborate method than just calculating the means of each bin. For instance, something similar to a Savitzky-Golay smoothing would be nice.
> > >
> > > Thanks for your help!
> > >
> > > Matthias
> > >
> > > =============================
> > > Code example of my current approach:
> > > =============================
> > >
> > > % raw frequency response (before smoothing):
> > > f = 10:10:40000;                        % frequency values of loudspeaker frequency response
> > > Y = 10 + rand(size(f));         % frequency response (absolute values of Fourier transporm)
> > >
> > > % 1/24 octave smoothing:
> > > sm = 1/24;                                                              % smoothing band width (fraction of an octave)
> > > n  = floor ( log2(f(end)/f(1)) / sm );  % number of frequency bins for avaraging
> > > YY = repmat (NaN,1,n); ff = repmat (NaN,size(YY)); % initialize vectors that will contain the smoothed data
> > > f1 = f(1);      % lower end of first frequency bin
> > > for i = 1:n     % loop through frequency bins and calculate means of each bin
> > >         f2 = f1*2^sm;   % upper end of frequency bin
> > >         ff(i) = sqrt(f1*f2); % center frequency of current bin
> > >         j = find (f >= f1 & f < f2); % find data in frequency bin f1...f2
> > >         if any (j) % compute mean
> > >                 YY(i) = mean (Y(j));
> > >         end
> > >         f1 = f2; % prepare for next iteration
> > > end
> > >
> > > % remove missing / NaN entries:
> > > i = find (~isnan(YY));
> > > ff = ff(i); YY = YY(i);
> > >
> > > plot (f,Y,'k-','linewidth',1 , ff,YY,'r-','linewidth',5);
> > > legend ('Raw','Smoothed 1/24 octave’)
> > >
> > > =============================
> > >
> > > _______________________________________________
> > > Help-octave mailing list
> > > address@hidden
> > > https://lists.gnu.org/mailman/listinfo/help-octave
> >
> > The signal package has sgolayfilt, there is also the smooth function.

I know. But that does not solve my question related to the logarithmic distribution of the data when applying octave-band smoothing.

Matthias