On Mon, 07 Nov 2011 17:48:51 -0800, Carnë Draug <carandraug+...@gmail.com> wrote:
> On 7 November 2011 22:42, Jeffrey Cunningham <jeff...@jkcunningham.com> >> >> Are you sure it is a bug? I am assuming that you want the filter to be >> normalized the same way as Matlab's fir1 function, and they say: >> >> "By default, the filter is scaled so that the center of the first >> passband >> has a magnitude of exactly 1 after windowing." >> >> In other words, they are normalizing so the *frequency domain* passband >> comes out to 1. The d.c. output level divided by the bandwidth will be >> (roughly) 1. > > I asked in ##matlab and it seems it returns exactly 1 there. > > Carnė Right. I don't know what I was thinking of: The dc level is the sum of the weights. But I'm not seeing a problem. Here's a low-pass fir1 filter: b = fir1(48,[0,0.35]); sum(b) => 1.005 freqz shows unity gain in the passband. And I can drive it with a unit gain d.c. voltage and after it stops ringing I get unity output. c=filter(b,1,ones(520,1))(end) ==> 1.005 I think I see what his trouble is. He doesn't have nearly enough coefficients for a low-pass filter that narrow. Its a bad fit to an impulse-function. octave:17> version ans = 3.2.4 octave:19> sum(fir1(240,0.01)) ans = 1.1822 octave:20> sum(fir1(512,0.01)) ans = 1.0804 octave:21> sum(fir1(10000,0.01)) ans = 0.99934 Jeff ------------------------------------------------------------------------------ RSA(R) Conference 2012 Save $700 by Nov 18 Register now http://p.sf.net/sfu/rsa-sfdev2dev1 _______________________________________________ Octave-dev mailing list Octave-dev@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/octave-dev
