Butterworth filter cut-off frequency

Hello,

i wrote this program:

fp=500;     %f trans
fs=2000;    %f stopband
ap=3;       %transition atten
as=22;      %stopband atten

nfp = fp/(FS1/2);
nfs = fs/(FS1/2);

[n,Wn] = buttord(nfp,nfs,ap,as) ;
[b,a] = butter(n,Wn);
[bt,at] = lp2lp(b,a,Wo)

fvtool(b,a)

And the characteristic is: http://i.imgur.com/UShdzql.png

But i would like it to go trough the blue point (exactly [fp,ap] point) instead of green and red - so the "critical point" is near the transition band.

Is there a easy way to modify the code to accomplish it?

I would also need it to work, if "ap" is not necessary 3dB, but e.g. 1dB.

[Star Strider]

Need ‘FS1’ and ‘Wo’.

I’m guessing FS1=1E+4 but I have no idea what ‘Wo’ may be.

[raf]

soryy, ignore lp2lp() line

here is corrected code

clc;clear,close all;

FS1=10000;

fp=500;     %f trans
fs=2000;    %f stopband
ap=3;       %transition atten
as=22;      %stopband atten

nfp = fp/(FS1/2);
nfs = fs/(FS1/2);

[n,Wn] = buttord(nfp,nfs,ap,as) ;
[b,a] = butter(n,Wn);

fvtool(b,a)

Answer 1

[Star Strider]

It may not be possible to design a Butterworth filter with that much precision. You have to make some tradeoffs to get about 3 db attenuation at 500 Hz. Consider Chebychev and fir filters as well, if you need more definite passbands and stopbands.

Changing to these selected parameters works about as well as anything else I’ve tried:

fp = 230;
n  =   3;
ap =   2;
as =  40;

You probably will need to experiment yourself to get the exact characteristics you want. This tends to be heuristic at some point.

I also refer you to the yulewalk function. It lets you design filters with specific passband and stopband characteristics.