Butterworth filter and normalised frequency

Hi, I have a little problem with a B filter, I want to compare the frequency response of the analog and digital filter, so I tried with this

if true

    >> Rp=1;
    >> Rs=80;
    >> nupass=20000;
    nustop=22000;
    >> Ts=1/44100;
    >> nucontpass=1/pi/Ts*tan(pi*Ts*nupass);
    >> nuconstop=1/pi/Ts*tan(pi*Ts*nustop);
    >> nu=0:0.1:1/2/Ts;
    >> nucont=0:0.1:1000;
    [n,wn]=buttord(2*pi*nucontpass,2*pi*nuconstop, Rp, Rs,'s');
    [b,a]=butter(n,wn);
    [B,A]=bilinear(b,a,1/Ts);
    >> Hc=freqs(b,a,2*pi*nucont);
    >> Hb=freqz(B,A,nu,1/Ts);
    >> figure, plot(nu,abs(Hb)),hold on,plot(nucont,abs(Hc),'g--');
  end

But the problem come of wn which is not normalised so tried to do Wn=2wn/Ts but i have but the result is always superior at 1. Have you got any solution to fix this problem? Thanks

Answer 1

[Rick Rosson]

   Wn = 2*wn*Ts;

[jack]

Thankls but it's an error in my previous post but i have tried Wn = 2*wn*Ts; so it's not the problem