# Digital Butterworth filter design error

I am trying to design a digital ButterWorth filter for the given specifications.

rp=3;
rs=15;
FS=1;
wp=0.5*pi;
ws=0.75*pi;
pwp=2*FS*tan(wp/2);
pws=2*FS*tan(ws/2);
[n,wn]=buttord(pwp,pws,rp,rs,'s')
[b,a]=butter(n,wn,'s');
[bn,an]=bilinear(b,a,FS);%error
freqz(bn,an,512,FS);


If I design a filter manually i am getting different values for bn and an but Matlab gives different answer. Can anybody please help me?

• Do you get the same filter order? Are both filters meeting the specification? Jan 22, 2015 at 14:51

I cannot run your code because I have not matlab on this pc, but I can try to give you some advice.

The first thing I will check is the way you defined the cut-off frequencies. The Matlab function uses normalized frequencies (look at the examples here http://it.mathworks.com/help/signal/ref/buttord.html), which means you have to define it in this way: lets assume you want to cut your signal at 5 Hz and your sampling frequency is $Fs$

cut_freq = 5;
Wn = cut_freq/(Fs/2);


Of course if you have a band-pass filter you have to define 2 cut-off frequencies and therefore $Wn$ will be a vector of 2 elements.

Hope this helps. As soon as I get my laptop I will try to run your code and see what happens.

• thanks for your advice.I checked the examples.. But there ,they are directly designing digital filter...But what i am trying to do here is given the analog specifications of a butterworth filter, designing digital butterworth filter using Binilear Tranformation teqniq..For more clarity i will post original Q. below Nov 22, 2014 at 15:29
• Design a digital filter,given the equavalent analog filter with following specifications Passband ripple=3.01 db Passband edge=500 hz Stopband attenuation =15 db Stopband edge=750 hz Sample rate=2000 hz, and design using Bilinear Tranformation. Nov 22, 2014 at 15:34
• Sorry I did not specify, my answer was related to the first part of your code where you used the 'buttord()' function. There I do not understand the way you are defining the frequencies 'pwp=2*FStan(wp/2); pws=2*FStan(ws/2); [n,wn]=buttord(pwp,pws,rp,rs,'s')'
– Rhei
Nov 22, 2014 at 15:35
• To be more clear, $pwp$ and $pws$ are not normalized frequencies, but buttord() asks for normalized values of frequencies. So I think the problem could be in the way you are using the buttord() function
– Rhei
Nov 22, 2014 at 15:37