2
$\begingroup$

I have to design 2 IIR bandpass filter with the following specifications:

Sampling frequency 1000 Hz Pass-band 50 Hz to 200 Hz Order 6 using bilinear transform and impulse invariance with Butterworth prototype analogue filter.

I'm confused, because the two methods I've got for producing the filters have different results.

fs = 1000;
fNq = fs/2;
f1 = 50/fNq;
f2 = 200/fNq;
[z,p,k] = buttap(6);
[A,B,C,D] = zp2ss(z,p,k);
[zd,pd,kd] = bilinear(z,p,k,fs);
Bw = f2-f1;
Wo = sqrt(f1*f2);
[At,Bt,Ct,Dt] = lp2bp(A,B,C,D,Wo,Bw);
[b,a] = ss2tf(At,Bt,Ct,Dt);

[Ad,Bd,Cd,Dd] = bilinear(At,Bt,Ct,Dt,fs);
[bz, az] = ss2tf(Ad,Bd,Cd,Dd);
fvtool(bz,az);

(I think that) The above code creates an analogue prototype, transforms to a state-space form, applies the bilinear transform, transforms the low-pass filter to a band-pass filter, transforms to a transfer function and the plots it.

My problem is that it is noticably different to:

[num,den] = butter(6, [f1, f2], 's');
[B,A] = bilinear(num, den, fs);
fvtool(B, A);

I have similar problems with the impulse invariance method.

Could someone please give me a hint as to what I'm doing wrong?

$\endgroup$
4
$\begingroup$

The issue is resolved: the butter() function automatically does bilinear transform, you just need to specify the frequencies normalised to the Nyqyist freq.

$\endgroup$
  • $\begingroup$ Please give this answer the check mark if it really solved your problem. Thanks! $\endgroup$ – Peter K. Oct 4 '16 at 11:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.