Le 25/02/2016 18:10, grivet a écrit :
Le 24/02/2016 21:40, Serge Steer a écrit :
Le 24/02/2016 11:30, grivet a écrit :
I appreciate your help; however, neither suggestion works: I still
get the same error message.
The similar line
[frq,repf]=repfreq(hz,0.01,0.49);
has no problem
You are right the problem is exactly for the frequency 0.5 besause
exp(2*%pi*%i*0.5) -> - 1. + 1.225D-16i
and hz.num has 2 zeros very near -1
I checked the hz value with Matlab, the results are the same. So it
seems that hz is ok. So it is probabily a probleme due to floating point
computations procucing a zero value instead of a very small one.
Serge
please can you save the hz value using the Scilab save function and
send the file?
Serge
Voila le code:
//filtre Butterworth
Order = 2; // The order of the filter
Fs = 1000; // The sampling frequency
Fcutoff = 40; // The cutoff frequency
// We design a low pass Butterworth filter
hz = iir(Order,'lp','butt',Fcutoff/Fs/2,[0.1 0.1]);
// We compute the frequency response of the filter
[frq,repf]=repfreq(hz,0,0.5);
[db_repf, phi_repf] = dbphi(repf);
// And plot the bode like representation of the digital filter
subplot(2,1,1);
plot2d(Fs*frq,db_repf);
xtitle('Obtained Frequency Response (Magnitude)');
subplot(2,1,2);
plot2d(Fs*frq,phi_repf);
xtitle('Obtained Frequency Response (Phase in degree)');
iir est sauvegardé dans "svgd_iir" joint (binaire).
Cordialement,
JP Grivet
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