Re: [Scilab-users] bug in function iir ?
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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Re: [Scilab-users] bug in function iir ?
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
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
svgd_iir
Description: Binary data
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Re: [Scilab-users] bug in function iir ?
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 please can you save the hz value using the Scilab save function and send the file? Serge Your problem arises because one frequency value you ask for corresponds exactly to a zero of hz.num log(roots(hz.num))/(2*%pi) so you want to compute the gain in dB of a zero value which is -inf To avoid such problem you can let repfreq to do the frequency discretization. [frq,repf]=repfreq (hz) or equivalently [frq,repf]=repfreq (hz,0,0.5) in this case the discretization uses varying frequency step Serge Le 23/02/2016 14:41, grivet a écrit : Le 23/02/2016 14:21, Serge Steer a écrit : Please can you give more details : value of Order and Fcutoff/Fs/2 and what you are doing with hz (because iir does not call dbphi) Serge ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] bug in function iir ?
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 Your problem arises because one frequency value you ask for corresponds exactly to a zero of hz.num log(roots(hz.num))/(2*%pi) so you want to compute the gain in dB of a zero value which is -inf To avoid such problem you can let repfreq to do the frequency discretization. [frq,repf]=repfreq (hz) or equivalently [frq,repf]=repfreq (hz,0,0.5) in this case the discretization uses varying frequency step Serge Le 23/02/2016 14:41, grivet a écrit : Le 23/02/2016 14:21, Serge Steer a écrit : Please can you give more details : value of Order and Fcutoff/Fs/2 and what you are doing with hz (because iir does not call dbphi) Serge ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
Re: [Scilab-users] bug in function iir ?
Your problem arises because one frequency value you ask for corresponds
exactly to a zero of hz.num
log(roots(hz.num))/(2*%pi)
so you want to compute the gain in dB of a zero value which is -inf
To avoid such problem you can let repfreq to do the frequency
discretization.
[frq,repf]=repfreq (hz)
or equivalently
[frq,repf]=repfreq (hz,0,0.5)
in this case the discretization uses varying frequency step
Serge
Le 23/02/2016 14:41, grivet a écrit :
Le 23/02/2016 14:21, Serge Steer a écrit :
Please can you give more details :
value of Order and Fcutoff/Fs/2
and what you are doing with hz (because iir does not call dbphi)
Serge
I am just running the example found in "how to design an elliptic
filter". Here is the code:
Order= 2; // The order of the filter
Fs = 1000; // The sampling frequency
Fcutoff = 40;// The cutoff frequency
// We design a low pass elliptic filter
hz = iir (Order,'lp','ellip',[Fcutoff/Fs/2 0],[0.1 0.1]);
// We compute the frequency response of the filter
[frq,repf]=repfreq (hz,0:0.001:0.5);
[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)');
with 'ellip' replaced by 'butt', and [Fcutoff/Fs/2 0] replaced
byFcutoff/Fs/2.
I am beginning to use digital filters to treat some data. As my first
step, I try to run the examples in the help,how to design an elliptic
filter (using Scilab 5.5.1, Win7-64). This works . However, when I
select a Butterworth filter:
hz = iir(Order,'lp','butt',Fcutoff/Fs/2,[0.1 0.1]);
I get this error message:
Singularité de la fonction log ou tan.
at line 6 of function dbphi called by :
[db_repf, phi_repf] = dbphi(repf);
What did I miss ?
No bugs have been reported for function dbphi, but three similar
bugs are listed for iir.
Any suggestion welcome.
JPGrivet
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[Scilab-users] bug in function iir ?
Hello, I am beginning to use digital filters to treat some data. As my first step, I try to run the examples in the help,how to design an elliptic filter (using Scilab 5.5.1, Win7-64). This works . However, when I select a Butterworth filter: hz = iir(Order,'lp','butt',Fcutoff/Fs/2,[0.1 0.1]); I get this error message: Singularité de la fonction log ou tan. at line 6 of function dbphi called by : [db_repf, phi_repf] = dbphi(repf); What did I miss ? No bugs have been reported for function dbphi, but three similar bugs are listed for iir. Any suggestion welcome. JPGrivet ___ users mailing list users@lists.scilab.org http://lists.scilab.org/mailman/listinfo/users
