Effacer les filtres
Effacer les filtres

Bode different phase issue

14 vues (au cours des 30 derniers jours)
Bertug
Bertug le 21 Oct 2022
Modifié(e) : Bertug le 23 Oct 2022
Hello,
I have a homework. I have a transfer function which is related to butterworth low pass filter. Teacher asks me phase and amplitude response, and bode diagram separately. I use freqs(num,den) for the phase and amplitude response. I use bode(sys) for the bode diagram. When I compare the outputs, amplitude responses are quite similar. But phase responses are quite different. You can see the outputs and my codes.
Here is my code:
>> wp = 1;
>> ws = 1.2;
>> Rp = 0.2;
>> Rs = 25;
>> [n, Wn] = buttord(wp,ws,Rp,Rs,'s');
>> [num, den] = butter(n, Wn, 's');
>> sys = tf(num, den);
Continuous-time transfer function.
>> bode(sys)
>> freqs(num, den)
bode(sys) output:
freqs(num, den) output:
Is it necessary to see them similar or same? Could you please explain me?
Thanks in advance.

Réponses (1)

Star Strider
Star Strider le 21 Oct 2022
In the freqs call, the phase is wrapped. Unwrap it and freqs matches bode
wp = 1;
ws = 1.2;
Rp = 0.2;
Rs = 25;
[n, Wn] = buttord(wp,ws,Rp,Rs,'s');
[num, den] = butter(n, Wn, 's');
sys = tf(num, den);
sys
sys = 5.373 ----------------------------------------------------------------------------------------------------------------------------------------------------------------------- s^25 + 17.03 s^24 + 145.1 s^23 + 821.6 s^22 + 3471 s^21 + 1.164e04 s^20 + 3.215e04 s^19 + 7.51e04 s^18 + 1.509e05 s^17 + 2.639e05 s^16 + 4.054e05 s^15 + 5.504e05 s^14 + 6.626e05 s^13 + 7.087e05 s^12 + 6.734e05 s^11 + 5.675e05 s^10 + 4.225e05 s^9 + 2.763e05 s^8 + 1.574e05 s^7 + 7.708e04 s^6 + 3.191e04 s^5 + 1.089e04 s^4 + 2949 s^3 + 595.6 s^2 + 80 s + 5.373 Continuous-time transfer function.
figure
bode(sys)
figure
freqs(num,den, 1E+2)
Ax2 = get(subplot(2,1,2));
Line = Ax2.Children;
Ax2.Children.YData = rad2deg(unwrap(deg2rad(Line.YData)));
Helpful functions for this are unwrap, deg2rad and rad2deg.
.

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