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I am following the process that is described in this question: Transfer function of second order notch filter , I want to create a notch filter with the band suppressed equal to $f_c = 4000$ Hz, so using $\omega_n= f_c / f_s $, ($f_s = 48000$), I obtained the $\omega_n = \frac{\pi}{6}$, then using the exact same formula, with $a =0.8$. The pole-zero graph I get has the zero in $1$ and a pole in $0.8$, is this correct?? I am getting the half of the filter since the filter is centered in $0$ and not in $4000$ Hz.

As far as I know it must be centered in the wn I have (based on $f =4000$ Hz), but I am not sure why it is centered in $0$, or how to center it in the desired frequency. I get a pole zero graph like this one, with zeros in $1$ and poles in $0.8$.

enter image description here

Please advice

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  • $\begingroup$ Which formula did you use? Did you see that the first formula in the accepted answer is a filter with a notch at DC? Could it be that you used that formula? $\endgroup$ – Matt L. Nov 27 '18 at 16:07
  • $\begingroup$ Related: this answer $\endgroup$ – Matt L. Nov 27 '18 at 16:59
  • $\begingroup$ Hello Matt I used H(z) = k * (z^2-2cos(wn)z +1) / (z^2-2acos(wn)z +a^2) $\endgroup$ – ImperialF7 Nov 27 '18 at 17:09
  • $\begingroup$ How can you then get only one zero and one pole? That equation clearly has two poles and two zeros. $\endgroup$ – Matt L. Nov 27 '18 at 17:15
  • $\begingroup$ If you get real-valued double zeros and poles then $\cos(\omega_n)$ must be equal to $1$, which can't be the case with $\omega_n=\pi/6$. $\endgroup$ – Matt L. Nov 27 '18 at 17:17
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This Octave / Matlab code gives you a 2nd ord notch filter at $\omega_n = \pi/6$

r = 0.99;          % notch radius (closer to 1 stiffer)
wn = pi/6;         % notch radian frequency...  

% Create the 2nd order NOTCH filter coefficients b() and a()
b = r*conv([1,  -exp(j*wn)],[1,  -exp(-j*wn)]);      
a = conv([1,  -r*exp(j*wn)],[1,  -r*exp(-j*wn)]);    

figure,freqz(b,a,2048);

with the following result:

enter image description here

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