3
$\begingroup$

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

Improve this question
$\endgroup$
6
  • $\begingroup$ 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, 2018 at 16:07
  • $\begingroup$ Related: this answer $\endgroup$
    – Matt L.
    Nov 27, 2018 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, 2018 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, 2018 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, 2018 at 17:17

1 Answer 1

Reset to default
2
$\begingroup$

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

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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