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I am studying a Notch filter. I found its Transfer Function (in Z) and its poles. How can I find the frequency of theses poles (which corresponds to the cut-off frequency)?

Thank you very much for your help

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    $\begingroup$ Welcome to DSP.SE! A notch filter doesn't have a "cut-off frequency" as such. It has a single frequency that it aims to nullify. Is that the frequency you are trying to find? $\endgroup$ – Peter K. Nov 4 '15 at 20:30
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Usually a discrete-time notch filter is implemented by an IIR (infinite impulse response) filter, which has a numerator as well as a denominator polynomial. The notch frequency - the frequency where the input signal should be completely suppressed - is determined by the numerator polynomial, i.e. by the zeros of the filter's transfer function, not (directly) by its poles. Of course, a practical design will also place the poles close to the notch-frequency, but the actual notch is implemented by a zero on the unit circle of the complex plane, i.e. by the numerator polynomial.

The numerator of a second order notch filter looks like this:

$$N(z)=z^2-2\cos(\omega_0)z+1\tag{1}$$

You can verify that this polynomial has two complex conjugate zeros at $z=e^{j\omega_0}$ and at $z=e^{-j\omega_0}$, so the notch frequency is given by the normalized angular frequency $\omega_0$.

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