# Magnitude response of mirrored (with respect to unit circle) poles and zeros

I just want to check that my understanding about the following paragraph from Optical Filter Design and Analysis by Christi K. Madsen, Jian H. Zhao is correct:

A filter’s magnitude response is equal to the modulus of its transfer function, $$|H(z)|$$, evaluated at $$z = e^{j\omega}$$. Based on the pole/zero representation of $$H(z)$$, only the distance of each pole and zero from the unit circle, i.e. $$|e^{j\omega}-z_m|$$ or $$|e^{j\omega}-p_n|$$, affects the magnitude response. Consequently, a zero that is located at the mirror image position about the unit circle, i.e. $$1/z_m^*$$, cannot be differentiated from $$z_m$$ based on the magnitude response.

Let us assume a filter with just one zero $$c_1=|c_1|e^{j\phi}$$ so that its transfer function is: $$H(z)=1-c_1 z^{-1}=1-\lvert c_1\rvert e^{j\phi} z^{-1}$$ Its mirrored zero is $$c_2=1/c_1^*=\lvert c_1\rvert^{-1}e^{j\phi}$$ and its transfer function is $$H_m(z)=1-c_2 z^{-1}=1-\lvert c_1\rvert^{-1}e^{j\phi} z^{-1}$$ Evaluating the transfer function at $$z=e^{j\omega}$$ and computing $$H(e^{j\omega})H^*(e^{j\omega})$$ we get the power response: $$\big\lvert H(e^{j\omega)}\big\rvert^2=\big\lvert 1-\lvert c_1\rvert e^{j(\phi-\omega)}\big\rvert^2=1 +\lvert c_1\rvert^2-2\lvert c_1\rvert\cos(\phi-\omega)$$ and $$\big\lvert H_m(e^{j\omega})\big\rvert^2=\bigg\lvert 1-\frac{1}{\lvert c_1\rvert}e^{j(\phi-\omega)}\bigg\rvert^2=1 +\frac{1}{\lvert c_1\rvert^2}-2\frac{1}{\lvert c_1\rvert}\cos(\phi-\omega)$$

$$\big\lvert H_m(e^{j\omega})\big\rvert^2$$ can also be expressed as: $$\big\lvert H_m(e^{j\omega})\big\rvert^2=\frac{1}{\lvert c_1\rvert^2}\Big[1 +\lvert c_1\rvert^2-2\lvert c_1\rvert\cos(\phi-\omega)\Big]=\frac{1}{\lvert c_1\rvert^2}\lvert H(e^{j\omega)}\rvert^2$$ which is just the power response of the first filter scaled by a factor of $$\lvert c_1\rvert^{-2}$$. Is this analysis right?

You're right, the contribution of a zero $$z_0=re^{j\phi}$$ to the squared magnitude response is

$$\big|e^{j\omega}-re^{j\phi}\big|^2=1-2r\cos(\omega-\phi)+r^2\tag{1}$$

From $$(1)$$ it is clear that another zero resulting in the same frequency dependence as in $$(1)$$ must have the same phase angle $$\phi$$. Assuming $$z_1=Re^{j\phi}$$ we get

$$\big|e^{j\omega}-Re^{j\phi}\big|^2=1-2R\cos(\omega-\phi)+R^2=R^2\left(1-\frac{2}{R}\cos(\omega-\phi)+\frac{1}{R^2}\right)\tag{2}$$

Comparing $$(1)$$ and $$(2)$$ we get $$R=1/r$$ and a squared magnitude scaling of $$R^2=1/r^2$$. Consequently, the other zero $$z_1$$ is given by

$$z_1=\frac{e^{j\phi}}{r}=\frac{1}{z_0^*}\tag{3}$$

and

$$\big|e^{j\omega}-z_0\big|=r\left|e^{j\omega}-\frac{1}{z_0^*}\right|\tag{4}$$