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If a second order system has 2 poles/zeros that are conjugate symmetric, how does this affect the phase response? I know that if there are 4 zeros/poles that are conjugate reciprocals, then it is a zero phase system.

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If you have conjugate symmetric poles and zeros then the coefficients of the system (i.e., the filter coefficients, or, equivalently, the coefficients of the corresponding difference equation) are real-valued. E.g., if the zeros are $z_0$ and $z_0^*$ then the numerator polynomial of the transfer function $H(z)$ is (up to a scaling factor)

$$N(z)=(z-z_0)(z-z_0^*)=z^2-(z_0+z_0^*)z+|z_0|^2=z^2-2\text{Re}(z_0)z+|z_0|^2\tag{1}$$

Note that the coefficients in $(1)$ are all real-valued. The same is true for the denominator polynomial. So the frequency response (i.e., the transfer function evaluated at $z=e^{j\omega}$) can be written as

$$H(e^{j\omega})=\frac{b_0e^{2j\omega}+b_1e^{j\omega}+b_2}{e^{2j\omega}+a_1e^{j\omega}+a_2}\tag{2}$$

with all coefficients $a_i$ and $b_i$ being real-valued. Consequently, the following equation must hold:

$$H(e^{j\omega})=H^*(e^{-j\omega})\tag{3}$$

If you write $H(e^{j\omega})$ in polar form $H(e^{j\omega})$=$M(\omega)e^{j\phi(\omega)}$ with magnitude $M(\omega)$ and phase $\phi(\omega)$ you get

$$M(\omega)e^{j\phi(\omega)}=M(-\omega)e^{-j\phi(-\omega)}\tag{4}$$

From $(4)$ it follows that for conjugate symmetric poles and zeros the magnitude of the frequency response must be an even function of frequency, and the phase must be an odd function of frequency, i.e., $M(\omega)=M(-\omega)$ and $\phi(\omega)=-\phi(-\omega)$.

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  • $\begingroup$ Whoa! Couldn't have asked for a better answer. Could you please elucidate on the conjugate symmetry equality $(3)$? I am trying to substitute $e^{-j\omega}$ in $(1)$ but i don't see it getting equal to $(2)$ $\endgroup$ – Rithesh Nov 10 '15 at 13:08
  • $\begingroup$ @Rithesh: If you conjugate (2), the coefficients don't change (because they're real-valued), just the sign in the exponent changes, so you get (for the numerator) $b_0e^{-2j\omega}+b_1e^{-j\omega}+b_2$. Now if you take $-\omega$ instead of $\omega$ you get back the original expression for $H(e^{j\omega})$, yielding the equality (3). $\endgroup$ – Matt L. Nov 10 '15 at 13:14
  • $\begingroup$ Danke schon @Matt ! $\endgroup$ – Rithesh Nov 10 '15 at 13:19

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