The zeros of the transfer function
$H(z)$ of a linear-phase filter lie in specific configurations.
We can write the symmetry condition
$$h(n)=h(N-1-n)$$ in the
$Z$ domain. Taking the
$Z$ -transform of both sides gives
$H(z)=z^{-(N-1)}H(\frac{1}{z})$
Recall that we are assuming that
$h(n)$ is real-valued. If
${z}_{0}$ is a zero of
$H(z)$ ,
$$H({z}_{0})=0$$ then
$$H(\overline{{z}_{0}})=0$$ (Because the roots of a polynomial with real coefficients
exist in complex-conjugate pairs.)
Using the symmetry condition,
, it follows that
$$H({z}_{0})=z^{-(N-1)}H(\frac{1}{{z}_{0}})=0$$ and
$$H(\overline{{z}_{0}})=z^{-(N-1)}H(\frac{1}{\overline{{z}_{0}}})=0$$ or
$$H(\frac{1}{{z}_{0}})=H(\frac{1}{\overline{{z}_{0}}})=0$$
If
${z}_{0}$ is a zero of a (real-valued) linear-phase filter, then so
are
$\overline{{z}_{0}}$ ,
$\frac{1}{{z}_{0}}$ , and
$\frac{1}{\overline{{z}_{0}}}$ .
Zeros locations
It follows that
generic zeros of a linear-phase filter exist in sets of 4.
zeros on the unit circle (
${z}_{0}=e^{i{}_{0}}$ ) exist in sets of 2. (
${z}_{0}\neq (1)$ )
zeros on the real line (
${z}_{0}=a$ ) exist in sets of 2. (
${z}_{0}\neq (1)$ )
zeros at 1 and -1 do not imply the existence of zeros at
other specific points.
Zero locations: automatic zeros
The frequency response
${H}^{f}()$ of a Type II FIR filter always has a zero at
$=\pi $ :
$$h(n)$$
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