What is importance of the conjugate reciprocal and none reciprocal zeros in FIR Type 2 Filters? In DSP class we have this slide:



So i like to know why this classification is so imoortant?

I have seen this post :


  Conjugate reciprocal pairs of zeros and poles in ...
  
  Assuming the impulse response h[n]h[n] of an FIR filter is real for
  all nn,
  
  Why are zeros and poles in FIR design found in reciprocal and
  conjugate pairs?Is the assumption necessary for this phenomenon to
  take place?


But i dont get my answer?( mybe because of being newbi in this field)
I would like to provide a short answer.

Consider

$$ h[n] = \sum_{k=0}^{M} b_k \delta[n-k] ~~~~\longleftrightarrow ~~~~H(z) =  \sum_{k=0}^{M} b_k z^{-k} $$

Left: LTI FIR filter impulse response coefficients $b_k$, and right: its Z-Transform.

If the FIR filter is real; i.e., coefficients $b_k$ are real, then the Z-transform is a polynomial of real coefficients, and the roots of such polynomials are either real complex-conjugate pairs. This explains why a real FIR filter's Z-domain poles &amp; zeros are in conjugate pairs when they are complex.

To explain why, also, they are in reciprocals, we need to consider linear phase property (without losing generality) : $h[n] = h[-n]$; then since the Z-transform of $h[-n]$ is $H(1/z)$; we have the following :

$$ h[n] = h[-n]  ~~~\implies ~~~ H(z) = H(1/z) $$

And this means that if $z_0$ is a zero (or pole) of $H(z)$; i.e., $H(z_0) = 0$ then its reciprocal $1/z_0$ is also a zero of $H(z)$.

To sum up; for a real and linear phase FIR filter, if $z_0$ is a zero of $H(z)$ then $z_0^*$, $1/z_0$ and $1/z_0^*$ are also zeros; hence zeros and poles are in pairs of four when they are complex.