# 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 & 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.

This gets into standard properties of polynomials which all FIR filters are represented by. But to help give you some insight consider the following by understanding additional filters beyond linear phase filters: the minimum phase filter, the maximum phase filter and the reverse filter:

A minimum phase filter will have it's dominant taps earliest in the filter thus providing the dominant portions of the input signal first, so has minimum delay and the phase response deviates the minimum possible for that number of taps and frequency response). If you solve for the roots (the zeros) of such a filter you can confirm the property that all the zeros will be inside the unit circle. This is a property of polynomials with leading coefficients.

A maximum phase filter will have it's dominant taps latest in the filter thus resulting in the longest delay and have the largest delay and largest phase excursion for that numbrer of taps and frequency response. If you solve for the roots (the zeros) of such a filter you can confirm the property that all the zeros will be outside the unit circle. This is a property of polynomials with trailing coefficients.

A reverse filter is formed by flipping the coefficients of a filter from end to start and will have the exact same magnitude response but a very different phase response. The reverse of a minimum phase filter is a maximum phase filter!

Look at this graphic below to give further insight-- the frequency response can be formed using phasors as I have done by replacing $$z^{-1}$$ with a phasor that rotates clockwise as we sweep the frequency from $$0$$ to $$2\pi$$ (The frequency response is found from the z-transform by replacing z with $$e^{j\omega}$$. So a filter with coefficients 1, -0.5 would have the response $$1-0.5e^{-j\omega}$$ which the plot on the left is showing as magnitude and phase on a complex plane, while the reverse filter with coefficients -0.5, 1 would have the response give by $$-0.5+1e^{-j\omega}$$ which is the plot of the right. Notice how the magnitude response of each is identical but how small the phase deviates on the plot on the left and how large is the overall phase in comparison for the plot on the right.

To cascade two filters, you convolve their coefficients. If you cascade a filter with its reverse you will always get a linear phase filter since such a convolution results in symmetric coefficients:

Indeed in the example above, convolving $$[1, -0.5]$$ with $$[-0.5, 1]$$ results in $$[-.5, 1.25, -.5]$$

Therefore a linear phase filter with zeros not on the unit circle can be factored into a minimum phase filter with zeros inside the unit circle and maximum phase filter with zeros outside the unit circle (You can follow the same logic for the cascade of a filter with its reverse for zeros directly on the unit circle).