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When we say that a Generalized Linear Phase System must satisfy the pole zero plot with the condition that a complex zero not on the unit circle exist's in a pair of 4. Then I understand that I need the conjugate pairs to have real coefficients. Why do we need the complex conjugate reciprocal pairs and what role do they play for the FIR filter? I am aware that we use complex conjugate reciprocal pairs in case of all pass systems but how does that relate to the Linear Phase System if it even does.

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The impulse response of a linear phase FIR filter is odd or even symmetrical. This has as a consequence that if $z_0$ is a zero, $1/z_0^*$ must also be a zero. If in addition to the linear phase property you also have real-valued coefficients, complex zeros always occur in complex conjugate pairs. So if a real-valued linear phase filter has a complex root $z_0$, you automatically get $z_0^*$, $1/z_0$, and $1/z_0^*$ as roots.

Complex roots on the unit circle just occur in complex conjugate pairs (because $z_0=1/z_0^*$, and real-valued roots just occur in pairs $z_0$ and $1/z_0$, unless they're on the circle (i.e., $z_0=1$ or $z_0=-1$).

In sum, for real-valued linear phase FIR filters, there are four types of "clusterings" of roots:

  1. complex roots with $|z_0|\neq 1$ occur in groups of four: $z_0$, $z_0^*$, $1/z_0$, $1/z_0^*$
  2. complex roots on the unit circle occur as complex conjugate pairs $z_0$ and $z_0^*$
  3. real roots with $|z_0|\neq 1$ occur in pairs as $z_0$ and $1/z_0$
  4. real roots on the unit circle ($z_0=1$ or $z_0=-1$ ) don't imply another root

Mathematically, for a real-valued length $N$ FIR filter the linear phase condition in the frequency domain is

$$H(z)=\pm z^{-(N-1)}H\left(\frac{1}{z}\right)\tag{1}$$

where $N$ is the filter length (number of taps). From $(1)$ it is clear that if $z_0$ is a zero of $H(z)$ then also $1/z_0$ must be a zero.

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  • $\begingroup$ I understand you completely. The thing is I missed a word in my question my fault entirely. I wanted to know just like you said "Linear phase FIR filter must have a complex conjugate reciprocal zero for each zero not on the unit circle". Why do we have this reciprocal conjugate zero? What effect do I see in the frequency response because of it? Cause in all pass we have a zero and a pole at the complex conjugate reciprocal location which gives us unity mag response. $\endgroup$ Jan 13 at 21:44
  • $\begingroup$ @PowerSurge: It's caused by the symmetry of the filter coefficients. Note that for an allpass filter the numerator and denominator polynomials are mirrored versions of each other. That's why you have a zero at $1/p^*$ where $p$ is a pole. Same for FIR filters, just that the polynomial is a mirror of itself. That's why you get another zero at $1/z_0^*$ for each zero at $z_0$. $\endgroup$
    – Matt L.
    Jan 13 at 22:08

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