Question

Consider a finite impulse response filter with unit-sample response h(n) of length (2N+1). If h(n) is real and even, show that the zeros of the system function occur in mirror-image pairs about the unit circle; i.e., if H(z)=0 for z=rhoeʲθ, then H(z)=0, also for z=(1/rho)eʲθ.

Answer 1

The zeros of H(z) of a finite impulse response filter with a real and even unit-sample response h(n) occur in mirror-image pairs about the unit circle because the conjugate reciprocal of any zero z is also a zero when H(z) has real coefficients, which is the case due to h(n) being real and even.

Step-by-step explanation:

To show that zeros of the system function H(z) of a finite impulse response filter with real and even unit-sample response h(n) occur in mirror-image pairs about the unit circle, we begin with the fact that h(n) is real and even. This means that h(n) = h(-n) for all n and, hence, the z-transform H(z) is real-valued on the unit circle (i.e., for |z|=1).

Let's consider that H(z)=0 for z=ρeˆθ, where ρ and θ are real numbers. Due to h(n) being even, H(z) can be represented as a sum of cosines which implies that if z is a zero, then its complex conjugate z* is also zero, which in our case would be ρe-jθ since the complex conjugate mirrors the angle θ across the real axis.

The property that needs to be proved is that if H(z)=0 for z=ρejθ, then H(z)=0 for z=1/ρejθ. This comes from the conjugate reciprocal property of polynomials with real coefficients which states that if a complex number is a zero then its conjugate reciprocal (1/ρ)ejθ is also a zero. The filter coefficients being real ensures that the polynomial representing H(z) only has real coefficients, which lets us apply this property to conclude the desired result: zeros occur in pairs such that if z=ρejθ is a zero, so is z=(1/ρ)ejθ, which shows the mirror-image pairing about the unit circle.

Answer 2

A real and even finite impulse response filter will have a z-transform that is also real and even, leading to zeros of the system function occurring in symmetric pairs with respect to the real axis and in mirror-image pairs about the unit circle.

Step-by-step explanation:

Finite Impulse Response (FIR) Filters and Zeros Position

For a finite impulse response filter that has a unit-sample response h(n) which is real and even, the filter's length is given by (2N+1). We know that if the filter's response h(n) is real and even, then its z-transform, H(z), is also real and even in the z-domain. The real and even properties of H(z) suggest that the zeros of the system function must be symmetric with respect to the real axis.

If a zero occurs at z = rho ejθ, then due to the even symmetry, a zero must also occur at z = rho e-jθ, since a real and even function has this hermitian symmetry property. Additionally, due to the nature of the FIR filter's polynomial in z, the reciprocal of these zeros z = 1/rho e±jθ will also be zeros of the system function H(z). This shows that zeros occur in mirror-image pairs about the unit circle.