Final answer:
To find the minimum-phase system Hmin(z), we need to factorize the transfer function H(z) into its zeros and poles. The all-pass system consists of the remaining zeros and poles.
Step-by-step explanation:
To find the minimum-phase system Hmin(z), we need to factorize the transfer function H(z) into its zeros and poles. The minimum-phase system consists of the poles and zeros that are inside the unit circle in the z-plane. In this case, the poles at z = 2 and z = -2 are inside the unit circle, so we only keep them. The zeros at z = -0.31 and z = -0.64 are outside the unit circle, so we remove them. Therefore, Hmin(z) is given by Hmin(z) = (1)(1 - 4z^2)(1).
To find the all-pass system Hap(z), we need to divide H(z) by Hmin(z). The all-pass system consists of the remaining zeros and poles. In this case, the remaining zero is at z = -0.31 and the remaining pole is at z = 0.64. Therefore, Hap(z) is given by Hap(z) = (1 + 0.31)(1 + 0.64z^(-2)).
So the factorization is H(z) = Hmin(z) * Hap(z).
To find Hmin(z) and Hap(z) for the discrete-time system H(z), one must decompose H(z) into factors and categorize them based on their phase properties, with Hmin(z) including factors with zeros inside the unit circle and Hap(z) containing factors representing poles and zeros on the unit circle.
The question involves finding a minimum-phase system, Hmin(z), and an all-pass system, Hap(z), such that when multiplied together, they give the original transfer function H(z). To do so, one must first decompose the original transfer function into factors and sort them based on whether they have poles or zeros inside the unit circle (for minimum-phase) or on the unit circle (for all-pass). The minimum-phase system should have all its zeros and poles inside the unit circle, while the all-pass system will have its zeros and poles mirrored across the unit circle.
To find Hmin(z), identify factors of H(z) that have their zeros inside the unit circle. Likewise, for Hap(z), identify factors with zeros on the unit circle and ensure that for every zero at z=a, there is a corresponding pole at z=1/a, maintaining the all-pass property.
In this case, zeros at z=-0.31 and z=0.64 are inside the unit circle, and so they will be part of Hmin(z). The term (1 - 4z^2) will contribute to Hap(z), as it represents zeros on the unit circle at Z=±1/2 and corresponding poles at Z=±2. Therefore, Hmin(z) will include the terms (1 + 0.31) and (1 + 0.64z^-2), and Hap(z) will include the term (1 - 4z^2) as well as (1 - 0.25z^-2) to ensure the system remains all-pass.