113_1_chapter14-posted - All-Pass Systems An all-pass...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: All-Pass Systems An all-pass system will be defined as a causal and stable LTI system that has unit magnitude response over the entire frequency range [-, ]. The condition of causality means that its impulse response sequence is a right-sided sequence so that the ROC of its z-transform is the outside of a circular region. Now, by the assumption of stability, we know that the ROC must include the unit circle. Therefore, the poles of such an all-pass system have to lie inside the unit circle. The simplest examples of an all-pass system are H(z) = 1 or H(z) = z -k (i.e., pure delays). More generally, a first-order all-pass rational transfer function takes the form H(z) = 1 - a z z -1 - a = , -1 1 - az z-a |a| < 1. 149 That is, it has a single pole inside the unit circle (at the point a) and a single zero outside the unit circle (at the point 1/a ). The magnitude response of the above first-order system can be easily seen to be unity. Indeed e-j - a e-j - a 1 - a ej = ej j = =1 |H(ej )| = j - a e e -a ej - a since the expressions in the numerator and denominator are conjugates of each other. The phase response, on the other hand, can be verified to be H(ej ) = - - 2 arctan r sin( - ) 1 - r cos( - ) where we introduced the polar representation for a, a = rej . The associated group delay is obtained by differentiation, which leads to () = 1 - r2 . 1 + r 2 - 2r cos( - ) Now since r < 1, we find that the group delay of an all-pass section is necessarily nonnegative at all . The transfer functions of rational all-pass systems of higher orders have the form H(z) = where A(z) is a polynomial in z, A(z) = z N + 1 z N -1 + 2 z N -2 + . . . + N with roots inside the unit circle, and where A# (z) denotes the conjugate reversal polynomial that is defined as A# (z) = z N A 1 z = N z N + . . . + 2 z 2 + 1 z + 1. A# (z) A(z) That is, the coefficients of A(z) are conjugated and reversed in order. It is an easy exercise to verify that such transfer functions have unit magnitude response. Indeed, H(ej ) = ejN e-jN + 1 e-j(N -1) + . . . + N 1 + 1 ej + . . . + N ejN = ejN j(N -1) + . . . + jN + ej(N -1) + . . . + + 1 e e N 1 N which allows us to conclude that |H(ej )| = 1 for all . Note also the useful property that if A(z) has a zero at zo then A# (z) has a zero at 1/zo . In particular, since all the zeros of A(z) are assumed to be inside the unit circle, then the zeros of A# (z) will be outside the unit circle. Minimum Phase Systems A rational minimum phase system will be defined as causal LTI system with rational transfer function whose poles and zeros are all inside the unit circle. The condition of causality means that its impulse response sequence is a right-sided sequence so that the ROC of the transfer function is the outside of a circular region. Now 150 All-Pass and Minimum-Phase Systems Chapter 14 since the poles are, by definition, inside the unit circle, it follows that the ROC must include the unit circle so that a rational minimum-phase system is necessarily BIBO stable. Such a system also has a stable inverse since its zeros are also inside the unit circle. For example, 1 z-2 H(z) = z-1 4 is minimum phase, while H(z) = is not. z-2 z-1 4 A Fundamental Decomposition Not every stable causal system is all-pass. Also, not every stable causal system is minimum phase. But every stable causal system can be expressed as the product of a minimum phase system and an all-pass system. Indeed, let H(z) = N (z) D(z) denote an arbitrary proper transfer function of a causal stable system. We can factor N (z) as the product of two polynomials, N (z) = N1 (z)N2 (z), with N1 (z) having all its zeros inside the unit circle and N2 (z) having all its zeros outside the unit circle. Then we can write # N1 (z)N2 (z) N2 (z) H(z) = . # D(z) N2 (z) minimum phase More compactly, we write H(z) = Hmin (z) Hap (z) This also shows that H(z) and Hmin (z) have the same magnitude response. Moreover, H () = min () + ap () . all-pass ...
View Full Document

Ask a homework question - tutors are online