12 (z )2 − Write the inverse system • = (z )2 Hinv(z) − 12 (z 4)2 − = Double poles at 4, hence the poles are outside the unit circle and inverse z system is not stable and causal Unsolved ex:7.35(b) Find the transfer function of an inverse system and determine whether it can be both causal and stable z2 − 81 100 = 1 H(z) z2 − = 1 Hinv(z) z2 − 10081 z2 − 109 = Double at , hence the poles are inside the unit z circle and inverse system is stable and causal
(z − 12 )2 • Write the inverse system Hinv(z) = (z − 12 )2 (z − 4)2 Double poles at z = 4, hence the poles are outside the unit circle and inverse system is not stable and causal Unsolved ex:7.35(b) Find the transfer function of an inverse system and determine whether it can be both causal and stable z2 − 81 H(z) = 100 z2 − 1 Hinv(z) = z2 − 1 z2 − 10081 Double at z = 109 , hence the poles are inside the unit circle and inverse system is stable and causal