5 1 7 1 21 41 (1 (1 + − = 2 + 2 z−1) 2 z−1) H(z) − Write in product form H(z) • + (1 + z−1)(1 z−1)+ 1 1 5 1 2(1 − (1 z−1)+ z−1) − 7 1 H(z)= 2 4 2 4 2 2 12 14 (1 + − z−1)(1 z−1) After simplification we get • 1 (1 + 2z−1) − z−1)(1 H(z) 8 = 12 14 (1 + z−1)(1 z−1) − H1(z) We know = , so we can write Hinv(z) • = Hinv(z) 12 (1 + z−1)(1 − 14 18 (1 z−1) − + 2z−1) z−1)(1 18 Poles are at = and = so the system can not be both stable z z −2, and causal 1.8.6 Unsolved examples from [2] Unsolved ex:7.34(a) Is the system (i)causal and stable (ii)minimum phase • 2z + 3 = H(z) 165 + z2 z − H(z)
H(z) = 2 + 5 1 − 7 1 2 (1 − 21 z−1) 2 (1 + 41 z−1) • Write H(z) in product form 2(1 − 1 z−1)(1 + 1 z−1)+ 5 (1 + 1 z−1)+ 7 (1 − 1 z−1) H(z)= 2 4 2 4 2 2 (1 − 12 z−1)(1 + 14 z−1) • After simplification we get (1 − 1 z−1)(1 + 2z−1) H(z) = 8 (1 − 12 z−1)(1 + 14 z−1) • We know Hinv(z) = H1(z) , so we can write Hinv(z) = (1 − 12 z−1)(1 + 14 z−1) (1 − 18 z−1)(1 + 2z−1) Poles are at z = 18 and z = −2, so the system can not be both stable and causal 1.8.6 Unsolved examples from [2] Unsolved ex:7.34(a) • Is the system (i)causal and stable (ii)minimum phase H(z) = 2z + 3 z2 + z − 165 H(z)
U8-L47_Solved-Examples-on-Causality-stability-and-Inverse-systems.docx