Figure 6.3 illustrates the use of residuez (§H.5) for performing a partial fraction expansion on the transfer function
B = [1 0 0 0.125]; A = [1 0 0 0 0 0.9^5]; [r,p,f] = residuez(B,A) % r = % 0.16571 % 0.22774 - 0.02016i % 0.22774 + 0.02016i % 0.18940 + 0.03262i % 0.18940 - 0.03262i % % p = % -0.90000 % -0.27812 - 0.85595i % -0.27812 + 0.85595i % 0.72812 - 0.52901i % 0.72812 + 0.52901i % % f = [](0x0) |