in terms of partial fraction expansion H(z) • 1 −3 = + H(z) 21 41 + 1 1 − z−1 z−1 Ex3: Stable and causal system Now write the impulse response for a causal and stable system • 1 1 = )nu[n] + ( )nu[n] h[n] −3( − Inverse system Impulse response (hinv) of an inverse system satisfies • = [n] hinv[n] h[n] ∗ where is the impulse response of a system to be inverted h[n] Take inverse on both sides gives z-transform • = 1 Hinv(z)H(z) = 1 Hinv(z) H(z) The transfer function of an LTI inverse system is the inverse of the transfer function of the system that we desire to invert If we write the pole-zero form of as H(z) ˜ 1 p M−p = k=1 H(z) (1 bz− − kN=−1l z−l (1 dkz−1) −
• H(z) in terms of partial fraction expansion H(z) = −3 + 1 1 + 21 z−1 1 − 41 z−1 Ex3: Stable and causal system • Now write the impulse response for a causal and stable system h[n] = −3( − 1 )nu[n] + ( 1 )nu[n] Inverse system • Impulse response (hinv) of an inverse system satisfies hinv[n] ∗ h[n] = [n] where h[n] is the impulse response of a system to be inverted • Take inverse z-transform on both sides gives Hinv(z)H(z) = 1 Hinv(z) = 1 H(z) The transfer function of an LTI inverse system is the inverse of the transfer function of the system that we desire to invert If we write the pole-zero form of H(z) as ˜ p M−p 1 H(z) = bz− k=1 (1 − z−l kN=−1l (1 − dkz−1)
U8-L47_Solved-Examples-on-Causality-stability-and-Inverse-systems.docx