The pole outside the unit circle gives left sided sequence • Ex1(a): Stable system Impulse response is • = )nu[n] + 2(0.9e h[n] j 4 2(0.9e− )nu[n] j 4 3(−2)nu[−n 1] − − n)u[n] = 4(0.9)n cos( h[n] 4 3(−2)nu[−n 1] − − Ex1(b): Causal system For causal system: all the poles must contribute the right sided terms, • = )nu[n] + 2(0.9e h[n] j 4 )nu[n] 2(0.9e− j 4 + 3(−2)nu[n] n)u[n] = + 4(0.9)n cos( 3(−2)nu[n] h[n] 4 Ex1(c): Causal and stable system For causal and stable system: all the poles must be inside the unit circle We have one pole at = is outside the unit circle, hence the z −2, system can not be both stable and causal system Ex2: Recursive system
• The pole outside the unit circle gives left sided sequence Ex1(a): Stable system • Impulse response is h[n] = 2(0.9e j 4 )nu[n] + 2(0.9e− j 4 )nu[n] − 3(−2)nu[−n − 1] h[n] = 4(0.9)n cos( n)u[n] 4 − 3(−2)nu[−n − 1] Ex1(b): Causal system • For causal system: all the poles must contribute the right sided terms, h[n] = 2(0.9e j 4 )nu[n] + 2(0.9e− j 4 )nu[n] + 3(−2)nu[n] h[n] = 4(0.9)n cos( n)u[n] + 3(−2)nu[n] 4 Ex1(c): Causal and stable system For causal and stable system: all the poles must be inside the unit circle We have one pole at z = −2, is outside the unit circle, hence the system can not be both stable and causal system Ex2: Recursive system