Figure 1.52: Location of poles in Example 1 this system be causal and stable? 2 2 3 j j 1 0.9e 4 1 0.9e− 1 + − − 4 2z−1 = + + z−1 z−1 H(z) j j = Solution: The system has poles at z e = 4 = • and and z e− z 4 −2 Ex1(a): Stable system The location of poles in the z-plane • Ex1(a): Stable system For stable system: the ROC must include the unit circle Two inside the unit circle contribute right sided terms
Figure 1.52: Location of poles in Example 1 this system be causal and stable? H(z) = 2 + 2 + 3 1 − 0.9e j 4 z−1 1 − 0.9e− j 4 z−1 1 + 2z−1 • Solution: The system has poles at z = e j 4 and z = e− j 4 and z = −2 Ex1(a): Stable system • The location of poles in the z-plane Ex1(a): Stable system For stable system: the ROC must include the unit circle Two inside the unit circle contribute right sided terms
U8-L47_Solved-Examples-on-Causality-stability-and-Inverse-systems.docx