02_Discrete_Time_Signals_Systems_Release

Curtin university australia 2 51 yh leung 2005 2006

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Curtin University, Australia 2-51 YH Leung (2005, 2006, 2007, 2012, 2016) 15. (a) 1 1 2 3 2 1 ( ) , 2 z z z H z z - - - - - = > (b) ( ) ( ) ( ) 2 2 1 5 5 2 [ ] 2 [ ] [ ] n n h n u n u n = - - (c) ( ) ( ) ( ) 2 2 1 5 5 2 [ ] 2 [ 1] [ ] n n h n u n u n = - - - - - 16. (a) Cannot be determined (b) Cannot be determined (c) False (d) True 17. (a) 1 2 z > (b) ( ) H z is stable (c) 1 1 2 1 1 1 2 ( ) z z X z - - - - = 18. (a) 1 2 1 (1 ) az - - (b) 1 3 2 (1 ) az - - (c) 1 ( 1)! (1 ) k k az - - -
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Curtin University, Australia 2-52 YH Leung (2005, 2006, 2007, 2012, 2016) 19. 1 1 1 1 ( ) 1 1 N N a a X z b z b z - - = + + - - It follows ( ) X z can be written, for appropriate values of 1 c , , 1 N c - , as follows 1 ( 1) 1 2 1 1 1 1 1 2 ( ) (1 )(1 ) (1 ) N N N c c z c z X z b z b z b z - - - - - - - + + + = - - - Suppose the numerator can be factored such that, with no loss of generality, ( ) X z has pole-zero cancellation at 1 z b = . In other words, we can write ( ) X z as follows. 1 1 ( 2) 1 1 1 2 1 1 1 1 2 3 (1 )(1 ) ( ) (1 )(1 ) (1 ) N N c b z d z d z X z b z b z b z - - - - - - - - - + + + = - - - Given the above expression for ( ) X z , suppose we now wish to perform a PFE. The residue 1 a is then given by 1 1 1 1 ( 2) 1 1 1 1 2 1 1 1 1 2 3 (1 )(1 ) (1 ) ( ) 0 (1 ) (1 ) N N z b z b c b z d z d z a b z X z b z b z - - - - - - - - = = - + + + = - = = - - which contradicts the original assumption that 1 0 a ¹ . In other words, ( ) X z cannot exhibit any pole-zero cancellations.
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