630_rec_1_sol - ENEE630 ADSP RECITATION 1 w/ solution...

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Unformatted text preview: ENEE630 ADSP RECITATION 1 w/ solution September 10/11, 2009 1. Define H ( z ) = ∑ + ∞ n =-∞ h ( n ) z- n . (1). If h ( n ) = a n u ( n ), what is H ( z )? (2). If h ( n ) =- a n u (- n- 1), what is H ( z )? (3). Given H ( z ) = 3 1- . 5 z- 1 + 1 1+0 . 75 z- 1 + 1 1- 2 z- 1 , what are the possible h ( n )? Discuss whether the systems are causal or stable. Solution: (1) By definition. H ( z ) = ∑ + ∞ n =0 a n z- n = 1 1- az- 1 with R.O.C | z | > | a | . (2) By definition. H ( z ) = ∑- 1 n =-∞- a n z- n = 1 1- az- 1 with R.O.C | z | < | a | . (3) Four possible cases. i. | z | < . 5, h ( n ) =- 3(0 . 5) n u (- n- 1)- (- . 75) n u (- n- 1)- 2 n u (- n- 1). ii. 0 . 5 < | z | < . 75, h ( n ) = 3(0 . 5) n u ( n )- (- . 75) n u (- n- 1)- 2 n u (- n- 1). iii. 0 . 75 < | z | < 2, h ( n ) = 3(0 . 5) n u ( n ) + (- . 75) n u ( n )- 2 n u (- n- 1). iv. | z | > 2, h ( n ) = 3(0 . 5) n u ( n ) + (- . 75) n u ( n ) + 2 n u ( n ). Only the fourth case is causal by definition. Only the third case is stable since the unit circle | z | = 1 lies inside the R.O.C. Note: 1. R.O.C is important. For a right-sided sequence as in (1), R.O.C is outside; for a left-sided sequence as in (2), R.O.C is inside. 2. The general condition of a system being stable is that the unit circle ( | z | = 1) lies within the R.O.C. However, since usually we are interested in the causal system (must be right-sided), itsthe R....
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This note was uploaded on 10/31/2010 for the course EE 630 taught by Professor Wu during the Spring '10 term at Aarhus Universitet, Aarhus.

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630_rec_1_sol - ENEE630 ADSP RECITATION 1 w/ solution...

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