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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 rightsided sequence as in (1), R.O.C is outside; for a leftsided 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 rightsided), 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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