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Homework_7_solution

# Homework_7_solution - 1 EE113 Homework 7 Problem 10.13...

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1 EE113 Homework 7 Problem 10.13 Determine all possible sequences x ( n ) with z –transform X ( z )= z - 2 z 2 - z + 1 2 Solution: Note that X ( z ) can be rewritten as X ( z )= 2 z - 3 · z sin ( π 4 ) z 2 - 2 1 2 z cos ( π 4 ) + 1 ( 2) 2 Therefore, we can obtain a right-hand side sequence as x ( n )= parenleftbigg 1 2 parenrightbigg n - 4 sin parenleftBig π 4 ( n - 3) parenrightBig u ( n - 3) or a left-hand side sequence as x ( n )= parenleftBig 2 parenrightBig - n +5 sin parenleftBig π 4 ( - n +3) parenrightBig u ( - n +3) November 8, 2011 DRAFT

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2 Problem 10.17 Let X ( z )= 1 z - 1 2 · 1 z 2 +1 , | z | > 1 Decide whether x ( n ) is a causal sequence without inverting the transform. Solution: Since the ROC of the z -transform is outside of the unit disk, we conclude that it is a causal sequence. November 8, 2011 DRAFT
3 Problem 10.21 Use the z -transform technique to evaluate the convolution parenleftbigg 1 2 parenrightbigg n - 1 u ( n - 2) bracketleftBigg 1+ parenleftbigg 1 3 parenrightbigg n +1 bracketrightBigg u ( n ) Solution: The z -transform of x ( n )= 1 2 · ( 1 2 ) n - 2 u ( n - 2) is X ( z )= z - 2 2 · z z - 1 2 , | z | > 1 2 The z -transform of w ( n )= 1 3 · ( 1 3 ) n u ( n ) is W ( z )= 1 3 · z z - 1 3 , | z | > 1 3 Since y ( n )= x ( u ) [ u ( n )+ w ( n )]= x ( u ) ⋆ u ( n )+ x ( n ) ⋆ w ( n ) the z -transform of y ( n ) can be expressed as Y ( z )= X ( z ) · U ( z )+ X ( z ) · W ( z ) = z - 2 2 · z z - 1 2 · z z - 1 + z - 2 2 · z z - 1 2 · 1 3 · z z - 1 3 = 1 2 bracketleftbigg A z - 1 2 + B z - 1 bracketrightbigg + 1 6 bracketleftbigg C z - 1 2 + D z - 1 3 bracketrightbigg , | z | > 1 where A = - 2 , B =2 , C =6 , D = - 6 Then, Y ( z ) is Y ( z )= - 1 z - 1 2 + 1 z - 1 + 1 z - 1 2 - 1 z - 1 3 = 1 z - 1 - 1 z - 1 3 , | z | > 1 Thus, we can get y ( n ) as y ( n )= u ( n - 1) - parenleftbigg 1 3 parenrightbigg n - 1 u ( n - 1)= bracketleftBigg 1 - parenleftbigg 1 3 parenrightbigg n - 1 bracketrightBigg u ( n - 1) November 8, 2011 DRAFT

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4 Problem 10.27 Consider a causal system that is described by the difference equation y ( n )= 5 6 y ( n - 1) - 1 6 y ( n - 2)+ x ( n - 2) , y ( - 2)=0 , y ( - 1)=1 Determine its complete response to the sequence x ( n )=( n - 1) parenleftbigg 1 4 parenrightbigg n - 2 u ( n - 1) Solution: We shall find the complete response sequence y ( n ) by the zero-input and zero-state responses. First of all, the general solution for the homogeneous equation can be written as y h ( n )= C 1 parenleftbigg 1 2 parenrightbigg n + C 2 parenleftbigg 1 3 parenrightbigg n Substituting the initial conditions y ( - 2)=0 and y ( - 1)=1 yields C 1 = 3 2 , C 2 = - 2 3 Hence, the zero-input response sequence can be written as y zi ( n )= bracketleftbigg 3 2 parenleftbigg 1 2 parenrightbigg n - 2 3 parenleftbigg 1 3 parenrightbigg n bracketrightbigg u ( n +2) The transfer function of the system is H ( z )= Y ( z ) X ( z ) = z - 2 1 - 5 6 z - 1 + 1 6 z - 2 = 1 ( z - 1 2 ) · ( z - 1 3 ) , | z | > 1 2 The z -transform of the input sequence x ( n )=4( n - 1) ( 1 4 ) n - 1 u ( n - 1) can be expressed as X ( z )=4 z - 1 1 4 z ( z - 1 4 ) 2 = 1 ( z - 1 4 ) 2 , | z | > 1 4 Thus, the z
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Homework_7_solution - 1 EE113 Homework 7 Problem 10.13...

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