Homework_7_solution

Homework_7_solution - 1 EE113 Homework 7 Problem 10.13...

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Unformatted text preview: 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 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 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...
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This note was uploaded on 01/24/2012 for the course EE 113 taught by Professor Walker during the Fall '08 term at UCLA.

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Homework_7_solution - 1 EE113 Homework 7 Problem 10.13...

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