This preview shows pages 1–5. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 righthand side sequence as x ( n ) = parenleftbigg 1 √ 2 parenrightbigg n 4 sin parenleftBig π 4 ( n 3) parenrightBig u ( n 3) or a lefthand 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 ztransform is outside of the unit disk, we conclude that it is a causal sequence. November 8, 2011 DRAFT 3 Problem 10.21 Use the ztransform 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 ztransform 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 ztransform 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 ztransform 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 zeroinput and zerostate 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 zeroinput 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...
View
Full
Document
This note was uploaded on 01/24/2012 for the course EE 113 taught by Professor Walker during the Fall '08 term at UCLA.
 Fall '08
 Walker

Click to edit the document details