This preview shows pages 1–4. 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 4 Problem 4.7 Consider the moving average system with exponential weighting y ( n ) = 1 M + 1 M summationdisplay k =0 λ k x ( n − k ) where  λ  < 1 . Is the system linear? causal? timeinvariant? stable? Solution: Yes, it is linear, causal, timeinvariant, and stable. The reason is (a) for linearity: for the input sequence αx 1 ( n ) + βx 2 ( n ) , where α and β are two constant scalars, the output sequence y ( n ) should be y ( n ) = 1 M + 1 M summationdisplay k =0 λ k [ αx 1 ( n − k ) + βx 2 ( n − k )] = α M + 1 M summationdisplay k =0 λ k x 1 ( n − k ) + β M + 1 M summationdisplay k =0 λ k x 2 ( n − k ) = αy 1 ( n ) + βy 2 ( n ) where y k ( n ) is the output sequence for the input sequence x k ( n ) . (b) for causality: the output y ( n ) only depends on the input { x ( n ) ,x ( n − 1) ,...,x ( n − M ) } for any n . (c) for timeinvariability: for the input sequence x 1 ( n ) is a shifted version of x ( n ) ,i.e.,x 1 ( n ) = x ( n − m ) for any constant integer m , the output sequence y 1 ( n ) should be y 1 ( n ) = 1 M + 1 M summationdisplay k =0 λ k x 1 ( n − k ) = 1 M + 1 M summationdisplay k =0 λ k x ( n − m − k ) = y ( n − m ) which is the shifted version of y ( n ) by the same shifts m as the input sequence x 1 ( n ) . (d) for stability: if the input sequence x ( n ) is bounded by  x ( n )  ≤ B x < ∞ , then the output sequence y ( n ) can be bounded as  y ( n )  = vextendsingle vextendsingle vextendsingle 1 M + 1 M summationdisplay k =0 λ k x ( n − k ) vextendsingle vextendsingle vextendsingle ≤ 1 M + 1 M summationdisplay k =0  λ k  ·  x ( n − k )  ≤ 1 M + 1 M summationdisplay k =0  λ  k · B x = B x ( M + 1) 1 −  λ  M +1 1 −  λ  October 25, 2011 DRAFT 2 Problem 4.9 True or False: (a) A relaxed system is linear. (b) A linear system is relaxed. Solution: (a) A relaxed system is not necessarily linear. Here is a counterexample: y ( n ) = x 2 ( n ) u ( n ) (b) A linear system is relaxed. This can be shown by contradiction. Say there is a linear, but not relaxed system S such that y ( n ) = S [ x ( n )] where x ( n ) is the input sequence and y ( n ) is the output sequence. There should be a time instant n o such that the input sequence x ( n ) = 0 if n ≤ n o but the output sequence y ( n o ) negationslash = 0 . Then it can be verified that the linearity does not hold at the time instant n o because S [ αx ( n o )] = S [ α × 0] = S [0] = S [ x ( n o )] = y ( n o ) negationslash = αy ( n o ) for a nonzero constant scalar α October 25, 2011 DRAFT 3 Problem 4.11 The response of a linear timeinvariant system to x ( n ) = u ( n ) is y ( n ) = 0 . 5 n u ( n ) . Find its response to δ ( n ) ....
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