Homework_4_solution(1)

Homework_4_solution(1) - 1 EE113 Homework 4 Problem 4.7...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the 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? time-invariant? stable? Solution: Yes, it is linear, causal, time-invariant, 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 time-invariability: 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 time-invariant 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.

Page1 / 12

Homework_4_solution(1) - 1 EE113 Homework 4 Problem 4.7...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online