Homework_4_solution(1)

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

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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 ) ....
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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_4_solution(1) - 1 EE113 Homework 4 Problem 4.7...

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