20112ee113_1_Hw2-Solution-Updated

# 20112ee113_1_Hw2-Solution-Updated - Spring 2011 EE113:...

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EE113: Digital Signal Processing Spring 2011 Prof. Mihaela van der Schaar (Instructor) Homework #2 Solutions Prepared by Khoa T Phan and Yuanzhang Xiao (TAs) Problem 4.4. a) True. However, y ( n ) is not time-invariant. b) False. y ( n ) = x (2 n ) is composed of all the even numbered samples of x ( n ). Shifting the input by one sample will change the output to all odd numbered samples of x ( n ), and consequently, the system cannot be time-invariant. c) False. The system reverses time, and hence cannot be causal. d) False. If the sequence x ( n ) is bounded, so is the product of a ±nite number of samples. The system is nonlinear, non-causal, and time-varying. Problem 4.8. a) y ( n ) = ln[ | x ( n ) | + 1] is nonlinear, time-invariant, causal, BIBO stable, and relaxed. We have ln[ | x 1 ( n ) | +1]+ln[ | x 2 ( n ) | +1] n = ln[ | x 1 ( n )+ x 2 ( n ) | +1] for two (arbitrary) input sequences x 1 ( n ) and x 2 ( n ): non-linear We have ln[ | x ( n k ) | + 1] = y ( n k ): time-invariant The system is causal since the output at time n depends on the input at time n only. If | x ( n ) | ≤ A for all n for some number A then y ( n ) ln[ A + 1] = B for all n : BIBO stable. The system is relaxed since the output will be zero as long as the input is zero. b) y ( n ) = y ( n 1) + x ( n ), y ( 1) = 0 is linear, not time-invariant, non-causal, not stable (The input x ( n ) = u ( n ) has output as y ( n ) = n + 1 for all n 0. Hence lim n →∞ y ( n ) = and the system is not BIBO stable), and not relaxed. Consider input x ( n ) = u ( n + 2). Then y ( n ) = n + 1 ,n 0 0 ,n = 1 1 ,n ≤ − 2 (1) 1

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Then, lim n →∞ y ( n ) = , hence, the system is not BIBO stable. Since
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## This note was uploaded on 10/07/2011 for the course EE 113 taught by Professor Walker during the Spring '08 term at UCLA.

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20112ee113_1_Hw2-Solution-Updated - Spring 2011 EE113:...

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