ee20-hw03-f09-sol - EECS 20N: Structure and Interpretation...

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Unformatted text preview: EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences UNIVERSITY OF CALIFORNIA BERKELEY Problem Set 3 SOLUTIONS HW 3.1 ( F 1 ( x ))( n ) = x ( n ) (a) cannot be memoryless. ( F 1 ( x ))( n ) does not depend only on x at time n , but on x at time n . (b) cannot be causal. For example, ( F 1 ( x ))( 1) = x (1) Output at n = 1 depends on input at n = 1 F 1 not causal F 1 not memoryless. (c) The system must be linear ( F 1 ( x 1 ))( n ) = x 1 ( n ) ( F 1 ( x 2 ))( n ) = x 2 ( n ) x 3 = a ( x 1 )( n ) + bx 2 ( n ) ( F 1 ( x 3 ))( n ) = a ( x 1 )( n ) + bx 2 ( n ) ? = a ( F 1 ( x 1 ))( n ) + b ( F 1 ( x 2 ))( n ) = a ( x 1 )( n ) + bx 2 ( n ) TRUE! (d) The system cannot be time-invariant Let y ( n ) = ( F 1 ( x ))( n ) y ( n ) = x ( n ) y ( n N ) = x ( ( n N )) n, let x ( n ) = x ( n N ) ( F 1 ( x ))( n ) = y ( n ) = x ( n ) = x ( n N ) (e) Notice that the output signal F 1 ( x ) depends upon the values of the input signal at different points in discrete-time. In other words, if we were to feed the bounded signal x to the system F 1 , such that | x ( n ) | B for all n ( B R ), then we are assured that the output signal y ( n ) is also bounded by B , since it depends upon the values of x at different points in discrete-time. Thus, the system is BIBO stable . ( F 2 ( x ))( t ) = x ( e | t | ) (a) cannot be memoryless; See part (b). 1 (b) cannot be causal; Let t = 1 ( F 2 ( x ))( 1) = x ( e | 1 | ) = x ( 1 e ) non causal not memoryless. Note that at t = 1 , ( F 2 ( x ))( t ) depends not only on x at t = 1 , but at t = 1 e , which is at a future time with respect to 1 . (c) The system must be linear ( F 2 ( x 1 ))( t ) = x 1 ( e | t | ) ( F 2 ( x 2 ))( t ) = x 2 ( e | t | ) x 3 = a ( x 1 )( t ) + bx 2 ( t ) ( F 2 ( x 3 ))( t ) = ax 1 ( e | t | ) + bx 2 ( e | t | ) ? = a ( F 2 ( x 1 ))( n ) + b ( F 2 ( x 2 ))( n ) TRUE ! (d) The system cannot be time-invariant Let y ( t ) = ( F 2 ( x ))( t ) y ( t ) = x ( e | t | ) y ( t T ) = x ( e | t T | ) t, let x ( t ) = x ( t T ) ( F 2 ( x ))( t ) = y ( t ) = x ( e | t | ) = x ( e | t | T ) (e) The system is BIBO stable , by same reasoning as given for F 1 . ( F 3 ( x ))( n ) = x (2 n ) (a) cannot be memoryless. Consider the input signal x ( n ) = ( n ) + ( n 2) . In this case, the output signal is given by y ( n ) = ( n ) + ( n 1) . (Why?) Then, even though x (0) = x (2) , y (0) negationslash = y (2) , and so the system is not memoryless....
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This note was uploaded on 12/24/2011 for the course ELECTRICAL 20 taught by Professor Babak during the Spring '11 term at University of California, Berkeley.

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ee20-hw03-f09-sol - EECS 20N: Structure and Interpretation...

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