hw03-f06-sol

# hw03-f06-sol - EECS 20N Structure and Interpretation of...

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EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences U NIVERSITY OF C ALIFORNIA B ERKELEY Problem Set 3 SOLUTIONS HW 3.1 ( F 1 ( x ))( n ) = x ( - n ) (a) cannot be memoryless; (b) cannot be causal. For example, ( F 1 ( x ))( - 1) = x (1) Output at t = - 1 depends on input at t = 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 ˆ x ( n ) = x ( n - N ) ( F 1 x ))( n ) = ˆ y ( n ) = ˆ x ( - n ) = x ( - n - N ) ( F 2 ( x ))( t ) = x ( e -| t | ) (a) cannot be memoryless; (b) cannot be causal; Let t = - 1 ( F 2 ( x )( - 1) = x ( e -|- 1 | ) = x ( 1 e ) non causal not memoryless. (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 ! 1

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(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 ˆ x ( t ) = x ( t - T ) ( F 2 x ))( t ) = ˆ y ( t ) = ˆ x ( e -| t | ) = x ( e -| t | - T ) ( F 3 ( x ))( n ) = x (2 n ) (a) cannot be memoryless; (b) cannot be causal. Let t = 1 ( F 3 ( x )(1) = x (2) peeks ahead non causal not memoryless. (c) The system must be linear ( F 3 ( x 1 )( n ) = x 1 (2 n ) ( F 3 ( x 2 )( n ) = x 2 (2 n ) x 3 = a ( x 1 )( n ) + bx 2 ( n ) ( F 3 ( x 3 )( n ) = a ( x 1 )(2 n ) + bx 2 (2 n ) ? = a ( F 3 ( x 1 ))( n ) + b ( F 3 ( x 2 )( n ) TRUE ! (d) The system cannot be time-invariant Let y ( n ) = ( F 3 ( x ))( n ) y ( n ) = x (2 n ) y ( n - N ) = x (2( n - N )) n, let ˆ x ( n ) = x ( n - N ) ( F 3 x ))( n ) = ˆ y ( n ) = ˆ x (2 n ) = x (2 n - N ) ( F 4 ( x ))( t ) = x ( t/ 2) (a) cannot be memoryless; (b) cannot be causal. Let t = - 1 ( F 4 ( x )( - 1) = x ( - 1 2 ) peeks ahead non causal not memoryless. 2
(c) The system must be linear ( F 4 ( x 1 )( t ) = x 1 ( t/ 2) ( F 4 ( x 2 )( t ) = x 2 ( t/ 2) x 3 = a ( x 1 )( t ) + bx 2 ( t ) ( F 4 ( x 3 )( t ) = a ( x 1 )( t/ 2) + bx 2 ( t/ 2) ? = a ( F 4 ( x 1 ))( t ) + b ( F 4 ( x 2 )( t ) TRUE ! (d) The system cannot be time-invariant Let y ( t ) = ( F 4 ( x ))( t ) y ( t ) = x ( t/ 2) y ( t - T ) = x (( t - T ) / 2) t, let ˆ x ( t ) = x ( t - T ) ( F 4 x ))( t ) = ˆ y ( t ) = ˆ x ( t/ 2) = x ( t/ 2 - T ) ( F 5 ( x ))( t ) = 1 C t -∞ x ( τ ) d τ , C (a) cannot be memoryless; (b) must be causal. Let x 1 , x 2 [ R R ] be input signals such that x 1 ( τ ) = x 2 ( τ ) , τ t ( F 5 ( x 1 ))( t ) = 1 c t -∞ x 1 ( τ ) d τ = 1 c t -∞ x 2 ( τ ) d τ ( F 5 ( x 1 ))( t ) = ( F 5 ( x 2 ))( t ) Since t was selected to be arbitrary equality ( t )

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