hw04-f06-sol

# hw04-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 4 SOLUTIONS HW 4.1 (1) (a) The system cannot be memoryless because same input values are mapped into different output values depending if their time stamp was positive or negative (b) The system must be causal. Consider x 1 ( n ) and x 2 ( n ) such that x 1 ( n ) = x 2 ( n ) , n n 0 , then for the definition of the system we will also have y 1 ( n ) = y 2 ( n ) , n n 0 , hence it is causal. (c) The system must be linear. Consider ˆ x ( n ) = ax 1 ( n ) + bx 2 ( n ) ... (d) The system is not time invariant. Consider ˆ x ( n ) = x ( n N ) , it follows ˆ y ( n ) = braceleftBigg ˆ x ( n ) n < 0 x ( n ) n 0 = braceleftBigg x ( n N ) n < 0 + x ( n N ) n 0 On the other hand: y ( n N ) = braceleftBigg x ( n N ) n N < 0 + x ( n N ) n N 0 which is different from ˆ y . (2) (a) The system must be memoryless and the function f is exactly the ceiling function f ( x ) = x . (b) The system must be causal, because it is memoryless and memoryless causal . (c) The system cannot be linear. Consider ˆ x ( t ) = ax 1 ( t ) + bx 2 ( t ) , then F 2 x )( t ) = ax 1 ( t )+ bx 2 ( t ) which is different from a x 1 ( t ) + b x 2 ( t ) . (Assume for instance that a = b = 1 and that for a given time t 0 , we have x 1 ( t 0 ) = x 2 ( t 0 ) = 1 2 , in this case ax 1 ( t ) + bx 2 ( t ) ⌉ negationslash = a x 1 ( t ) + b x r ( t ) ).) (d) The system must be time invariant. (3) (a) The system cannot be memoryless. Consider N < t 0 < N +1 where N is a Natural number, y ( t 0 ) = x ( N ) , hence it is not memoryless. (b) The system must be causal, because it is looking at current or past values of the input. (c) The system must be linear. Consider ˆ x ( t ) = ax 1 ( t ) + bx 2 ( t ) , then F 3 x )( t ) = ax 1 ( t ) + bx 2 ( t ) = aF 3 ( x 1 )( t ) + bF 3 ( x 2 )( t ) .

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