problem 2

problem 2 - ECE 101 - Linear Systems, Winter 2009 Problem...

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ECE 101 - Linear Systems, Winter 2009 Problem Set # 2 Solutions (send comments/questions to kquest@ucsd.edu) Problem 1 (1.27(c,e), 1.28(c,d) OW2) 1.27(c) y ( t ) = Z 2 t -∞ x ( τ ) (1) Memoryless - no. Because of the integral, y ( t ) depends on time values of x other than t . (2) Time invariant - no Let x 1 ( t ) = x ( t - t 0 ), and x 1 ( t ) y 1 ( t ). Then y 1 ( t ) = R 2 t -∞ x ( τ - t 0 ) Now let y 2 ( t ) = y ( t - t 0 ), and x 2 ( t ) y 2 ( t ). Then, y 2 ( t ) = R 2( t - t 0 ) -∞ x ( τ ) . Using τ 0 = τ + 2 t 0 , we get y 2 ( t ) = R 2 t -∞ x ( τ 0 - 2 t 0 ) 0 . Thus, y 1 ( t ) 6 = y 2 ( t ). (3) Linear - yes. Let x A ( t ) = Ax 1 ( t ) + Bx 2 ( t ), and x A ( t ) y A ( t ). Then y A ( t ) = R 2 t -∞ ( Ax 1 ( τ ) + Bx 2 ( τ )) . Now let y B ( t ) = Ay 1 ( t ) + By 2 ( t ), and x B ( t ) y B ( t ). It follows that y B ( t ) = A R 2 t -∞ x 1 ( τ ) + B R 2 t -∞ x 2 ( τ ) . Thus, y A ( t ) = y B ( t ). (4) Causal - no. The upper limit on the integral is at a future time. (5) Stable - no. Assume | x ( t ) | = B for all t , where B is a finite, positive- definite number. It follows that y ( t ) = R 2 t -∞ B dτ = B R 2 t -∞ = 1.27(e) y ( t ) = 0 x ( t ) < 0 x ( t ) + x ( t - 2) x ( t ) 0 (1) Memoryless - no. Because of the term x ( t - 2), y ( t ) depends on time values of x other than t . 1
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(2) Time invariant - yes Let x 1 ( t ) = x ( t - t 0 ). Then y 1 ( t ) = 0 x ( t - t 0 ) < 0 x ( t - t 0 ) + x ( t - t 0 - 2) x ( t - t 0 ) 0 Now let y 2 ( t ) = y ( t - t 0 ). Then, y 2 ( t ) = 0 x ( t - t 0 ) < 0 x ( t - t 0 ) + x ( t - t 0 - 2) x ( t - t 0 ) 0 Thus, y 1 ( t ) = y 2 ( t ). (3) Linear - no. Let
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problem 2 - ECE 101 - Linear Systems, Winter 2009 Problem...

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