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hw3sol - Instructor Christopher Wayne Walker Ph.D TA...

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Instructor: Christopher Wayne Walker, Ph.D. TA: Federico Cattivelli EE132A Winter 2008 EE 132A Homework 3 Solutions Problem 1. Proakis & Salehi 2.16 parts 2 and 13. Solution: 2) Nolinear, if we multiply the input by a constant -1, the output does not change. In a linear system the output should be scaled by -1. 13) Linear, we can write the output of this feedback system as y ( t ) = x ( t ) + y ( t 1) = summationdisplay n =0 x ( t n ) Then for x ( t ) = αx 1 ( t ) = βx 2 ( t ) y ( t ) = summationdisplay n =0 [ αx 1 ( t n ) + βx 2 ( t n )] = α summationdisplay n =0 x 1 ( t n ) + β summationdisplay n =0 x 2 ( t n ) = αy 1 ( t ) + βy 2 ( t ) Problem 2. Proakis & Salehi 2.24 parts 1 and 9. Solution: 1) The invariant: The response to x ( t t 0 ) is 2 x ( t t 0 ) + 3 which is y ( t t 0 ). 9) Time-invariant system. Writing y ( t ) as n =0 x ( t n ) we get y ( t t 0 ) = summationdisplay n =0 x ( t t 0 n ) = T [ x ( t t 0 )] Problem 3. Proakis & Salehi 5.6. Solution: 1) X can take four different values: 0 if no head shows up, 1 if only one head shows up in the four flips of the coin, 2 for two heads, and 3 if the outcome of each flip is head. 2) X follows the binomial distribution with n = 3. Thus P ( X = k ) = parenleftbigg 3 k parenrightbigg p k (1 p ) 3 - k for 0 k 3 0 otherwise 3) F X ( k ) = k summationdisplay m =0 parenleftbigg 3 m parenrightbigg p m (1 p ) 3 - m 1
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Instructor: Christopher Wayne Walker, Ph.D. TA: Federico Cattivelli EE132A Winter 2008 Hence F X ( k ) = 0 k < 0 (1 p ) 3 k = 0 (1 p ) 3 + 3 p (1 p ) 2 k = 1 (1 p ) 3 + 3 p (1 p ) 2 + 3 p 2 (1 p ) k = 2 (1 p ) 3 + 3 p (1 p ) 2 + 3 p 2 (1 p ) + p 3 = 1 k = 3 1 k > 3 A plot of F X ( k ) is shown in Figure 1. 0
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