HW4 - ECE-2523 Homework 4 Solutions February 15, 2009 3.3...

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ECE-2523 Homework 4 Solutions February 15, 2009 3.3 (a) x a [ n ] = α | n | , 0 < | α | < 1, n = sign ( n ) n = ( - n ,n < 0 n ,n 0 X ( z ) = - 1 X n = -∞ α - n z - n + X n =0 α n z - n The first sum is - 1 X n = -∞ α - n z - n = X m =1 α m z m = X m =0 α m z m - 1 = 1 1 - αz - 1 , provided | αz | < 1 or | z | < 1 | α | = αz 1 - αz So X ( z ) = αz 1 - αz + 1 1 - αz - 1 , | z | < 1 α ∩ | z | > α = αz 1 - αz + z z - α = z (1 - α 2 ) (1 - αz )( z - α ) , α < | z | < 1 α 1
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(b) x [ n ] is a pulse train of length N. X ( z ) = N - 1 X n =0 z - n = 1 - z - N 1 - z - 1 = z N - 1 z N - 1 ( z - 1) ,z 6 = 0 The reason the ROC doesn’t depend on 1 is due to a pole/zero cancellation. The zeros are the roots of z N - 1 = 0 or z N = 1 z N = e j 2 π z k = e j 2 π N k k = 0 , ..., N - 1 Poles are the roots of z N - 1 ( z - 1) = 0, or z = 1, z = 0 of order N - 1 (c) Let y [ n ] = x b [ n ] * x b [ n ] and Y ( z ) = X b ( z ) 2 . Then we have
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HW4 - ECE-2523 Homework 4 Solutions February 15, 2009 3.3...

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