# A7_SOL - Signals and Systems Fall 2013 Solution to Homework...

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Signals and Systems Fall 2013 Solution to Homework Assignment #7 1. (a) F ( z ) 2 3 z 1 6 z 3 4 z 7 Taking the inverse z-transform:     2 3 1 6 3 4 7 f k k k k k (b) F ( z ) ) )( 1 ( 1 2 T e z z z , T 2 1   1 1 1 z F z z z e Using partial-fraction expansion, we have   1 1 2 1 1 1 1 1 1 1 1 1 z z z F z A B C z z e z z e A B C z Ae B C e Ce z z e   Equating the numerator of the fraction F ( z ): 1 1 1 1 1 2 1 0 1 1 1 1 1 A e A B C Ae B C e B e e C e Ce   This yields: 1 1 1 2 1 ( ) 1 1 1 z e z F z e e z e z e       Taking the inverse z-transform by looking up from table, we obtain: 1 1 2 1 ( ) ( ) ( ) ( ) 1 1 k s s e f k u k e u k e k e e Note that you can use the technique introduced in class to find A,B,C (see solution to (e)). (c)   6 6 6 7 1 1 1 1 1 ( 1) 1 1 1 1 z z F z z z z z z z z z z z z Taking the inverse z-transform yields:   6 ( ) 1 6 7 s s s s f k E u k u k u k u k

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(d)    2 1 1 3 2 1 2 F z z z z z z z Analyzing F ( z ) as following:     2 1 1 2 1 2 2 3 2 1 2 z z z F z A B C z z z z A B C z A B C z C z z   Equating the numerator of the fraction F ( z ): 0 2 2 3 1 3 2 2 1 1 2 A B C A A B C B C C   This yields: 3 1 ( ) 2 1 2 2 2 z z F z z z   Taking the inverse z-transform by looking up from table, we obtain: 3 1 ( ) 2 ( ) 2 ( ) ( ) 2 2 k s s f k u k u k k   Note that you can use the technique introduced in class to find A,B,C (see solution to (e)).
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