SOL08 - Signals and Systems Fall 2012 Solution to Homework...

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Signals and Systems Fall 2012 Solution to Homework Assignment #8 1. (a) (E + 0.5){ y ( k )} = 0, y (0) = 2. ( ) ( ) 1 0.5 0 y k y k + + = Taking the z-transform, we have: ( ) ( ) ( ) ( ) ( ) 0 0.5 0 2 0.5 0 z Y z y Y z zY z z Y z + = + = ( ) 2 0.5 ZI z Y z z = + It then follows that: ( ) ( ) ( ) ( ) 2 0.5 k ZI s y k y k u k = = (b) (E 2 3E + 2){ y ( k )} = u s ( k ), y (0) = 2, y (1) = 1. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 0 1 3 0 2 3 2 1 3 0 S S z Y z y z y z Y z y Y z U z z z Y z U z z y z z y + = + = + + ( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( ) 2 2 2 2 2 2 2 2 1 3 2 3 2 1 3 2 3 2 2 6 2 1 2 1 2 1 2 5 2 1 2 1 ZIR ZSR ZIR ZSR ZIR ZSR z z z z Y z z z z z z z z z z z z z z z z z z z z z z z z z = + + + + + = + + = +       Using partial-fraction expansion, we have: ( ) ( )( ) ( ) 2 2 2 1 2 1 1 ZSR z z z z Y z A B C z z z z z = = + + Multiply ( ) ZSR Y z by ( ) 2 z and set 2 z = , we have: ( ) 2 2 2 1 2 1 A A = × = Multiply ( ) ZSR Y z by ( ) 2 1 z and set 1 z = , we have: ( ) 1 1 1 1 2 C C = × ⇒ =− Then let z = ∞ , we have: 0 1 A B B = + =
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