# soln4 - EECS 451 SOLUTIONS TO PROBLEM SET #4 1. y (0) = y...

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EECS 451 SOLUTIONS TO PROBLEM SET #4 1. y (0) = y ( - 1)+ y ( - 2)+ x (0) = 0+0+1 = 1 . y (2) = y (1)+ y (0) = 1+1 = 2. Clearly y (1) = y (0) + y ( - 1) + x (1) = 1 + 0 + 0 = 1 . y (3) = y (2) + y (1) = 2 + 1 = 3. works. Take 2-sided Z : H ( z ) = z 1 H ( z ) + z 2 H ( z ) + 1 H ( z ) = 1 1 z 1 z 2 (see p.202). Partial fraction: H ( z ) = φ/ 5 1 φz 1 - φ / 5 1 φ z 1 h ( n ) = 1 5 ( φ n +1 - ( φ ) n +1 ) u ( n ) where φ = 1+ 5 2 = 1 . 618=golden ratio and φ = 1 5 2 = - 1 φ = 1 - φ (see why?) Note: For n 10 or so, h ( n ) 1 5 φ n +1 h ( n +1) h ( n ) φ ; hence ”golden ratio.” 2b. Y ( z ) - 1 . 5( z 1 Y ( z ) + 1) + 0 . 5( z 2 Y ( z ) + 0 + 1 z 1 ) = 0 Y ( z ) = 1 . 5 0 . 5 z 1 1 1 . 5 z 1 +0 . 5 z 2 Y ( z ) = 2 1 z 1 - 0 . 5 1 0 . 5 z 1 y ( n ) = 2 u ( n ) - 0 . 5(0 . 5) n u ( n ) = (2 - (0 . 5) n +1 ) u ( n ). 2c.

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## This note was uploaded on 12/10/2009 for the course EECS 451 taught by Professor Andrewyagle during the Spring '08 term at University of Michigan.

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soln4 - EECS 451 SOLUTIONS TO PROBLEM SET #4 1. y (0) = y...

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