2 - 2/10/10 ECE 483 Homework #2 Solution Problem 1. Problem...

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Unformatted text preview: 2/10/10 ECE 483 Homework #2 Solution Problem 1. Problem 2-19 (a)-(c). Note that the results should be in terms of the known constants b 2 , g 1 , g 2 , g 3 , g 4 (forget about the A , A * , p and p * in the problem formulation). Solution: (a) The difference equations are f 1 ( k ) = g 1 f 1 ( k- 1)- g 2 f 2 ( k- 1) + g 3 e ( k ) f 2 ( k ) = g 1 f 2 ( k- 1) + g 2 f 1 ( k- 1) + g 4 e ( k ) y ( k ) = b 2 e ( k ) + f 2 ( k- 1) . (b) Taking the z-transform of the above equations, we have F 1 ( z ) = g 1 z- 1 F 1 ( z )- g 2 z- 1 F 2 ( k ) + g 3 E ( z ) (1) F 2 ( z ) = g 1 z- 1 F 2 ( z ) + g 2 z- 1 F 1 ( z ) + g 4 E ( z ) (2) Y ( z ) = b 2 E ( z ) + z- 1 F 2 ( z ) . (3) From (1), we can solve for F 1 ( z ) as F 1 ( z ) =- g 2 z- 1 F 2 ( k ) + g 3 E ( z ) 1- g 1 z- 1 . Plugging this into (2), we get F 2 ( z ) = g 4 z 2- g 1 g 4 z + g 2 g 3 z ( z- g 1 ) 2 + g 2 2 E ( z ) , which, together with (3), yields the transfer function Y ( z ) E ( z ) = b 2 + z- 1 g 4 z 2- g 1 g 4 z + g 2 g 3 z ( z- g 1 ) 2 + g 2 2 = b 2 + g 4 z- g 1 g 4 + g 2 g 3 ( z- g 1 ) 2 + g 2 2 . (4) (c) After plotting the equivalent signal flow graph (omitted), we find that there are three loops, with gains L 1 = g 1 z- 1 , L 2 = g 1 z- 1 , and L 3 = z- 1 g 2 z- 1 (- g 2 ) =- g 2 2 z- 2 , with the first two being nontouching. Thus, the graph determinant iswith the first two being nontouching....
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2 - 2/10/10 ECE 483 Homework #2 Solution Problem 1. Problem...

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