solution_Exam_F09

solution_Exam_F09 - 1 Exact Solution The delay through each...

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1 Exact Solution The delay through each path can be written as follows D 1 = X p i + 3 f + 2 f 1 (1) D 2 = X p j + 3 f + 2 f 2 (2) D 3 = X p k + 3 f + 2 f 3 (3) where f,f 1 ,f 2 ,f 3 are the stage eﬀort of the common path and each of the indivdual paths after the branch. Let the input device capcitance of paths 1,2,3 be xC inv ,yC inv ,zC inv respec- tively. f 1 = p ( g 1 g 2 h 1 h 2 ) = s ± g 1 g 2 600 x ² (4) f 2 = p ( g 3 g 4 h 3 h 4 ) = s ± g 3 g 4 1000 y ² (5) f 3 = p ( g 5 g 6 h 5 h 6 ) = s ± g 5 g 6 500 z ² (6) f = 3 p (2 g 7 g 8 g 9 h 7 h 8 h 9 ) = 3 p (2 g 7 g 8 g 9 ( x + y + z )) (7) Since this is a constraint n three vairables the minimum delay will occur when ∂D 1 ∂x + ∂D 1 ∂y + ∂D 1 ∂z = 0 (8) ∂D 2 ∂x + ∂D 2 ∂y + ∂D 2 ∂z = 0 (9) 1

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∂D 3 ∂x + ∂D 3 ∂y + ∂D 3 ∂z = 0 (10) This should intuitvely remind you of a gradient descent, since a sum of partial derivatives of three orthogonal coordinates is a gradient. We know that ∂D i ∂x + ∂D i ∂y + ∂D i ∂z = 3 ± ∂f ∂x + ∂f ∂y + ∂f ∂z ² + 2 ± ∂f i ∂x + ∂f i ∂y + ∂f i ∂z ² . Computing the derivatives we have ∂f ( x,y,z ) = 1 3 3 p 2 g 7 g 8 g 9 ( x + y + z ) - 2 3 (11) ∂f 1 ∂x = - 1 2 p 600 g 1 g 2 x - 3 2 (12) ∂f 2 ∂y = - 1 2 p 1000 g 3 g 4 y - 3 2 (13) ∂f 3 ∂z = - 1 2 p 500 g 5 g 6 z - 3 2 (14) The other derivatives all compute to zero. Substituting the derivative in the minimum delay constraint we get 600 g 1 g 2 3 2 g 7 g 8 g 9 x - 3 2 = 3 ( x + y + z ) - 2 3 (15) 1000 g 3 g 4 3 2 g 7 g 8 g 9 y - 3 2 = 3 ( x + y + z ) - 2 3 (16) 500 g 5 g 6 3 2 g 7 g 8 g 9 z - 3 2 = 3 ( x + y + z ) - 2
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This note was uploaded on 01/16/2012 for the course ECE 223 taught by Professor Luketheogarajan during the Fall '11 term at UCSB.

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solution_Exam_F09 - 1 Exact Solution The delay through each...

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