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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 (4) f 2 = p ( g 3 g 4 h 3 h 4 ) = s g 3 g 4 1000 y (5) f 3 = p ( g 5 g 6 h 5 h 6 ) = s g 5 g 6 500 z (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 ∂f i ∂x + ∂f i ∂y + ∂f i ∂z . 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 3 (17) Since the right-hand side of the equations above are the same we get the following relationships y - 3 2 = 600 g 1 g 2
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solution_Exam_F09 - 1 Exact Solution The delay through each...

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