sol6

# sol6 - Chapter 6 6.1 w En y w 1 y 1 y 2 y 3 y 7 y 6 y 5 y 4...

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Unformatted text preview: Chapter 6 6.1. w En y w 1 y 1 y 2 y 3 y 7 y 6 y 5 y 4 w 2 f 1 w 1 w 2 w 3 6.2. w En y w 1 y 1 y 2 y 3 y 7 y 6 y 5 y 4 w 2 f 1 w 1 w 2 w 3 6.3. 1 1 1 1 1 1 1 1 1 1 1 1 w 1 w 2 w 3 f 1 1 1 1 1 f w 1 w 2 w 3 + w 2 w 3 f w 3 w 1 w 2 6-1 6.4. 1 1 1 1 1 1 1 1 1 1 1 1 w 1 w 2 w 3 f 1 1 1 1 1 f w 1 w 2 w 3 w 2 w 3 + f w 3 w 1 w 2 6.5. The function f can be expressed as f = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 Expansion in terms of w 1 produces f = w 1 ( w 2 + w 3 ) + w 1 ( w 2 w 3 ) The corresponding circuit is f w 2 w 1 w 3 6.6. The function f can be expressed as f = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 Expansion in terms of w 2 produces f = w 2 ( w 3 ) + w 2 ( w 1 ) The corresponding circuit is f w 2 w 3 w 1 6-2 6.7. Expansion in terms of w 2 gives f = w 2 (1 + w 1 w 3 + w 1 w 3 ) + w 2 ( w 1 w 3 + w 1 w 3 ) = w 1 w 2 w 3 + w 1 w 2 w 3 + w 2 + w 1 w 2 w 3 + w 1 w 2 w 3 Further expansion in terms of w 1 gives f = w 1 ( w 2 w 3 + w 2 w 3 + w 2 ) + w 1 ( w 2 w 3 + w 2 w 3 + w 2 ) = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 Further expansion in terms of w 3 gives f = w 3 ( w 1 w 2 + w 1 w 2 + w 1 w 2 + w 1 w 2 ) + w 3 ( w 1 w 2 + w 1 w 2 + w 1 w 2 + w 1 w 2 ) = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 6.8. Expansion in terms of w 1 gives f = w 1 w 2 + w 1 w 3 + w 1 w 2 Further expansion in terms of w 2 gives f = w 2 ( w 1 w 3 ) + w 2 ( w 1 + w 1 + w 1 w 3 ) = w 1 w 2 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 Further expansion in terms of w 3 gives f = w 3 ( w 1 w 2 + w 1 w 2 + w 1 w 2 + w 1 w 2 ) + w 3 ( w 1 w 2 + w 1 w 2 ) = w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 + w 1 w 2 w 3 6.9. Proof of Shannon’s expansion theorem f ( x 1 , x 2 , ..., x n ) = x 1 · f (0 , x 2 , ..., x n ) + x 1 · f (1 , x 2 , ..., x n ) This theorem can be proved using perfect induction , by showing that the expression is true for every possible value of x 1 . Since x 1 is a boolean variable, we need to look at only two cases: x 1 = 0 and x 1 = 1 . Setting x 1 = 0 in the above expression, we have: f (0 , x 2 , ..., x n ) = 1 · f (0 , x 2 , ..., x n ) + 0 · f (1 , x 2 , ..., x n ) = f (0 , x 2 , ..., x n ) Setting x 1 = 1 , we have: f (1 , x 2 , ..., x n ) = · f (0 , x 2 , ..., x n ) + 1 · f (1 , x 2 , ..., x n ) = f (1 , x 2 , ..., x n ) This proof can be performed for any arbitrary x i in the same manner....
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## This note was uploaded on 02/16/2011 for the course EE 4745 taught by Professor Rai during the Spring '07 term at LSU.

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sol6 - Chapter 6 6.1 w En y w 1 y 1 y 2 y 3 y 7 y 6 y 5 y 4...

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