Lecture 10

# Lecture 10 - ECE52 Spring 11 Lecture 10 2/4/11 Midterm 2...

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1 ECE52 Spring 11 Lecture 10 2/4/11 Midterm 2 weeks away Next Wednesday: class 1:15-2:30 Next Friday: no class

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2 A five-variable Karnaugh map. x 1 x 2 x 3 x 4 00 01 11 10 1 1 1 1 1 1 00 01 11 10 x 1 x 2 x 3 x 4 00 01 11 10 1 1 1 1 1 1 1 00 01 11 10 f 1 x 1 x 3 x 1 x 3 x 4 x 1 x 2 x 3 x 5 + + = x 5 1 = x 5 0 =
3 From intuition to algorithms Intuition was “ to find as few as possible and as large as possible groups of 1s that cover all cases where the function has a value of 1 .” We can get away with this for small functions representable via a K-map; the idea is sound for complex functions but the strategy needs to be automated.

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4 Terms Literal : each appearance of a variable, in true or uncomplemented form, in a product term is called a “literal”. x 1 x 2 ’x 3 has 3 literals. (will also be true for sum terms!) Implicant : for SOP, any product term is an implicant. Minterm : for an n-variable function, an implicant with n literals is a minterm.
5 Three-variable function f ( x 1 , x 2 , x 3 ) = Σ m (0, 1, 2, 3, 7) has 11 implicants. 11 implicants: 5 minterms, 2 prime (minimal) implicants, and 4 more non-minimal, non-minterm implicants x 1 x 2 x 3 1 1 1 1 x 1 0 0 1 0 00 01 11 10 0 1 x 2 x 3

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6 Three-variable function f ( x 1 , x 2 , x 3 ) = Σ m (0, 1, 2, 3, 7) has 11 implicants. 11 implicants: 5 minterms, 2 prime (minimal) implicants, and 4 more non-minimal, non-minterm implicants x 1 x 2 x 3 1 1 1 1 x 1 0 0 1 0 00 01 11 10 0 1 x 2 x 3
7 Terms, continued Prime implicant : an implicant that cannot be combined into another implicant that has fewer literals. Restated, it is impossible to delete any literal and still have a valid implicant. Cover : Any subset of the implicants that includes all “1”s in the map is a cover. Equivalent to a valid SOP expression. Each unique cover corresponds to a different way to implement the function. The set of all minterms comprise a cover The set of all prime implicants comprise a cover

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8 Three-variable function f ( x 1 , x 2 , x 3 ) = Σ m (0, 1, 2, 3, 7) has 11 implicants. Valid covers include
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## This note was uploaded on 07/05/2011 for the course ECE 52 taught by Professor Dr.jonathanboard during the Spring '11 term at Duke.

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Lecture 10 - ECE52 Spring 11 Lecture 10 2/4/11 Midterm 2...

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