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Chapter 3. Logic Gates & Boolean Algebra. Algebra Example. Simplify this expression:. ACD + ABCD. Factor out common variables:. CD(A + AB). Apply theorem 15a:. CD(A + B) or ACD + BCD. DeMorgan’s Theorems.
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Chapter 3 Logic Gates & Boolean Algebra
Algebra Example Simplify this expression: ACD + ABCD Factor out common variables: CD(A + AB) Apply theorem 15a: CD(A + B) or ACD + BCD
DeMorgan’s Theorems Demorgan’s Theorems allow us to break up a “bar”which is over an entire expression: (X + Y) = X ● Y (X ● Y) = X + Y
DeMorgan’s Theorem Example (A B + C) Simplify this expression: Apply Demorgan’s: (A B) ●C Apply Demorgan’s: (A + B) ●C (notice importance of keeping the parentheses!) Cancel double bars: (A + B) ●C or AC + BC
DeMorgan’s Theorem Example (A + C) ( B + D) Simplify this expression: Apply Demorgan’s: (A + C) + (B + D) Apply Demorgan’s: (A ●C) + (B ● D) Cancel double bars: AC + BD
DeMorgan’s Theorem Example (A + BC) (D + EF) Simplify this expression: Apply Demorgan’s: (A + BC) + (D + EF) Apply Demorgan’s: (A ● BC) + (D ● EF) Apply Demorgan’s: (A ● (B+C)) + (D ● (E+F)) Distribute: AB + AC + DE + DF
Alternate Gate Representations (X + Y) = X ● Y X Y X + Y X Y X Y X ● Y = X + Y
Alternate Gate Representations NOR Gates NAND Gates
NAND Gates are Universal Gates X X ● X = X X X X Y XY X ● Y XY X Y X Y X + Y
NAND Gates are Universal Gates X Y X ● Y = X + Y = X + Y X + Y X Y X Y X ● Y X Y
Using NAND Gates to Simplify 2 CHIPS: 7408 = AND 7432 = OR
Using NAND Gates to Simplify 1 CHIP: 7400 = NAND