Sum of Minterms and Product of Maxterms, Study notes for Digital Logic Design and Programming

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Sum of Minterms and Product of Maxterms
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Preview1 page / 1  Example. Express the Boolean function F = x + yz as a sum of minterms. Solution:F = x + yz = x + (yz) AND (multiply) has a higher precedence than OR (add) = x(y+y')(z+z') + (x+x')yz expand 1st term by ANDing it with (y + y’)(z + z’), and 2nd term with (x + x’) = x y z + x y z' + x y' z + x y' z' + x y z + x' y z = m7 + m6 + m5 + m4 + m3 = Σ(3, 4, 5, 6, 7) sum of 1-minterms

Example. Express the Boolean function F = x + yz as a product of maxterms. Solution: First, we need to convert the function into the product-of-OR terms by using the distributive law as follows: F = x + yz = x + (yz) AND (multiply) has a higher precedence than OR (add) = (x + y) (x + z) use distributive law to change to product of OR terms = (x + y + zz') (x + yy' + z) expand 1st term by ORing it with zz', and 2nd term with yy' = (x + y + z) (x + y + z') (x + y + z) (x + y' + z) = M0 • M1 • M2 = Π(0, 1, 2) product of 0-maxterms

Example. Express F ' = (x + yz)' as a sum of minterms. Solution: F' = (x + yz)' = (x + (yz))' AND (multiply) has a higher precedence than OR (add) = x' (y' + z') use dual or De Morgan’s Law = (x' y') + (x' z') use distributive law to change to sum of AND terms = x' y' (z + z') + x' (y + y') z' expand 1st term by ANDing it with (z + z'), and 2nd term with (y + y') = x' y' z + x' y' z' + x' yz' + x' y' z' = m1 + m0 + m2 = Σ(0, 1, 2) sum of 0-minterms

Example. Express F ' = (x + yz)' as a product of maxterms. Solution: F' = (x + yz)' = (x + (yz))' AND (multiply) has a higher precedence than OR (add) = x' (y' + z') use dual or De Morgan’s Law = (x' + yy' + zz') (xx' + y' + z') expand 1st term by ORing it with yy' and zz', and 2nd term with xx' = (x' + y + z) (x' + y + z') (x' + y' + z) (x' + y' + z') (x + y' + z') (x' + y' + z') = M4 • M6 • M5 • M7 • M3

= Π(3, 4, 5, 6, 7) product of 1-maxtermsF and F' are shown in the following truth table:

xyz Minterms Maxterms FF ' 0 0 0 m0=x' y' z' M0=x + y + z 0 1 0 0 1 m1=x' y' zM1=x + y + z' 0 1 0 1 0 m2=x' y z' M2=x + y' + z 0 1 0 1 1 m3=x' y zM3=x + y' + z' 1 0 1 0 0 m4=xy' z' M4=x' + y + z 1 0 1 0 1 m5=xy' zM5=x' + y + z' 1 0 1 1 0 m6=xyz' M6=x' + y' + z 1 0 1 1 1 m7=xyzM7=x' + y' + z' 1 0