Page 62 - DCAP108_DIGITAL_CIRCUITS_AND_LOGIC_DESIGNS

Unit 4: Minimization of Boolean Algebra

All other cells contain 1s because any input values other than ((0,0,0) for (A + B + C) yields 1s Notes
upon evaluation.
Referring to the Figure 4.3, the procedure for placing a maxterm in the K-map is:
• Identify the Sum term to be mapped
• Write corresponding binary numeric value
• Form the complement
• Use the complement as an address to place a 0 in the K-map
• Repeat for other maxterms (Sum terms within Product-of-Sums expression).

Figure 4.4: Maxterm Schematic

Another maxterm A’ + B’ + C’ is shown. Numeric 000 corresponds to A’ + B’ + C’. The complement
is 111. Place a 0 for maxterm (A’ + B’ + C’) in this cell (1,1,1) of the K-map as shown above.
Why should (A’ + B’ + C’) cause a 0 to be in cell 111? When A’ + B’ + C’ is (1’ + 1’ + 1’), all 1s in,
which is (0 + 0 + 0) after taking complements, we have the only condition that will give us a 0.
All the 1s are complemented to all 0s, which is 0 when ORed.
Figure 4.5: Maxterms Schematic

Out = (A +B+C)(A +B+C)
Maxterm = (A +B+C) Maxterm = (A +B+C)
Numeric = 1 1 1 Numeric = 1 1 0
Complement = 0 0 0 Complement = 0 0 1
BC
A 00 01 11 10
0 0 0 1 1
1 1 1 1 1
For a two-variable expression, the minterms and maxterms are as follows:
Table 4.1: Two-variable Expression of Minterms and Maxterms

X Y Minterm Maxterm
0 0 X′.Y′ X + Y
0 1 X′.Y X + Y′
1 0 X.Y′ X′ + Y
1 1 X.Y X′ + Y′

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