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Solved Examples · Example 12

Q.Convert Y=AC+BC‾Y = AC + B\overline{C} into its canonical SOP form

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[!TLDR]

AC=ABC+AB‾CAC = ABC + A\overline{B}C and BC‾=ABC‾+A‾BC‾B\overline{C} = AB\overline{C} + \overline{A}B\overline{C}, so the canonical SOP is Y=A‾BC‾+AB‾C+ABC‾+ABC=∑m(2,5,6,7)Y = \overline{A}B\overline{C} + A\overline{B}C + AB\overline{C} + ABC = \sum m(2,5,6,7).

In this Karnataka II PUC Electronics digital-logic example, a canonical (standard) SOP requires each product term (minterm) to contain all input variables, either in true or complemented form. Convert each term by ANDing it with (X+X‾)(X + \overline{X}) for every missing variable XX, then expand (A is the MSB, C the LSB).

Term 1 - ACAC is missing B, so AND it with (B+B‾)(B + \overline{B}):

AC=AC(B+B‾)=ABC+AB‾C=m7+m5.AC = AC(B + \overline{B}) = ABC + A\overline{B}C = m_7 + m_5.

Term 2 - BC‾B\overline{C} is missing A, so AND it with (A+A‾)(A + \overline{A}):

BC‾=BC‾(A+A‾)=ABC‾+A‾BC‾=m6+m2.B\overline{C} = B\overline{C}(A + \overline{A}) = AB\overline{C} + \overline{A}B\overline{C} = m_6 + m_2.

Combining the four minterms (all distinct, none repeated): …

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