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Question Bank (3 marks) · Q9

Q.Convert A+BC+ A‾B\overline{A}B into its canonical SOP and write the expression in min term designation.

Karnataka PUCTextbookNumeric· 3mImportance★★★★★est
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[!TLDR]

Y=A+BC+A‾B=A+B=Σm(2,3,4,5,6,7)Y=A+BC+\overline{A}B=A+B=\Sigma m(2,3,4,5,6,7) over the three variables A, B, C.

Step 1 — simplify: Using the absorption identity A+A‾B=A+BA+\overline{A}B=A+B,

Y=A+BC+A‾B=(A+A‾B)+BC=(A+B)+BC=A+B(1+C)=A+B.Y = A + BC + \overline{A}B = (A+\overline{A}B) + BC = (A+B) + BC = A + B(1+C) = A+B.

So the function reduces to Y=A+BY = A+B, but it is a three-variable function (A, B, C).

Step 2 — expand each term to the missing variables (canonical SOP):

  • A=A(B+B‾)(C+C‾)=ABC+ABC‾+AB‾C+AB‾ C‾A = A(B+\overline{B})(C+\overline{C}) = ABC + AB\overline{C} + A\overline{B}C + A\overline{B}\,\overline{C}.
  • B=B(A+A‾)(C+C‾)=ABC+ABC‾+A‾BC+A‾BC‾B = B(A+\overline{A})(C+\overline{C}) = ABC + AB\overline{C} + \overline{A}BC + \overline{A}B\overline{C}.

Collecting the distinct minterms:

Y=A‾BC‾+A‾BC+AB‾ C‾+AB‾C+ABC‾+ABC.Y = \overline{A}B\overline{C} + \overline{A}BC + A\overline{B}\,\overline{C} + A\overline{B}C + AB\overline{C} + ABC. …

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