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Question Bank (5 marks) · Q5

Q.Convert Y = AB‾+BC+BD‾A\overline{B} + BC + B\overline{D} into its canonical SOP form and write the truth table for the corresponding min term designation.

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[!TLDR]

Expanding every term to four variables gives Y=Σm(4,6,7,8,9,10,11,12,14,15)Y=\Sigma m(4,6,7,8,9,10,11,12,14,15).

Variables: A, B, C, D (weights 8 4 2 1).

Step 1 — expand each term to the missing variables:

  • AB‾=AB‾(C+C‾)(D+D‾)A\overline{B} = A\overline{B}(C+\overline{C})(D+\overline{D}) → minterms 8 (1000), 9 (1001), 10 (1010), 11 (1011).
  • BC=(A+A‾)BC(D+D‾)BC = (A+\overline{A})BC(D+\overline{D}) → minterms 6 (0110), 7 (0111), 14 (1110), 15 (1111).
  • BD‾=(A+A‾)B(C+C‾)D‾B\overline{D} = (A+\overline{A})B(C+\overline{C})\overline{D} → minterms 4 (0100), 6 (0110), 12 (1100), 14 (1110).

Step 2 — collect the distinct minterms: {8,9,10,11} ∪ {6,7,14,15} ∪ {4,6,12,14} = {4, 6, 7, 8, 9, 10, 11, 12, 14, 15}. Hence Y=Σm(4,6,7,8,9,10,11,12,14,15)Y=\Sigma m(4,6,7,8,9,10,11,12,14,15).

Step 3 — truth table (Y = 1 for the minterms above):

mABCDY
000000
100010
200100
300110
401001

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