For F=ABC+ A BC+A BC+ ABC+ A B C, the equivalent product of sums is

$(A+\bar B+\bar C)(A+B+C)(\bar A+\bar B+\bar C)$

$(\bar A+B+\bar C)(A+\bar B+\bar C)(A+B+C)$

$(\bar A+\bar B+C)(A+B+\bar C)(A+B+C)$

GATE 2015 EE (EE2) – Question 23

Analog and Digital Electronics · VCOs and timers, combinatorial and sequential logic circuits, multiplexers, demultiplexers · 1 mark · Multiple choice

For $F=ABC+\bar A\bar BC+A\bar BC+\bar ABC+\bar A\bar B\bar C$, the equivalent product of sums is

$(A+\bar B+C)(\bar A+B+C)(\bar A+\bar B+C)$
$(A+\bar B+\bar C)(A+B+C)(\bar A+\bar B+\bar C)$
$(\bar A+B+\bar C)(A+\bar B+\bar C)(A+B+C)$
$(\bar A+\bar B+C)(A+B+\bar C)(A+B+C)$

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Show answer and explanation

Correct answer: (A) $(A+\bar B+C)(\bar A+B+C)(\bar A+\bar B+C)$

Explanation

The product terms make F=1 at binary indices 7,1,5,3,0. Thus its zero indices are 2,4,6, and the canonical POS is ΠM(2,4,6).

Maxterm 2 is (A+B̄+C), maxterm 4 is (Ā+B+C), and maxterm 6 is (Ā+B̄+C). Their product is exactly **A**. A maxterm uses an uncomplemented variable for a zero bit and a complemented variable for a one bit.

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