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GATE 2018 EC – Question 19

Digital Circuits · Combinatorial Circuits · 1 mark · Multiple choice

A function $F(A,B,C)$ defined by three Boolean variables $A$, $B$ and $C$ when expressed as sum of products is given by

$$F=\bar A\cdot\bar B\cdot\bar C+\bar A\cdot B\cdot\bar C+A\cdot B\cdot\bar C$$

where, $\bar A$, $\bar B$, and $\bar C$ are the complements of the respective variables. The product of sums (POS) form of the function F is

  1. $F=(A+B+C)\cdot(A+B+\bar C)\cdot(A+\bar B+\bar C)$
  2. $F=(\bar A+\bar B+\bar C)\cdot(\bar A+B+\bar C)\cdot(A+\bar B+\bar C)$
  3. $F=(A+B+\bar C)\cdot(A+\bar B+\bar C)\cdot(\bar A+B+C)\cdot(\bar A+B+\bar C)\cdot(\bar A+\bar B+\bar C)$
  4. $F=(\bar A+\bar B+C)\cdot(\bar A+B+C)\cdot(A+\bar B+C)\cdot(A+B+\bar C)\cdot(A+B+C)$

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

Explanation

The minterms of $F$ are $000,010,110$, i.e. $m_0,m_2,m_6$. So $F$ is 0 for $m_1,m_3,m_4,m_5,m_7$, and the POS is the product of the maxterms $(A+B+\bar C)$, $(A+\bar B+\bar C)$, $(\bar A+B+C)$, $(\bar A+B+\bar C)$ and $(\bar A+\bar B+\bar C)$, which is option C.