UPSC 2019 Maths Optional Paper 2 Q8a — Step-by-Step Solution

15 marks · Section B

Boolean algebra · Numerical Analysis · asked 15× in 13 yrs · Read the full method →

Question

Given the Boolean expression $X=AB+ABC+A\bar B\bar C+A\bar C$

(i) Draw the logical diagram for the expression.
(ii) Minimize the expression.
(iii) Draw the logical diagram for the reduced expression.

Technique

Absorption laws $AB+ABC=AB$ and $A\bar B\bar C+A\bar C=A\bar C$ ; factor $A$ to get $A(B+\bar C)$ ; confirm via K-map / truth table.

Solution

Setup. $X=AB+ABC+A\bar B\bar C+A\bar C$ over inputs $A,B,C$ .

Step (i) — Logic diagram of the original expression

Four AND gates feeding one OR gate:

 A,B ─────────────►[AND]──► AB ───────┐
 A,B,C ───────────►[AND]──► ABC ──────┤
 A,B',C' ─────────►[AND]──► AB'C' ────┤►[OR]──► X
 A,C' ────────────►[AND]──► AC' ──────┘

(Inverters produce $\bar B$ and $\bar C$ for the third and fourth product terms.)

Step (ii) — Minimize

Algebraically:

X=AB+ABC+A\bar B\bar C+A\bar C.

$AB+ABC=AB(1+C)=AB$ (absorption).
$A\bar B\bar C+A\bar C=A\bar C(\bar B+1)=A\bar C$ (absorption).

So $X=AB+A\bar C=A(B+\bar C)$ .

Answer

\boxed{\;X=A(B+\bar C)=AB+A\bar C.\;}