UPSC 2024 Maths Optional Paper 2 Q6b — Step-by-Step Solution

15 marks · Section B

Boolean algebra · Numerical Analysis · asked 15× in 14 yrs · Read the full method →

Question

Draw the logical circuit for $Y=AB\overline C+B\overline C+\overline A B$ . Also obtain the output $Y$ for the bit sequences $A=10001111$ , $B=00111100$ , $C=11000100$ .

Technique

Simplify via $AB\overline C+B\overline C=B\overline C$ ; draw the circuit for $Y=B\overline C+\overline A B$ ; evaluate bit-by-bit.

Solution

Step 1 — Simplify

Y=AB\overline C+B\overline C+\overline A B=B\overline C(A+1)+\overline A B=B\overline C+\overline A B=B(\overline C+\overline A).

Simplified: $Y=\overline A B+B\overline C$ .

Step 2 — Circuit

The circuit for $Y=\overline A B+B\overline C$ requires:

NOT gates on $A$ (giving $\overline A$ ) and on $C$ (giving $\overline C$ ).
AND gate 1: inputs $\overline A,B$ ; output $\overline A B$ .
AND gate 2: inputs $B,\overline C$ ; output $B\overline C$ .
OR gate: inputs $\overline A B$ , $B\overline C$ ; output $Y$ .

Step 3 — Truth table for given sequences

Pos	$A$	$B$	$C$	$\overline A B$	$B\overline C$	$Y$
1	1	0	1	0	0	0
2	0	0	1	0	0	0
3	0	1	0	1	1	1
4	0	1	0	1	1	1
5	1	1	0	0	1	1
6	1	1	1	0	0	0
7	1	0	0	0	0	0
8	1	0	0	0	0	0

Answer

\boxed{\;Y=\overline A B+B\overline C;\qquad Y\text{-sequence}=00111000.\;}