UPSC 2014 Maths Optional Paper 2 Q5d — Step-by-Step Solution

10 marks · Section B

Logic gates and truth tables · Numerical Analysis · asked 2× in 13 yrs · Read the full method →

Question

Use only AND and OR logic gates to construct a logic circuit for the Boolean expression $z=xy+uv$ .

Technique

Direct realisation of sum-of-products Boolean expression using AND gates for product terms and OR gate for the sum.

Solution

The expression $z=xy+uv$ is the OR of two AND products. It requires:

AND gate 1: inputs $x,y$ → output $xy$ .
AND gate 2: inputs $u,v$ → output $uv$ .
OR gate: inputs $xy,\,uv$ → output $z=xy+uv$ .

Circuit diagram

   x ----+
         |---[AND]----+
   y ----+            |
                      +---[OR]---- z
   u ----+            |
         |---[AND]----+
   v ----+

ASCII rendering:

Two AND gates compute the two product terms in parallel.
One OR gate combines them.

Total gate count: 2 AND gates + 1 OR gate = 3 gates.

Truth table verification

$x$	$y$	$u$	$v$	$xy$	$uv$	$z=xy+uv$
$0$	$0$	$0$	$0$	$0$	$0$	$0$
$0$	$0$	$1$	$1$	$0$	$1$	$1$
$1$	$1$	$0$	$0$	$1$	$0$	$1$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$0$	$1$	$0$	$1$	$0$	$0$	$0$
…	…	…	…	…	…	…

(All 16 rows of the 4-variable truth table can be checked; $z=1$ iff either pair is $11$ .)

Answer

\boxed{\;z=xy+uv\text{ is realised by 2 AND gates feeding into 1 OR gate.}\;}

$x$	$y$	$u$	$v$	$xy$	$uv$	$z=xy+uv$
$0$	$0$	$0$	$0$	$0$	$0$	$0$
$0$	$0$	$1$	$1$	$0$	$1$	$1$
$1$	$1$	$0$	$0$	$1$	$0$	$1$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$0$	$1$	$0$	$1$	$0$	$0$	$0$
…	…	…	…	…	…	…

$x$	$y$	$u$	$v$	$xy$	$uv$	$z=xy+uv$
$0$	$0$	$0$	$0$	$0$	$0$	$0$
$0$	$0$	$1$	$1$	$0$	$1$	$1$
$1$	$1$	$0$	$0$	$1$	$0$	$1$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$0$	$1$	$0$	$1$	$0$	$0$	$0$
…	…	…	…	…	…	…

$x$	$y$	$u$	$v$	$xy$	$uv$	$z=xy+uv$
$0$	$0$	$0$	$0$	$0$	$0$	$0$
$0$	$0$	$1$	$1$	$0$	$1$	$1$
$1$	$1$	$0$	$0$	$1$	$0$	$1$
$1$	$1$	$1$	$1$	$1$	$1$	$1$
$0$	$1$	$0$	$1$	$0$	$0$	$0$
…	…	…	…	…	…	…