An X-OR gate has following truth table:
It is represented by following logic relation
`y=barA.B+A.barB`
Build this gate using AND, OR and NOT gates.
`|{:(A,B,Y),(0,0,0),(0,1,1),(1,0,1),(1,1,0):}:|`

Explanation

Given, the logical relation for the given truth table is
`Y = bar(A).B + A. bar(B) = Y_(1) + Y_(2)`
When `Y_(1) = A.B " and " Y_(2) = A.bar(B)`
`Y_(1)` can be obtained as output of AND gate `I` for which one Input is of A through NOT gate and another input is of `B. Y_(2)` can be obtained as output of AND gate `II` for which one input is of A and other input is of B through NOT gate.
Now `Y_(2)` can be obtained as output from OR gate, where, `Y_(1) " and Y_(2)` are input of OR gate. Thus, the given table can be obtained from the logic circuit given below.