Question

The output Y of the logic circuit given below is:-

A. \[1\]
B. \[0\]
C. \[X\]
D. \[\overline X \]

Answer 1

Hint: Here in this question, we need to determine the output of the logic circuit. The given circuit is a combination of the two gates, first is a NOT gate, and the second is the XOR gate, the output from the NOT gate will be one of the inputs for the XOR gate in the circuit, using the TRUTH TABLE we will find the result.

Complete step by step solution:
In the given circuit we can see the first logical gate is the NOT gate and the second logical gate is the XOR gate. The input \[\overline X \] to the logical circuit is first divided by the parallel circuit, one going direct to the XOR gate and another through the NOT gate.
When a binary input is given to a NOT gate then the signal is INVERTED so the signal \[\overline X \] will become \[X\].

Now, we will find the output Y of the XOR gate, when its inputs are \[\overline X \] and \[X\], to find the same we will draw a truth table-

\[\overline X \]	\[X\]	Output \[y = \overline X \oplus X\]
0	0	0
0	1	1
1	0	1
1	1	0

Now from the above table we will write the equation for the high output signals, we have
\[y = X \oplus \overline X \\y = X\overline X + \overline X X\]
On further solving this, we have
\[y = XX + \overline X \overline X \]
Now, by using the inverse law \[A + \overline A = 1\], we can write-
\[y = X + \overline X \,\\ \therefore y = 1\]
Therefore, we get the output Y of the logic circuit as \[1\].

Hence, option A is correct.

Note: Always note that a XOR gate gives a true output when the number of logical inputs is odd, \[y = A\overline B + \overline A B\] and \[y = A\overline {BC} + \overline A B\overline C + \overline {AB} C + ABC\].