Q.What is an X-OR gate? Realize the logic expression Y= A⊕B⊕C using 2 input two X-OR gate.

An exclusive-OR (X-OR) gate is a two-input logic gate whose output is HIGH only when the two inputs are unequal.

[!ANSWER]

An X-OR gate gives $Y = A\oplus B = \overline{A}B + A\overline{B}$; $Y=A\oplus B\oplus C$ is realised by cascading two 2-input X-OR gates — the first forms $A\oplus B$ and the second X-ORs that result with $C$.

[!TLDR]

An X-OR gate outputs 1 when its inputs differ; a three-variable X-OR is built by cascading two 2-input X-OR gates.

X-OR gate: The exclusive-OR gate is a two-input gate whose output is HIGH (1) only when the two inputs are at different logic levels, and LOW (0) when they are equal. Its Boolean equation is $Y = A\oplus B = \overline{A}B + A\overline{B}$.

Truth table:

Realising $Y = A\oplus B\oplus C$ with two 2-input X-OR gates: Since the X-OR operation is associative, the three-input function is grouped as $Y = (A\oplus B)\oplus C$.

• X-OR gate 1: inputs $A$ and $B$ → output $P = A\oplus B$.
• X-OR gate 2: inputs $P$ and $C$ → output $Y = P\oplus C = (A\oplus B)\oplus C = A\oplus B\oplus C$.

The output is 1 whenever an odd number of the three inputs are 1, which is exactly the parity/odd-function realised by the cascade.

[!ANSWER]

The X-OR gate gives output 1 only for unequal inputs, $Y=\overline{A}B+A\overline{B}$. Two cascaded 2-input X-OR gates realise $Y=A\oplus B\oplus C$: gate 1 gives $A\oplus B$, and gate 2 X-ORs it with $C$.