Q.Construct an X-OR gate using only NAND gates and write the Boolean expression at the output of each gate.

An X-OR gate can be built from four 2-input NAND gates, since NAND is a universal gate.

[!ANSWER]

Using four NAND gates: G1 = $\overline{AB}$; G2 = $\overline{A\cdot G1}$; G3 = $\overline{B\cdot G1}$; G4 = $\overline{G2\cdot G3}=A\overline{B}+\overline{A}B=A\oplus B$.

[!TLDR]

Four 2-input NAND gates realise the X-OR function $Y=A\oplus B$.

Construction: NAND is a universal gate, so the X-OR gate is built from four NAND gates connected as follows.

• Gate G1 — both inputs $A, B$: $G1 = \overline{AB}$.
• Gate G2 — inputs $A$ and $G1$: $G2 = \overline{A\cdot \overline{AB}} = \overline{A(\overline{A}+\overline{B})} = \overline{A\overline{B}} = \overline{A}+B$.
• Gate G3 — inputs $B$ and $G1$: $G3 = \overline{B\cdot \overline{AB}} = \overline{B(\overline{A}+\overline{B})} = \overline{\overline{A}B} = A+\overline{B}$.
• Gate G4 — inputs $G2$ and $G3$: $G4 = \overline{G2\cdot G3} = \overline{(\overline{A}+B)(A+\overline{B})}$.

Simplifying the final output: $(\overline{A}+B)(A+\overline{B}) = \overline{A}\,\overline{B}+AB$, so $G4 = \overline{\overline{A}\,\overline{B}+AB} = \overline{A}B+A\overline{B} = A\oplus B$.

Thus the network output equals the X-OR function.

[!ANSWER]

With four NANDs: $G1=\overline{AB}$, $G2=\overline{A\cdot G1}=\overline{A}+B$, $G3=\overline{B\cdot G1}=A+\overline{B}$, and $G4=\overline{G2\cdot G3}=A\overline{B}+\overline{A}B=A\oplus B$.