A two-bit comparator gives an active-HIGH output, Y, if two two-bit words, A and B, are the same. Give Y in fundamental sum of products form and then use Boolean algebra to show that Y=(A0⊕B0)+(A1⊕B1)

Question

A two-bit comparator gives an active-HIGH output, Y, if two two-bit words, A and B, are the same. Give Y in fundamental sum of products form and then use Boolean algebra to show that [latex]Y=\overline{(A_{0}⊕B_{0})+(A_{1}⊕B_{1})} [/latex]

Accepted Answer

[latex]Y=\overline{A_{0}} \overline{A_{1}} \overline{B_{0}} \overline{B_{1}} + \overline{A_{0}} A_{1} \overline{B_{0}} B_{1}+A_{0} \overline{A_{1}} B_{0} \overline{B_{1}}+A_{0} A_{1} B_{0} B_{1}[/latex]

[latex]=\overline{A_{0}} \overline{B_{0}} (\overline{A_{1}} \overline{B_{1}}+A_{1} B_{1})+A_{0} B_{0}(\overline{A_{1}} \overline{B_{1}}+A_{1} B_{1})[/latex]

[latex]=(\overline{A_{0}} \overline{B_{0}}+A_{0} B_{0})·(\overline{A_{1}} \overline{B_{1}}+A_{1} B_{1})[/latex]

[latex]=(\overline{(A_{0}⊕B_{0})}·\overline{(A_{1}⊕B_{1})})[/latex]

[latex]=\overline{(A_{0}⊕B_{0})+(A_{1}⊕B_{1})}[/latex] De Morgan's theorem

Question 4.7: A two-bit comparator gives an active-HIGH output, Y, if two ...

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