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The following configuration of gates is equivalent to |
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Answer» `HAND` Output of `G_(1)=BAR(A.B)` Output of `G_(3)` `=(A+B).bar(A.B)=(A+B).(bar(A)+bar(B))` If `A=1` and `B=1`, then `A+B=1` `A+B=0` So, output is zero, If `A=0` and `B=0`, then `A+B=0` `bar(A)+bar(B)=1+1=1` So, output is zero. Clearly, the given combination is `XOR` gate. |
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