A) OR
B) XOR
C) NAND
D) NOR
Correct Answer: A
Solution :
The given combination can be represented as All the gates in the combination are NAND gates. Output of gate-1,\[{{Y}_{1}}=\overline{A\cdot A}\] Output of gate-2,\[{{Y}_{2}}=\overline{B\cdot B}\] Output of gate-3,\[Y=\overline{{{Y}_{1}}\cdot {{Y}_{2}}}\] \[\therefore \] \[Y=\overline{\overline{A\cdot B}\cdot \overline{B\cdot B}}\] ?. (i) By Demorgans theorem, we have \[\overline{A\cdot A}=\overline{A}\] and \[\overline{B\cdot B}=\overline{B}\] Therefore, Eq. (i) becomes, \[Y=\overline{\overline{A}\cdot \overline{B}}\] Again from Demograns theorem \[\overline{\overline{A}\cdot \overline{B}}=\overline{\overline{A}}+\overline{\overline{B}}\] \[\therefore \] \[Y=\overline{\overline{A}}+\overline{\overline{B}}=A+B\] \[(as\,\,\overline{\overline{A}}=A)\] Hence, the combination behaves as\[OR\]gate.You need to login to perform this action.
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