The truth table for the following logic circu | vedclass

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(B) The given circuit consists of two $AND$ gates,one $OR$ gate,and one $NOT$ gate (forming a $NOR$ gate at the output). The inputs to the first $AND$

Vedclass · Accepted Answer

$A, B, Y$$0, 0, 1$$0, 1, 0$$1, 0, 0$$1, 1, 1$

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(B) The given circuit consists of two $AND$ gates,one $OR$ gate,and one $NOT$ gate (forming a $NOR$ gate at the output). The inputs to the first $AND$ gate are $A$ and $\bar{B}$. The inputs to the second $AND$ gate are $\bar{A}$ and $B$.The output of the $OR$ gate is $(A \cdot \bar{B} + \bar{A} \cdot B)$.The final output $Y$ is the inversion of this,so $Y = \overline{(A \cdot \bar{B} + \bar{A} \cdot B)}$.Using De Morgan's theorem,$Y = \overline{(A \cdot \bar{B})} \cdot \overline{(\bar{A} \cdot B)} = (\bar{A} + B) \cdot (A + \bar{B})$.Expanding this,$Y = \bar{A} \cdot A + \bar{A} \cdot \bar{B} + B \cdot A + B \cdot \bar{B} = 0 + \bar{A} \cdot \bar{B} + A \cdot B + 0 = A \cdot B + \bar{A} \cdot \bar{B}$.This is the Boolean expression for an $XNOR$ gate.$A$$B$$Y$$0$$0$$1$$0$$1$$0$$1$$0$$0$$1$$1$$1$

The truth table for the following logic circuit is ...... .

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