Which of the following truth tables correctly represents the final...

The circuit consists of three logic gates:

1. A NAND gate with inputs $A$ and $B$. Its output is $Y_1 = \overline{A \cdot B}$.

2. An AND gate with inputs $Y_1$ and $B$. Its output is $Y_2 = Y_1 \cdot B = (\overline{A \cdot B}) \cdot B$.

3. A NOR gate with both inputs connected to $Y_2$. This acts as a NOT gate. Its output is $Y = \overline{Y_2 + Y_2} = \overline{Y_2}$.

Now, simplify the Boolean expression for $Y$:

$Y_2 = (\overline{A \cdot B}) \cdot B$

Using De Morgan's law, $\overline{A \cdot B} = \bar{A} + \bar{B}$.

$Y_2 = (\bar{A} + \bar{B}) \cdot B = \bar{A} \cdot B + \bar{B} \cdot B$

Since $\bar{B} \cdot B = 0$, we have $Y_2 = \bar{A} \cdot B$.

The final output is $Y = \overline{Y_2} = \overline{\bar{A} \cdot B}$.

Using De Morgan's law again:

$Y = \overline{\bar{A}} + \bar{B} = A + \bar{B}$

Constructing the truth table for $Y = A + \bar{B}$:

For $A = 0, B = 0 \Rightarrow Y = 0 + \bar{0} = 0 + 1 = 1$

For $A = 0, B = 1 \Rightarrow Y = 0 + \bar{1} = 0 + 0 = 0$

For $A = 1, B = 0 \Rightarrow Y = 1 + \bar{0} = 1 + 1 = 1$

For $A = 1, B = 1 \Rightarrow Y = 1 + \bar{1} = 1 + 0 = 1$

This matches the truth table given in option (A).

Answer:

A	B	Y
0	0	1
0	1	0
1	0	1
1	1	1