Output of Logic Circuit for Four Bit Binary Inputs

To determine the output $Y$ of the logic circuit, we first find the Boolean expression for $Y$ in terms of inputs $A$ and $B$.

1. Analysis of the Circuit:

   • Input $A$ passes through a NOT gate, giving the output $\overline{A}$.
   • Input $B$ is directly fed into the logic gate.
   • The two inputs, $\overline{A}$ and $B$, enter a NAND gate.

2. Boolean Expression: The output of the NAND gate is given by: $Y = \overline{\overline{A} \cdot B}$

   Using De Morgan's Law ($\overline{X \cdot Y} = \overline{X} + \overline{Y}$), we simplify the expression: $Y = \overline{\overline{A}} + \overline{B} = A + \overline{B}$

3. Bitwise Calculation: Given inputs: $A = 1101 \quad \text{and} \quad B = 1010$

   Taking the bitwise NOT of $B$: $\overline{B} = 0101$

   Now, performing the bitwise OR operation ($Y = A + \overline{B}$):

   $$\begin{array}{rccccccc} A &=& 1 & 1 & 0 & 1 \\ \overline{B} &=& 0 & 1 & 0 & 1 \\ \hline Y = A + \overline{B} &=& 1 & 1 & 0 & 1 \end{array}$$

Thus, the output is $Y = 1101$.

Correct Answer: Option A ($Y = 1101$)