Four NOR gates are connected as shown in figure. The truth table for the given figure is:

We need to determine the correct truth table for a digital combinational circuit consisting of four interconnected NOR gates.

1. Understand the Operation of a NOR Gate

A standard NOR gate takes two inputs, sums them together, and then inverts the result. The Boolean output expression for any two inputs $$X$$ and $$Y$$ passing through a NOR gate is:

$$\text{Output} = \overline{X + Y}$$

Furthermore, if the two inputs of a NOR gate are shorted/tied together to receive the same signal $$X$$, it functions as a NOT gate:

$$\overline{X + X} = \bar{X}$$

2. Track the Logic States Through the Circuit

Let's analyze the output of each gate stage by stage, using inputs $$A$$ and $$B$$:

• Stage 1 (Input Level):
  • Gate 1 (Top-Left): Both of its inputs are tied together to signal $$A$$. It acts as a NOT gate:

    $$\text{Output}_1 = \bar{A}$$

  • Gate 2 (Bottom-Left): Both of its inputs are tied together to signal $$B$$. It also acts as a NOT gate:

    $$\text{Output}_2 = \bar{B}$$



• Stage 2 (Middle Level):
  • Gate 3: This gate receives the outputs from the first two gates ($$\bar{A}$$ and $$\bar{B}$$) as its inputs. Applying the NOR operation:

    $$\text{Output}_3 = \overline{\bar{A} + \bar{B}}$$

    Using De Morgan's Law ($$\overline{\bar{A} + \bar{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}}$$), this simplifies directly into an AND gate operation:

    $$\text{Output}_3 = A \cdot B$$



• Stage 3 (Final Output Level):
  • Gate 4 (Rightmost): The final NOR gate has its inputs tied together and receives the signal from Gate 3 ($$A \cdot B$$). Because its inputs are shorted, it performs a NOT operation on its input:

    $$Y = \overline{\text{Output}_3} = \overline{A \cdot B}$$

3. Determine the Final Logic Operation

The simplified Boolean expression for the final output is:

$$Y = \overline{A \cdot B}$$

This expression represents the standard logic operation of a NAND gate.

4. Construct the Truth Table

Let's evaluate the output $$Y$$ for all four possible combinations of the binary inputs $$A$$ and $$B$$:

• If $$A = 0$$ and $$B = 0$$:

  $$Y = \overline{0 \cdot 0} = \bar{0} = 1$$

• If $$A = 0$$ and $$B = 1$$:

  $$Y = \overline{0 \cdot 1} = \bar{0} = 1$$

• If $$A = 1$$ and $$B = 0$$:

  $$Y = \overline{1 \cdot 0} = \bar{0} = 1$$

• If $$A = 1$$ and $$B = 1$$:

  $$Y = \overline{1 \cdot 1} = \bar{1} = 0$$

Final Answer: The circuit behaves as a NAND gate, yielding an output of $$1$$ for all input states except when both inputs are $$1$$.