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Question 19

The logic performed by the circuit shown in figure is equivalent to

Solution

$$\text{Outputs of the first two NOR gates with tied inputs: } Y_a = \overline{a + a} = \bar{a}, \quad Y_b = \overline{b + b} = \bar{b}$$

$$\text{Output of the final NOR gate: } Y = \overline{Y_a + Y_b} = \overline{\bar{a} + \bar{b}}$$

$$\text{Applying De Morgan's Law: } Y = \overline{\bar{a}} \cdot \overline{\bar{b}} = a \cdot b \implies \text{AND gate}$$

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