The problem asks us to analyze a given logic circuit with inputs $A$ and $B$. (i) We need to write the logic expression for the output. (ii) We need to construct a truth table for the expression. (iii) We need to identify a single equivalent logic gate. (iv) We need to determine which basic gates are formed together to realize the equivalent logic gate we identified in part (iii).

1. Problem Description

The problem asks us to analyze a given logic circuit with inputs $A$ and $B$.

(i) We need to write the logic expression for the output.

(ii) We need to construct a truth table for the expression.

(iii) We need to identify a single equivalent logic gate.

(iv) We need to determine which basic gates are formed together to realize the equivalent logic gate we identified in part (iii).

2. Solution Steps

(i) Logic expression:

The circuit consists of two NOT gates connected to inputs $A$ and $B$, respectively, followed by an OR gate.

The output of the NOT gate connected to $A$ is $\overline{A}$.

The output of the NOT gate connected to $B$ is $\overline{B}$.

The OR gate takes $\overline{A}$ and $\overline{B}$ as inputs and produces the output $\overline{A} + \overline{B}$.

Therefore, the logic expression for the output is $Output = \overline{A} + \overline{B}$.

(ii) Truth table:

We need to construct a truth table for $Output = \overline{A} + \overline{B}$.

| A | B | $\overline{A}$ | $\overline{B}$ | $\overline{A} + \overline{B}$ |

|---|---|------------------|------------------|-----------------------------|

| 0 | 0 | 1 | 1 | 1 |

| 0 | 1 | 1 | 0 | 1 |

| 1 | 0 | 0 | 1 | 1 |

| 1 | 1 | 0 | 0 | 0 |

(iii) Equivalent logic gate:

The truth table is the same as the truth table of a NOR gate applied to inputs $A$ and $B$. DeMorgan's law states that $\overline{A} + \overline{B} = \overline{A \cdot B}$. The right-hand side represents the negation of $A$ AND $B$, which is a NAND gate.

We can also look at the truth table for NAND gate:

| A | B | $A \cdot B$ | $\overline{A \cdot B}$ |

|---|---|---------------|-----------------------------|

| 0 | 0 | 0 | 1 |

| 0 | 1 | 0 | 1 |

| 1 | 0 | 0 | 1 |

| 1 | 1 | 1 | 0 |

So the single equivalent gate is the NAND gate.

(iv) Basic logic gates:

The equivalent logic gate is a NAND gate. A NAND gate can be constructed using basic gates: AND and NOT.

3. Final Answer

(i) $Output = \overline{A} + \overline{B}$

(ii)

| A | B | $\overline{A}$ | $\overline{B}$ | $\overline{A} + \overline{B}$ |

|---|---|------------------|------------------|-----------------------------|

| 0 | 0 | 1 | 1 | 1 |

| 0 | 1 | 1 | 0 | 1 |

| 1 | 0 | 0 | 1 | 1 |

| 1 | 1 | 0 | 0 | 0 |

(iii) NAND gate

(iv) AND gate and NOT gate