The problem consists of three parts: (A) Complete a logic circuit and write the Boolean expression for the circuit. The bicycle engine starts (F=1) only if the rider wears a helmet (A=1), inserts the key (B=1), and releases the bicycle stand (C=1). Releasing the bicycle stand means C=0. (B) Complete the truth table for the Boolean expression $F = (A+B) \overline{C}$. (C) Convert the octal number $657_8$ to a hexadecimal number.

1. Problem Description

The problem consists of three parts:

(A) Complete a logic circuit and write the Boolean expression for the circuit. The bicycle engine starts (F=1) only if the rider wears a helmet (A=1), inserts the key (B=1), and releases the bicycle stand (C=1). Releasing the bicycle stand means C=

0. (B) Complete the truth table for the Boolean expression $F = (A+B) \overline{C}$.

(C) Convert the octal number $657_8$ to a hexadecimal number.

2. Solution Steps

(A) (i) Completing the logic circuit:

The problem states that the engine starts (F=1) only when A=1, B=1, and C=

0. This implies that the output F should be 1 only when A and B are 1, and C is

0. The AND gate takes A and B as inputs. So, the output of the AND gate will be A AND B.

Since we need A AND B AND (NOT C), the gate X should be a NOT gate. Therefore, the input C to the NOT gate becomes $\overline{C}$.

The output of the AND gate (A AND B) is connected to the input of gate Y. Also, $\overline{C}$ is connected to the input of gate Y. Thus, gate Y is also an AND gate with the inputs (A AND B) and $\overline{C}$. Therefore, the circuit consists of an AND gate for A and B, a NOT gate for C, and an AND gate for (A AND B) and $\overline{C}$.

(ii) Writing the Boolean expression F:

The output F is the result of ANDing (A AND B) with (NOT C).

So, $F = (A \cdot B) \cdot \overline{C}$ or $F = (A \land B) \land \overline{C}$.

However, the problem states that $F = (A+B) \overline{C}$. So the first gate must be an OR gate.

Therefore the logic gate connecting A and B should be an OR gate and the logic gate connecting $\overline{C}$ and $A+B$ should be an AND gate.

$F = (A + B) \cdot \overline{C}$

or

$F = (A \lor B) \land \overline{C}$

(B) Completing the truth table for $F = (A+B) \overline{C}$:

| A | B | C | A+B | $\overline{C}$ | F = (A+B) $\overline{C}$ |

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

| 0 | 0 | 0 | 0 | 1 | 0 |

| 0 | 0 | 1 | 0 | 0 | 0 |

| 0 | 1 | 0 | 1 | 1 | 1 |

| 0 | 1 | 1 | 1 | 0 | 0 |

| 1 | 0 | 0 | 1 | 1 | 1 |

| 1 | 0 | 1 | 1 | 0 | 0 |

| 1 | 1 | 0 | 1 | 1 | 1 |

| 1 | 1 | 1 | 1 | 0 | 0 |

(C) Converting $657_8$ to hexadecimal:

First, convert the octal number to decimal:

$657_8 = (6 \times 8^2) + (5 \times 8^1) + (7 \times 8^0) = (6 \times 64) + (5 \times 8) + (7 \times 1) = 384 + 40 + 7 = 431_{10}$

Now, convert the decimal number 431 to hexadecimal:

Divide 431 by 16:

$431 \div 16 = 26$ remainder 15 (F)

$26 \div 16 = 1$ remainder 10 (A)

$1 \div 16 = 0$ remainder 1 (1)

Reading the remainders from bottom to top, we get $1AF_{16}$

3. Final Answer

(A) (i) Gate X is a NOT gate, and Gate Y is an AND gate.

(ii) $F = (A+B) \overline{C}$

(B)

| A | B | C | F |

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

| 0 | 0 | 0 | 0 |

| 0 | 0 | 1 | 0 |

| 0 | 1 | 0 | 1 |

| 0 | 1 | 1 | 0 |

| 1 | 0 | 0 | 1 |

| 1 | 0 | 1 | 0 |

| 1 | 1 | 0 | 1 |

| 1 | 1 | 1 | 0 |

(C) $657_8 = 1AF_{16}$