1. **Problem Statement:** We have three logic circuits (Questions 4, 5, and 6) involving NOT, AND, OR, and XOR gates. We need to find the Boolean expressions and truth tables for each. --- ### Question 4 2. **Boolean Expression:** - Input B goes to a NOT gate: $\overline{B}$ - Input A goes to the AND gate along with $\overline{B}$: $A \cdot \overline{B}$ - Input C goes directly to the AND gate output and also to the OR gate. - The output of the AND gate and input C go to the OR gate: $Y = (A \cdot \overline{B}) + C$ 3. **Truth Table:** | A | B | C | $\overline{B}$ | $A \cdot \overline{B}$ | $Y = (A \cdot \overline{B}) + C$ | |---|---|---|--------------|---------------------|-----------------------------| | 0 | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 0 | 0 | 0 | 0 | | 0 | 1 | 1 | 0 | 0 | 1 | | 1 | 0 | 0 | 1 | 1 | 1 | | 1 | 0 | 1 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 0 | 0 | | 1 | 1 | 1 | 0 | 0 | 1 | --- ### Question 5 4. **Boolean Expression:** - Input A goes to NOT gate: $\overline{A}$ - $\overline{A}$ and B go to AND gate: $\overline{A} \cdot B$ - B goes to OR gate directly - Outputs of AND gate and B go to OR gate: $Y = (\overline{A} \cdot B) + B$ 5. **Simplification:** Since $B + (\overline{A} \cdot B) = B$, the output simplifies to $Y = B$ 6. **Truth Table:** | A | B | $\overline{A}$ | $\overline{A} \cdot B$ | $Y = (\overline{A} \cdot B) + B$ | |---|---|--------------|---------------------|-----------------------------| | 0 | 0 | 1 | 0 | 0 | | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 0 | 1 | --- ### Question 6 7. **Boolean Expression:** - Left XOR gate: $X_1 = A \oplus B$ - Bottom-left AND gate: $A \cdot B$ - C input splits to two AND gates: one is the right AND gate, which also takes input from bottom-left AND gate output. - Right XOR gate inputs: $X_1$ and $C$, output $X = X_1 \oplus C$ - Right AND gate inputs: $(A \cdot B)$ and $C$, output $D = (A \cdot B) \cdot C$ - OR gate inputs: $X$ and $D$, output $Y = X + D$ 8. **Truth Table:** | A | B | C | $A \oplus B$ | $A \cdot B$ | $X = (A \oplus B) \oplus C$ | $D = (A \cdot B) \cdot C$ | $Y = X + D$ | |---|---|---|-------------|------------|--------------------------|-----------------------|------------| | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | | 0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 | | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 0 | | 1 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | | 1 | 1 | 1 | 0 | 1 | 1 | 1 | 1 | --- **Summary:** - Q4: $Y = (A \cdot \overline{B}) + C$ - Q5: $Y = B$ - Q6: $X = (A \oplus B) \oplus C$, $Y = X + (A \cdot B \cdot C)$ These expressions and truth tables fully describe the circuits.