1. **State De Morgan's law.** De Morgan's laws are two transformation rules that are useful in Boolean algebra and set theory. They state: $$\overline{A \cdot B} = \overline{A} + \overline{B}$$ $$\overline{A + B} = \overline{A} \cdot \overline{B}$$ This means the complement of a product is the sum of the complements, and the complement of a sum is the product of the complements. 2. **Define the term "minterm".** A minterm is a product (AND) term in Boolean algebra that includes all variables in the function, each appearing exactly once, either complemented or uncomplemented. It represents a single combination of variables that makes the function true. 3. **Convert (231.25)₁₀ to binary number.** Step 1: Convert integer part 231 to binary: $231 \div 2 = 115$ remainder $1$ $115 \div 2 = 57$ remainder $1$ $57 \div 2 = 28$ remainder $1$ $28 \div 2 = 14$ remainder $0$ $14 \div 2 = 7$ remainder $0$ $7 \div 2 = 3$ remainder $1$ $3 \div 2 = 1$ remainder $1$ $1 \div 2 = 0$ remainder $1$ Reading remainders from bottom to top: $11100111_2$ Step 2: Convert fractional part 0.25 to binary: Multiply by 2: $0.25 \times 2 = 0.5$, bit = 0 Multiply fractional part 0.5 by 2: $0.5 \times 2 = 1.0$, bit = 1 Fractional binary is $0.01$ Final binary number: $231.25 = 11100111.01_2$ 4. **Use 1's complement technique to perform $(1001)_2 - (1111)_2$.** Step 1: Find 1's complement of subtrahend (1111): Complement of $1111$ is $0000$ Step 2: Add minuend and 1's complement of subtrahend: $1001 + 0000 = 1001$ Step 3: Since no carry out, take 1's complement of result: Complement of $1001$ is $0110$ So, the result is $-0110_2$ (negative sign shows subtraction result is negative). 5. **Minimize the Boolean function:** $$F= (A + B' + C')(A + B' + C)(A + B + C')$$ Use distributive and absorption laws: Notice common term $A$ in all; rewrite: Since $A$ appears in all sums, expand combination only for the parts without $A$ would be unnecessary; instead simplify each term: Since $A + B' + C'$ and $A + B' + C$ differ by $C$ only, their AND simplifies: $$(A + B' + C')(A + B' + C) = A + B'$$ (Using consensus theorem) Now multiply with the third term: $$(A + B')(A + B + C')$$ Distribute: $= A(A + B + C') + B'(A + B + C')$ $= A + B' A + B' B + B' C'$ Simplify terms: $B' A$ is $A B'$, $B' B = 0$, so: $= A + A B' + 0 + B' C'$ Since $A + A B' = A$ (absorption law), Final simplified function: $$F = A + B' C'$$ 6. (i) **Implement NAND using NOR gate:** NAND operation is the complement of AND: $\overline{AB}$ Using NOR gates, NAND can be implemented by applying De Morgan's law: $$\overline{A B} = \overline{\overline{\overline{A} + \overline{B}}}$$ So, - First NOR gate to get $\overline{A}$: NOR A with A - Second NOR gate to get $\overline{B}$: NOR B with B - Third NOR gate to get $\overline{A} + \overline{B}$ NORed: output is $\overline{\overline{A} + \overline{B}} = A B$ - NAND is complement of AND, so feeding the AND output to NOT (NOR gate with same input) gives NAND. (ii) **Implement XOR gate using NAND gate:** XOR can be expressed as: $$A \oplus B = (A \cdot \overline{B}) + (\overline{A} \cdot B)$$ Using NAND gates only, XOR can be constructed as: $$X = ((A \text{ NAND } B) \text{ NAND } (A \text{ NAND } (A \text{ NAND } B))) \text{ NAND } ((B \text{ NAND } (A \text{ NAND } B)) \text{ NAND } (B \text{ NAND } (A \text{ NAND } B)))$$ This is a standard NAND gate XOR implementation. 7. Circuit design for counting number of 1's in 3-bit input A, B, C with outputs X and Y: 7a. **Complete the truth table:** | A | B | C | Number of 1's | X | Y | |---|---|---|---------------|---|---| | 0 | 0 | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 1 | 0 | 1 | | 0 | 1 | 0 | 1 | 0 | 1 | | 0 | 1 | 1 | 2 | 1 | 0 | | 1 | 0 | 0 | 1 | 0 | 1 | | 1 | 0 | 1 | 2 | 1 | 0 | | 1 | 1 | 0 | 2 | 1 | 0 | | 1 | 1 | 1 | 3 | 1 | 1 | 7b. **Expression for X and Y:** X represents the high bit and Y the low bit in binary count of 1's. From the table: - $X = 1$ when number of 1's is 2 or 3. Boolean expression: $$X = (A \cdot B) + (B \cdot C) + (A \cdot C)$$ - $Y = 1$ when number of 1's is odd (1 or 3), which is the XOR of inputs: $$Y = A \oplus B \oplus C$$ 7c. **Minimize X and Y using 3-variable Karnaugh Map:** - For $X$: KMap for $X$ shows 1s in cells with 2 or 3 ones. Simplified expression using KMap is: $$X = AB + BC + AC$$ (This is already minimal.) - For $Y$: $Y$ is the parity of inputs, the XOR of three variables. Minimal form: $$Y = A \oplus B \oplus C$$ 7d. **Implement circuit:** - $X$ can be implemented using three 2-input AND gates and one 3-input OR gate (or chained OR gates): $X = (A \cdot B) + (B \cdot C) + (A \cdot C)$ - $Y$ can be implemented with two XOR gates: First XOR: $D = A \oplus B$, Second XOR: $Y = D \oplus C$ ---