Digital Electronics CIE-1 Exam Preparation

Here is a complete and concise answer sheet for your Digital Electronics CIE-1 preparation based on your questions:

1. Design Half and Full Subtractor with Truth Table

Half Subtractor

• Inputs: A, B
• Outputs: Difference (D), Borrow (Bo)
• Logic Equations:
  • Difference (D) = A ⊕ B
  • Borrow (Bo) = A' · B

Full Subtractor

• Inputs: A, B, B_in (Borrow In)
• Outputs: Difference (D), Borrow Out (B_o)
• Logic Equations:
  • D = A ⊕ B ⊕ B_in
  • B_o = B · B_in + B · A' + A' · B_in

2. Working of Adder and Subtractor

Adder Circuits

• Half Adder: Adds two bits A and B.
  • Sum = A ⊕ B, Carry = A · B
• Full Adder: Adds three bits A, B, and C_in.
  • Sum = A ⊕ B ⊕ C_in, C_out = AB + AC_in + BC_in

Subtractor Circuits

• Half Subtractor: Subtracts B from A.
  • Difference = A ⊕ B, Borrow = A' · B
• Full Subtractor: Subtracts B and B_in from A.
  • Difference = A ⊕ B ⊕ B_in, Borrow = B · B_in + A' · B + A' · B_in

3. Logic Gates with Truth Table

(Ensure you draw the standard symbols for AND, OR, NOT, NAND, NOR, XOR, and XNOR alongside their respective truth tables.)

4. Universal and Exclusive Gates

• Universal Gates:
  • NAND and NOR gates are called universal gates because you can construct all other logic gates (AND, OR, NOT) using only NAND or only NOR gates.
• Exclusive Gates:
  • XOR (Exclusive OR): The output is 1 only when the inputs are different.
  • XNOR (Exclusive NOR): The output is 1 only when the inputs are the same.

5. Number System Conversions and Solved Problems

b. Convert (26.25)₁₀ to Binary

• Integer Part: 26 ÷ 2 = 13 (R: 0), 13 ÷ 2 = 6 (R: 1), 6 ÷ 2 = 3 (R: 0), 3 ÷ 2 = 1 (R: 1), 1 ÷ 2 = 0 (R: 1) → 11010
• Fractional Part: 0.25 × 2 = 0.5 (Integer: 0), 0.5 × 2 = 1.0 (Integer: 1) → .01
• Answer: (11010.01)₂

c. Convert (155C.125)₁₆ to Decimal

• Integer Part (155C): (1 × 16³) + (5 × 16²) + (5 × 16¹) + (12 × 16⁰) = 4096 + 1280 + 80 + 12 = 5468
• Fractional Part (0.125): (1 × 16^-1) + (2 × 16^-2) + (5 × 16^-3) = 1/16 + 2/256 + 5/4096 = 0.0625 + 0.0078125 + 0.0012207 ≈ 0.0715
• Answer: ≈ 5468.0715₁₀

d. Find 2's Complement of (1001001)

• Step 1 (1's Complement): Invert the bits → 0110110
• Step 2: Add 1 to the LSB → 0110110 + 1 = 0110111
• Answer: 2’s complement = 0110111

6. Simplify the Expression: (A + B)(B + C)

Using Boolean Algebra laws:

• = AB + AC + BB + BC
• = AB + AC + B + BC (Since B · B = B)
• = B + AB + BC + AC (Rearranging)
• = B(1 + A + C) + AC (Since 1 + A + C = 1)
• = B(1) + AC
• Answer: B + AC

7. DeMorgan’s Theorem

1. Theorem 1: (A · B)' = A' + B' (The complement of a product is equal to the sum of the complements.)
2. Theorem 2: (A + B)' = A' · B' (The complement of a sum is equal to the product of the complements.)

These theorems are essential for simplifying Boolean expressions by converting AND operations to OR operations (and vice versa) using inversions.

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