Jawaharlal Nehru Technological University Hyderabad B.Tech II DIGITAL LOGIC AND PULSE CIRCUITS Previous Question Paper (November/December - 2020 - R16)

i) For binary 110001.1010010 to hexadecimal, group bits in fours from the radix point: (0011)(0001).(1010)(0010) = 31.A2₁₆. For decimal, calculate the sum of powers: 1×2⁵ + 1×2⁴ + 0×2³ + 0×2² + 0×2¹ + 1×2⁰ + 1×2^-1 + 0×2^-2 + 1×2^-3 + 0×2^-4 + 0×2^-5 + 1×2^-6 = 32 + 16 + 1 + 0.5 + 0.125 + 0.015625 = 49.640625₁₀.

ii) For (423.25)₁₀ to binary, convert integer 423: 423/2 = 211 rem 1, 211/2 = 105 rem 1, 105/2 = 52 rem 1, 52/2 = 26 rem 0, 26/2 = 13 rem 0, 13/2 = 6 rem 1, 6/2 = 3 rem 0, 3/2 = 1 rem 1, 1/2 = 0 rem 1. Binary is 110100111. Convert fractional 0.25: 0.25×2 = 0.5 (0), 0.5×2 = 1.0 (1). Binary is 110100111.01₂. For hexadecimal, group bits: (0001)(1010)(0111).(0100) = 1A7.4₁₆.

A gate is universal if any Boolean function can be implemented using only that gate. NAND gates are universal because:

1. NOT: Connect both inputs of a NAND gate together.
2. AND: Pass the NAND output through a NOT gate (NAND followed by NAND).
3. OR: Invert inputs using NAND gates and pass to a NAND gate (De Morgan's Law).

NOR gates are universal because:

1. NOT: Connect both inputs of a NOR gate together.
2. OR: Pass the NOR output through a NOT gate (NOR followed by NOR).
3. AND: Invert inputs using NOR gates and pass to a NOR gate (De Morgan's Law).

Since AND, OR, and NOT operations can be realized, any logic circuit can be constructed using exclusively NAND or NOR gates.