Binary Conversions and K-Map Boolean Simplification

Sure! Here are the conversions you requested:

1. (10000101)₂ to hexadecimal (base 16):
   • Group the binary digits in sets of four from the right: 1000 0101
   • (1000)₂ = 8, (0101)₂ = 5
   • Therefore, (10000101)₂ = (85)₁₆.
2. (AA)₁₆ to decimal (base 10):
   • A = 10 in decimal, so (AA)₁₆ = 10 × 16¹ + 10 × 16⁰ = 160 + 10 = 170.
   • Therefore, (AA)₁₆ = (170)₁₀.
3. (5.5)₁₀ to binary (base 2):
   • The integer part: 5 = (101)₂.
   • The fractional part: 0.5 = (0.1)₂.
   • Combine them: (5.5)₁₀ = (101.1)₂.
4. 2's complement of (01001)₂:
   • Invert the digits: (01001)₂ → (10110)₂.
   • Add 1: (10110)₂ + (00001)₂ = (10111)₂.
   • Therefore, the 2's complement of (01001)₂ is (10111)₂.

Simplify F(A, B, C, D) with a K-Map

To simplify the Boolean expression F(A, B, C, D) = Σ m(0, 2, 3, 8, 10, 11, 12, 14) using a Karnaugh map (K-map), follow these steps.

Step 1: Create the K-map

A 4-variable K-map has 16 cells corresponding to the minterms. Arrange AB as rows in Gray order (00, 01, 11, 10) and CD as columns in Gray order (00, 01, 11, 10).

        CD
      00  01  11  10
    +-----------------
  00 |  1   0   1   1   | A=0, B=0
  01 |  0   0   0   0   | A=0, B=1
  11 |  1   0   0   1   | A=1, B=1
  10 |  1   0   1   1   | A=1, B=0
  

Step 2: Fill in the K-map

From the given minterms, place 1s in the respective cells:

• m(0) = 1
• m(2) = 1
• m(3) = 1
• m(8) = 1
• m(10) = 1
• m(11) = 1
• m(12) = 1
• m(14) = 1

Step 3: Identify groups

Make groups of 1, 2, 4, 8 ... (prefer largest groups, overlaps allowed). A minimal covering can be achieved with three groups of four (overlapping):

1. Group 1 (4 cells): m(2), m(3), m(10), m(11) → simplifies to B' C (B = 0, C = 1).
2. Group 2 (4 cells): m(8), m(10), m(12), m(14) → simplifies to A D' (A = 1, D = 0).
3. Group 3 (4 cells): m(0), m(2), m(8), m(10) → simplifies to B' D' (B = 0, D = 0).

Step 4: Write the simplified expression

From the groups above the expression is:

F(A, B, C, D) = B' C + B' D' + A D'

This can be factored as:

F(A, B, C, D) = B'(C + D') + A D'

Both forms are equivalent; the factored form is compact and highlights shared terms.

Step 5: Implement using logic gates

To implement F(A, B, C, D) = B'(C + D') + A D':

• NOT gates: produce B' and D' (also A', C' only if needed elsewhere).
• OR gate: compute (C + D').
• AND gates:
  • AND B' with (C + D') to get B'(C + D').
  • AND A with D' to get A D'.
• OR gate: combine B'(C + D') and A D' to get F.

Summary of required gates

• NOT gates for B and D (and for any other inverted signals you choose to generate)
• One OR gate for (C + D')
• Two AND gates for B'(C + D') and A D'
• One OR gate to combine the two AND outputs

If you prefer the sum-of-products expansion, use three AND gates for B'C, B'D', A D' and one OR gate to combine their outputs.

Let me know if you want the circuit diagram, truth table, or further simplification (e.g., using NAND/NOR-only implementations).