Haynes Vaughn
08/10/2024 · High School

a) \( \mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})=\sum(0,1,2.3,4,6,7,8,9,10,15) \) b) \( \mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})=\sum(0,4,6,8,9,11,11,14,15) \)

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To analyze the given functions \( F(A, B, C, D) \), we will first convert the minterms into their binary representations and then derive the corresponding Boolean expressions. ### Part a: Given: \[ F(A, B, C, D) = \sum(0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 15) \] **Step 1: Convert minterms to binary** - \( 0 \) = \( 0000 \) - \( 1 \) = \( 0001 \) - \( 2 \) = \( 0010 \) - \( 3 \) = \( 0011 \) - \( 4 \) = \( 0100 \) - \( 6 \) = \( 0110 \) - \( 7 \) = \( 0111 \) - \( 8 \) = \( 1000 \) - \( 9 \) = \( 1001 \) - \( 10 \) = \( 1010 \) - \( 15 \) = \( 1111 \) **Step 2: Identify the combinations** - From the binary representations, we can group the minterms based on common variables. **Step 3: Write the Boolean expression** Using the minterms, we can derive the expression: \[ F(A, B, C, D) = A'B'C'D' + A'B'C'D + A'B'CD' + A'B'CD + A'BC'D' + A'BC'D + AB'C'D' + AB'C'D + AB'CD + ABC'D + ABCD \] This can be simplified using Karnaugh maps or Boolean algebra. ### Part b: Given: \[ F(A, B, C, D) = \sum(0, 4, 6, 8, 9, 11, 14, 15) \] **Step 1: Convert minterms to binary** - \( 0 \) = \( 0000 \) - \( 4 \) = \( 0100 \) - \( 6 \) = \( 0110 \) - \( 8 \) = \( 1000 \) - \( 9 \) = \( 1001 \) - \( 11 \) = \( 1011 \) - \( 14 \) = \( 1110 \) - \( 15 \) = \( 1111 \) **Step 2: Identify the combinations** - Similar to part a, we can group these minterms based on common variables. **Step 3: Write the Boolean expression** Using the minterms, we can derive the expression: \[ F(A, B, C, D) = A'B'C'D' + A'B'CD' + A'BC'D' + AB'C'D' + AB'C'D + AB'CD + ABC'D + ABCD \] This can also be simplified using Karnaugh maps or Boolean algebra. ### Summary - For part a, the function is derived from the minterms \( 0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 15 \). - For part b, the function is derived from the minterms \( 0, 4, 6, 8, 9, 11, 14, 15 \). Both functions can be further simplified using Karnaugh maps or Boolean algebra techniques. If you need the simplified forms or further assistance, please let me know!

Quick Answer

For part a, the function \( F(A, B, C, D) \) is derived from the minterms \( 0, 1, 2, 3, 4, 6, 7, 8, 9, 10, 15 \) and for part b, the function is derived from the minterms \( 0, 4, 6, 8, 9, 11, 14, 15 \).
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