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    Logic Circuits for Boolean Expressions

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    1. Using Boolean algebra, reduce the following Boolean expression to its simplest form and implement it using your method of choice.

    F = A'B'D + ABC' + AB'CD + BC + A'CD + BCD

    2. Using a Karnaugh map (K-map), reduce the following Boolean to its simplest form and implement it using SOP (Sum of Products)

    F = B'C'D + B'D + ABC + BC'D' +ABD' + A'B'D

    3. Using a Karnaugh map (K-map), reduce the following Boolean to its simplest form and implement it using POS (Products of Sum)

    F(A, B, C, D) = ∑(0,2,3,5,7,8,9,10,11,13,15)

    Show karnaugh map steps and details.

    © BrainMass Inc. brainmass.com December 24, 2021, 8:22 pm ad1c9bdddf
    https://brainmass.com/math/boolean-algebra/logic-circuits-boolean-expressions-270115

    SOLUTION This solution is FREE courtesy of BrainMass!

    Please see attached for the full solution to the given problems.

    1. Using Boolean algebra, reduce the following Boolean expression to its simplest form and implement it using your method of choice.

    F = A'B'D + ABC' + AB'CD + BC + A'CD + BCD
    Solution: We first group the terms with common logic:
    F = A'B'D + B(AC'+C) + CD(AB'+A') + BCD
    Then we apply Boolean algebra laws to simplify the highlighted terms:
    F = A'B'D + B(A+C) + CD(A+B') + BCD
    Then we distribute the terms:
    F = A'B'D + AB + AC + ACD + B'CD + BCD
    Then we group terms common logic:
    F = A'B'D + AB + AC + ACD + CD(B'+ B)
    Then we apply Boolean algebra laws to simplify the highlighted terms:
    F = A'B'D + AB + AC + ACD + CD
    Then we group terms common logic:
    F = A'B'D + AB + AC + CD(A+1)
    Then we apply Boolean algebra laws to simplify the highlighted terms to get the final answer of:
    F = A'B'D + AB + AC + CD
    Using K-map and SOP, we also get the same answer:
    ABCD 00 01 11
    10
    00 0 1
    1 0
    01 0
    0 1
    1
    11 1 1 1 1
    10 0 0 1 0

    F = A'B'D + AB + AC + CD

    2. Using a Karnaugh map (K-map), reduce the following Boolean to its simplest form and implement it using SOP (Sum of Products)

    F = B'C'D + B'D + ABC + BC'D' +ABD' + A'B'D
    Solution: Using K-map and SOP, we have:
    A'B' A'B
    AB AB'
    C'D' 0
    1 1 0
    C'D 1 0 0 1
    CD 1 0 1
    1
    CD' 0 0 1 0

    Grouping the 1's, the final expression in SOP is:
    F = B'D + BC'D' + ABC

    3. Using a Karnaugh map (K-map), reduce the following Boolean to its simplest form and implement it using POS (Products of Sum)

    F(A, B, C, D) = ∑(0,2,3,5,7,8,9,10,11,13,15)

    Solution: Using K-map and POS, we have:
    ABCD 00
    01 11
    10
    00 0 1 0 0
    01 1 0
    0 1
    11 1
    0 0 1
    10 0 0 0 0

    The product of sum terms are placed in the K-map as 0's. Grouping the 0's, the final expression in POS is:
    F = (B'+D')(A'+B)(C'+D')(B+D).

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 8:22 pm ad1c9bdddf>
    https://brainmass.com/math/boolean-algebra/logic-circuits-boolean-expressions-270115

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