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Explain the three Boolean algebra laws using ORing and ANDing, and explain logic gates in terms of the number system.

Computer Science

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Here are the answers for questions 2, 3, and 4:

2.

a) Explain the three Boolean algebra laws you know using ORing and ANDing system.

  1. Commutative Law: This law states that the order of operands does not affect the result of an operation.
    • For ORing: A+B=B+AA + B = B + A
    • For ANDing: A⋅B=B⋅AA \cdot B = B \cdot A
  2. Associative Law: This law states that the grouping of operands does not affect the result of an operation when performing the same operation multiple times.
    • For ORing: (A+B)+C=A+(B+C)(A + B) + C = A + (B + C)
    • For ANDing: (A⋅B)⋅C=A⋅(B⋅C)(A \cdot B) \cdot C = A \cdot (B \cdot C)
  3. Distributive Law: This law describes how operations interact with each other. It allows for the expansion or factorization of Boolean expressions.
    • A⋅(B+C)=(A⋅B)+(A⋅C)A \cdot (B + C) = (A \cdot B) + (A \cdot C)
    • A+(B⋅C)=(A+B)⋅(A+C)A + (B \cdot C) = (A + B) \cdot (A + C)

b) Explain logic gate in terms of number system.

A logic gate operates within the binary number system, which uses only two digits: 0 and 1. These digits represent the two possible states of an electrical signal (e.g., 0 for low voltage/off, 1 for high voltage/on). Logic gates take one or more binary inputs and produce a single binary output based on a specific Boolean function, effectively performing arithmetic or logical operations on these binary numbers.

3. Simplify the following Boolean algebra.

i) A+AˉB+A+BA + \bar{A}B + A + B

  1. Apply the Idempotence Law (X+X=XX+X=X) for A+AA+A.

    A+AˉB+A+B=A+AˉB+BA + \bar{A}B + A + B = A + \bar{A}B + B
  2. Apply the Absorption Law (X+XˉY=X+YX + \bar{X}Y = X+Y).

    A+AˉB+B=A+BA + \bar{A}B + B = A + B

    The simplified expression is A+B\boxed{A+B}.

    ii) Z=AB+A(B+C)+B(B+C)Z = AB + A(B+C) + B(B+C)

  3. Apply the Distributive Law (X(Y+Z)=XY+XZX(Y+Z) = XY+XZ).

    Z=AB+AB+AC+BB+BCZ = AB + AB + AC + BB + BC
  4. Apply the Idempotence Law (X+X=XX+X=X and XX=XXX=X).

    Z=AB+AC+B+BCZ = AB + AC + B + BC
  5. Factor out B from terms containing B.

    Z=B(A+1)+AC+BCZ = B(A+1) + AC + BC
  6. Apply the Identity Law (X+1=1X+1=1).

    Z=B(1)+AC+BCZ = B(1) + AC + BC Z=B+AC+BCZ = B + AC + BC
  7. Factor out B again from terms containing B.

    Z=B(1+C)+ACZ = B(1+C) + AC
  8. Apply the Identity Law (1+X=11+X=1).

    Z=B(1)+ACZ = B(1) + AC Z=B+ACZ = B + AC

    The simplified expression is B+AC\boxed{B+AC}.

    iii) F=A+AˉBC+AˉBˉCF = A + \bar{A}BC + \bar{A}\bar{B}C

  9. Factor out AˉC\bar{A}C from the last two terms.

    F=A+AˉC(B+Bˉ)F = A + \bar{A}C(B + \bar{B})
  10. Apply the Complement Law (B+Bˉ=1B + \bar{B} = 1).

    F=A+AˉC(1)F = A + \bar{A}C(1)
  11. Apply the Identity Law (X⋅1=XX \cdot 1 = X).

    F=A+AˉCF = A + \bar{A}C
  12. Apply the Absorption Law (X+XˉY=X+YX + \bar{X}Y = X+Y).

    F=A+CF = A + C

    The simplified expression is A+C\boxed{A+C}.

    4. Form the truth table using the Boolean expression of the following:

    i) F=A+AˉBC+AˉBˉCF = A + \bar{A}BC + \bar{A}\bar{B}C
    From 3(iii), we simplified this expression to F=A+CF = A+C. We will use this simplified form to construct the truth table.

    ACF = A+C
    000
    011
    101
    111

    ii) Z=AB+A(B+C)+B(B+C)Z = AB + A(B+C) + B(B+C)
    From 3(ii), we simplified this expression to Z=B+ACZ = B+AC. We will use this simplified form to construct the truth table.

    ABCACZ = B+AC
    00000
    00100
    01001
    01101
    10000
    10111
    11001
    11111
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