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    Find the logic equation for the truth table given: A B Y 0 0 0 0 1 1 1 0 0 1 1 1
    Question

    Find the logic equation for the truth table given:

    A
    B
    Y
    0
    0
    0
    0
    1
    1
    1
    0
    0
    1
    1
    1

    A.

    Y=AˉB(A+Bˉ)Y=\bar{A}B*(A+\bar{B})​​

    B.

    Y=AˉBˉ+ABˉY=\bar{A}\bar{B}+A\bar{B}​​

    C.

    Y=(AˉBˉ)BY= (\bar{A} \cdot \bar{B})*B​​

    D.

    Y = B

    Correct option is D

    From the truth table: when B = 0, Y = 0; when B = 1, Y = 1, regardless of A.
    Thus Y follows B exactly—so Y = B.
    Equivalently, in canonical SOP: Y=m(1,3)=AˉB+AB=B(Aˉ+A)=BY=\sum m(1,3)=\bar A B + A B = B(\bar A + A)=B​.
    A K-map would also group the two 1s in the B = 1 column, yielding Y = B.
    Algebraically, option (b) simplifies to AˉBˉ+ABˉ=Bˉ(Aˉ+A)=Bˉ\bar{A}\bar{B}+A\bar{B}=\bar{B}(\bar{A}+A)=\bar{B}​, which is not B.
    Option (a) is AˉB(A+Bˉ)\bar{A}B(A+\bar{B})​, which is impossible to satisfy (requires A = 0 and A = 1 or B = 1 and B = 0 simultaneously), hence always 0.
    Option (c) is (AˉBˉ)B=Aˉ(BˉB)=0(\bar{A}\bar{B})B=\bar{A}(\bar{B}B)=0​, also always 0.
    Therefore, only (d) matches the truth table behavior.
    Important Key Points

    1. Identity behavior: When a truth table shows Y identical to B across all rows, the logic equation is simply Y = B.
    2. Independence from A: If outputs do not change with A, A is a don’t-care input for the function.
    3. Simplification check: Use basic identities—BˉB=0,X+Xˉ=1\bar{B}B=0, X+\bar{X}=1​, and factoring—to test alternatives quickly.
    4. Design implication: Implementing Y = B needs just a wire (or buffer), minimizing gates, delay and power.
    5. Verification tip: Compare column-by-column: if two columns match exactly, they are logically equivalent.
    6. Robustness: Expressions that simplify to constants (0/1) cannot fit a non-constant truth table; eliminate them first.

    Knowledge Booster

    • Why (a) is wrong: AˉB(A+Bˉ)=AˉBA+AˉBBˉ=0+0=0\bar{A}B(A+\bar{B}) = \bar{A}B\cdot A + \bar{A}B\cdot\bar{B} = 0 + 0 = 0​; contradicts rows where B = 1 but Y = 1.
    • Why (b) is wrong: AˉBˉ+ABˉ=Bˉ(Aˉ+A)=Bˉ\bar{A}\bar{B}+A\bar{B} = \bar{B}(\bar{A}+A) = \bar{B}​; this outputs 1 when B = 0, the opposite of Y = B.
    • Why (c) is wrong: (AˉBˉ)B=Aˉ(BˉB)=0(\bar{A}\bar{B})B = \bar{A}(\bar{B}B) = 0​, a constant zero function.
    • Extra tip: If the truth table equals B, alternative equivalent forms include Y=(A+Aˉ)B=BY=(A+\bar{A})B=B​ and Y=B1Y=B\cdot 1​; both reaffirm the identity.

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