The output of the combinational circuit given below is

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The output of the combinational circuit given below is

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GATE EC 2016 Official Paper: Shift 1

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A+B+C
A(B+C)
B(C+A)
C(A+B)

Answer 1

Gate EC 2016 paper 1 Images-Q17.1

$Y = A B C \oplus (A B \oplus B C) = (A B C) (A B ⊙ B C) + (A B \oplus B C) (\overset{―}{A B C})$

Consider $A B ⊙ B C = (A B) (B C) + (\overset{―}{A B}) (\overset{―}{B C})$

$\begin{array}{l} = A B C + (\bar{A} + \bar{B}) (\bar{B} + \bar{C}) = A B C + \bar{A} \bar{B} + \bar{B} + \bar{A} \bar{C} + \bar{B} \bar{C} \\ = A B C + \bar{A} \bar{C} + \bar{B} = \bar{A} \bar{C} + (B + \bar{B}) (A C + \bar{B}) = \bar{A} \bar{C} + A C + \bar{B} \\ (A B C) (A B ⊙ B C) = (A B C) (\bar{A} \bar{C} + A C + \bar{B}) = A B C \end{array}$

Consider $\overset{―}{A B C} (A B \oplus B C) = A B (\overset{―}{B C}) + \overset{―}{A B} (B C) = A B (\bar{B} + \bar{C}) + (\bar{A} + \bar{B}) B C = \bar{A} B C + A B \bar{C}$

$\begin{array}{l} ∴ Y = A B C + \bar{A} B C + A B \bar{C} = B C + A B \bar{C} = B (C + \bar{C} A) \\ = B (C + \bar{C}) (C + A) = B (C + A) \end{array}$

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The output of the combinational circuit given below is

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