Q.

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To obtain the given truth table, following logic gate should be placed at G :

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a

NOR Gate

b

NAND Gate

c

OR Gate

d

AND Gate

answer is A.

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Detailed Solution

From the circuit:

  • The first AND gate receives inputs AA' (NOT AA) and BB.
  • The second AND gate receives inputs AA and BB' (NOT BB).

Thus:

  • The output of the first AND gate is ABA' \cdot B,
  • The output of the second AND gate is ABA \cdot B'.

Let's say G is NOR gate.

These two outputs are then fed into the NOR gate GG, which performs:

Y=A'B+AB'¯

 

Truth Table Derivation

Now, calculate the values step by step:

AABBAA'BB'ABA' \cdot BABA \cdot B'AB+ABA' \cdot B + A \cdot B'Y=A'B+AB'¯
00110001
01101010
10010110
11000001
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