Question No: 1 Following logic circuit consist of eight gates reducethis circuit

Question

Question No: 1

Following logic circuit consist of eight gates reducethis circuit to six gates such that its out put may notchange.                                                                              Marks: 10

Question No: 2

Following is the karnaugh map of 4 variables A,B,C and Dsolve this into minimal Boolean expression                                                                                            Marks:5

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Question No: 3

Simplify the following expression and implement theresulting expression using NAND gates only

A’(A + B) + (B + AA)(A +B’):                                                                  Marks: 5

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Explanation / Answer

1. I can't see the picture so Ican't answer that 2. By the map we know that this function is =m(0,1,2,5,13,15,8,9,10) (Anywhere the function is 1). We canthen simplifiy it like this: 0000 0001 0010 1000 0101 1001 1010 1101 1111 Simplifies to 000- 00-0 -000 0-01 -001 100- 10-0 -010 -101 1-01 11-1 Which simplifies down to: -00- -0-0 --01 11-1 Therefore the function is simplified down to F(A,B,C,D) =(B'C') + (B'D') + (C'D) + (ABD) 3. With distribution we get: A'A + BA' + AB + BB' + AA + AB' = 0 + A'B + AB + 0 + A + AB' = A'B + AB + A + AB' = A'B + A(B + B' + 1) = A'B + A Which is simply A' NAND B' Hope this helps. 1. I can't see the picture so Ican't answer that 2. By the map we know that this function is =m(0,1,2,5,13,15,8,9,10) (Anywhere the function is 1). We canthen simplifiy it like this: 0000 0001 0010 1000 0101 1001 1010 1101 1111 Simplifies to 000- 00-0 -000 0-01 -001 100- 10-0 -010 -101 1-01 11-1 Which simplifies down to: -00- -0-0 --01 11-1 Therefore the function is simplified down to F(A,B,C,D) =(B'C') + (B'D') + (C'D) + (ABD) 3. With distribution we get: A'A + BA' + AB + BB' + AA + AB' = 0 + A'B + AB + 0 + A + AB' = A'B + AB + A + AB' = A'B + A(B + B' + 1) = A'B + A Which is simply A' NAND B' Hope this helps.

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