(a) A 0 OHH 0 1 1 B 0 1 1 0 Y HOHE 0 1...

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(a) A 0 OHH 0 1 1 B 0 1 1 0 Y HOHE 0 1 1 1 (b) A 0 0 DOOHHHH B с BOOHHOONE ok 0 1 1 1 1 1 1 1 0 0 1 0 1 0 0 1 0 0 -OTOTOT 0 (c) Y A В 1 0 0 0 HOOOooo. 0 0 0 0 1 1 BOO с oko 0 1 Y 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 HC 1 1 0 0 1 0 1 1 1 0 1 1 1 (d) A </c 0 0 0 0 0 ooo 0 0 BOOO 0 0 0 1 1 0 1 0 1 1 0 C D Good 1 9OHO 0 1 1 0 1 1 0 HOO Y 0 1 0 0 1 1 0 1 1 0 0 1 1 0 10 0 1 10 1 0 10 1 1 11 0 0 11 0 1 11 1 0 0 0 0 1 0 0 0 0 1 (e) A B 0 00000 с 0 0 0 1 0 1 0 0 looooOLLE 0 0 0 1 0 1 0 1 0 1 1 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 1 1 1 D Y 1 0 0 1 1 0 0 1 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 0 0 1 0 1 1 For each truth table (a-e) you will receive the following extra credit points: +1 pt for producing a correct and fully simplified Boolean expression +1 pt for drawing a correct digital circuit (using two-input AND, OR and NOT gates) 1 1 Please show all work, particularly in the simplification. You can use Boolean Algebra and/or a Karnaugh Map - though just a reminder - the Karnaugh Map will only do the DCI (Distributive, Complement, Identity) and Idempotency properties and it is possible you can go further. To get the point on the Boolean expression, it should be simplified such that it uses the minimum number of transistors possible (NOT gates have 2, AND and OR gates have 6). (a) A 0 OHH 0 1 1 B 0 1 1 0 Y HOHE 0 1 1 1 (b) A 0 0 DOOHHHH B с BOOHHOONE ok 0 1 1 1 1 1 1 1 0 0 1 0 1 0 0 1 0 0 -OTOTOT 0 (c) Y A В 1 0 0 0 HOOOooo. 0 0 0 0 1 1 BOO с oko 0 1 Y 1 1 0 1 0 1 1 0 0 0 0 0 1 0 0 HC 1 1 0 0 1 0 1 1 1 0 1 1 1 (d) A </c 0 0 0 0 0 ooo 0 0 BOOO 0 0 0 1 1 0 1 0 1 1 0 C D Good 1 9OHO 0 1 1 0 1 1 0 HOO Y 0 1 0 0 1 1 0 1 1 0 0 1 1 0 10 0 1 10 1 0 10 1 1 11 0 0 11 0 1 11 1 0 0 0 0 1 0 0 0 0 1 (e) A B 0 00000 с 0 0 0 1 0 1 0 0 looooOLLE 0 0 0 1 0 1 0 1 0 1 1 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 1 1 1 1 D Y 1 0 0 1 1 0 0 1 0 1 1 0 0 0 1 1 0 1 0 0 1 0 1 0 0 1 0 1 1 For each truth table (a-e) you will receive the following extra credit points: +1 pt for producing a correct and fully simplified Boolean expression +1 pt for drawing a correct digital circuit (using two-input AND, OR and NOT gates) 1 1 Please show all work, particularly in the simplification. You can use Boolean Algebra and/or a Karnaugh Map - though just a reminder - the Karnaugh Map will only do the DCI (Distributive, Complement, Identity) and Idempotency properties and it is possible you can go further. To get the point on the Boolean expression, it should be simplified such that it uses the minimum number of transistors possible (NOT gates have 2, AND and OR gates have 6).

Answer rating: 100% (QA)

The image shows five truth tables labeled a through e Each table corresponds to a different Boolean function of variables A B C and D where some table... View the full answer