Question

5. Realize the Boolean expression Y = A B C + Ā B C + ABC using NOR gates only. You may assume that the inputs are available both in their true form and complement form, i.e., input A is available both as A and A and similarly the inputs B and C. Repeat the problem but this time using NAND gates only.

Answer 1

Using nor Gales only.. у = ABC +Ā BCA ĀBE Yo & Yn & Yg Ya ABC using Demorgan's low A BC A ti té | ABC = (Ã + B + E) (Авс Y,= Āt Bra] A+Btc A + Ŝ t Y =Ā BC Ã + B+ca A+ BIC Ā BE = Ã +Bēs ĀtBtc Y₃ A + B + 7 + A+B+@+ A+Btc yo Y,+ Y + Y Y*%+ Detetle А (A) ) y[A+B+ 2) + (A+B+ 7) + (A+B+C) А B

늘 B ㄹ 3 30 C 쓰 는 4월 . 야 늘

Using NAND Gates:- Y - ABC+ ARC+ ĀBO Y = ABC (ABC) Y = Ã BC = A B A B C) ĀBC = (ABC) Yay = Y,+Y₂+ Y₃ Yital Demorgan's low

А А B се NOR Gates Do y Dabei B 13 clo DO* Dry NAMD Gates