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Simplify the Boolean expression F = C(B + C)(A + B + C).
Ans. Simplification of the given Boolean Expression F = C (B +C) (A+B+C) given asF = C (B+C) (A+B+C)= CB + CC [(A+B+C)]= CB + C [(A+B+C)] (Q CC = C)= CBA + CBB + CBC + CA + CB + CC= ABC + CB + CB + CA + CB + CC (Q CBB =CB & CBC = CB)= ABC + CB + CA + C (Q CB+CB+CB = CB; CC = C)= ABC + BC + C (1+A)= ABC + BC + C (Q 1+A = 1)= ABC + C (1+B)= ABC + C (Q 1+B = 1)= C (1+AB) = C {Q (1+AB)=1}
Q. F'(a,b,c,d) = (a + b + d')(a + c' + d')(a + b' + c') d'(a,b,c,d) = (a + b' + c + d')(a' + c' + d')(a' + b + d) Find the simplified function F and implement it using NOR Ga
Add 648 and 487 in BCD code. Ans. In BCD Code, addition of 648 and 487: 6 4 8 = 0 1 1 0 0 1 0 0 1 0 0 0 4 8 7 = 0 1 0 0 1 0 0 0 0 1 1 1 ---------------------
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