Given f1, f3 and f in canonical sum of products form (in decimal) for the circuit
A)m(4, 6) B) m(4, 8) C) m(6, 8) D) m(4, 6, 8)
(A) A
(B) B
(C) C
(D) D
Answer: (C)
Explanation: From logic diagram we have f=f1.f2+f3
f=m(4,5,6,7,8).f2+m(1,6,15)—-(1)
from eq(1) we need to find such f2 so that we can get f=m(1,6,8,15)
eq(1) says we can get m(1,6,15) from f3 ,so only 8 left
now from option (a,b,d) we get (4,6), (4,8) ,(4,6,8) respectively for m(4,5,6,7,8)f2 which is not required as m4 is undesired.
But option (C) m(4,5,6,7,8)(6,8)+(1,6,15)
- m(6,8)+m(1,6,15)
- m(1,6,8,15)an
Option (C) is correct.
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