For the function : f(A,B,C,D) = A\'BCD+ A\'B + ACD\'+ BC What would be the canon
ID: 643965 • Letter: F
Question
For the function:
f(A,B,C,D) = A'BCD+ A'B + ACD'+ BC
What would be the canonical sum of products expansion?
Question 5 options:
a)
f(A,B,C,D)= A'BC'D'+A'BC'D+A'BCD'+ A'BCD + AB'CD'+ABCD'+ ABCD
b)
f(A,B,C,D)= A'BC'D'+A'BC'D+A'BCD'+ A'BCD + ABC'D'
c)
f(A,B,C,D)= A'BC'D'+C'D+A'BCD'+ A'BCD + ABC'D'+ BC'D+BCD'+ CD
d)
The function is already expressed in canonical form
a)
f(A,B,C,D)= A'BC'D'+A'BC'D+A'BCD'+ A'BCD + AB'CD'+ABCD'+ ABCD
b)
f(A,B,C,D)= A'BC'D'+A'BC'D+A'BCD'+ A'BCD + ABC'D'
c)
f(A,B,C,D)= A'BC'D'+C'D+A'BCD'+ A'BCD + ABC'D'+ BC'D+BCD'+ CD
Explanation / Answer
f(A,B,C,D) = A'BCD+ A'B + ACD'+ BC
canonical form should contain all the variables in each term
Also we know that x + x' = 1
f(A,B,C,D) = A'BCD+ A'B + ACD'+ BC
f(A,B,C,D) = A'BCD + (A'B .1 .1) + (A . 1 . CD') + (1.BC . 1)
= A'BCD + A'B .(C+C') .(D+D') + A . (B+ B') . CD' + (A+A').BC . (D+D')
= A'BCD + A'BCD +A'BC'D +A'BCD'+A'BC'D' + ABCD'+AB'CD' + ABCD+ABCD'+A'BCD+A'BCD'
= A'BCD + A'BC'D + A'BCD' + A'BC'D' + ABCD' + AB'CD' + ABCD
hence the answer is A