Discussion in 'Homework Help' started by ann, Feb 22, 2008.
Can this boolean algebra be reduced further? f=/cd+/ab+/a/c+a/bcd
Double NOT the function (this doesn't change the function), so: f=//(/cd+/ab+/a/c+a/bcd).
Now think about DeMorgan's theorems.
Have a go and post up your efforts, I'll give you some feedback.
I did the boolean algebra again this time using k maps and realized I had made a mistake in reducing my equation. The reduced one now is /CD+/AB+/A/C+A/BD and I don't think this can be reduced further. Can it?
Sorry I missed this one. Feel free to PM to remind me in future.
Without your working out I cannot give you any feedback on your answer, but I simplified it to:
f = (ABCD)'
(Note I use ' to denote NOT/INVERTED)
I used the two techniques I told you about above. How?
f = (CD)' + (AB)' + (A'B') + (AB'CD)
Double NOT (this doesn't change the function):
f = ((CD)' + (AB)' + A'B' + AB'CD)''
Take out a common factor in the last two terms:
f = ((CD)' + (AB)' + B'( A' + ACD))''
Apply DeMorgan's theorem:
f = ((CD)''(AB)''(B'(A' + ACD))')'
(CD)'' = CD; (AB)'' = AB; so:
f = (CDAB(B'(A' + ACD))')'
Apply DeMorgan's theorem on (B'(A' + ACD))':
f = (CDAB(B'' + (A' + ACD)'))'
B'' = B
Apply DeMorgan's theorem on (A' + ACD)':
f = (CDAB(B + (A''(ACD)')))'
A'' = A
Apply DeMorgan's theorem on (ACD)':
f = (CDAB(B + (A(A' + C' + D'))))'
Multiply out (A(A' + C' + D'):
f = (CDAB(B + (AA' + AC' + AD')))'
AA' = 0
f = (CDAB(B + AC' + AD'))'
Multiply everything out:
f = (CDABB + CDABAC' + CDABAD')'
BB = B; CC' = 0; DD' = 0; therefore
f = (CDAB)' = (ABCD)'
Not easy eh?!