Why the D' (bolded)? It doesn't follow on further down your calculation.could some1 please help me?
i would like to know if:
ABCD + ABCD + ABCD + ABCD
can be simplified to
AB
what i did was:
ABCD + ABCD (B+B') = 1
= AC' (B+B' + D+D') (D+D') =1
so this = AC'D'
Where does the B come from in the first expression: AC' (bolded)? Other than this it seems pretty good.then
ABCD + ABCD
= ABC (D+D') (D+D') =1
so this = ABC
then
AC' + ABC
= AB (C'+C) (C+C')= 1
so this = AB
is what i have done here correct?
=
I have just tried this and got this to simplify to A(B + C')So you have AC' + ABC
Can I ask which theorum you have used to simplify this to AB?
You can't make that assumption because B is not a common factor in the first expresion.So you have AC' + ABC
(C' + C) = 1
so u would then be left with A + AB ?
A + AB would simplify to A, not AB. Ref: http://www.allaboutcircuits.com/vol_4/chpt_7/5.html at the topwhich could be simplified to AB ?
im kinda lost now
The expression there can be simplified further for use in a NAND implementation.ok
so the answer... is it in its simplest form or is there anything else that could be done
(A(B'C(B'C'D)')')''
There is possibly a simplification for (B'C)'(B'CD')', but I just can't see it now.(A(B'C)'(B'CD')')''
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by Jake Hertz
by Luke James
by Luke James
by Jake Hertz