digital-logic
if I have a truth table for ABC and output Z
ABC Z
000 1
001 1
111 1
110 1
SOP=A'B'C'+A'B'C+ABC+ABC'
the most straight forward way will be 3 AND,1 NAND,1 OR
.But, I was told this can be done with 3 NANDS…
I tried simplified it, I got A'B'+AB…which can be done with 3 NANDS
but,is it make sense to ignore input c?
Can someone help?
Thanks
Going by the problem definition directly:
D and B turn ON: B•D
A turns OFF while C turns ON: ~A•C
A and B turn ON and the rest turn OFF: This condition is already covered by condition 4.
C and D turn OFF: ~C•~D
all switches turn ON: This condition is already covered by 1.
So your final function is: B•D + ~A•C + ~C•~D
3 ands, 2 ors.
You've got a good start. Just re-distribute the second term over the first:
(A + B)(AB)' = A(AB)' + B(AB)'
And then apply De Morgan to the whole thing:
A(AB)' + B(AB)' = ((A(AB)')'(B(AB)')')'
The Boolean expression gets to be a little hard to read, but it translates to the following circuit:
simulate this circuit – Schematic created using CircuitLab
Note that the exact same network works if all of the NAND gates are changed to NOR gates, except that you get an XNOR gate — the output is high if the inputs are equal.
Best Answer
It makes sense, because if you look at the function, there are pairs of terms in which C appears both negated and not; so it has no influence on the output; and, if you look at the truth table, the same applies.
$$ {\lnot C} + {C} = 1 $$ and $$ AB \cdot (C+ \lnot C)=AB \cdot 1 = AB $$
So you can just discard C.