So, in other words, you have X = (AB + BCD + AC)' ?Hi I've got this question to draw a circuit diagram for X=AB+BCD+AC (the whole expression is complemented) so do i need to draw an invertor after every input?
The bit about the DeMorgan's theorem is the a bit I don't really get. This is a Homework question that I got, and correct me if I'm wrong the question doesn't ask to simplify the circuit so should I break up the expression as DeMorgan's theorm. Or does it make sense to draw the logic diagram as it is and just put an inverter after the final OR gate?So, in other words, you have X = (AB + BCD + AC)' ?
There are many logic circuit that implement this. As Alec_t indicated, it depends on what gates you have available. Not only what the underlying function happens to be (AND vs NAND, for instance), but also any limitations on how many inputs each gate has.
As to the question of whether you need to draw an inverter after every input, that depends on how you manipulate the circuit. For instance:
Y = (A+B)' = (A')(B')
So in that case the answer is yes, but it is not a complete answer -- you must also swap the underlying logic function from OR to AND. This is the heart of DeMorgan's theorem.
However, you can't just blindly apply this to multi-term inputs such as
Y = (AB + C)' = (A')(B')(C') -- this is WRONG.
It is term by term
Y = (AB + C)' = ((AB)')(C') = (A' + B')(C') = A'C' + B'C'
Depends on the context of the assignment. If this homework follows a bunch of coursework having to do with simplifying logic equations or applying DeMorgan's, then you will probably be in for a surprise if you just implement it as is. In general, homework should be done in a manner compatible with the course material is it related to.The bit about
The bit about the DeMorgan's theorem is the a bit I don't really get. This is a Homework question that I got, and correct me if I'm wrong the question doesn't ask to simplify the circuit so should I break up the expression as DeMorgan's theorm. Or does it make sense to draw the logic diagram as it is and just put an inverter after the final OR gate?