Combinational Logic + Boolean equation help

On April 21 2008 11:38 fight_or_flight wrote:
1) Yes

2) See above post, just factoring.

3) So this is an AND gate with inversions on all inputs and outputs?

Basically, an AND gate with inverted inputs is the same as a NOR by demorgan's theorem A'B' = (A+B)'

Then you have a NOR with an inverted output which is the same as OR.

On April 21 2008 11:38 Raithed wrote:
I really don't see how BA + BC + C = BA + C(B + 1) as factoring, I mean..

x^2 + 4x + 4 would be (x+2)(x+2) ...

Can you explain further, I really don't see it.

On April 21 2008 11:40 Muirhead wrote:
In general C(A+B)=CA+CB for any three numbers A,B,C. This is called the distributive law of multiplication.

Plug in A=1, and you get C(1+B)=C(1)+CB=C+CB

On April 21 2008 11:46 Raithed wrote:

Oh, wow, nice explanation. But what if you don't know the answer, would you always plug A = 1?

On April 21 2008 11:43 Raithed wrote:

For number 3, so it's still (AB)'? Because there's an inverter in the input which makes it:

A' B' = AB? It doesn't work that way?

On April 21 2008 11:53 fight_or_flight wrote:

Does not work that way. Look at the truth tables:

out = A'B'

A B out
0 0 1
0 1 0
1 0 0
1 1 0

----------------------------

out = AB

A B out
0 0 0
0 1 0
1 0 0
1 1 1

------------------------------

Therefore you conclude that A'B' = NOR. (look up demorgan's theorem)