Question about logic gates

WBahn

Joined Mar 31, 2012
29,976
no im not sure
Then you probably need to go back and read through the basic intro material to digital logic again -- there's probably other things you've missed.

The overbar is one way of indicating logical inversion (or NOT).

\(
\text{Y\;=\;\bar{A}\;=\;NOT(A)\;=\;A'}
\)

If A is True, then Y is False, while if A is False then Y is True.
 

WBahn

Joined Mar 31, 2012
29,976
Also, note that X and Y below are NOT equal.

\(
\text{X = \bar{A}\.\bar{B}}
\text{Y = \bar{A\.B}}
\)

Because of this, and because it is easy for adjacent overbars to accidentally be joined, either by the person reading it or by the software rendering it, it is best to use parens to make it very clear what the groupings are.

\(
\text{X = \(\bar{A}\)\(\bar{B}\)}
\text{Y = \bar{\(AB\)}}
\)
 

WBahn

Joined Mar 31, 2012
29,976
is this the same as my (AB) bar question?
that would be A?
y does your have 2 solid bars while mine is 1 solid and 1 split?
Yes, A double bar is just A.

The solid versus split is critical.

If you have AB with a solid overbar, it means that you AND the signals A and B together and THEN negate the result.

If you have AB with a split overbar, it means that you negate A and B separately and THEN you AND the results.
 
Last edited:

WBahn

Joined Mar 31, 2012
29,976
so then would my (AB) with a solid and split just be AB?
NO! The difference is CRITICAL!

It is the difference between saying that the bomb will explode anytime that [(light A is on) AND (light B is on)] is False and saying that the bomb will explode anytime that [(light A is NOT on) AND (light B is NOT on)] is True. In the latter case, it's pretty clear that bomb will only explode if both lights are off, so that as long as one of the lights is on, the bomb blows up. But in the former case, both lights have to be on in order for the bomb to detonate.

The overbar is an implicit pair of parentheses -- everything under the bar is evaluated and then THAT result is negated.
 

Thread Starter

TL314

Joined Mar 29, 2015
64
im going to try and attempt my hw problem, i know im just trying random things but I think this would work.
A AND gate gives me AB
A NAND gates gives me AB(bar)
A OR gates gives me (A+B)
so if I have those 3 hooked to a OR gate my output would be AB+AB(bar)+(A+B) right?
Snapchat--4481435906559487436.jpg
 

WBahn

Joined Mar 31, 2012
29,976
im going to try and attempt my hw problem, i know im just trying random things but I think this would work.
A AND gate gives me AB
A NAND gates gives me AB(bar)
A OR gates gives me (A+B)
so if I have those 3 hooked to a OR gate my output would be AB+AB(bar)+(A+B) right?
View attachment 84090
Yes, but what is it you are trying to accomplish?

In your original post you said you were trying to solve an equation that had three Boolean terms? What does "solving" mean for this homework?

I had assumed that it probably meant simplifying that expression into something that had fewer terms.
 

WBahn

Joined Mar 31, 2012
29,976
Draw a circuit that implements the X and the Y that were presented in Post #7.

Draw the truth table for both.
 

Thread Starter

TL314

Joined Mar 29, 2015
64
I have to create a circuit that satisfies that equation. So I need OR gates to sum and ill sum the 3 individual gates to give a single output C. So in my attempt I did a AND a NAND and a OR summed through a OR gate so I think it would give me that equation but i could be completely wrong
 

WBahn

Joined Mar 31, 2012
29,976
It will give you a solution, but you can achieve the same solution (i.e., the same function) with a single two-input logic gate.
 

WBahn

Joined Mar 31, 2012
29,976
1 gate will give me that solution? how?
There are several ways to see how. Let's work through a couple of them.

The gold standard for digital logic is the truth table.

So draw the truth table for that function. Since it only has two inputs, there are only four rows. Once you do that, see if you recognize the logic function that is implemented.
 
Top