Boolean simplification help please.

[cvriv]

Oh my goodness. How do you get AB'+BC' from AB'+AC'+BC'. Please. I was given an assignment and from a logic circuit I was about to write an initial expression and from there I simplified it down to AB'+BC'. I created a truth table with the simplified expression. Using the truth table I made a Karnaugh Map and from the K-Map I got AB'+AC'+BC'. Ugh!

Using Multsim I checked everything and it's all right! The truth tables are the same for both expressions. I can't figure out for the life of me how the K-Map expression is reduced to what I got initially.

Please help.

[Paul Rose]

The AC' term is redundant.

The AC' is only true when A,B,C = 1,0,0 or 1,1,0

In the larger expression, the 1,0,0 case is already covered by the AB' term,
and the 1,1,0 case is already covered by the BC' term

[Paul Rose]

Also, here is what I come up with for a Karnaugh map ( courtesy of Notepad  )

      BC
   00 01 10 11
  |--|--|--|--|
 0| 0| 0| 1| 0|
A |--|--|--|--|
 1| 1| 1| 1| 0|
  |--|--|--|--|
 

This looks a lot like AB' + BC' to me

Is it possible that your original truth table was wrong?

[rstofer]

Very carefully lay out the K map with AB on the left edge and C across the top

Make sure the map looks like:

AB    C
     0 1
---------
00   0 0
01   1 0 
11   1 0
10   1 1



When you review the map, you will see the BC' and AB' cover the 4 squares of the original function and the mapping of AC' is redundant.

Note:
Paul's ASCII art is better than mine...

[cvriv]

I attached my work. I don't understand how it's redundant. Maybe I need to take a break and then try again. Like I said, I checked it with Multisim as well. The truth table is right. Maybe I screwed up the K-Map?

[cvriv]

I'm lost. I got the 3 term expression from the k-map. My k-map looks like the two maps you two posted. Not exactly sure how you two set up your k-maps either.

[Paul Rose]

The map you drew looks fine.

Now, draw a 1x2 "circle" around the two 1's in the bottom row, this is your AB' term.
And draw a 2x1 "circle" around the middle two 1's in the left ( C' ) column, this is your BC' term

You could draw an additional 2x1 "circle" around the bottom two 1's in the left column, but you don't need to, because those 1's have already been included in your other circles.  This additional circle would be the redundant AC' term.

[rstofer]

Right, you only need to cover the 1s, not link them all together.  The two circles Paul noted on your K map is all it takes.

[cvriv]

Alright. Gotcha:) I think if I remember correctly... my professor had us grabbing all possible groups. I don't know. Thanks!!!!

[Paul Rose]

my professor had us grabbing all possible groups.

Just a guess, your prof may have mentioned using the K-map for hazard analysis.  Linking unconnected circles and including the redundant term can prevent some kinds of basic hazard ( more than one input changing at a time, not in perfect sync, the output can go through an intermediate state ).   If you were paying attention at just the wrong time, you may have gotten the idea that linking the circles was required.

[cvriv]

my professor had us grabbing all possible groups.

Just a guess, your prof may have mentioned using the K-map for hazard analysis.  Linking unconnected circles and including the redundant term can prevent some kinds of basic hazard ( more than one input changing at a time, not in perfect sync, the output can go through an intermediate state ).   If you were paying attention at just the wrong time, you may have gotten the idea that linking the circles was required.

As i don't always pay attention to him, you're probably right. I learn mostly from the book and as I haven't had much time to really read through the book, I have to go back and read the book again. This book we're using is so damn wordy too. I find myself skimming though this book more than any other book I own.

[pitagoras]

The question is where do you get AC' from in the kmap?

[Ratch]

Oh my goodness. How do you get AB'+BC' from AB'+AC'+BC'. Please. I was given an assignment and from a logic circuit I was about to write an initial expression and from there I simplified it down to AB'+BC'. I created a truth table with the simplified expression. Using the truth table I made a Karnaugh Map and from the K-Map I got AB'+AC'+BC'. Ugh!

Using Multsim I checked everything and it's all right! The truth tables are the same for both expressions. I can't figure out for the life of me how the K-Map expression is reduced to what I got initially.

Please help.

Trivial! Who needs a K-map when the function can be reduced by simply applying Boolean algebra?

AC'=AB'C'+ABC'

So therefore, AB'+AC'+BC' = AB'+AB'C'+BC'+ABC' = AB'(1+C)+BC'(1+A) = AB'+BC'

Ratch