karnaugh map problem

Thread Starter

poindexter

Joined Aug 26, 2006
10
Hi folks,I'm stuck on one particular problem and have wasted nearly a whole day on it.I'm sure with a little help I can get a breakthrough.Her it is.Simplify this expression using a karnaugh map:AC[B+A(B+C')].Hope someone can help me Thanx.Poindexter
 

Papabravo

Joined Feb 24, 2006
13,543
You just work from the inside out using:
AA = A and
CC' = 0
Rich (BB code):
AC[B + A(B + C')]    =
AC[B + AB + AC']     =
ACB + ACAB + ACAC'   =
ACB + ACB  + 0       =
ACB
Not much of a Karnaugh map problem after all -- is it?
 

Thread Starter

poindexter

Joined Aug 26, 2006
10
I'm supposed to used a k map to simplify it so Im not sure if we've done it right.But i did learn something from you .thanx for the time to help
 

Papabravo

Joined Feb 24, 2006
13,543
As you correctly pointed out, the beginning of a Karnaugh map problem is a sum of products expression. If getting to that point leaves you with only one term then the map doesn't do much to help. It is also not the only technique that can be used. It works best with three or four variables. For a large number of variables other techniques need to be used because the maps loose their intuitive value.
 

Thread Starter

poindexter

Joined Aug 26, 2006
10
I'm supposed to use the karnaugh map to simplify.But it seems that getting to the sum of products already simplifies it and eliminates the need to use the karnaugh map.I dunno what the solution is so will just submit it and let the teacher decide what it is they want.What other problem are you talking about?
 

Papabravo

Joined Feb 24, 2006
13,543
I'm supposed to use the karnaugh map to simplify.But it seems that getting to the sum of products already simplifies it and eliminates the need to use the karnaugh map.I dunno what the solution is so will just submit it and let the teacher decide what it is they want.What other problem are you talking about?
I could have sworn there was another problem involving x's and y's in your original post, but I don't see it now.
 

terrytapp

Joined Mar 30, 2006
15
You just work from the inside out using:
AA = A and
CC' = 0
Rich (BB code):
AC[B + A(B + C')]    =
AC[B + AB + AC']     =
ACB + ACAB + ACAC'   =
ACB + ACB  + 0       =
ACB
Not much of a Karnaugh map problem after all -- is it?
My question after reading this post is this.
I can make a Karnaugh map from a truth table and reduce everything.
(easy enough) BUT how do you do it backwards.
Say you have the answer of the sum of your product A C [ (B+A) (B+C) ]
AS Poindexter originally posted, how do you build up from that? I am not clear on this.

_______A ___B __C


A

B
_
A B
___
A B
_
A B

I mean how do you graph it out?
 

jpates

Joined May 15, 2007
1
Hi folks,I'm stuck on one particular problem and have wasted nearly a whole day on it.I'm sure with a little help I can get a breakthrough.Her it is.Simplify this expression using a karnaugh map:AC[B+A(B+C')].Hope someone can help me Thanx.Poindexter
I know it is a bit late...and I am really new to this material, but maybe I can help. The other post about solving this one algebraically is definitely a whole lot easier, but since you had to solve using a K-Map, here ya go.

To solve this using a Karnaugh Map, step 1 would be to create the Truth Table.
(using a bit of logic, you can see if A or C is 0, then the whole answer is 0, so you only need to solve for 101 and 111)

*Attaching a Word document since it is a pain to format on this forum.

From there, you can see that the answer is ABC from the map...and the previously posted algebraic solution.
 

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Papabravo

Joined Feb 24, 2006
13,543
No to both. A Karnaugh map with a single term will have a single 1. It is both trivial and useless. The whole purpose of a Karnaugh map is to spot overlapping terms which result in simplified solution. No matter how you try to dress this pig there is no point in using a Karnaugh map for a single term boolean equation.
 
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