I'm stuck on boolean algebra, it has 4 variables

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Discussion Overview

The discussion revolves around solving a boolean algebra problem involving four variables to design an integrated circuit (IC) that outputs a 1 when a number is greater than 9. Participants explore truth tables, Karnaugh maps (k-maps), and simplification techniques.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a truth table and seeks help with the next steps in the boolean algebra process.
  • Another suggests using the most significant bit (MSB) in conjunction with preceding bits to derive a solution.
  • A participant claims to have simplified the equation using k-maps, arriving at z = AB + AB'C, but questions the redundancy of variables.
  • Some participants argue that the inclusion of certain variables adds unnecessary complexity to the equation.
  • There is a discussion about the proper use of k-maps, with one participant asserting that not overlapping groups leads to redundancies in the solution.
  • A later reply provides a visual representation of the k-map and explains how certain combinations lead to parts of the solution, specifically highlighting the roles of AB and AC.
  • Another participant expresses confusion over their professor's advice against overlapping in k-maps, contrasting it with the suggestions from others in the thread.

Areas of Agreement / Disagreement

Participants express differing views on the use of overlaps in k-maps and the necessity of certain variables in the boolean expression. There is no clear consensus on the best approach to simplifying the equation.

Contextual Notes

Some participants mention specific conditions regarding the arrangement of terms in k-maps and the implications of overlaps, indicating that the discussion may depend on individual interpretations of k-map methodology.

mr_coffee
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Hello everyone, I'm designing a IC that will result in a 1 output if the number is > 9. So I wrote out the truth table and or'ed all the min terms and got the following:
http://img282.imageshack.us/img282/482/lastscan45vd.jpg I'm stuck now! any help would be great!
 
Last edited by a moderator:
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AND the MSB with each of the two preceeding bits and then OR that.

(AB)+(AC) will yield a 1 when the output is greater than 9.
 
Thanks for the reply but i finally figured out how k-maps work, and when i threw that big mess in the k-map it reduced down to:
z = AB + AB'C
note; B' = B complimented
here is a picture:
http://img43.imageshack.us/img43/3353/w0t1bp.jpg
 
Last edited by a moderator:
You don't need the B! though. You're adding complexity by having it.
 
I don't understand how the kmap didn't find the simplest forum of the equation i thought it always does, I didn't overlap any 1's in my k-map so no variables should be redudnat should they?
 
mr_coffee said:
I don't understand how the kmap didn't find the simplest forum of the equation i thought it always does, I didn't overlap any 1's in my k-map so no variables should be redudnat should they?

You have it backward. It's because you didn't use all possible overlaps that you have redundancies. Go back this time and group your "1's" so that each group is as large as possible, taking every overlap you can get, but the smallest number of groupings possible. (In this case there are obviously only two groupings needed, so there will be two 'AND' terms.) When you do that you will see the answer.

KM

By the way: your drawing was hard to see.
 
mr_coffee said:
I don't understand how the kmap didn't find the simplest forum of the equation i thought it always does, I didn't overlap any 1's in my k-map so no variables should be redudnat should they?

You do get the right answer if you arrange things logically:

Code:
______________________________ 
|  \cd                        |
|ab \ ________________________|
|     | cd  | cd! | c!d | c!d!|
|______________________________
|  ab |  1  |  1  |  1  |  1  |
|______________________________
| ab! |  1  |  1  |  0  |  0  |
|______________________________
| a!b |  0  |  0  |  0  |  0  | 
|______________________________
|a!b! |  0  |  0  |  0  |  0  |
_______________________________

You see AB is one part of the solution because it works for all combinations of CD. If you look in the columns you will see AC is the other solution because the solution AC does not depend on B at all.
 
Thanks so much guys, my professor so told us that we should not overlap, infact we should avoid overlapping, what the heck is he talking about! I ended up with AB+AC like u guys said!
 

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