Simplifying an expression

Discussion in 'Homework Help' started by eagle136, Aug 25, 2009.

1. eagle136 Thread Starter New Member

Aug 25, 2009
3
0
I have an expression, I know the simplified answer but I would like to know the simplified answer has been arrived at. I'd like to use the example with a class I have to teach next year - thrown in right at the deep end with a topic I have never covered before ( I know, state of the education system and all taht ....)

The expression is:
Q=(A'BC)+(ABC')+(ABC)

Simplified this becomes Q = AB + BC (or even Q=B(C+A))

but which rules are used to arrive at this? How does one determine which rules are used where?

Sep 25, 2008
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3. eagle136 Thread Starter New Member

Aug 25, 2009
3
0
Hi,
Thanks for the pointers:

Original:
Q=(A'BC)+(ABC')+(ABC)

? I can see that the expression has been expanded but I do not know how to get to this as being the way in ...
Q=(A'BC)+(ABC')+(ABC)+(ABC)

Rewrite the terms to facilitate applying subsequent laws. Commutative law permits this.
Q=(ABC) + (A'BC) + (ABC') + (ABC)

Associative law
Q=(A+A')(BC) + (AB)(C'+C)

Identity
Q=(BC) + (AB)

How did I do?

I did a bit of digging myself and cam up with an alternative - the answer is correct (according to the truth table) - any comment on my application?

Original:
Q=(A'BC)+(ABC')+(ABC)

BC is common in the 1st and 3rd terms, so this can be factored out:
rewrite to make clearer:
(ABC)+(A'BC)+(ABC')
Q = BC(A+A') + (ABC')

Apply identity theorem to A+A':
Q = BC + (ABC')

Factor out the B using associative property:
Q = B(C+AC')

Use absorbtion property on (C+AC') which through association is the same as C+C'A thus C+A
Q=B(C+A)

Any comments? I know that Q=B(C+A) and Q=(BC)+(AB) are equivalent from building the truth table for each, but which is the more correct answer?

Thanks

4. Ratch New Member

Mar 20, 2007
1,068
3
eagle136,

The best way to simplify a Boolean expression is to expand it out to its minterms, and then use a Karnaugh map or a tabulation method such as the Quine-McCluskey operation. The above mentioned methods systematically reduce the Boolean expression if it can be reduced. You should download the program at http://forum.allaboutcircuits.com/showthread.php?t=22347&highlight=boolean, which will give the minterms of a Boolean expression. If the minterms of the original expression and its simplification are the same, then your simplification does not have any errors.

Ratch

5. eagle136 Thread Starter New Member

Aug 25, 2009
3
0
Hi Ratch,
Thanks for the reminder. I remember doing K-Maps at Uni ages ago, our class syllabus does not include them so I had thought I should steer clear, however, having spent an evening with them again (ah memories) they seem SO much simpler than tryign to work out which law/property to apply where.

One question however, the syllabus bangs on about de Morgan's laws and students will have to be able to simplify using this. Do K-Maps help here as well? Is De Morgan used to turn a SOP into POS (and vice versa) and then use the K-Map?

6. Ratch New Member

Mar 20, 2007
1,068
3
eagle136,

Certainly, here's how. Using your example, Q = A'BC+ABC'+ABC ,
the minterms are 3,6,7 . The missing minterms of 0,1,2,4,5 are the complement of your expression or Q' . You can easily see that if you make a truth table. Next, plot the 5 minterms of 0,1,2,4,5 on a K-map to get a reduction of Q' = B'+A'C' . Then use DeMorgan's theorem to easily get Q = B(A+C), which is what you started out with. This was a trivial example, but it illustrates the method. Search the archives for "Boolean" to find other problems and resolutions.

This link can give you some tips.

This link shows another POS to SOP conversion.