# Boolean Algebra - Simplification

Discussion in 'Homework Help' started by D0wnl04d3r, Dec 30, 2009.

1. ### D0wnl04d3r Thread Starter New Member

Dec 30, 2009
3
0
Hi guys...

I'm having real trouble simplifying the following equation, so that a minimal number of gates are used...
I'm very new to this, I have no previous knowledge of electronics and have just started a degree...I just cannot understand this concept at all!

here is the question...

Simplify the following function, such that a minimal number of gates are used:
(D.B'+A'C).(C+A')

Do I need to create a truth table or Karnaugh graph from this? is it possible to simplify without therse? I've tried browsing through the site but find it all very confusing!

thanks

2. ### Ratch New Member

Mar 20, 2007
1,068
3
D0wnl04d3r,

A degree in what? How much was taught to you already? Is this part of a course you are taking? It find it worrisome that you are trying to simplify a Boolean expression, and you don't understand the concept.

Did you see the post from just three days ago? http://forum.allaboutcircuits.com/showthread.php?t=32040 . What do you find confusing? We can answer specific questions, but we cannot start you from ground zero. You have to do some preliminary studying on your own.

Ratch

3. ### D0wnl04d3r Thread Starter New Member

Dec 30, 2009
3
0
it's a degree in IT but more than half of the course is electronics based. My lecturers so far have touched on Truth Tables & Karnaugh Graphs but there is no maths module as such within my course so the boolean algebra side I am finding very difficult...

for staters I dont understand how to expand the current expression, so that the data can be input into the Karnaugh map

Last edited: Dec 30, 2009
4. ### Ratch New Member

Mar 20, 2007
1,068
3
D0wnl04d3r,

First of all, get rid of the dots between the logic variables and the terms. They are redundant and ridiculous.

(D.B'+A'C).(C+A') = (DB' + A'C)(C +A')

(DB' + A'C)(C +A') = B'CD + A'B'D + A'C + A'C

= B'CD + A'B'D + A'C ; Remove redundant term

= (A' + A)B'CD + A'B'(C' + C)D + A'(B' + B)C

= A'B'CD + A'B'CD + A'B'C'D + A'B'CD + A'B'C + A'BC

= A'B'CD + A'B'CD + A'B'C'D + A'B'CD + A'B'C(D' + D) + A'BC(D' + D)

= A'B'CD + A'B'CD + A'B'C'D + A'B'CD + A'B'CD' + A'B'CD) + A'BCD' + A'BCD

= minterms = Ʃ(3,11,1,11,2,3,6,7) = Ʃ(1,2,3,6,7,11) ; Remove redundant minterms

From there you can use a Karnaugh map to simplify it.

You can obtain the minterms a lot faster if you use the program from the following link.

Ratch

5. ### D0wnl04d3r Thread Starter New Member

Dec 30, 2009
3
0
okay thankyou, i've done that now, need to have a more in depth look at the karnaugh map now, to see how I can group the 1's. thanks ever so much for your help!

Well i've finally come up with an expression of :

A'(B'+C)+CD

is anyone able to verify whether or not this is correct?

Last edited: Dec 31, 2009
6. ### Ratch New Member

Mar 20, 2007
1,068
3
D0wnl04d3r,

No one should have to do so. You should be able to do it yourself. Did you download that program from the link, read the instructions, and plug in your K-map simplification? Did the program spit out the same minterms as you started with? If not, you did not intrepret the K-map correctly.

Ratch