# Digital Electronics

Discussion in 'Homework Help' started by coo0004, Dec 6, 2006.

1. ### coo0004 Thread Starter New Member

Dec 6, 2006
4
0
hi guys
i am new to this forum and i am currently undertakin an electronic engineering degree and i thought it would be a good idea to join, anyway at the moment i am having a big problem with boolean algebra and all i can find on the internet is simple stuff. anyway if anyone can help me figure all this out it would be much appreciated.

for example:
__ _ _
(ABC + ABC + ABC + ABC)(A+B)

and

___ _ _ _ __ _ _
ABC + ABC + ABC + ABC + ABC + ABC

thanks heaps guys

2. ### coo0004 Thread Starter New Member

Dec 6, 2006
4
0
Sorry guys disregard those last problems

these are the correct ones, (' represents not)

A'B'C' + A'BC' + A'BC + AB'C' + AB'C + ABC'

and

(AB'C' + AB'C + ABC + ABC')(A+B)

thanks again

3. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
What are you trying to do, simplify the expressions?

Dave

4. ### coo0004 Thread Starter New Member

Dec 6, 2006
4
0
yeh i was trying to minimise the equation. any help is appreciated

5. ### m4yh3m Senior Member

Apr 28, 2004
186
42
I always threw those things into Electronics Workbench and let it simplify it

6. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
The first one A'B'C' + A'BC' + A'BC + AB'C' + AB'C + ABC':

Take out common factors B'C' ; BC' ; C

B'C'(A'+A) + BC'(A'+A) + C(A'B+AB')

(A'+A) = 1

Therefore:

B'C' + BC' + C(A'B+AB')

Take out common factor C'

C'(B'+B) + C(A'B+AB')

(B'+B) = 1

C' + C(A'B+AB')

The second one will use the same techniques as the first but look at taking out common factors AB and AB' from the first bracket, don't multiple out the two brackets.

If you want further advice, or wish to run your answer past us, post it up and I'll be glad to check it. I am also moving this to the Homework Help forum.

Dave

7. ### dina Member

Nov 12, 2006
19
0
Hi
here's a simplification for the first problem:
C' + A'B + AB'
for the second one:
it's simply A

8. ### Dave Retired Moderator

Nov 17, 2003
6,960
145
Do you care to embelish on how you simplied the first expression to C' + A'B + AB' - where did the C go?

Dave

9. ### dina Member

Nov 12, 2006
19
0
hi Dave...
i used the karnaugh map to simplify both the problems
if u need a help with karnaugh maps i'll be glad to help

10. ### beck Member

Nov 27, 2006
26
0
yup,
K-Maps are the best.

11. ### dragan733 Senior Member

Dec 12, 2004
152
0
A'+AB=A'+B
A'(B+B')+AB
A'B+A'B'+AB+A'B
A'B=A'B+A'B
A'(B+B')+B(A+A')=A'+B

12. ### Sonofjustice New Member

Jul 3, 2008
3
0
Yeah There Is Also This Formula Saying X'+XY= X'+Y

13. ### Ratch New Member

Mar 20, 2007
1,068
3
Dave,

Sure, use the identity X +X'Y = X+Y where X=C', X'=C, and Y=A'B+AB'

Proof of The Identity:

The dual of X +X'Y is X(X'+Y) = XY

The dual of XY is X+Y, so the identity is proved. Other ways exist to prove the identity. Ratch

14. ### Ratch New Member

Mar 20, 2007
1,068
3
beck,

Nope, not when the number of variables is greater than 5 or 6. It blows your mind to try to interpret that many variables variables on a K-map. Try:

P(A,B,C,D,E,F) = SOP(6,9,13,18,19,25,27,29,41,45,57,61)

Tabulation methods like the Quine-McCluskey method guarantee the best simplification and are applicable to computer operations.

The answer to the first problem I got from the Quine-McCluskey method was A'B+AB'+C' .

For the second problem, by inspection it is easily seen that
(AB'C' + AB'C + ABC + ABC')(A+B)=(AB'C' + AB'C + ABC + ABC') .

Then = (AB'C' + AB'C + ABC + ABC')= AB'(C'+C)+AB(C+C') = AB'+AB = A(B+B') = A

Ratch

Last edited: Jul 20, 2008