# Digital Electronics

#### coo0004

Joined Dec 6, 2006
4
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

#### coo0004

Joined Dec 6, 2006
4
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

#### Dave

Joined Nov 17, 2003
6,969
What are you trying to do, simplify the expressions?

Dave

#### coo0004

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

#### m4yh3m

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

Joined Nov 17, 2003
6,969
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
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

#### dina

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

#### Dave

Joined Nov 17, 2003
6,969
Hi
here's a simplification for the first problem:
C' + A'B + AB'
Do you care to embelish on how you simplied the first expression to C' + A'B + AB' - where did the C go?

Dave

#### dina

Joined Nov 12, 2006
19
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

#### beck

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

#### dragan733

Joined Dec 12, 2004
152
Do you care to embelish on how you simplied the first expression to C' + A'B + AB' - where did the C go?

Dave
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

#### Sonofjustice

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

#### Ratch

Joined Mar 20, 2007
1,070
Dave,

Do you care to embelish on how you simplied the first expression to C' + A'B + AB' - where did the C go?
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

#### Ratch

Joined Mar 20, 2007
1,070
beck,

yup,
K-Maps are the best.
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

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