Notation difference

E

Thread Starter

eastmus

Joined Dec 28, 2008

9

Dec 28, 2008

• #1

hoomun said:

hi, im supose to complete the design of " A'B'CD'+A'B'CD+A'BC'D'+A'BC'D+AB'C'D+AB'CD'+AB'CD+ABCD " or simplifed ver of it " A'BC'+AB'D+B'C+ACD " with a multiplexer and then build the circut at uni any idea on how to do this? ( i have multisim software aswell )

cheers

Click to expand...

could someone explain me how "NOT" works with A'B' etc...

im only familiar with the lines on top of the letter like



and how would u write the above boolean algebra in AB'CD' with those ' (quotes)


thx!

 

beenthere

Joined Apr 20, 2004

15,819

Dec 28, 2008

• #2

Wlecome to AAC. We would like you to always open a new topic for each question.

The thread you posted in was this one - http://forum.allaboutcircuits.com/showthread.php?t=492.

 

E

Thread Starter

eastmus

Joined Dec 28, 2008

9

Dec 28, 2008

• #3

Thanks and I appologize for asking in the older topic.

I'm trying to learn from someone else's post about mux but I dont understand his boolean algebra.

 

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silvrstring

Joined Mar 27, 2008

159

Dec 28, 2008

• #4

The example you posted above might be typed here as X = ((A+B')+C'D)'

The apostrophes are bars.

 

E

Thread Starter

eastmus

Joined Dec 28, 2008

9

Dec 29, 2008

• #5

mmh then there is still a small confusion to me.

If the expression reads C'D'

will this mean its one long bar over C and D or a small bar on C and a small one on D.
The difference is that Demorgan can be applied.
Where a long bar over C and D will simplify it to C'+D'

Thanks

 

S

silvrstring

Joined Mar 27, 2008

159

Jan 1, 2009

• #6

Okay. Using DeMorgan's theorem, we can type it as

C'D' = (C + D)'

and your last sentence would read (CD)' = C'+D'

Happy New Year, and take care!

 

E

Thread Starter

eastmus

Joined Dec 28, 2008

9

Jan 1, 2009

• #7

ty happy new year

i understand it now