Boolean Logic

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ckaiser813

Joined Jan 21, 2009
17
Just checking if I'm right with these?

Given F=A'B+(C'+E)(D+F'), use de Morgan's theorem to find F'

I got F=A+B'(CE')+(D'F)????


Simplify the following: {[(AB)'C]'D}'

I got A'+(B'+C')D


Pretty tough for me
 

mkbutan

Joined Sep 30, 2008
299
Just checking if I'm right with these?

Given F=A'B+(C'+E)(D+F'), use de Morgan's theorem to find F'

I got F=A+B'(CE')+(D'F)????


Simplify the following: {[(AB)'C]'D}'

I got A'+(B'+C')D


Pretty tough for me


pl explane your steps for the five input F
 

Ratch

Joined Mar 20, 2007
1,070
ckaiser813,

You need to read this thread and download this program.

http://forum.allaboutcircuits.com/showthread.php?t=22347&highlight=BOOLEAN

When you do plug in the Boolean expression, A'B+(C'+E)(D+F'), into the program, you will obtain the following minterms:

0 2 4-7 10 14-32 34 36-39 42 46-48 50 52-55 58 62-63

The negative of that expression should have the missing minterms of the above expression, which are:

1 3 8-9 11-13 33 35 40-41 43-45 49 51 56-57 59-61

Since your answer, A+B'(CE')+(D'F) does not these minterms, that is a sure indication that your result is wrong. Try again and use the program to check your new result.

The program gives the minterms for ((AB)'C)'D as 1 5 9 13 15

Your answer, A'+(B'+C')D, has minterms 0-7 9 11 13 , so again you have the wrong answer. Get the idea?

Ratch
 
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