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Boolean algebra problem

  • Thread starter mr_coffee
  • Start date
  • #1
1,629
1
Boolean algebra problem!!

Hello everyone 'im designing a IC that will tell output a 1 if the number is > 9. so I wrote this boolean expression but i'm stuck on reducing it.
B' = B compelemnted
AB'CD' + AB'CD + ABC'D' + ABC'D + ABCD' + ABCD
AB'(CD' + CD) + AB(C'D' + C'D + CD' + CD)
any hints? thanks!
 

Answers and Replies

  • #2
1,482
3
re

If my karnaugh map is not rusty, i got

AB + AC
 
  • #3
*ab'cd' + Ab'cd + Abc'd' + Abc'd + Abcd' + Abcd
*ab'c(d'+d) +abc'(d'+d) + Abc(d'+d)-----> (d+d') =>1
*ab'c + Abc' + Abc
*a(b'c+bc') +abc
*a[b Xor C + Bc]
 
  • #4
424
3
The answer that 'waht' got is the correct one (a mapping easily shows that). I don't think though that I should do the actual Boolean equation for you, but I'll give you a couple of hints. Group as many of the "AND" terms together with (X + X') terms, to eliminate as much as you can to start. Then remember that you can take something like: "XYZ +........." and make it "XYZ + XYZ + ......". In other words, writing a term twice in an equation doesn't change anything. It can make it easier to work out. If you try these, you should get the answer to your original equation. Then let us see it and we'll check it for you.

KM
 

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