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Is This Logic Diagram Correct?

  1. Apr 14, 2013 #1
    The boolean expression I am working with is:

    ((ab)'(b'c)' + a'bpc') where ' is the NOT symbol.

    I have the following logic circuit diagram. Is it correct?








     

    Attached Files:

  2. jcsd
  3. Apr 14, 2013 #2

    cepheid

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    Welcome to PF!

    The diagram is confusing a bit, because one of the a lines is very close to one of the b lines. However it looks correct to me.
     
  4. Apr 14, 2013 #3
    Ok, I know exactly which line you are talking about; I thought the same thing. I'll change that around. Thanks!
     
  5. Apr 14, 2013 #4

    cepheid

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    I think you can simplify this alot using DeMorgan's and other things. For instance, take the first input to the OR gate:

    (ab)' * (b'c)'

    Using DeMorgan's theorem (xy)' = x' + y', you can expand each thing in parentheses:

    = (a' + b') * (b + c') [eq. 1]

    and now using the distributive property of AND operator (which I believe is present for Boolean algebra?)

    = a'b + a'c' + b'b + b'c'

    and using the associative property:

    a'(b + c') + 0 + b'c'

    Anything OR'd with 0 is just itself, so this becomes:

    a'(b + c') + b'c' [eq. 2]

    I'm not sure that this is any simpler in this case, but the point is that you could, in principle, try to minimize the number of gates. I wrote a little program to generate the truth tables for eq. 1 and eq. 2 and they came out the same.
     
    Last edited: Apr 14, 2013
  6. Apr 14, 2013 #5
    Alright, variable p is missing from that equation, but I get the idea. On that note, is this the correct truth table outputs:

    1
    1
    0
    0
    1
    1
    1
    1
    1
    1
    0
    0
    0
    0
    0
    0
     
  7. Apr 14, 2013 #6
    Ahh, the a' a' are adjacent so it is absorbed, got ya
     
  8. Apr 14, 2013 #7
    So the equation should be a'(b + c') + b'pc'
     
  9. Apr 14, 2013 #8
    which gives me three and gates and two or gates
     
  10. Apr 14, 2013 #9

    cepheid

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    Well, I was ONLY considering the (ab)'(b'c)' in the original equation, as an example. I wasn't trying to simplify the whole thing.

    Here's what I get:

    Code (Text):


    a   |   b   |   c   |   p   |   f
    0   |   0   |   0   |   0   |   1
    0   |   0   |   0   |   1   |   1
    0   |   0   |   1   |   0   |   0
    0   |   0   |   1   |   1   |   0
    0   |   1   |   0   |   0   |   1
    0   |   1   |   0   |   1   |   1
    0   |   1   |   1   |   0   |   1
    0   |   1   |   1   |   1   |   1
    1   |   0   |   0   |   0   |   1
    1   |   0   |   0   |   1   |   1
    1   |   0   |   1   |   0   |   0
    1   |   0   |   1   |   1   |   0
    1   |   1   |   0   |   0   |   0
    1   |   1   |   0   |   1   |   0
    1   |   1   |   1   |   0   |   0
    1   |   1   |   1   |   1   |   0

     
     
  11. Apr 14, 2013 #10
    Sorry, I wasn't complaining. :) Alright, that's the same as what I have.
     
  12. Apr 14, 2013 #11
    The equation breaks down to a'b + b'c' + a'bp, I believe.
     
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