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Building a logic circuit from a truth table

  1. Feb 10, 2015 #1
    P.S I know I am not supposed to use attachment but there are no way for me to input the table in a clean way... same for the logic circuit drawing. Would anyone mind helping me?

    1. The problem statement, all variables and given/known data


    upload_2015-2-10_19-19-20.png


    2. Relevant equations

    AND gates require both input to be True to get a true ouput
    NOT gates are just the opposite of what you input in them ( 1 -- > 0 and 0 --> 1)
    OR gates require at least one input to be true to create a true output

    3. The attempt at a solution


    I can't seem to understand why the answer key uses 4 AND gates with 3 output. Like the one below :

    upload_2015-2-10_19-21-48.png

    It just seems very... weird that there are only 4 a inputs where as in the truth table there are 8 inputs in the truth table. Anybody can help me understand that?
     
  2. jcsd
  3. Feb 10, 2015 #2

    berkeman

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    Staff: Mentor

    Have you drawn a Karnaugh Map for that truth table? Can you share it with us? :-)
     
  4. Feb 10, 2015 #3
    Oh, well I know how, I just didn't try because this was a set of problems before it was taught, but I'll do that right now. ^^
     
  5. Feb 10, 2015 #4

    Mark44

    Staff: Mentor

    There are four inputs in the truth table -- A, B, C, and F -- not eight.

    Edit: Make that three inputs -- A, B, and C --and one output -- F.
     
    Last edited: Feb 11, 2015
  6. Feb 10, 2015 #5
    So I'm guessing the first thing to do is reduce the truth table either using theorems or karnaugh map right? ( hence why you asked)... If so then I got the following

    FAB 00 01 11 10
    C
    0 0 1 0 1
    1 1 0 1 0
    and then I group together if I remember correctly and that allows me to cancel the input that "changes" if the input is the same ( I don't know if I'm clear? :/). But none of them are similar.

    Edit: Sorry I didn't express myself right... I meant 8 possibilities out of 3 inputs ( 2^3)
    Edit2: Can't seem to format this table right ughhhh
     
  7. Feb 11, 2015 #6

    berkeman

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    Staff: Mentor

    Try using the "code" tags to preserve spaces for formatting. Like this:

    Code (Text):

    A B C D
    0 1 0 1
     
     
  8. Feb 11, 2015 #7
    Code (Text):

     
      AB      00  01  11  10
    C
    0          0   1   0   1
    1          1   0   1   0

     
    Thanks!!
     
  9. Feb 11, 2015 #8

    berkeman

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    Staff: Mentor

    Wow, doesn't look like there are any simplifications. That's why there are 4 separate minterms that you OR together to get the output.
     
  10. Feb 11, 2015 #9

    NascentOxygen

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    Staff: Mentor

    @MarcL Has your question now been answered?

    As an aside, if you were allowed to use any gates of your choosing, you'd just need a pair of 2-input exclusive-ORs.

    F = A ⊕ (B⊕C)
     
  11. Feb 11, 2015 #10

    berkeman

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    Staff: Mentor

    Nice! Too bad he can't use XORs. I'll remember that trick! :-)
     
  12. Feb 15, 2015 #11

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    You can do nice K map tables in Latex:$$
    \begin{array}{|c|c|c|c|c|}
    \hline C|AB & 00 & 01 & 11 & 10\\
    \hline 0&0&1&0&1\\
    \hline 1 & 1 & 0 & 1 & 0\\
    \hline
    \end{array}$$
     
    Last edited: Feb 15, 2015
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