1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Logic gates - Electronics

  1. Feb 14, 2012 #1

    Femme_physics

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Given the function:


    http://img26.imageshack.us/img26/9718/elel0.jpg [Broken]

    A) Write the truth table of the function F (A, B, C, D)

    B) Present the function F (A, B, C, D) via Karnaugh Map

    C) Express the function F as the sum of multiplications with minimum literals

    D) Realize the minimized function F via logic gates


    3. The attempt at a solution

    I just wanna see if I got it right :)
    http://img84.imageshack.us/img84/1940/elel1.jpg [Broken]

    http://img577.imageshack.us/img577/1851/elel2.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 14, 2012 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    See your gate arrangement--you have used two NOT gates to twice produce B'. This is unnecessary duplication.

    I can't say much about your Karnaugh map, I need to revise that topic myself. :( But I can see that your equation F= AB + B'C + B'C'D' does not match your truth table. Isn't it supposed to??
     
    Last edited: Feb 14, 2012
  4. Feb 14, 2012 #3

    NascentOxygen

    User Avatar

    Staff: Mentor

    let's take as correct, your equation F= AB + B'C + B'C'D'

    take out as a factor B' --> F = AB + B'(C + C'D')

    consider that last term, B'(C + C'D')

    when C is true, the bracketed term evaluates as true
    when C is false, the bracketed term evaluates as D

    so I think there should be some algebra reduction that allows you to make this

    F = AB + B'(C + D')
     
  5. Feb 14, 2012 #4

    NascentOxygen

    User Avatar

    Staff: Mentor

    Here's how to go about demonstrating this. After you've been shown once, you'll probably be able to figure it out for yourself thereafter.

    let's focus on the term in brackets,
    C + ¬C¬D

    EDITED:
    Consider two basic logic relations:
    you can OR anything with TRUE and it's still TRUE
    you can AND anything with TRUE and it doesn't change its value
    = C ( 1 + ¬D) + ¬C¬D

    remove the brackets
    = C + C¬D + ¬C¬D

    take out a common factor ¬D
    = C + ¬D (C + ¬C)

    what's in brackets evaluates as always TRUE, so simplifies to
    = C + ¬D
     
    Last edited: Feb 14, 2012
  6. Feb 14, 2012 #5

    Femme_physics

    User Avatar
    Gold Member

    I never heard this "Or everything" rule in our Boolean algebra. Could our teacher only want us to use the basic list he gave us?

    Point taken.


    I don't know how to relate truth tables to functions, only functions to Karnaugh Maps and Karnaugh Maps to truth tables. Is it even possible?
     
    Last edited: Feb 14, 2012
  7. Feb 14, 2012 #6

    I like Serena

    User Avatar
    Homework Helper

    I believe that expression does match the truth table.



    Your truth table is fine, your Karnaugh table is fine, your function is fine and your circuit is fine. :smile:


    Btw, you can construct a truth table from a function.
    Just start with every combination of A, B, C, and D, which you already have in your current truth table.
    If you want, introduce a couple of intermediary results.
    Then calculate the result of the function for each combination of A, B, C, and D.



    Which basic list did he give you?
    It should include that (1 + a) is always true, that is, it is equal to 1.



    Btw, can't you put in a drawing of something? Anything? A beetle would do. :wink:
     
    Last edited: Feb 14, 2012
  8. Feb 14, 2012 #7

    NascentOxygen

    User Avatar

    Staff: Mentor

    Oops, oops, oops! :redface: :redface: I left out half the explanation in that step.

    I meant to also add:
    You can AND anything with TRUE and you don't change its value.

    Sorry for the oversight. :cry:
     
    Last edited: Feb 14, 2012
  9. Feb 14, 2012 #8

    I like Serena

    User Avatar
    Homework Helper

    Uhh :uhh:... where's the typo?
     
  10. Feb 14, 2012 #9

    NascentOxygen

    User Avatar

    Staff: Mentor

    Fixed now.
     
  11. Feb 15, 2012 #10

    NascentOxygen

    User Avatar

    Staff: Mentor

    I overlooked the distinction between Ø and O so was just a little puzzled. Now I recognize you used Ø for your "don't care" states. So all is correct.

    F = AB + B'(C + D')
    ⇔ F = AB + B'C + B'D'
     
  12. Feb 16, 2012 #11

    Femme_physics

    User Avatar
    Gold Member

    :) Thank you!
     
  13. Feb 16, 2012 #12
    The simplification in the equation could have also been obtained by grouping the four corners on the Karnaugh map instead of just boxes 0 and 8.

    I didn't see anyone else mention it, so I thought I would throw that out there :smile:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Logic gates - Electronics
  1. Logic Gates (Replies: 1)

  2. Logic Gates (Replies: 8)

  3. Logic Gates (Replies: 26)

  4. Logic Gates Algebra (Replies: 15)

Loading...