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!

Implementing boolean functions with decoder and external gate

  1. Oct 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Design an combinational circuit using a decoder and external gates defined by the boolean functions F1, F2, F3(see picture)

    2. Relevant equations

    3. The attempt at a solution


    I'm quite confused as to the exact method in doing this. I understand that a decoder takes n inputs and produces 2^n outputs. The combination of the n inputs correspond to binary numbers, whatever binary numbers the inputs make, the corresponding output line will be 1. For example if x=0, y=0, z=0, that repersents binary 0, so D0 will be 1.

    My confusion arises when there are minterms in F1,F2,F3 that only have 2 variables. Do I just set the third, non present variable to 0? Any ideas?
  2. jcsd
  3. Oct 17, 2016 #2
    I went ahead and implemented F1 to my understanding.


    I got F1 = Σm(0,5).
    The answers given to us is F1 = Σ(0,5,7)
    Any ideas

    Also, is it better to write Σm or is just Σ fine?
  4. Oct 17, 2016 #3
    I was always told that with decoders that each output equation contains all of the input variables but since the one of the minterms in the F1 output equation doesn't contain all variables, how do I deal with that?
  5. Oct 18, 2016 #4


    User Avatar
    Science Advisor

    The missing variable is a 'Don't Care.' That means the result of the minterm is the same regardless of the value of the missing variable.
  6. Oct 18, 2016 #5
    I ended up connecting D0, D5 and D7 to an OR gate which basically resulted in the y term being cancelled out by simplification.
  7. Oct 18, 2016 #6
    Also how did you figure out that it's a don't care term without simplification
  8. Oct 18, 2016 #7


    User Avatar
    Science Advisor

    F1 = ## \bar X \bar Y \bar Z##
    ##X Z##​

    The equation for F1 says there are two different conditions, either of which can satisify it:
    • X, Y, Z are all False
    • X and Z are both True
    Restating it; For F1 to be satisified, it is sufficient that both X and Z are True, nothing else matters and there is no reason to include anything else in the minterm; all other variables are extraneous.

    By the same token, F1 can also be satisified if X, Y, Z are all False.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Implementing boolean functions with decoder and external gate