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Simplying the circuit into Boolean expressions [2 Questions]

  1. Feb 2, 2013 #1
    Question 1: http://imgur.com/Px0n152
    Question 2: http://imgur.com/tZ0beF6


    For Question 1, I got:

    (AB.CD) + [(CD)' . (AB)'] with a "." = multiplication and ' = complement.
    into (AB.CD) + [(CD+AB)']

    I"m not sure how to simplify this further.

    Question 2 is really long and I would appreciate help. I got:
    A' + (D'+C') + A + (A' + C'+ D') + B' .D ' + (D+B)' = F

    I'm not even sure if this is correct, and these two questions are due as homework tonight.

    Any help would be much appreciated, thank you.
     
  2. jcsd
  3. Feb 2, 2013 #2

    NascentOxygen

    User Avatar

    Staff: Mentor

    Hi lunapt. At any line in your working you yourself can check that you are proceeding correctly by drawing up a truth table for that line. Compare it with the truth table for the original logic, and they must agree at all times, right up to including your answer. :smile:

    By not providing your earlier working, you don't allow us to identify whether the line you show has been arrived at correctly or not, or even whether it is your own effort. So can you show how you obtained that line?

    Good luck! The simplified expression could not be much simpler. :approve:

    BTW, it is best to present only one question, as it is likely that the help you receive with it will enable you to go on and solve other similar ones.
     
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