Homework - Boolean Simplification

Discussion in 'Homework Help' started by SaadMo, Sep 15, 2014.

  1. SaadMo

    Thread Starter New Member

    Sep 15, 2014
    2
    1
    Hi guys, I am new to the forums but have benefited from the sites tutorials.

    I have a hw problem: To simplify (a+b+c')(a'b'+c)

    If someone could work it out step by step, how they would do it, I would really appreciate it. What I typically do is take the dual of the function, then start working on the a'+b'c side by using x+x' to make sure all the terms have the variables.

    I end up with a'bc+a'b'c'+ab'c+a'b'c, but I am not sure how to proceed from here, or even if I did the earlier steps correctly.
     
  2. JoyAm

    Member

    Aug 21, 2014
    112
    4
    Have you considered using a Karnaugh map ? I believe its the best way to solve stuff like that.
    I got a'c+ab'+b'c ( i havent worked on it lately so i may be wrong)
     
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  3. MrChips

    Moderator

    Oct 2, 2009
    12,415
    3,354
    To simplify (a+b+c')(a'b'+c)

    multply the terms (a'b'+c) by a, then by b and by c'

    and then simplify the SOP (sums of products).

    Yes, Karnaugh mapping is a very useful tool for simplifying boolean logic. However, it does not help to reduce the expression in this example.
     
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  4. dalam

    Member

    Aug 9, 2014
    58
    6
    SaadMo likes this.
  5. SaadMo

    Thread Starter New Member

    Sep 15, 2014
    2
    1
    Thanks guys, I appreciate the help.

    I actually used a karnaugh map to solve it as that is way easier for me, but the HW required just striaght algebraic manipulation.
     
    JoyAm likes this.
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