Homework Problems

Discussion in 'Homework Help' started by ahmadrashid17, Sep 27, 2007.

  1. ahmadrashid17

    Thread Starter New Member

    Sep 27, 2007
    3
    0
    First of all, i would like to thanks for the developer of this website especially the forum section. I am new to digital logics and design. Our teacher has given homework to simplify the booleon algebric expression to specific literals; i tried alot and able to solve just one out of them. Can someone help me and solve this for me.
    I am trying for last one week, though i know and learnt the laws like commutative, distributive, assosiative, but still i can not find the way to proceed. The one i solved is:

    A: Simplify the expession to the said literals:

    1. ABC + A'B'C + A'BC + ABC' + A'B'C' to five literals
    =>ABC + ABC' + A'B'C + A'B'C' + A'BC
    =>AB(C+C') + A'B'(C+C') + A'BC
    =>AB(1) + A'B'(1) + A'BC
    =>AB + A'B' + A'BC
    =>AB + A'(B' + BC)
    =>AB + A'(B' + B)(B' + C)
    =>AB + A'(1)(B' + C)
    =>AB + A'(B' + C) ------------- Answer

    Hereby u are request to help me in question 2 to 8 which are as following:

    2. BC + AC' + AB + BCD to four literals
    3. [(CD)' +A]' + A + CD + AB
    4. (A + C + D)(A + C + D')(A + C' + D)(A + B')

    B: First take the complement and then reduce the expression to minimum literals possible:

    1. (BC' + A'D)(AB' + CD')
    2. B'D + A'BC' + ACD + A'BC
    3. [(AB)'A][(AB)'B]
    4. AB' + C'D'


    I will highly appriciate your kind help. thanks
     
  2. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
    6
    Yes!
    No!

    Please list the work you yourself have done so far on the questions you have trouble with. We will gladly assist, suggest, teach, point out, encourage, and re-phrase. But will never do your work for you.
     
  3. ahmadrashid17

    Thread Starter New Member

    Sep 27, 2007
    3
    0
    Ok here are my try for part B:

    1. [(BC' + A'D)(AB' + CD')]' i took the whole complemet
    (BC' + A'D)' + (AB' + CD')'
    (BC')'.(A'D)' + (AB')'.(CD')'
    (B' + C)(A + D') + (A' + B)(C' + D)
    B'A + B'D' + AC + CD' + A'C' + A'D + BC' + BD

    Now i cant go furture can you help me....

    2. [B'D + A'BC' + ACD + A'BC]'
    (B'D)'(A'BC')'(ACD)'(A'BC)'
    (B + D')(A + B' + C)(A' + C' + D')(A + B' + C')
    Then i got hanged

    3. [((AB)'A)((AB)'B)]'
    ((AB)'A)' + ((AB)'B)'
    AB + A' + AB + B'
    AB + B' +AB + B'
    (A + B')(B + B') + (A + A')(A' + B)
    (A + B')(1) + (1)(A' + B)
    A + B' + A' + B
    A + A' + B + B'
    1 + 1
    1


    4. [AB' + C'D']'
    (AB')'(C'D')'
    (A' + B)(C + D)

    Could some one help me in Q 1 and 2 of part B
     
  4. ahmadrashid17

    Thread Starter New Member

    Sep 27, 2007
    3
    0
    And for part A Q2, 3, and 4.... JUST GIVE ME A STARTING HINT for two or three steps... then i will try my self the rest
     
  5. adn07

    Member

    Sep 25, 2007
    17
    0
    the best soultion to your problem is to read this book :
    fundamentals of logic design by Charles H.Roth,Jr.
    go to chapter 2 and 3 and in chapter for you will find applications of boolean algebra ...
     
  6. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    A point about taking the compliment is that you have to take the double compliment to ensure the function remains the same. Consider the following:

    - You have a function A (very simple I know)

    - If you take a single compliment you end up with NOT A, which is not the same as A. You have changed the function and therefore the simplification will be different.

    - If you take a double the compliment you end up with NOT NOT A, which is just A. You haven't changed the function, however you now have a compliment with which to use De-Morgans theorems.

    With that in mind, have another look at your answers to section B. Also you should look at the relevant sections in the e-book for Boolean identities:

    http://www.allaboutcircuits.com/vol_4/chpt_7/3.html

    http://www.allaboutcircuits.com/vol_4/chpt_7/4.html

    http://www.allaboutcircuits.com/vol_4/chpt_7/5.html

    Dave
     
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