Implementing a function.

Discussion in 'Homework Help' started by cazz0026, Sep 7, 2012.

  1. cazz0026

    Thread Starter New Member

    Sep 6, 2012
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    Hi all,

    I have this question for my homework.

    Implement the following function u sing a 4-to-16 decoder with active low outputs and any additional logic which might be necessary.

    F(a,b,c,d) = ab'c + bcd + ac'd + (abd)'

    I started to tackle this question by standardizing the function but when I used De Morgan's theoreum on (abd)' , I realized that I already had a b' in the function.

    (abd)' = a' + b' + d'
    ab'c = ab'cd + ab'cd'

    When standardizing b' , two of the minterms will be ab'cd and ab'cd'. I'm confused about this can you help me please?
     
  2. WBahn

    Moderator

    Mar 31, 2012
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    What is (ab'cd + ab'cd) equal to? In otherwords, what is any expression OR'ed with itself equal to? Similarly, what is any expression AND'ed with itself equal to?
     
  3. cazz0026

    Thread Starter New Member

    Sep 6, 2012
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    Nono I need to standardise the function in order to make a truth table. The variable ab'c is standardized to ab'cd + ab'cd'. Those two variables replace the ab'c in the function.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    I get what you are trying to do. But it seems you are getting stuck on how to handle repeated terms, so I asked a couple of questions aimed at helping you see how to handle repeated terms.

    Let's say you f(w,x,y,z) = wxy + wxz'

    When you expand these into full terms, you get:

    f(w,x,y,z) = wxy(z+z') + wx(y+y')z'

    f(w,x,y,z) = wxyz + wxyz' + wxyz' + wxy'z'

    You've got a repeated term. But, like I asked in the last post, what do you get when you OR two identical terms together?

    f(w,x,y,z) = wxyz + (wxyz' + wxyz') + wxy'z'

    What can the expression in parentheses be reduced to?
     
  5. cazz0026

    Thread Starter New Member

    Sep 6, 2012
    7
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    Oh I understood your question now;;

    When they are OR'ed together you get the same variable.. :)

    Thank you :)
     
  6. cazz0026

    Thread Starter New Member

    Sep 6, 2012
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    0
    Since they are active low outputs, I connect the variables in the truth table which are '0' to Vcc and those variables which are '1' to ground?
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    It says they are active low outputs, not active low inputs.

    My guess is that the inputs are active hi, otherwise they probably would have said active low logic (implying both inputs and outputs are active low).

    There's always room for interpretation differences, but I read the statement about having a 4-to-16 decoder as meaning that you present a four-bit active hi binary value to the inputs and one of the outputs goes high while the others go low. So when you say that you have active low outputs, the only difference I see is that the one output that went high now goes low and the others all go high.

    So you want to take a subset of the 16 outputs and produce an high output any time any one of the outputs in that subset is low.
     
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