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General or sumarize a MUX 2-1 only using NAND gate.

Discussion in 'Electronics Homework Help' started by julia, Apr 7, 2013.

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  1. julia

    julia

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    Mar 12, 2013
    Hey everyone! Please guide or give me some instruct how to general or sumarize a MUX 2-1 only using NAND gate, thank very much, I have try as follow:
    S = NAND.E0) + NAND or S = NAND + NAND.E1) or S = NAND[(NOT.E0) . (C.E1)] or S = NAND[(C.E0) . (NOTE1)], I do it true or false, Please give me some advice.
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    At least Lee Yeon-Hee isn't a porn star I guess.
     
  3. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    What does all this mean? That's not the standard way to describe a boolean function - at least not that I knew.

    If I were to apply a NAND function to two variables (inputs), it would come down to
    Y= not(A *B) where A, B are the inputs, Y is the output, * is the AND function and not is the not function.

    To set up a multiplexer you have 3 inputs and 1 output:
    A, B are the inputs to be multiplexed
    Y is the output
    S is the selector. S=1 -> Y=A, S=0 -> Y=B (or vice versa, it doesn't matter much)
    Therefore Y=S*A + (not S)*B where + is the OR function.
    Now you apply boolean logic to transform this equation into one containing only * and not, then you have a multiplexer made from NAND gates only. Although I strongly think that this should have been part of your class, I'll link you to this text that gives you information on boolean simplification.

    You could also use Karnaugh maps for this task. This is a quick way to find the basic boolean equations. You still have to do the minimizing or transformation into logic types as required (here: NAND).
     
  4. Harald Kapp

    Harald Kapp Moderator Moderator

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    Nov 17, 2011
    Btw: you never finished your other thread "Design a NOT gate by using Diode and Resistance"
     
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