Digital Electronics help

Discussion in 'Homework Help' started by aniskazi, May 6, 2014.

  1. aniskazi

    Thread Starter Member

    May 3, 2014
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    Design a logic circuit of electronic lock with inputs P,Q,R so that output S is HIGH whenever Q=R. Using NAND gates.
    P Q R S
    0 0 0 1
    0 0 1 0
    0 1 0 0
    0 1 1 1
    1 0 0 1
    1 0 1 0
    1 1 0 0
    1 1 1 1

    I made the eq as : P' Q' R' +P ' Q R + P Q'R '+ PQR ... ' = compliment taking QR common
    QR( P'+P) + Q'R'(P'+P)
    QR + Q'R'

    is this eq: correct?
     
    Last edited: May 6, 2014
  2. tshuck

    Well-Known Member

    Oct 18, 2012
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    If the output doesn't depend on one of the inputs, the simplified equation shouldn't have it, right?

    Your truth table has some places where Q=R, yet S=0...

    Edit: The OP has come back and modified the post, so the answer provided is not what was originally there....
     
    Last edited: May 9, 2014
  3. aniskazi

    Thread Starter Member

    May 3, 2014
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    yes edited the post sorry about that are you talking about the P input? that it shouldn't be in the circuit?
     
  4. tshuck

    Well-Known Member

    Oct 18, 2012
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    I'm saying that if the output is only dependent on Q and R, it stands to reason that the function that describes the output should only consist of Q and R, in a reduced form.
     
  5. aniskazi

    Thread Starter Member

    May 3, 2014
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    okay so if I only include Q and R in it so it will become with the equations as:

    Q' R' + QR + Q' R' + QR

    This will be the eq....
     
  6. tshuck

    Well-Known Member

    Oct 18, 2012
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    I'm not suggesting that you simply omit the P, but rather that, as a check, your final equation cannot have a P variable.

    I would suggest either posting your K-map or Boolean algebraic steps so we can comment...
     
  7. aniskazi

    Thread Starter Member

    May 3, 2014
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    Got rid of the P in the final eq:
    P' Q' R' +P ' Q R + P Q'R '+ PQR ... ' = compliment taking QR common
    QR( P'+P) + Q'R'(P'+P)
    QR + Q'R'

    correct my truth table with the eq and the final eq is this correct? I'm trying to make the truth table will show you when it's done :)
     
  8. tshuck

    Well-Known Member

    Oct 18, 2012
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    There you go.

    You should learn to check your answers by comparing your simplified results against the original, that way, you needn't ask if you are right, but you'll know it.
     
  9. aniskazi

    Thread Starter Member

    May 3, 2014
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    you mean to say that the eq that I got I should put the inputs and verify the results right? By the way made the circuit check it out please :)

    http://postimg.org/image/tx0s7njqb/
     
  10. tshuck

    Well-Known Member

    Oct 18, 2012
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    Precisely. If the functionality is any different, it is, by definition, not a simplified function. So, ensuring the same output for the same inputs will allow you to check the results yourself.

    Looks good.

    Another way to check this is to map out the truth table (like above).

    It is typically encouraged for you to upload images to this forum as links to other website will eventually break. You can do that by clicking the paperclip icon while posting.
     
    aniskazi likes this.
  11. full

    Member

    May 3, 2014
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    I think the answer is true:

    [​IMG]
     
  12. tshuck

    Well-Known Member

    Oct 18, 2012
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    Hi full,

    While I'm sure you can do your own homework, doing other people's homework does not benefit them.

    Please do not post answers as we try to teach people here and it's probably safe to reason the OP has seen many problems worked before. Showing the OP one more problem is unlikely to give them the clarity that comes from struggling through and grasping a problem one step at a time.
     
  13. full

    Member

    May 3, 2014
    225
    2
    ok ,I'm sorry,

    I new in this forum
     
  14. tshuck

    Well-Known Member

    Oct 18, 2012
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    No worries.

    Just keep your answers to your threads.:)
     
    brahms likes this.
  15. aniskazi

    Thread Starter Member

    May 3, 2014
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    I had one more question in my mind that if I have got a circuit in which only AND gates have been used and I have to remake it using only NAND gates than for every one AND gate I have to use TWO NAND gates is that true?
     
  16. tshuck

    Well-Known Member

    Oct 18, 2012
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    To implement the exact approach, yes, however, there are typically ways to reduce the logic and implement a complemented function to realize a smaller number of NAND gates than a 1 to 1 AND gate logic implementation.

    In other words, reexamining the NAND gate version can usually reduce the number of gates used when converting to a NAND only implementation.
     
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