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Circuit Realization of XOR function with Three inputs

  1. Dec 13, 2016 #1
    1. The problem statement, all variables and given/known data
    This is from a past exam paper for logic design in my course. I have an exam coming up and would love to know how to solve this one.

    Develop a circuit realization of the XOR function with three inputs. You may use AND, OR and NOT-gates with not more than two inputs.

    Either draw the circuit or (for simplicity) write down the exact equivelant Boolean Expression with correctly positioned parenthesis

    2. Relevant equations
    DeMorgans Law...I think...

    3. The attempt at a solution
    I have not really gotten anywhere with this one.
    NAND and NOR functions with three inputs worked out fine for me, but not this one.

    For DeMorgans you have to:
    1. Change all variables to their complements.
    2. Change all AND operations to ORs.
    3. Change all OR operations to ANDs.
    4. Take the complement of the entire expression.
    So how does this one work with XOR?
    (A xor B xor C) = (A'B + AB') AND C - this seems off..
    I am not even sure if im moving in the right direction. If anyone could lay out the steps in how to do this one I would appreciate it.
     
  2. jcsd
  3. Dec 13, 2016 #2

    berkeman

    User Avatar

    Staff: Mentor

    Welcome to the PF.

    I would start with a K-map of the XOR function of 3 inputs, and then use a SOP representation to convert the K-map into the equivalent circuit using 2-input gates. Can you give that a try and show us your work?
     
  4. Dec 13, 2016 #3
    Let me give it a go. I am currently in the process of learning K maps so im still not familiar with how to do them properly.
    I will get back to you when I have tried working it out.

    Have to head out for a while so I can't do it right now. Thanks =]
     
  5. Dec 13, 2016 #4
    I have added an image to show what I have currently done. What I am struggling with is the Kmap.
    I see that you are supposed to group 1's together to get a minimised expression. SSOP.
    Even for SOP, how do I actually get that from this map?
    79d475885a2edf31e13023946ae318df.png
     
  6. Dec 14, 2016 #5
    Now I feel silly looking at this.

    So I was close to the answer already but kept moving off in completely the wrong direction most of the time. So what I did was just, got the SOP minterms from the truth table. Applied the associative law on the boolean expression. And that gave me the answer I was looking for. Tested it out in LogiSim and it works just great. Thanks for the help, I clearly didnt get enough sleep yesterday. But after some rest I took another look and solved it.

    150907e72fde7ba2ab1f0daff73fa38e.png
     
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