Engineering Circuit Realization of XOR function with Three inputs

AI Thread Summary
The discussion centers around developing a circuit realization of a three-input XOR function using basic logic gates. The initial poster struggles with the problem and expresses confusion over applying DeMorgan's Law and creating the correct Boolean expression. A suggestion is made to use a Karnaugh map (K-map) to derive the simplified expression, which the poster admits they are not yet familiar with. After further attempts and applying the associative law to the Boolean expression derived from the truth table, the poster successfully finds the solution and confirms it works in LogiSim. The conversation highlights the importance of understanding K-maps and Boolean algebra in logic design.
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Homework Statement


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.
[/B]
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

Homework Equations


DeMorgans Law...I think...

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 I am moving in the right direction. If anyone could lay out the steps in how to do this one I would appreciate it.
 
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Na0s said:

Homework Statement


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.
[/B]
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

Homework Equations


DeMorgans Law...I think...

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 I am moving in the right direction. If anyone could lay out the steps in how to do this one I would appreciate it.
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?
 
Let me give it a go. I am currently in the process of learning K maps so I am 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 =]
 
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
 
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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