Circuit Realization of XOR function with Three inputs

In summary, the XOR function can be realized with three inputs using AND, OR and NOT-gates. To find the SOP minimization for this function, one needs to find the truth table of the boolean expression and apply the associative law.
  • #1
Na0s
3
0

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.
 
Physics news on Phys.org
  • #2
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?
 
  • #3
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 =]
 
  • #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
 
  • #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
 

1. What is an XOR function with three inputs?

An XOR (Exclusive OR) function with three inputs is a logic gate that outputs a 1 (true) if exactly one of the three inputs is 1. If two or all three inputs are 1, the output will be 0 (false). This function is also known as a three-input binary adder.

2. How is an XOR function with three inputs realized in a circuit?

An XOR function with three inputs can be realized using a combination of AND, OR, and NOT gates. First, the three inputs are fed into two 2-input XOR gates, producing two intermediate outputs. These outputs are then combined with a 3-input OR gate, with the final output being the result of the XOR function with three inputs.

3. What is the truth table for an XOR function with three inputs?

The truth table for an XOR function with three inputs is:

A B C Output
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 1
1 0 1 0
1 1 0 0
1 1 1 1

4. What are some applications of an XOR function with three inputs?

An XOR function with three inputs has several applications in digital electronics, including error detection and correction in data transmission, encryption algorithms, and generating parity bits in computer memory systems.

5. Can an XOR function with three inputs be realized using only two-input XOR gates?

Yes, an XOR function with three inputs can be realized using only two-input XOR gates by cascading two of the gates together. However, this may not be the most efficient or compact circuit design, as using a combination of AND, OR, and NOT gates may require fewer overall gates.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
4
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
10
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
26K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
7K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
24K
  • Engineering and Comp Sci Homework Help
Replies
9
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
13
Views
4K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
6K
Back
Top