Hi, just like some feedback if I've tackled a problem from my homework from the right angle in my introductory computer engineering unit.

We've been given the problem to design a 4-bit even-parity generator using only AND, OR and NOT gates.

The logic diagram I have come up with is:

And here is my truth table:

Does this look correct? Any help will be appreciated! This is my first assignment, and we have to implement it on a breadboard in class and I'm a little hesitant at how many gates my circuit requires so I have a feeling I'm on the wrong path.

 

I am just now taking digital logic, but, for what its worth, given the gate restrictions, I don't see any simplifications. Well, you do realize you only need four NOT gates, right?

 

Ah, yes. Thank you! I've now got it down to 4 NOT gates and a bit more confident of actually implementing this now.

 

This is a good site to learn how to derivative the XOR function. (it's simpler then doing derivatives in mathematics)

http://www.play-hookey.com/digital/xor_function.html

basically it should be 3 times that one shown on the website.

 

I couldn't get it to upload in Logisim but the attached is pdf and jpeg of the 4 bit even parity generator. It should be correct.

 

Thanks Arbitrator, that's definitely a more simplified version then my attempts. However before I move on I really want to get to that answer by myself.

My boolean expression that I ink out from my k-map is ABCD' + ABC'D + AB'CD + AB'C'D' + A'B'CD' + A'B'C'D

Is there a further simplification rule that I am missing?

 

this has labels on inputs and output.

 

i do it now , i think there should be 8 Boolean expressions wait 5 minutes

 

yes i got a 8 expressions of groups of a,b,c,d . i don't know if you can group anything from k-map since they all diagonal.
i got A'B'CD' + A'BC'D' + ABCD' + ABC'D + AB'CD + AB'C'D' + A'BCD + A'B'C'D

how did you simplify out AB'C'D' and A'BCD?

 

Yeap, that's the expression I've been getting too. In the interest of actually learning this, what method do I use to get the expression used in your circuit diagram?

 

i got it from the simplest XOR expression then just changed gates :S that doesn't help you much

 

we did this in a lab a few weeks back, there is no way to use k-maps to simplify Boolean. You can only simplify this further if you use Boolean logic to get XOR expressions by combining (in pairs) A'B'CD'+A'B'C'D to get A'B'(C XOR D) - eventually it is a simple implementation with XOR gates only, if XORs are unavailable, they can be implemented with regular gates, as is done in the diagram posted above.

 

we did this in a lab a few weeks back, there is no way to use k-maps to simplify Boolean. You can only simplify this further if you use Boolean logic to get XOR expressions by combining (in pairs) A'B'CD'+A'B'C'D to get A'B'(C XOR D) - eventually it is a simple implementation with XOR gates only, if XORs are unavailable, they can be implemented with regular gates, as is done in the diagram posted above.

i understand it now you explained it like that.