Reduce Circuit

Thread Starter

Damon G

Joined Feb 11, 2018
4
Hi Everyone. I have a Digital Logic class that I take at college. My professor gave us a problem to work on, its not for a grade but its just for us to make sure we understand what he has been teaching us.
I am pretty sure I have everything right up to the XNOR gate.. My issue is I cant figure out what the output will be from the XNOR gate. I researched the internet multiple sites , and I cant seem to get what the output will be.

If you look on my paper that I drew on with pen, I complimented (A'+b)(C)' to AB'+C' and I wasnt sure if that is correct if you would compliment before it goes into the XNOR
Also the (BC)' to B'+C' if that is the same and it is the other input into the XNOR


Not sure if I clarified this enough.. any help would be great./ Thank you.
 

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dl324

Joined Mar 30, 2015
10,933
Welcome to AAC!

Thank you for posting your work; you'd be surprised how many forget to do that.

This is incorrect: ( EDIT: corrected for posterity.)
upload_2018-2-11_10-57-39.png

It might be simpler if you treat the XNOR as an XOR and then invert the output.
 
Last edited:

Thread Starter

Damon G

Joined Feb 11, 2018
4
Thanks for the response. That part you just showed me, was actually something I found on the internet..
It was here :
https://steemit.com/logic/@drifter1/logic-design-from-function-to-circuit-using-multi-input-gates

A XNOR Gate can be found by searching for (A⊕B)' or AXNORB directly that is rare. But, we can again search for the following inside of our function:


(A⊕B)' = AB + A'B' (pretty easy one)

I seen this answer some where else also so that is why I jotted it down to the side of my paper to remember.

As for what you said about Invert the XOR out put.. this is where I am having an issue..
I understand the Truth Tables for every gate just fine.. So XOR is basically inputs the same means its False/ zero/ off , and if both input are opposite then out put will be True/one/ on. COmpliment that would just be the opposite of what I said..
But that is not the part that is giving me an issue.. its I dont understand how to combine the two inputs and know what the output will be.

on the piece of paper I put down something down I am sure is wrong..

If we look at the two inputs going to the XNOR gate it is

(A'+B)(C)' XNOR (BC)'
This is where I am stuck trying to figure the output.. I dont know how to get a result from the two inputs. I can do it for AND NAND OR NOR, Also I know how to XOR and XNOR two binary bits, but this is more than just XOR XNOR a bit with another bit. ..
 

WBahn

Joined Mar 31, 2012
25,904
Welcome to AAC!

Thank you for posting your work; you'd be surprised how many forget to do that.

This is incorrect:
View attachment 145699

It might be simpler if you treat the XNOR as an XOR and then invert the output.
What's wrong with that. Unless I'm misinterpreting what appear to be dots around the first term (and that I'm assuming are just extraneous marks on the paper), it is correct.

An XNOR is also known as an equality gate and outputs a high if both inputs are the same.
 

WBahn

Joined Mar 31, 2012
25,904
Thanks for the response. That part you just showed me, was actually something I found on the internet..
It was here :
https://steemit.com/logic/@drifter1/logic-design-from-function-to-circuit-using-multi-input-gates

A XNOR Gate can be found by searching for (A⊕B)' or AXNORB directly that is rare. But, we can again search for the following inside of our function:


(A⊕B)' = AB + A'B' (pretty easy one)

I seen this answer some where else also so that is why I jotted it down to the side of my paper to remember.

As for what you said about Invert the XOR out put.. this is where I am having an issue..
I understand the Truth Tables for every gate just fine.. So XOR is basically inputs the same means its False/ zero/ off , and if both input are opposite then out put will be True/one/ on. COmpliment that would just be the opposite of what I said..
But that is not the part that is giving me an issue.. its I dont understand how to combine the two inputs and know what the output will be.

on the piece of paper I put down something down I am sure is wrong..

If we look at the two inputs going to the XNOR gate it is

(A'+B)(C)' XNOR (BC)'
This is where I am stuck trying to figure the output.. I dont know how to get a result from the two inputs. I can do it for AND NAND OR NOR, Also I know how to XOR and XNOR two binary bits, but this is more than just XOR XNOR a bit with another bit. ..
Then just get rid of the XOR or XNOR and replace them with what they are equivalent to, namely (A⊕B) = AB' + A'B and (A⊕B)' = AB + A'B'.

So

(A'+B)(C)' XNOR (BC)' = [(A'+B)(C)'][(BC)'] + [(A'+B)(C)']'[(BC)']'
 

WBahn

Joined Mar 31, 2012
25,904
Are you familiar with bubble logic?

You can swap an AND with an OR, and vice versa, as long as you invert ALL of the inputs and outputs. Doing so does not change the overall logic.

Bubble logic for XOR is that you can invert any even number of inputs and outputs.

A consequence of this is that you can swap an XOR with an XNOR, and vice versa, by inverting an odd number of inputs (since the inverted output makes the total even).

As a check, that means that to get the basic equation for a 2-input XNOR we invert one input, which makes it an equality gate.
 

Thread Starter

Damon G

Joined Feb 11, 2018
4
Are you familiar with bubble logic?

You can swap an AND with an OR, and vice versa, as long as you invert ALL of the inputs and outputs. Doing so does not change the overall logic.

Bubble logic for XOR is that you can invert any even number of inputs and outputs.

A consequence of this is that you can swap an XOR with an XNOR, and vice versa, by inverting an odd number of inputs (since the inverted output makes the total even).

As a check, that means that to get the basic equation for a 2-input XNOR we invert one input, which makes it an equality gate.
Thank you for responding.. Yes those were marks on the paper around the AB , my mistake for not having the paper clean from markings ..

I will figure this out , I just wanted to ensure that their was an equivalent to the XOR or XNOR , which you did in fact point out. With that new knowledge I can look at the problem again and work my way through it..
My last question would be ... Should I simplify the inputs that are going into the XNOR gate?

As they stand right now the top input is (A'+B)(C)' would the equivalent of that be AB'+C' ? I just wasnt sure if it was better to get rid of the one compliment bar, because its just going to get another one AFTER the XNOR gate..

And of course the lower input would also be (BC)' B'+C' Once this is answered I will take a shot at finishing this problem.. Thank you again.
 

WBahn

Joined Mar 31, 2012
25,904
Thank you for responding.. Yes those were marks on the paper around the AB , my mistake for not having the paper clean from markings ..

I will figure this out , I just wanted to ensure that their was an equivalent to the XOR or XNOR , which you did in fact point out. With that new knowledge I can look at the problem again and work my way through it..
My last question would be ... Should I simplify the inputs that are going into the XNOR gate?

As they stand right now the top input is (A'+B)(C)' would the equivalent of that be AB'+C' ? I just wasnt sure if it was better to get rid of the one compliment bar, because its just going to get another one AFTER the XNOR gate..

And of course the lower input would also be (BC)' B'+C' Once this is answered I will take a shot at finishing this problem.. Thank you again.
I think your being a bit sloppy in your notation.

(A'+B)(C)' is NOT the same as [(A'+B)(C)]', it is simply (A'+B)(C')

To answer your question, it is often good to simplify intermediate results as you go, but if an inverted quantity is going to pick up another inversion, then they will cancel out and it is better to not simplify them. So the answer is... it depends.
 

Thread Starter

Damon G

Joined Feb 11, 2018
4
I think your being a bit sloppy in your notation.

(A'+B)(C)' is NOT the same as [(A'+B)(C)]', it is simply (A'+B)(C')

To answer your question, it is often good to simplify intermediate results as you go, but if an inverted quantity is going to pick up another inversion, then they will cancel out and it is better to not simplify them. So the answer is... it depends.
oops sorry , I didnt have the brackets up.. never used them before. I am used with the bars overhead.. But I see the difference now you are pointing out. They way I had it (c)' was just complimenting the C... I thought i was complimenting the entire thing.. Now I see the use for Brackets.. Once again thank you. Also I see your point about the compliment just being negated anyway, so no need to do that extra work ..
 

dl324

Joined Mar 30, 2015
10,933
What's wrong with that. Unless I'm misinterpreting what appear to be dots around the first term (and that I'm assuming are just extraneous marks on the paper), it is correct.
You're right, it's correct. I should have looked at the truth table instead of trying to do the boolean algebra mentally.
 
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