Boolean Algebra problem , need help

Thread Starter

Petrix12

Joined Sep 22, 2018
4
Hi i rly need help I solved DNS for fisrt Y but for Z I cant figure out last step and with CNS I tried everything that came to my mind but I cant move a bit and i would like to ask you for help, y=(A nor B) xor (c xnor ~d) and z=(c xnor ~d) nand (b xor d)

I know final values from logisim but I need to knwo the steps
DNS:
Z=~B ~D + C D + B ~C
y=~A ~B ~C ~D + ~A ~B C D + B ~C D + B C ~D + A ~C D + A C ~D
CNS:
z=(B + C + ~D) (~B + ~C + D)
y=(A + B + C + ~D) (A + B + ~C + D) (~B + C + D) (~B + ~C + ~D) (~A + C + D) (~A + ~C + ~D)
 

WBahn

Joined Mar 31, 2012
24,854
What are "DNS" and "CNS"? I'm guessing that the first is probably disjunctive normal something and the latter is conjunctive normal something, but that's only a guess and engineering is not about guessing.

In order to help you find where you went wrong, you need to show your work so that we can see where you went, period. We are not mind readers.
 

Thread Starter

Petrix12

Joined Sep 22, 2018
4
What are "DNS" and "CNS"? I'm guessing that the first is probably disjunctive normal something and the latter is conjunctive normal something, but that's only a guess and engineering is not about guessing.

In order to help you find where you went wrong, you need to show your work so that we can see where you went, period. We are not mind readers.
sorry I wrote it wrong it is DNF and CNF , I spent like last 6 hours trying to figure it out.... and F means form, the problem is that i dont know where i went wrong I have several A4 full of calculations but it is all wrong, I need to know steps how to get from Y ad Z which is up to DNF and CNF form which is writen below.
 

WBahn

Joined Mar 31, 2012
24,854
sorry I wrote it wrong it is DNF and CNF , I spent like last 6 hours trying to figure it out.... and F means form, the problem is that i dont know where i went wrong I have several A4 full of calculations but it is all wrong, I need to know steps how to get from Y ad Z which is up to DNF and CNF form which is writen below.
Pick what you think is you best attempt at one of them and post it. We need a starting point.

At the very least, expand your starting expressions into ones that involve only the standard operations of NOT, AND, and OR -- and show your work in doing so. The start identifying the common elements of the various factors/terms. If you know about Karnaugh maps, that can help you identify them.
 

Papabravo

Joined Feb 24, 2006
12,514
Don't know why it took 6 hours. The top Google hits get you the definitions in less than 30 seconds. Maybe the online explanations were too obtuse. The solution you seek is the relentless application of De Morgan's theorem along with some basic identities.
 
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Thread Starter

Petrix12

Joined Sep 22, 2018
4
Pick what you think is you best attempt at one of them and post it. We need a starting point.

At the very least, expand your starting expressions into ones that involve only the standard operations of NOT, AND, and OR -- and show your work in doing so. The start identifying the common elements of the various factors/terms. If you know about Karnaugh maps, that can help you identify them.
DNF for Y i compled, I know how to put starting expresions into standart operations, but i got stuck for example here as you can see in atached file, it is very closed to be DNF of Z but i dont know what to do next. And most problematic thing for me is to do CNF from DNF, I tried various substitutions and so but none of them got me even close to result.
 

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Thread Starter

Petrix12

Joined Sep 22, 2018
4
Don't know why it took 6 hours. The top Google hits get you the definitions in less than 30 seconds. Maybe the online explanations were too obtuse. The solution you seek is the relentless application of De Morgan's theorem along with some basic identities.
I know definitions and theory, but if you think it is that easy show me how you go from DNF to CNF step by step
 

WBahn

Joined Mar 31, 2012
24,854
DNF for Y i compled, I know how to put starting expresions into standart operations, but i got stuck for example here as you can see in atached file, it is very closed to be DNF of Z but i dont know what to do next.
This might help you:

https://forum.allaboutcircuits.com/blog/boolean-logic-working-with-consensus-terms.663/

And most problematic thing for me is to do CNF from DNF, I tried various substitutions and so but none of them got me even close to result.
You might try looking at the following:

https://forum.allaboutcircuits.com/blog/boolean-logic-sop-and-pos-forms.583/

If some of the equations don't render properly, try a different browser. Chrome, in particular, made changes a while back that broke its ability to render MimeTex properly.
 

Papabravo

Joined Feb 24, 2006
12,514
I know definitions and theory, but if you think it is that easy show me how you go from DNF to CNF step by step
First of all do you understand how to look at a truth table and write, by inspection, the SOP and the POS expressions that represent the Boolean function represented by the truth table. If the answer is yes we can start with a simple example and work our way up to the more complicated ones.
 

Papabravo

Joined Feb 24, 2006
12,514
I know definitions and theory, but if you think it is that easy show me how you go from DNF to CNF step by step
OK
If we have a function y = ab' + a'b which is the familiar xor function expressed as SOP. We want to take the complement and apply DeMorgan's theorem twice. You will agree that taking the complement twice preserves the original function.

1: Take the complement of the original function and apply DeMorgan's Law, (ab' + a'b)' = (a' + b)(a + b')
2: Expand the expression, (a' + b)(a + b') = a'a + a'b' + ab + bb'
3: Drop terms that are identically 0, a'a + a'b' + ab + bb' = a'b' + ab
4: Take the complement again and apply DeMorgan's Law (a'b' + ab)' = (a + b)(a' + b')

The result is the POS expression which can be verified in numerous ways. You can simply do the indicated operation to go from POS to SOP.

(a + b)(a' + b') = aa' +ab' +a'b + bb'
Dropping the terms that are identically 0, and
aa' +ab' +a'b + bb' = ab' + a'b
 
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