boolean algebra help.

Thread Starter

norules44

Joined Oct 28, 2014
4
i did in the past boolean algebra but i forgot it, can any1 try to help me out on this?
problem 1:
F(wxyz)= wx+xyz+x'yz+x'yz'+[wx+w' (x+z)]
i did this but i got stacked.
F(wxyz)= wx+xyz+x'yz+x'yz'+[wx+w' (x+z)]
= wx+xyz+x'y(z+z')+[wx+w' (x+z)]
=wx+xyz+x'y+(w'+x')wx.wz
=wx+xyz+x'y

can any1 tell me if i made a mistake? if its going to simplify more?

also i got this one:
problem 2: f(a,b,c)= a'b'c+a'bc'+ab'c+abc'+abc i did this:
=a'b'c+a'bc'+ab'c+ab(c'+c)
=a'b'c+a'bc'+ab'c+ab
=a'b'c+a'bc'+ac+a
=a'b'c+a'bc'+a

can any1 tell me if i made a mistake? if its going to simplify more?

tnx a lot for reading!
 

WBahn

Joined Mar 31, 2012
29,978
That operation does the period represent?

Have you checked your work by seeing if the truth tables for the initial equation match your result? Have you tried to verify your design with a Karnaugh Map (assuming you have been introduced to those yet, if not don't worry).

And thank you for showing your work!

When checking for equality of two Boolean expressions, it is often a good first step to look for boundary cases. For instance, in your first one the first two terms are identical, so that means we can focus on whether the last term in your answer, x'y, covers all of the cases covered by the last four terms of the original problem that aren't covered by the first two, and no more. Look at the last term w'(x+z). If z=1, and w=0, then this term is 1 and the result is 1, regardless of the values of x or y. Does that agree with your proposed solution?

In your second one, your answer is saying that if a=1, then the output is 1 regardless of the values of b or c. So look at the original equation and see if you can violate this, namely see if you can force the output to be 0 even when a=1. If we set a=1, then we are left with the last three terms reduced to

f(1,b,c) = b'c + bc' + bc

Well, what if b=1 and c=0 ?

Look carefully at where you claimed that ab'c+ab = ac+a. Do you really believe this?
 

Thread Starter

norules44

Joined Oct 28, 2014
4
well tnx for your reply and ur time, but im not allowed to use karnaught maps yet, i know how to do them but im not allowed, also i know how to make a truth table and find when is 1 and when is 0, but i have to do this exercise and reduce it to the smallest possible equation, showing the way im doing it, yes u are right about the second problem but how do i proceed further?
f(a,b,c)= a'b'c+a'bc'+ab'c+abc'+abc i did this:
=a'b'c+a'bc'+ab'c+ab(c'+c)
=a'b'c+a'bc'+ab'c+ab
after a research which i did i found out that
if i got x+xy this equal to x+y
and x'+xy= x+y
and x+x'y= x'+y
but i dont know what to do now, i mean i got 3 terms
 
Last edited:

WBahn

Joined Mar 31, 2012
29,978
well tnx for your reply and ur time, but im not allowed to use karnaught maps yet, i know how to do them but im not allowed, also i know how to make a truth table and find when is 1 and when is 0, but i have to do this exercise and reduce it to the smallest possible equation, showing the way im doing it, yes u are right about the second problem but how do i proceed further?
Does that mean that you can't work the problem algebraically and then take out a sheet of paper and use a K-map to check your work?

I'm not talking about solving the problem using anything other than pure Boolean algebra -- I am talking about developing the ability to CHECK your work. Think about it, I doubt that coming to an on-line forum and asking if your work is correct is MORE acceptable than doing a Karnaugh map or a truth table to check it yourself!

As for proceeding, look at the part that I called into question and figure out what you did wrong and how to do it right? Feel free to post your musings on that part of it and we can delve into it more deeply if needed.

And I can't evaluate your first problem until you explain what operation you are using the period for (i.e., in wx.wz).
 

Thread Starter

norules44

Joined Oct 28, 2014
4
in the problem 1 i used de morgan's law in this part
F(wxyz)= wx+xyz+x'yz+x'yz'+[wx+w' (x+z)]
= wx+xyz+x'y(z+z')+[wx+w' (x+z)]
=wx+xyz+x'y+(w'+x')wx.wz
=wx+xyz+x'y


f(a,b,c)= a'b'c+a'bc'+ab'c+abc'+abc i did this:
=a'b'c+a'bc'+ab'c+ab(c'+c)
=a'b'c+a'bc'+ab'c+ab
after a research which i did i found out that
if i got x+xy this equal to x+y
and x'+xy= x+y
and x+x'y= x'+y
but i dont know what to do now, i mean i got 3 terms
 

WBahn

Joined Mar 31, 2012
29,978
I'll say it again: I CANNOT evaluate your work UNTIL you EXPLAIN what operation you are using the PERIOD for!

You have (w'+x')wx.wz

What does wx.wz mean?

This is a simple question. Why won't you answer it?
 

Thread Starter

norules44

Joined Oct 28, 2014
4
I'll say it again: I CANNOT evaluate your work UNTIL you EXPLAIN what operation you are using the PERIOD for!

You have (w'+x')wx.wz

What does wx.wz mean?

This is a simple question. Why won't you answer it?
i replied it i said i used the morgans law.




tnx a lot but anyway i dont know what to do to proceed more
 

djsfantasi

Joined Apr 11, 2010
9,156
Using DeMorgan's Law does not answer what the dot represents. Try describing the term in English, such as "w and x foo w and z", replacing "foo" with an English word.
 

WBahn

Joined Mar 31, 2012
29,978
i replied it i said i used the morgans law.




tnx a lot but anyway i dont know what to do to proceed more
So what does the dot represent?

If I tell you f(a,b) = a.b, what is the truth table?

DeMorgan's Theorems are either: ab = a' + b' or a + b = a'b'

The next step is to look carefully at your claim that x+xy equals x+y.
 

MrAl

Joined Jun 17, 2014
11,389
Hi,

You still need to define the dot. Saying it is DM theorem is not helping. You need to write it out completely because using a dot like that is too unconventional. Failure to write it out in FULL will result in you not getting an answer for that problem again.

Also, are you allowed to use exclusive or, ie "XOR", to simplify? This would help if you could because it can reduce the number of terms needed.
 

amilton542

Joined Nov 13, 2010
497
Boolean algebra of that complexity presumably implies you should be familiar with k-maps where that of which has already been bought to your attention in post #2. Just do them anyway, even if you're not allowed. If you're required to use Boolean algebra methods alone, do the k-map anyway and use it as an aid (provided you got it right and you're striving for the largest groupings of 2^n) in order to ensure you know what the final simplified logic should look like.
 
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