Hi everyone-

Well, another day, another homework assignment! This problem is simple, it has two parts and in each you're finding the complement of a function. I will post my attempts and would appreciate it if someone could tell me if I'm right or wrong. If I'm wrong, I'd appreciate the correct answer (with steps if possible).Thanks!!

Part a) Find the complement of AB'+A'B
(A'+B)*(A+B') is my answer

Part b) Find the complement of (V'W+X)Y+Z'
(V+W')*(X'+Y')*(Z) is my answer

Thanks in advance for your replies!!

 

The first one is correct, but it can be simplified further. Do the operations and remove the parentesis. The result is the complement of XOR, we did it in a previous post of yours.

The second one is wrong. Try not to rush and do one complementation at a time, and find the answer through multiple steps, not just one. Break the expression in terms and work with them.

For example, you should start like this:
((V'W+X)Y+Z')'=
((V'W+X)Y)'*(Z')'=
((V'W+X)'+Y')*Z=
...
Do you get the method? Take big chunks of the expression and work them gradually by following the simple rules: (ab)'=(a'+b') and (a+b)'=(a'b')

 

Hi Georacer:

Thanks for the reply!!

for part b, I worked it out like this. Is it correct?

(v'w+x)y+z'
((v'w+x)y)'+z'
((v'w+x)y)'*z''
((v'w+x)y)'*z
((v'w+x)'+y'*z
(v+w')*x'+y'*z
vx'+w'x'+zy' is my answer. Is it correct?

For part a, I did:
A'A+B'A+BA+B'B
I got this answer after doing the multiplication and removing the parenthesis. Is this correct? Does it need simplified more?

 

Your part a is wrong. It is A'A+B'A'+AB+B'B. Watch those typos, they cost and are hard to trace. Take your time from line to line.
Also A'A is an identity. If A=1 then A'=0 and therefore A'A=0. If A=0 then also A'A=0. As a result, your final expression is A'B'+AB (as we said in a previous thread of yours).

As for part b, well, you messed up again due to your haste. I will put the solution and your efforts back to back to see your mistakes.

Your attempt _____________________________________My attempt
​

(v'w+x)y+z' ____%You start by f, but we need f' ______________((v'w+x)y+z')'=
((v'w+x)y)'+z' __%This line is not a valid transition from the above
((v'w+x)y)'*z'' __%However, "magicaly", this one is correct ______((v'w+x)y)'z''=
((v'w+x)y)'*z ___%OK so far
((v'w+x)'+y'*z __%Even though you made the complementation __((v'w+x)'+y')z=
______________%correctly, you forgot the parenthesis and from
______________%now on you are on a downfall
(v+w')*x'+y'*z ________________________________________ ((v'w)'x'+y')z=
vx'+w'x'+zy' __________________________________________((v+w')x'+y')z

I don't see the need to simplify the expression more. As it is it's very "implementation friendly" and mirrors the actual gate arrangment.

 

Thanks for the detailed solution to part b! As for part A, are my steps right up until the typo? From where you corrected the typo,
A'A+B'A'+AB+B'B,
I guess you use the identity XX'=0 to eliminate the A'A term and the B'B term so then that leaves you with just the B'A'+AB.
Are my steps for part A correct?

 

Well, I corrected it just by knowing what the correct answer was. Write the steps for part a (in detail, so that you don't make any typographic mistakes) and we 'll discuss about them.

 

Hi-

Here is my detailed solution to this problem:

AB'+A'B (original function)

(A'+B)*(A+B') (got the complement of the function by interchanging all of the operators)

A'A+B'A'+BA+B'B (performed the multiplication/and operation)

A'A= 0 as does B'B (due to one of the Boolean identities)

0+B'A'+BA+0 (reduced function)

A'B'+AB (changed order of the variables so that they fit the Standard order of operations)

How do these steps look to you?

 

It is correct, but you still miss the point! If you want to improve at logic design, you need to go deeper than simply saying that you complemented the expression just by "interchanging all of the operators". That's not even a valid practice for more complex expressions!

Take your time! Do things step by step. What you should have done in order for this session to be really productive is this:

f=ab'+a'b
f'=(ab'+a'b)'
f'=(ab')'(a'b)'
f'=(a'+b)(a+b')
f'=a'a+a'b'+ab+bb'
f'=a'b'+ab

You have invested so much time in these forums, just to let it go to waste. Gain something out of it by, not just solving the exercises, but learning the prerequisites so that you don't need to come to the forum for help.

 

there's a table in this post on boolean algebra this might help you...Demorgan's theorem you should take a look it's what you are applying on your equation

http://forum.allaboutcircuits.com/showthread.php?t=3560

 

Thanks for the help everyone!!