Hi,

I am new to the forum so hello to all.

I am studying a BTEC in EE and going onto a HND hopefully next year. I am struggling a bit with reducing boolean algebra. I seem to be able to look at the same problem 3 times and come up with three different answers.

Can someone give me some hints. If some is NOT for example NOT A I write it as A'

I have this problem.... ABC + A'B'C' + ABC' + A'C'

this is one way I have "solved" it.

ABC + A'B'C' + ABC' + A'C'
ABC + A'B'C' + ABC' + A' . 1 (last C' cancelled against first C, turns into 1 (A+A' = 1))
AB + A'B'C' + C' + A' . 1 (cancelled first AB against second AB in third group)
AB + A'B'C' + A' . 1 (cancelled second C' as same as C' in second group)
AB + B'C' + A' . 1 (A' in second group removed as same as last A')
AB + B'C' + A' (A' . 1 = A')

Any help would be greatly appreciated!

Thanks!

 

I don't think you are allowed to do the 1st step of your logic.

Consider this sentence: F=ABC+A'C'
for C=1, F=AB
for C=0, F=A'

Now, if according to your reasoning I write F=ABC+A' then
for C=1, F=AB+A'
for C=0, F=A'

These two expressions aren't the same.

Make another effort. Look up in your textbook for the available Boolean identities.

 

I don't think you are allowed to do the 1st step of your logic.

Consider this sentence: F=ABC+A'C'
for C=1, F=AB
for C=0, F=A'

Now, if according to your reasoning I write F=ABC+A' then
for C=1, F=AB+A'
for C=0, F=A'

These two expressions aren't the same.

Make another effort. Look up in your textbook for the available Boolean identities.

Changing that method I come up with ABC + A'B'C'

Am I any closer?

 

One of the nice things about problems like this (and a surprisingly high fraction of problems in engineering in general), is that you can check the validity of your answer from the anwer itself.

In this case, you have the given function:

F = ABC + A'B'C' + ABC' + A'C'

You have "simplified" it (whatever that means) and come up with:

G = ABC + A'B'C'

Both represents functions with a total of eight possible input conditions, so the truth table for each is very tame. Write out the truth tables for each. If they match, the functions are equal, if they don't, they aren't. Figuring out how to get to G may or may not be tricky, but determining whether G is a valid answer is trivial and should ALWAYS be done whenever practical.

Look for pairs of terms that have common factors, expecially if the only differ in one term.

 

ABC+A'B'C'+ABC'+A'C' =

AB(C+C')+A'C'(B'+1) =

C+C' = 1 and B'+1 = 1 so the equation will reduce to

AB+A'C'

 

It was kind of you to provide a full solution, but for the threads in the Homework Help section, we prefer to offer hints and help for the inquirer to reach the final result by his own means. It is more educational that way.

 

It was kind of you to provide a full solution, but for the threads in the Homework Help section, we prefer to offer hints and help for the inquirer to reach the final result by his own means. It is more educational that way.

Oops....OK, I'll keep it in mind.

 

Oops....OK, I'll keep it in mind.

Live and learn. Welcome aboard!