# Homework

Discussion in 'Homework Help' started by stino, Mar 4, 2011.

1. ### stino Thread Starter New Member

Mar 4, 2011
1
0
Hi guys I need a little bit help with these two exercices, can someone just do and explain them to me please don't understand much.

Question 1​
Use only Boolean algebra to simplify the Boolean expression F. (First simplify F​
1 and F2, then simplify
F
1 · F2, showing all the steps. You need not provide the names of the Boolean rules that you apply.)
F
1 = x + x'(yw)' + w'y
F
2 = (y' + x')'

F(x, y, w) = F
1 F2.

Question 2 ​
Use a ​
Karnaugh map to find the simplest form of
H(A, B, C, D) = m
1 + m4 + m6 + m7 + m9 + m10 + m12 + m14.
Derive the terms of H directly from the
Karnaugh map without making use of algebraic manipulations
or truth tables. Clearly show the groupings.
Use exactly the same order for the variables as given in the following diagram:
C B A D​

Thank you!!!

2. ### Georacer Moderator

Nov 25, 2009
5,177
1,285
Which parts of the questions can you handle? Do you know the theoretical background behind them? Can you post your efforts for tackling the problem?

3. ### zaxisus New Member

Mar 12, 2011
1
0
Hi Georacer,

I'm stuck on the same question, but I've made some inroads. It's just the second part of question 1 (F1⋅F2) that I'm not 100% sure of...

Here are my steps:

F(x, y, w) = F1 ⋅ F2
= (x + x'y' + x'w' + w'y).(yx)
= xy + 0 + 0 + xyw'
= xy + xyw'
= xy

Does this look right? xy seems such an "easy" answer...

4. ### Georacer Moderator

Nov 25, 2009
5,177
1,285
It looks correct to me and I don't think it is easy in any way, just a typical Boolean logic class question.

If you have doubts, try to reach to the same result using a Karnaugh map.