Homework Problems

Joined Sep 27, 2007
3
First of all, i would like to thanks for the developer of this website especially the forum section. I am new to digital logics and design. Our teacher has given homework to simplify the booleon algebric expression to specific literals; i tried alot and able to solve just one out of them. Can someone help me and solve this for me.
I am trying for last one week, though i know and learnt the laws like commutative, distributive, assosiative, but still i can not find the way to proceed. The one i solved is:

A: Simplify the expession to the said literals:

1. ABC + A'B'C + A'BC + ABC' + A'B'C' to five literals
=>ABC + ABC' + A'B'C + A'B'C' + A'BC
=>AB(C+C') + A'B'(C+C') + A'BC
=>AB(1) + A'B'(1) + A'BC
=>AB + A'B' + A'BC
=>AB + A'(B' + BC)
=>AB + A'(B' + B)(B' + C)
=>AB + A'(1)(B' + C)
=>AB + A'(B' + C) ------------- Answer

Hereby u are request to help me in question 2 to 8 which are as following:

2. BC + AC' + AB + BCD to four literals
3. [(CD)' +A]' + A + CD + AB
4. (A + C + D)(A + C + D')(A + C' + D)(A + B')

B: First take the complement and then reduce the expression to minimum literals possible:

1. (BC' + A'D)(AB' + CD')
2. B'D + A'BC' + ACD + A'BC
3. [(AB)'A][(AB)'B]
4. AB' + C'D'

I will highly appriciate your kind help. thanks

thingmaker3

Joined May 16, 2005
5,073
Can someone help me
Yes!
and solve this for me.
No!

Please list the work you yourself have done so far on the questions you have trouble with. We will gladly assist, suggest, teach, point out, encourage, and re-phrase. But will never do your work for you.

Joined Sep 27, 2007
3
Ok here are my try for part B:

1. [(BC' + A'D)(AB' + CD')]' i took the whole complemet
(BC' + A'D)' + (AB' + CD')'
(BC')'.(A'D)' + (AB')'.(CD')'
(B' + C)(A + D') + (A' + B)(C' + D)
B'A + B'D' + AC + CD' + A'C' + A'D + BC' + BD

Now i cant go furture can you help me....

2. [B'D + A'BC' + ACD + A'BC]'
(B'D)'(A'BC')'(ACD)'(A'BC)'
(B + D')(A + B' + C)(A' + C' + D')(A + B' + C')
Then i got hanged

3. [((AB)'A)((AB)'B)]'
((AB)'A)' + ((AB)'B)'
AB + A' + AB + B'
AB + B' +AB + B'
(A + B')(B + B') + (A + A')(A' + B)
(A + B')(1) + (1)(A' + B)
A + B' + A' + B
A + A' + B + B'
1 + 1
1

4. [AB' + C'D']'
(AB')'(C'D')'
(A' + B)(C + D)

Could some one help me in Q 1 and 2 of part B

Joined Sep 27, 2007
3
And for part A Q2, 3, and 4.... JUST GIVE ME A STARTING HINT for two or three steps... then i will try my self the rest

Joined Sep 25, 2007
16
the best soultion to your problem is to read this book :
fundamentals of logic design by Charles H.Roth,Jr.
go to chapter 2 and 3 and in chapter for you will find applications of boolean algebra ...

Dave

Joined Nov 17, 2003
6,960
A point about taking the compliment is that you have to take the double compliment to ensure the function remains the same. Consider the following:

- You have a function A (very simple I know)

- If you take a single compliment you end up with NOT A, which is not the same as A. You have changed the function and therefore the simplification will be different.

- If you take a double the compliment you end up with NOT NOT A, which is just A. You haven't changed the function, however you now have a compliment with which to use De-Morgans theorems.

With that in mind, have another look at your answers to section B. Also you should look at the relevant sections in the e-book for Boolean identities: