I am having trouble simplifying the following Boolean Expression in Sum of Products form:

AB' (C+D') + AB'C'D

= AB'C + AB'D' + AB'C'D
= ?

Can anybody help me please?

Thanks in advance.

 

Originally posted by gino@Sep 17 2005, 04:35 AM
I am having trouble simplifying the following Boolean Expression in Sum of Products form:

AB' (C+D') + AB'C'D

= AB'C + AB'D' + AB'C'D
= ?

Can anybody help me please?

Thanks in advance.

[post=10401]Quoted post[/post]​

Go from step 1 and use DeMorgan's instead. Don't distribute the AB', that should put you on the right track.

 

Originally posted by Brandon@Sep 18 2005, 02:03 AM
Go from step 1 and use DeMorgan's instead. Don't distribute the AB', that should put you on the right track.

[post=10404]Quoted post[/post]​

I assume you apply DeMorgan's law to the first part of the equation?

e.g. [AB'(C+D')]'

I got the answer:

A' + B + C'D

Not sure if its correct.

Thanks for your assistance.

 

Originally posted by gino@Sep 17 2005, 04:35 AM
I am having trouble simplifying the following Boolean Expression in Sum of Products form:

AB' (C+D') + AB'C'D

= AB'C + AB'D' + AB'C'D
= ?

Can anybody help me please?

Thanks in advance.

[post=10401]Quoted post[/post]​

Start again and factor out the AB': AB' (C + D' + C'D). Do you know the rule A + A'B = A + B? See how you can use it to simplify the expression in parentheses? C + C'D = C + D. Then you have AB'(C + D + D'). Another basic rule: A + A' =1. This leads us to AB'(C + 1). And of course C + 1 = 1. Hence: AB' (C + 1) = AB'. The original expression thus simplifies to AB'. You could also use a four-variable Karnaugh map to quickly obtain the same result (maybe you haven't done K-maps yet...).

You can utilize another rule: A + AB = A. Can you see how this rule allows you to get to the result immediately?

 

I didn't know the rule A + A'B = A + B. Is this derived from some Boolean Identities?

Not sure about using A + AB = A.

I follow your workings though. Thanks so much.

And we haven't covered Kmaps yet.

 

Originally posted by gino@Sep 18 2005, 12:03 AM
I didn't know the rule A + A'B = A + B. Is this derived from some Boolean Identities?

Not sure about using A + AB = A.

I follow your workings though. Thanks so much.

And we haven't covered Kmaps yet.

[post=10420]Quoted post[/post]​

To understand A + AB = A, factor as is regular algebra: A + AB = A(1 + B), where 1 + B = 1 (if an OR gate has a high on one input and logic level B on the other input, the output is high, regardless of what B is). Therefore, A + AB = A(1 + B) = A(1) = A. The last part follows from the fact that when you have a high on one input of an AND and logic level A on the other input, the output is A, regardless of what A is.

Now extend the rule: A + AB + AC + AD + .... = A!!! You can keep factoring terms as above, one after another, and you end up with just A. So consider your original expression: AB'C + AB'D' + AB'C'D. Let X = AB': So, you have XC + XD' + XC'D. Just as A is the only survivor of the above expression, so X is all that survives of this one, where X is your AB', of course.

The other rule I showed you, A + A'B = A + B, is a bit more involved to derive, but I'm sure you'll see it in class soon enough! Good luck.

 

Can anyone plz help me out to simplify boolean function
1) (A + B)' + (A' + B')'

2) y(wz' + wz) + xy

3) xy + x'z + yz


plzz do help me as my exams are supposed to start n plzz do tell me any site wherein i can refer n learn more bout system architecture.
thanks in advance
shradha

 

3) xy + x'z + yz

You can you the consensus theorem to simplify this expression. YZ is a redundant term. In the terms XY + X'Z, ignore the Xs (since there is X + X') and look at the other variables. The redundant term for these two is YZ. Similarly, if you had AB + A'C, the redundant term would be BC. I have found that this theorem is quite helpful. You can also add redundant terms to any expression in order to simplify it. Sometimes adding a redundant term will allow you to cancel another term, and then you can also cancel the redundant term as well.

 

I'm trying to prove that the term AB is redundant in the function below.

F = BC + AB + AC'
F = A'BC + ABC + AB + AC' {multiplying (A' +A) by the first term}
F = A'BC + AB(C + 1) + AC'
F = A'BC + AB + AC'
F = B(A'C + A) + AC'
F = B(A + C) + AC'

I dont seem to be getting any closer, am I?

 

gino,

AB'C + AB'D' + AB'C'D = AB'C(D+D')+AB'D'(C+C')+AB'C'D

=AB'CD+AB'CD'+AB'CD'+AB'C'D'+AB'C'D

=AB'CD+AB'CD'+AB'C'D'+AB'C'D

11 10 8 9

So the SOP's are 8,9,10,11 . Now you can use a 4 variable K-map, or do as I did and use the Quine-McCluskey tabulation method.

The expression simplifies to AB'

You can tell that is the correct answer because no matter which of the 4 values you can assign to CD, the expression is always AB'. For instance C=0, D=0. The expression is 0+0+AB'+0 . Ratch

 

shradha,

(A + B)' + (A' + B')' = A'B'+AB by DeMorgan's theorem. It cannot be simplified further.

y(wz' + wz) + xy = wyz'+wyz+xy

= wyz'(x+x')+wyz(x+x')+xy(w+w')

=wxyz'+wx'yz'+wxyz+wx'yz+wxy+w'xy

=wxyz'+wx'yz'+wxyz+wx'yz+wxyz+wxyz'+w'xyz+w'xyz'

=wxyz'+wx'yz'+wxyz+wx'yz+w'xyz+w'xyz'

SOP'S = 14 10 15 11 7 6

Using a 4 variable K-map or the Quine McClusky tabulation method we get xy+wy .

xy + x'z + yz = xy(z+z')+x'z(y+y')+yz(x+x')

=xyz+xyz'+x'yz+x'y'z+xyz+x'yz

=xyz+xyz'+x'yz+x'y'z

SOP = 7 6 3 1

Using a 3 variable K-map or the Quine McCluskey tabulation method we get x'z+xy .

Ratch

 

Stacey,

I'm trying to prove that the term AB is redundant in the function below.

The best way is to simplify it.

F = A'BC + ABC + AB + AC' {multiplying (A' +A) by the first term}

Continuing further F = A'BC + ABC + ABC + ABC' + ABC'+AB'C'

= AB'C + ABC +ABC' + AB'C'

SOP = 5 7 6 4

Using a 3 variable K-map or the Quine-McCluskey tabulation method we get F = A . As you can see, no matter what values we assign to B or C, the value is always A .

I dont seem to be getting any closer, am I?

Now we have.

Ratch

 

I am having trouble simplifying the following Boolean Expression in Sum of Products form:

AB' (C+D') + AB'C'D

= AB'C + AB'D' + AB'C'D
= ?

Can anybody help me please?

Thanks in advance.

Doing it after your solution

Now,
AB'(D'+C'D)+AB'C [A+B'C=(A+B')(A+C)]
AB'[(D+D')(D'+C')]+AB'C
AB'[D'+C']+AB'C
AB'(C'+C)+AB'D'
AB'+AB'D'
AB'(1+D') [1+A'=A]
=AB'

 

Can anyone plz help me out to simplify boolean function
1) (A + B)' + (A' + B')'

2) y(wz' + wz) + xy

3) xy + x'z + yz


plzz do help me as my exams are supposed to start n plzz do tell me any site wherein i can refer n learn more bout system architecture.
thanks in advance
shradha

1) Remember that
A'+B'=(AB)'
(A+B)'=A'B'
Now
(A + B)' + (A' + B')'
[(A+B)(A'+B')]'
[AB+A'B]'
B'


2) y(wz' + wz) + xy
z+z'=1
yz+xy
y(x+y)


3) xy + x'z + yz
xy+x'z+(x+x')yz
x'z(1+y)+xy(1+z)
[1+x=1]
x'z+xy