Serial trouble with boolean algebra

• May 13, 2018

• #1

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kachuri

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Hi,

Here is the problem statement given in my book.



Here is the solution given in my book to this problem



Could you please help to understand that marked area ? Is the book solution correct ?
I could not understand that marked part . I'm stuck at that part. Need help

 

• May 13, 2018

• #2

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kachuri

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Re: Help needed to solve equation

sorry...understood that part. They are using the definition of "*" operator .

stuck with the next part with this



Note : Is it not possible to edit a post ? There seems no EDIT post button in the forum. How do you edit post ?

 

• May 13, 2018

• #3

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albbg

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Re: Help needed to solve equation

Using "_" to denote the upper bar, your part is _(_A*_B+A*B).
Now, applying De Morgan two times:

_(_A*_B+A*B) = _(_A*_B)*_(A*B)

_(_A*_B)*_(A*B) = (A+B)*(_A+_B)

now expanding:

(A+B)*(_A+_B) =A*_A+A*_B+_A*B+B*_B

But A*_A = 0 and B*_B = 0, then

(A+B)*(_A+_B) =A*_B+_A*B that is the result written into the book.

 

• May 13, 2018

• #4

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kachuri

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Re: Help needed to solve equation

albbg said:

Using "_" to denote the upper bar, your part is _(_A*_B+A*B).
Now, applying De Morgan two times:

_(_A*_B+A*B) = _(_A*_B)*_(A*B)

_(_A*_B)*_(A*B) = (A+B)*(_A+_B)

now expanding:

(A+B)*(_A+_B) =A*_A+A*_B+_A*B+B*_B

But A*_A = 0 and B*_B = 0, then

(A+B)*(_A+_B) =A*_B+_A*B that is the result written into the book.

Click to expand...

this is great. I understand this now.

Thanks for the help. You are very much helpful.


I have one more query.

Can I rearranged input values ? ...I am not sure if this is allowed ....could you please confirm ?

for example...

 

• May 13, 2018

• #5

FvM

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Re: Help needed to solve equation

Read about commutativity of Boolean operators. https://en.m.wikipedia.org/wiki/Boolean_algebra

 

• May 13, 2018

• #6

andre_luis

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Re: Help needed to solve equation

View attachment 146649

Click to expand...

Anyway, irrespective to the order that appears, you should realize that any variable multiplied to its inverse yields False (BB=0) as albbg said above, didn't you read?

 

• May 14, 2018

• #7

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kachuri

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Which Law is this?

Here are some simplification steps of this boolean expression given in my book.

I don't see any such Law in the book ... B+!BA = B+A

Could you please explain how we can write that red marker steps?

Is it correct ?

 

• May 14, 2018

• #8

andre_luis

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Re: Which Law is this ?

I don't see any such Law in the book ... B+!BA = B+A

Click to expand...

Make the K-map of the function for 2 inputs A and B and you will infer this more easily than would achieve by using algebraic boolean.

 

• May 14, 2018

• #9

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Akanimo

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Re: Which Law is this ?

Hi,

If you substitute values for B in B + !BA, you can visualize it as being equivalent to B + A.

Let's start with B=1, then we would have 1+0.A = 1. Here B prevails.

If we substitute B=0, then we get 0+1.A. Here A prevails.

So it is either A prevails or B prevails, hence B OR A.
Hope this helps.

 

• May 14, 2018

• #10

_Eduardo_

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Re: Which Law is this ?

B + !BA = B(1+A) + !BA = B + BA + !BA = B + (B+!B)A = B + A

But is more easy to see with kmap.

 

• May 15, 2018

• #11

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kachuri

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trouble with boolean algebra

Here are some steps given in my book to simplify a boolean expression.

I have some doubts on the red marked steps.

How come they put two same items there? Is it a printing mistake ?

Here is what I'm upto

Could you please tell by which Law book has written same items twice ?

 

• May 15, 2018

• #12

andre_luis

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Regardless of the utility of repeating the same term in this specific case, there is nothing to prevent you from doing so if you want to regroup miniterms into more simplified expressions; I assume you know that the sum of the same value in Boolean algebra is equal to itself.

Keep in mind that in K-map, it is usual to graphically enclose the same miniterms in different groups, and this is likely what may have been done in the book, better saying, could have been done just the same splitting procedure, but now performed in an algebraic fashion.

 

• May 15, 2018

• #13

ads-ee

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It was done so it's more obvious what the next operation is:

Code:

XY + X'Y' + X'YZ
XY + X'Y' +   (X'Y' + X'YZ) - extra parenthesis are mine
XY + X'Y' +   X'(Y' +YZ) - this can be reduced like the previous question you had on A + A'B

BTW that A + A'B is a variation on the Absorptive Law, A + AB = A