- #1
Master Boolean's Algebra with SOP: Simplifying F(X, Y) to XY' + XY
- Thread starter desmond iking
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- Algebra
Both of these terms have an X and a Y, so the first two terms should also have an X and a Y, but they don't. The first and third terms, X'Y' and XY, are not included in the final result because they have a different arrangement of X and Y.
![IMG_20141215_212410[1].jpg](/data/attachments/59/59697-3449da1a7e627dde7414800501b155b5.jpg){kind=small}
- #2
jedishrfu
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Use boolean algebra to reduce it. In boolean algebra X+X' = 1 and use algebra to extract common factors.
Is this a homework assignment?
Is this a homework assignment?
- #3
desmond iking
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no, i am studying myself during the holiday.jedishrfu said:
- #4
jedishrfu
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Did my advice make sense?
Boolean algebra rules are the same as ordinary algebra with the added concepts X=X'=1 and X.X'=0 and X=0 means X=false and X=1 means X=true. Engineers start with a truth table for a circuit, construct boolean expressions, reduce them down using boolean algebra ruls and then construct a circuit from the reduced expressions.
Boolean algebra rules are the same as ordinary algebra with the added concepts X=X'=1 and X.X'=0 and X=0 means X=false and X=1 means X=true. Engineers start with a truth table for a circuit, construct boolean expressions, reduce them down using boolean algebra ruls and then construct a circuit from the reduced expressions.
- #5
desmond iking
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i only managed to reduce it to XY +Y' +X'Yjedishrfu said:
here's my working:
(X'Y') + (X'Y) + (XY') + (XY)
=XY +Y'(X' +X ) +X'Y
= XY +Y' +X'Y
- #6
jedishrfu
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what about the 1st and 3rd term? they have a common Y.
- #7
desmond iking
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jedishrfu said:
continue form the previous post , i still didnt get the ans
XY +Y' +X'Y = Y(X +X') +Y' = Y+Y'
- #8
NascentOxygen
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Hi Desmond. No one can explain it. It's a misprint.desmond iking said:
It is just the exclusive OR function, X'Y + XY'. It doesn't simplify, except you can write it as X ⨁ Y.
Good luck with your self-study. http://thumbnails112.imagebam.com/37333/0363e9373324851.jpg
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- #9
desmond iking
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I noticed that theF(X , Y) is the outcome of the addition of the last two. *correct me if I m wrong* , why it only involve the addition of last two term? Why not also the first two terms??
- #10
NascentOxygen
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It's a misprint. F(X,Y) is true when either of these is true: X'Y or XY'desmond iking said:
FAQ: Master Boolean's Algebra with SOP: Simplifying F(X, Y) to XY' + XY
What is Master Boolean's Algebra with SOP?
Master Boolean's Algebra with SOP (Sum of Products) is a method used to simplify logic expressions in Boolean algebra. It involves breaking down a complex expression into smaller terms, adding them together, and then simplifying them using Boolean laws and identities.
What is F(X, Y) in the expression XY' + XY?
In the expression XY' + XY, F(X, Y) represents the Boolean function that we are trying to simplify. In this case, F(X, Y) is equal to XY' + XY.
How do you simplify F(X, Y) to XY' + XY using Master Boolean's Algebra with SOP?
To simplify F(X, Y) to XY' + XY, we first need to break down the expression into smaller terms using Boolean laws and identities. Then, we add together the terms that have the same variables. In this case, both terms have XY as a common factor. Finally, we can simplify the expression to XY' + XY using the distributive property.
Why is simplifying logic expressions important?
Simplifying logic expressions is important because it helps us understand and analyze complex systems and circuits. It also allows us to reduce the number of gates and inputs needed, which can save time and resources in designing and building electronic devices.
Can Master Boolean's Algebra with SOP be used for more complex expressions?
Yes, Master Boolean's Algebra with SOP can be used for more complex expressions with multiple variables and terms. The same principles of breaking down the expression, adding together like terms, and simplifying using Boolean laws and identities still apply, but may require more steps and careful organization of terms.