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2.1.6 Boolean Algebra

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Activity 2.1.6 Boolean Algebra
Introduction
Have you ever had an idea that you thought was so unique that when you told
someone else about it, you simply could not believe they thought you were wasting
your time? If so, you know how the mathematician George Boole felt in the 1800s
when he designed a math system that, at the time, had no practical application.
Today, however, his math system is the most important mathematical tool used in the
design of digital logic circuits. Boole introduced the world to Boolean algebra when
he published his work called “An Investigation of the Laws of Thought, on Which Are
Founded the Mathematical Theories of Logic and Probabilities.”
In the same way that normal algebra has rules that allow you to simplify algebraic
expressions, Boolean algebra has theorems and laws that allow you to simplify
expressions used to create logic circuits.
By simplifying the logic expression, we can convert a logic circuit into a simpler
version that performs the same function. The advantage of a simpler circuit is that it
will contain fewer gates, will be easier to build, and will cost less to manufacture.
In this activity you will learn how to apply the theorems and laws of Boolean algebra
to simplify logic expressions and digital logic circuits.
The moral of the story is to keep dreaming. Someday your grandchildren may be
using something that you’re thinking about right now. When your grandparents were
kids, do you think that they imagined someday that we would all have 10,000 songs
in our pockets or a telephone in our backpacks?
Equipment
● Paper & Pencil
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DE – Unit 2 – Lesson 2.1 – Activity 2.1.6 – Boolean Algebra – Page 1
Procedure
Using the theorems and laws of Boolean algebra, simplify the following logic
expressions. Note the Boolean theorem/law used at each simplification step. Be
sure to put your answer in Sum-Of-Products (SOP) form.
1.
F = A (A + AB)
F = AA + AB
F = 0 + AB
F = AB
2.
F = YZ (X + X) + XYZ
F = YZ (1) + XYZ
F = YZ + XYZ
F = Z(Y + YX)
F = Z(Y + X)
F = YZ + XZ
3.
F = J(K + K)
F = J(1)
F=J
4.
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DE – Unit 2 – Lesson 2.1 – Activity 2.1.6 – Boolean Algebra – Page 2
F = (1)(AB + ABC)
F = AB(1 + C)
F = AB(1)
F = AB
5.
F = XX + XY + YX + YY
F = X + XY + YX + 0
F = X + X (Y + Y)
F = X + X(1)
F=X+X
F=X
6.
F = JK + LJ + LK + JK
F = JK + LJ + LK
F = LJ + LK
F=L
7.
F = RS + RS + RT + SS + SU
F = RS + RT + 0 + SU
F = RS + RT + SU
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DE – Unit 2 – Lesson 2.1 – Activity 2.1.6 – Boolean Algebra – Page 3
8.
F = (1 + M)(1 + M)(N + M)
F = (1M)(1M)(N + M)
F = 1MN + M
F = MN
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DE – Unit 2 – Lesson 2.1 – Activity 2.1.6 – Boolean Algebra – Page 4
Almost as important as being able to use the laws of Boolean algebra (i.e.,
associative, commutative, or distributive) to simplify logic expressions, it is also
critical that you are able to identify them. Identify the law of Boolean algebra upon
which the following equalities are based.
1.
2.
= COMMUTATIVE
= ASSOCIATIVE
3.
= ASSOCIATIVE
4.
5.
= DISTRIBUTIVE
= DISTRIBUTIVE
Brew Circuit
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DE – Unit 2 – Lesson 2.1 – Activity 2.1.6 – Boolean Algebra – Page 5
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DE – Unit 2 – Lesson 2.1 – Activity 2.1.6 – Boolean Algebra – Page 6
Conclusion
1. Describe the process that you would use to simplify a logic expression using
Boolean algebra.
Use the Boolean Algebra to combine like terms, then to simplify the expression.
How do you know when you are finished simplifying and have arrived at the
simplest equation?
Once the expression is fully condensed, leaving no similar letters, there will be no
more Boolean theorems left.
2. Other than using Boolean algebra, how could you prove that two circuits are
equivalent?
Truth Tables are another way to prove that two circuits are equivalent.
3. If you worked for a company that manufactured the coffee vending machine that
used the poorly designed circuit, how much money would your new design save
the company annually if each GATE cost 15¢ and the company made 500,000
vending machines per year.
The company would be saving $450,000 annually.
As an experienced engineer, you earn $75 per hour. The total redesign took you
two hours (including a coffee break). What would the company’s
Rate-Of-Investment (ROI) be on your time?
Was it a good investment?
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DE – Unit 2 – Lesson 2.1 – Activity 2.1.6 – Boolean Algebra – Page 7
No
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DE – Unit 2 – Lesson 2.1 – Activity 2.1.6 – Boolean Algebra – Page 8
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