ee2 chapter8 applicationsof_booleanalgebra

Nov 20, 2012

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IT2001PAEngineering Essentials (2/2)

Chapter 8 - Applications of Boolean Algebra

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IT2001PA Engineering Essentials (2/2)

Lesson Objectives

Upon completion of this topic, you should be able to:

Apply Boolean algebra to solve combination logic of up to 2 variables.



Specific Objectives

Students should be able to : Determine the output logic expression from a given

logic circuit.

Use Boolean algebra to simplify logic expressions.

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Example 1

Write the Boolean equation for the circuit

Use DeMorgan’s Theorem and then Boolean Algebra rules to simplify the equation. Draw the simplified circuit.

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Example 1 (Solution)

Boolean equation at X, X = (AB)●B

Apply De Morgan’s theorem:

X = (A + B) ● B(Use parentheses, () to maintain proper grouping)

Apply Distributive Law:

X = AB + BB

Apply Boolean Algebra rule:

= AB + 0

= AB

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The simplified equation will be

X=AB

To draw the simplified circuit, first write down the inputs (A and B), and the output (X),

then insert the corresponding gates

Finally, join the gates.

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Example 2

Write the Boolean equation for the circuit.


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X = B●(A+C)+C, Simplifying equation:Boolean Equation at X: X = B●(A + C) + C

Apply Distributive Law: X = BA + BC + C

Since C is common for term 2 and term 3: X = BA + C●(B + 1)

Apply (B + 1) = 1: X = BA + C● 1

Apply C●1 = C: X = BA + C

Apply BA = AB: X = AB + C

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The simplified equation will be:

X = AB + C

To draw the simplified circuit, first write down the inputs (A, B and C),and the output (X),

then insert the corresponding gates.


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Example 3



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X = (A+B)●BC+A Simplifying equation:

Boolean Equation at X: X = (A+B) ● BC + A

Apply Distributive Law: X = ABC + BBC + A

Apply BBC = BC: X = ABC + BC+ A

Since BC is common for term 1 and term 2: X = BC (A +1) + A

Apply (A + 1) = 1: X = BC + A

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X = BC + A




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Example 4



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X = (A+B)●B+(B)+(BC)

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Simplifying equation:

Boolean Equation at X: X = (A + B)●B + B + BC

Apply Distributive Law: X = AB + BB + B + BC

Apply (BB = 0): X = AB + 0 + B + BC

Apply (AB + 0) = AB: X = AB + B + BC

Since B is common for term 1 and term 2:X = B●(A + 1) + BC

Apply (A + 1) = 1: X = B ● 1 + BC

Apply (B●1) = B: X = B + BC

Apply (B +BC) = B+C: X = B + C

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Example 4 (Solution) The simplified equation will be:

X = B + A




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Example 5



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X = AB●(C+D)●AB

Simplifying Equation:

Boolean Equation at X: X= AB●(C+D)●AB

Apply Demorgan’s Thereom: X= AB + (C+D) + AB

Apply C+D = C+D: X= AB + C+D + AB

Apply AB + AB = AB: X= AB + C+D

Apply AB = A + B: X= A + B + C+D

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X = A + B + C + D

To draw the simplified circuit, first write down the inputs (A, B,C and D), and the output (X),



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Summary

Steps taken to simplify a combination logic circuit: Write down the expression of a given Combination Logic

Circuit.

Simplify the expression using Boolean Algebra Theorem.

Draw the Logic Circuit using the simplified Logic Expression.

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