Boolean Algebra
9/15/09 - L5 Boolean Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
1
Class 5 outline
Boolean Algebra
Basic Boolean Equations
Multiple Level Logic Representation
Basic Identities
Algebraic Manipulation
Complements and Duals
Material from section 2-2 of text
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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History
George Boole
2 November 1815 Lincoln Lincolnshire, England
– 8 December 1864 Ballintemple, Ireland
Professor at Queens College, Cork, Ireland
“spring of 1847 that he put his ideas into the
pamphlet called Mathematical Analysis of
Logic.” from wikipedia.com
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Basic Boolean Equations
For the basic gates/functions
AND
3 input gate
4 input gate
OR
Z=AB
X=CDE
Y=FGHK
Z=A+B
Y=F+G+H+K
4 input gate
NOT
Z=A
Y = (F G H K)
9/15/09 - L5 Boolean
Algebra
actually 2 level logic
Copyright 2009 - Joanne DeGroat, ECE, OSU
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2 Level Logic
Consider the following logic equation
Z(A,B,C,D) = A B + C D
The Z(A,B,C,D) means that the output is a
function of the four variables within the ().
The AB and CD are terms of the expression.
This form of representing the function is an
algebraic expression.
For this function to be True, either both A AND
B are True OR both C AND D are True.
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Truth table expression
Just like we had the
truth tables for the
basic functions, we can
also construct truth
tables for any function.
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Examples of Boolean Equations
Some examples
F = AB + CD + BD’
Y = CD + A’B’
SUM = AB + A Cin + B Cin
P = A0A1A2A3A4B0B1B2B3B4 + …
Equations can be very complex
Usually desire a minimal expression
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Basic Identities of Boolean Algebra
1. X + 0 = X
3. X + 1 = 1
5. X + X = X
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Algebra
2. X · 1 = X
4. X · 0 = 0
6. X · X = X
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Basic Identities (2)
7. X + X’ = 1
9. (X’)’ = X
9/15/09 - L5 Boolean
Algebra
8. X · X’ = 0
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Basic Properties (Laws)
Commutative
10. X + Y = Y + X
Associative
12. X+(Y+Z)=(X+Y)+Z
Distributive
9/15/09 - L5 Boolean
Algebra
11. X · Y = Y · X
Associative
14. X(Y+Z) =XY+XZ
AND distributes over
OR
Commutative
13. X(YZ) = (XY)Z
Distributive
15. X+YZ=(X+Y)(X+Z)
OR distributes over
AND
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Basic Properties (2)
DeMorgan’s Theorem
Very important in simplifying equations
16. (X + Y)’ = X’ · Y’
17. (XY)’ = X’ + Y’
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Simplify, simplify
These properties (Laws and Theorems) can be used to
simplify equations to their simplest form.
Simplify
9/15/09 - L5 Boolean
Algebra
F=X’YZ+X’YZ’+XZ
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Affect on implementation
F = X’YZ + X’YZ’ + XZ
Reduces to F = X’Y + XZ
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Other examples
Examples from the text
1. X + XY = X·1 + XY = X(1+Y) = X·1 = X
2
14
3
2
2. XY+XY’ = X(Y + Y’) = X·1 = X
Use
Use
14
7
2
3. X+X’Y = (X+X’)(X+Y) = 1· (X+Y) = X+Y
9/15/09 - L5 Boolean
Algebra
Use
15
Copyright 2009 - Joanne DeGroat, ECE, OSU
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2
14
Further Examples
Examples from the text
4. X· (X+Y)=X·X+X·Y=
X+XY=X(1+Y)=X·1=X
Use
14
6
14
3
2
5. (X+Y) ·(X+Y’)=XX+XY’+XY+YY’=
X+XY’+XY+0=X(1+Y’+Y)=X·1=X
by a slightly different reduction
6. X(X’+Y) = XX’+XY = 0 + XY = XY
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Consensus Theorem
The Theorem gives us the relationship
XY + X’Z + YZ = XY + X’Z
Proof is on page 47.
Note that in doing the reduction the first step is to
and in a 1 to the YZ term. That 1 is in the form,
(X+X’).
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Application of Consensus Theorem
Consider (page 47 of text)
(A+B)(A’+C) = AA’ + AC + A’B + BC
= AC + A’B + BC
= AC + A’B
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Complement of a function
In real implementation sometimes the
complement of a function is needed.
Have
9/15/09 - L5 Boolean
Algebra
F=X’YZ’+X’Y’Z
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Duals
What is meant by the dual of a function?
The dual of a function is obtained by
interchanging OR and AND operations and
replacing 1s and 0s with 0s and 1s.
Shortcut to getting function complement
Starting with the equation on the previous slide
Generate the dual F=(X’+Y+Z’)(X’+Y’+Z)
Complement each literal to get:
F’=(X+Y’+Z)(X+Y+Z’)
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Getting XILINX software
Go to www.xilinx.com
In the upper line you have
Sign in
Language Documentation Downloads
Choose Downloads
On the right side of the page you will see
“Logic design tools”
Choose ISE WebPackTM
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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Class 5 assignment
Covered section 2-2
Problems for hand in
Problems for practice
2-7
2-2a,b,c 2-6b,c,d
Reading for next class: section 2-3
9/15/09 - L5 Boolean
Algebra
Copyright 2009 - Joanne DeGroat, ECE, OSU
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