Lecture 4: More Boolean Algebra - Wayne State University
Transcript of Lecture 4: More Boolean Algebra - Wayne State University
ENGIN112 L6: More Boolean Algebra September 15, 2003
Lecture 4: More Boolean Algebra
Syed M. Mahmud, Ph.D
ECE Department
Wayne State University
Original Source: Prof. Russell Tessier of University of Massachusetts
Aby George of Wayne State University
ENGIN112 L6: More Boolean Algebra September 15, 2003
Overview
° Expressing Boolean functions
° Relationships between algebraic equations, symbols, and truth tables
° Simplification of Boolean expressions
° Minterms and Maxterms
° AND-OR representations
• Product of sums
• Sum of products
ENGIN112 L6: More Boolean Algebra September 15, 2003
Conversion between Boolean Expression and Truth Table
Truth
Table
Boolean
Expression
ENGIN112 L6: More Boolean Algebra September 15, 2003
Truth Table to Expression
° What Boolean function is represented by the following Truth Table? Is it an AND or an OR function?
° The answer is AND because G1=1 only for one combination of x, y and z (x=1, y=1 and z=1).
• Therefore, G1 = xyz
x00001111
y00110011
z01010101
G100000001
ENGIN112 L6: More Boolean Algebra September 15, 2003
Truth Table to Expression
° What about G2 and G3?
° Both G2 and G3 represent AND operations because each one of these two functions is also equal to 1, only for one combination of x, y and z.
° G2=1 for x=1, y=1 and z=0, and G3=1 for x=0, y=1 and z=1.
• Therefore, G2 = xyz’ and G3 = x’yz
x00001111
y00110011
z01010101
G200000010
x00001111
y00110011
z01010101
G300010000
ENGIN112 L6: More Boolean Algebra September 15, 2003
Truth Table to Expression
° If G = G1+G2+G3, as shown in the following Truth Table, then
G = xyz + xyz’ + x’yz
x00001111
y00110011
z01010101
G100000001
G200000010
G300010000
G = G1+G2+G300010011
ENGIN112 L6: More Boolean Algebra September 15, 2003
Truth Table to Expression
° Converting a truth table to an expression
• Each row with output of 1 becomes a product term
• Sum all the product terms together.
x00001111
y00110011
z01010101
G00010011
xyz + xyz’ + x’yz
Any Boolean Expression can be
represented in sum of products form!
G = xyz + xyz' +x'yz
Product TermsSum of 3
Product Terms
ENGIN112 L6: More Boolean Algebra September 15, 2003
Implementation of the SOP Expression
SOP Expression: G = xyz + xyz' +x'yz
A SOP expression is implemented using one OR gate and
multiple AND gates.
This is not a minimum logic.
We will minimize later.
ENGIN112 L6: More Boolean Algebra September 15, 2003
Reducing the expression for G using Boolean Algebra
° G = xyz + xyz' +x'yz
° Step 1: Use Theorum 1 (a + a = a)
• So xyz + xyz’ + x’yz = xyz + xyz + xyz’ + x’yz
° Step 2: Use distributive rule a(b + c) = ab + ac
• So xyz + xyz + xyz’ + x’yz = xy(z + z’) + yz(x + x’)
° Step 3: Use Postulate 3 (a + a’ = 1)
• So xy(z + z’) + yz(x + x’) = xy.1 + yz.1
° Step 4: Use Postulate 2 (a . 1 = a)
• So xy.1 + yz.1 = xy + yz
° Therefore, G = xy + yz
ENGIN112 L6: More Boolean Algebra September 15, 2003
Reduced Hardware Implementation
° Reduced equation requires less hardware!
° Same function implemented!
x y z
x00001111
y00110011
z01010101
G00010011
G = xyz + xyz’ + x’yz = xy + yz
G
x x
xx
ENGIN112 L6: More Boolean Algebra September 15, 2003
Product of Sums (POS) Expression
Any Boolean Expression can be represented in Product
of Sums (POS) form. Let's find a POS expression for G.
We have shown how G was written in a
SOP expression.
Using the same technique we can write a
SOP expression for G'.
G' = x'y'z'+x'y'z+x'yz'+xy'z'+xy'z
Using DeMorgan's Theorem we can
convert the SOP expression for G' into a
POS expression for G
ENGIN112 L6: More Boolean Algebra September 15, 2003
Product of Sums (POS) Expression
G' = x'y'z'+x'y'z+x'yz'+xy'z'+xy'z
G = (G')' = (x'y'z'+x'y'z+x'yz'+xy'z'+xy'z)'
Using DeMorgan G=(x'y'z')'.(x'y'z)'.(x'yz')'.(xy'z')'.(xy'z)'
Using DeMorgan again
G=(x+y+z).(x+y+z').(x+y'+z).(x'+y+z).(x'+y+z')
Sum Terms
POS
Expression
ENGIN112 L6: More Boolean Algebra September 15, 2003
Summary of SOP and POS Expressions for G
SOP Expression:
G = xyz + xyz' +x'yz
POS Expression:
G=(x+y+z)(x+y+z')(x+y'+z)(x'+y+z)(x'+y+z')
Notice, the SOP expression has 3 Product terms and the POS
expression has 5 Sum terms.
There are 8 terms in both SOP and POS expressions together.
Why?
Because G is a function of 3 variable (x, y and z), and 3
variables have 8 different combinations.
• The 1's of G are used for SOP and the 0's of G are used
for POS.
ENGIN112 L6: More Boolean Algebra September 15, 2003
Implementation of the POS Expression
POS Expression:
G=(x+y+z)(x+y+z')(x+y'+z)(x'+y+z)(x'+y+z')
A POS expression is implemented using one AND gate and
multiple OR gates.
ENGIN112 L6: More Boolean Algebra September 15, 2003
Minterms and Maxterms
° The boolean variables and their complements are called Literals. For example, x is a literal, and y' is also a literal.
° The Product Terms of literals are called Minterms. The Sum Terms of literals are called Maxterms.
ENGIN112 L6: More Boolean Algebra September 15, 2003
Representing Functions with Minterms
° A SOP expression is a Sum of Minterms. The Minterms of a function correspond to the 1's of the function.
° Shorthand way to represent functions
x00001111
y00110011
z01010101
G00010011
G = x’yz + xyz’ + xyz
G = m3 + m6 + m7 = Σm(3, 6, 7)
ENGIN112 L6: More Boolean Algebra September 15, 2003
Representing Functions with Maxterms
° A POS expression is a Product of Maxterms. The Maxterms of a function correspond to the 0's of the function.
° Shorthand way to represent functions
G = (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z)(x’+y+z’)
G = M0M1M2M4M5 = ΠM(0,1,2,4,5)
x00001111
y00110011
z01010101
G00010011
ENGIN112 L6: More Boolean Algebra September 15, 2003
Canonical Forms
° The Minterm and Maxterm expressions are also called Canonical Forms
x00001111
y00110011
z01010101
G00010011
G = m3 + m6 + m7 = Σm(3, 6, 7)
G = M0M1M2M4M5 = ΠM(0,1,2,4,5)
Canonical Forms
ENGIN112 L6: More Boolean Algebra September 15, 2003
Conversion Between Canonical Forms
° Easy to convert between Minterm and Maxtermexpressions.
° Since 3 variables can take 8 combinations, if there are NMinterms in the expression for a function, then there will be 8-N Maxterms in the expression for the function.
° Similarly, since 4 variables can take 16 combinations, if there are N Minterms in the expression for a function, then there will be 16-N Maxterms in the expression for the function.
IF F = Σ m(1, 3, 6, 7, 10, 12, 15)
THEN F = Π M(0, 2, 4, 5, 8, 9, 11, 13, 14)
Example:
ENGIN112 L6: More Boolean Algebra September 15, 2003
Representation in Circuits
° All logic expressions can be represented in 2-level format
° Circuits can be reduced to minimal 2-level representation
° Sum of products representations are most common in industry.
ENGIN112 L6: More Boolean Algebra September 15, 2003
Summary
° Truth table, circuit, and boolean expression formats are equivalent
° Easy to translate truth table to SOP and POS representation
° Boolean algebra rules can be used to reduce circuit size while maintaining function
° All logic functions can be made from AND, OR, and NOT