Chapter 2: Boolean Algebra and Logic

Gates

CPIT 210

Outlines

1. Introduction2. Basic Boolean operations 3. Digital Logic Gates4. Universal Gates5. Boolean Expressions6. Boolean Algebra7. Simplification of Boolean Functions8. Canonical and Standard Forms

Introductionn In chapter 1, we’ve talked about how

arbitrary numbers can be represented using just the two binary values 1 and 0.

n Now we’ll interpret voltages as the logical values “true” and “false” instead. We’ll show:n How logical functions can be defined for

expressing computationsn How to build circuits that implement our functions

in hardware

Introductionn Earlier, we used electrical voltages to represent

two discrete values 1 and 0, from which binary numbers can be formed.

n It’s also possible to think of voltages as representingtwo logical values, true and false.

n For simplicity, we often still write digits instead:n 1 is truen 0 is false

n We will use this interpretation along with special operations to design functions and hardware for doing arbitrary computations.

Introduction

n Computers take inputs and produce outputs, just like functions in math!

n We can represent logical functions in two analogous ways too:n A finite, but non-unique Boolean expression. n A truth table, which will turn out to be unique and finite.

Basic Boolean operations

n There are three basic operations for logical values:

x y xy0 0 00 1 01 0 01 1 1

x y x+y0 0 00 1 11 0 11 1 1

x x’0 11 0

AND(product)oftwoinputs

OR(sum)oftwoinputs

NOT(complement)ononeinput

xy,orx•y x+y x’

Operation:

Expression:

Truthtable:

Digital Logic Gates

n A gate is an electronic device that produces a result based on two or more input values.n In reality, gates consist of one to six transistors,

but digital designers think of them as a single unit.

n Integrated circuits contain collections of gates suited to a particular purpose.

Digital Logic Gates

n The three simplest gates are the AND, OR, and NOT gates.

n They correspond directly to their respective Boolean operations, as you can see by their truth tables.

Digital Logic Gatesn Another very useful gate is the exclusive OR

(XOR) gate. n The output of the XOR operation is true only when

the values of the inputs differ.

Note the special symbol Åfor the XOR operation.

Digital Logic Gatesn The exclusive NOR (XNOR) gate. n The output of the XNOR operation is true only

when the values of the inputs similar.

x y XNOR0 0 10 1 01 0 01 1 1

xy

x Å yx � y

x y + x y

Digital Logic Gates

n NAND and NOR are two very important gates. Their symbols and truth tables are shown at the right.

L

J

Multiple Gates

n NAND and NOR are known as universal gates because they are inexpensive to manufacture and any Boolean function can be constructed using only NAND or only NOR gates.

n NAND and NOR are two very important gates. Their symbols and truth tables are shown at the right.

Universal Gates

n Universal Gates: A universal gate is a gate which can implement any Boolean function without need to use any other gate type.

n Proving NAND gate is universaln NAND gate is called universal gate

Universal Gates

n Proving NOR gate is universaln NOR gate is called universal gate

Universal Gates

Boolean Expressionsn We can use these basic operations to form more complex expressions:

f(x,y,z) = (x + y’)z + x’

n Some terminology and notation:n f is the name of the function.n (x,y,z) are the input variables, each representing 1 or 0. Listing the inputs is

optional, but sometimes helpful.n A literal is any occurrence of an input variable or its complement. The

function above has four literals: x, y’, z, and x’.n Precedence are important, but not too difficult.

n NOT has the highest precedence, followed by AND, and then OR.n Fully parenthesized, the function above would be kind of messy:

f(x,y,z) = (((x +(y’))z) + x’)

Boolean Expressionsn A truth table shows all possible inputs and outputs of a function.n Remember that each input variable represents either 1 or 0.

n Because there are only a finite number of values (1 and 0).n A function with n variables has 2n possible combinations of inputs.

n Inputs are listed in binary order—in this example, from 000 to 111.

x y z f(x,y,z)0 0 0 10 0 1 10 1 0 10 1 1 11 0 0 01 0 1 11 1 0 01 1 1 1

f(0,0,0) =(0+1)0+1 =1f(0,0,1) =(0+1)1+1 =1f(0,1,0) =(0+0)0+1 =1f(0,1,1) =(0+0)1+1 =1f(1,0,0) =(1+1)0+0 =0f(1,0,1) =(1+1)1+0 =1f(1,1,0) =(1+0)0+0 =0f(1,1,1) =(1+0)1+0 =1

f(x,y,z)=(x+y’)z+x’

Boolean Expressionsn Any Boolean expression can be converted into a circuit by combining basic

gates in a relatively straightforward way.n The diagram below shows the inputs and outputs of each gate.n The precedence are explicit in a circuit. We have to make sure that the

hardware does operations in the right order.

(x+y’)z+x’

Boolean Expressions

n After finding the circuit inputs and outputs, you can come up with either an expression or a truth table to describe what the circuit does.

n You can easily convert between expressions and truth tables.

Findthecircuit’sinputsandoutputs

FindaBooleanexpression

forthecircuit

Findatruthtableforthecircuit

Boolean Algebran Now we’ll learn how to how use Boolean algebra to simplify Booleans

expressions.n Last time, we saw this expression and converted it to a circuit:

(x+y’)z+x’Canwemakethiscircuit“better”?• Cheaper:fewergates• Faster:fewerdelaysfrominputstooutputs

Boolean Algebra

n Normal mathematical expressions can be simplified using the laws of algebra

n For binary systems, we can use Boolean algebra, which is superficially similar to regular algebra

n There are many differences, due ton having only two values (0 and 1) to work withn having a complement operationn the OR operation is not the same as addition

Boolean Algebran A Boolean algebra requires

n A set of elements B, which needs at least two elements (0 and 1)n Two binary (two-argument) operations OR and ANDn A unary (one-argument) operation NOTn The axioms below must always be true

1. x+0=x 2. x•1=x 3. x+1=1 4. x•0=0 5. x+x=x 6. x•x=x 7. x+x’=1 8. x•x’=0 9. (x’)’=x 10. x+y=y+x 11. xy=yx Commutative12. x+(y+z)=(x+y)+z 13. x(yz)=(xy)z Associative14. x(y+z)=xy+xz 15. x+yz=(x+y)(x+z) Distributive16. (x+y)’=x’y’ 17. (xy)’=x’+y’ DeMorgan’s

Boolean Algebran Here are some more useful laws. Notice the duals

again!

1. x+xy=x 4. x(x+y)=x2. xy+xy’=x 5. (x+y)(x+y’)=x3. x+x’y=x+y 6. x(x’+y)=xyxy+x’z+yz=xy+x’z (x+y)(x’+z)(y+z)=(x+y)(x’+z)

Boolean Algebran The complement of a function always outputs 0 where the original

function outputted 1, and 1 where the original produced 0.n In a truth table, we can just exchange 0s and 1s in the output

column(s)

f(x,y,z)=x(y’z’+yz)

x y z f(x,y,z)0 0 0 10 0 1 10 1 0 10 1 1 11 0 0 01 0 1 01 1 0 11 1 1 0

x y z f’(x,y,z)0 0 0 00 0 1 00 1 0 00 1 1 01 0 0 11 0 1 11 1 0 01 1 1 1

Boolean Algebra

n You can use DeMorgan’s law to keep “pushing” the complements inwards

f(x,y,z)=x(y’z’+yz)

f’(x,y,z)=(x(y’z’+yz))’=x’+(y’z’+yz)’=x’+(y+z)(y’+z’)

Simplification of Boolean Functionsn We can now start doing some simplifications

x’y’+xyz+x’y=x’(y’+y)+xyz [Distributive;x’y’+x’y=x’(y’+y)]=x’•1+xyz [Axiom7;y’+y=1]=x’+xyz [Axiom2;x’•1=x’]=(x’+x)(x’+yz) [Distributive]=1• (x’+yz) [Axiom7;x’+x=1]=x’+yz [Axiom2]

Simplification of Boolean Functions

n Here are two different but equivalent circuits.

n In general the one with fewer gates is “better”:n It costs less to buildn It requires less powern But we had to do some

work to find the second form

Canonical and Standard Forms

n Minterms and Maxterms

n Sum-of-Minterm (SOM) Canonical Form

n Product-of-Maxterm (POM) Canonical Form

n Representation of Complements of Functions

n Conversions between Representations

§ Literal:§ The appearance of a variable or its complement.

§ Product Term:§ one or more literals connected by AND operators.

§ Sum Term: § one or more literals connected by OR operators.

Canonical and Standard Forms

(1) x¢yz¢ + x¢yz + xy¢z¢ + xy¢z + xyz 5 sum terms, 15 literals

(2) x¢y + xy¢ + xyz 3 sum terms, 7 literals

(3) (x¢+y) ( x+y¢) (x+z) 3 product terms, 6 literals

(4) (x¢+y )( x+y¢) (y+z) 3 product terms, 6 literals

n Minterms are AND sum terms with every variable present in either true or complemented form.

n Example: Two variables (X and Y) produce2 x 2 = 4 combinations:

XY (both normal)XY’ (X normal, Y complemented)X’Y (X complemented, Y normal)X’Y’ (both complemented)

n Thus there are four minterms of two variables.

Canonical and Standard Forms

n Maxterms are OR product terms with every variable in true or complemented form.

n Example: Two variables (X and Y) produce2 x 2 = 4 combinations:

X+Y (both normal)X+Y’(x normal, y complemented)X’+Y(x complemented, y normal)X’+Y’(both complemented)

Canonical and Standard Forms

Thus there are four maxterms of two variables.

n Two variable minterms and maxterms.

n The minterm mi should evaluate to 1 for each combination of x and y.

n The maxterm is the complement of the minterm

x y Index Minterm Maxterm0 0 0 m0 = x y M0 = x + y0 1 1 m1 = x y M1 = x + y1 0 2 m2 = x y M2 = x + y1 1 3 m3 = x y M3 = x + y

Canonical and Standard Forms

Canonical and Standard Forms

M3 = x + y + zm3 = x y z3110M4 = x + y + zm4 = x y z4001M5 = x + y + zm5 = x y z5101M6 = x + y + zm6 = x y z6011

1

100y

1

000x

1

010z

M7 = x + y + zm7 = x y z7

M2 = x + y + zm2 = x y z2M1 = x + y + zm1 = x y z1M0 = x + y + zm0 = x y z0

MaxtermMintermIndex

Canonical and Standard Forms

n All variables should be present in a minterm or maxtermand should be listed in the same order (usually alphabetically)

n Example: For variables a, b, c:n Maxterms (a + b + c), (a + b + c) are in standard ordern However, (b + a + c) is NOT in standard order

(a + c) does NOT contain all variablesn Minterms (a b c) and (a b c) are in standard ordern However, (b a c) is not in standard ordern (a c) does not contain all variables

Sum-Of-Minterm (SOM)

x y z F Minterm0 0 0 00 0 1 1 m1 = x y z0 1 0 00 1 1 01 0 0 01 0 1 01 1 0 1 m6 = x y z1 1 1 1 m7 = x y z

F = m1 + m6 + m7 = ∑ (1, 6, 7) = x y z + x y z + x y z

Focus on the ‘1’ entries

+ a b c d

SOM Examplesn F (a, b, c, d) = ∑(2, 3, 6, 10, 11)n F (a, b, c, d) = m2 + m3 + m6 + m10 + m11

n G (a, b, c, d) = ∑(0, 1, 12, 15)n G (a, b, c, d) = m0 + m1 + m12 + m15

+ a b c da b c d + a b c d+ a b c d + a b c d

+ a b c da b c d + a b c d

Product-Of-Maxterm (POM)

x y z F Maxterm0 0 0 10 0 1 10 1 0 0 M2 = (x + y + z)0 1 1 11 0 0 0 M4 = (x + y + z)1 0 1 11 1 0 0 M6 = (x + y + z)1 1 1 1

Focus on the ‘0’ entries

F = M2·M4·M6 = ∏ (2, 4, 6) = (x+y’+z) (x’+y+z) (x’+y’+z)

n F (a, b, c, d) = ∏(1, 3, 6, 11)n F (a, b, c, d) = M1 · M3 · M6 · M11

n G (a, b, c, d) = ∏(0, 4, 12, 15)n G (a, b, c, d) = M0 · M4 · M12 · M15

POM Examples

(a+b+c+d) (a+b+c+d) (a+b+c+d) (a+b+c+d)

(a+b+c+d) (a+b+c+d) (a+b+c+d) (a+b+c+d)

Observations

n We can implement any function by "ORing" the mintermscorresponding to the ‘1’ entries in the function table. A mintermevaluates to ‘1’ for its corresponding entry.

n We can implement any function by "ANDing" the maxtermscorresponding to ‘0’ entries in the function table. A maxtermevaluates to ‘0’ for its corresponding entry.

n The same Boolean function can be expressed in two canonical ways: Sum-of-Minterms (SOM) and Product-of-Maxterms(POM).

n If a Boolean function has fewer ‘1’ entries then the SOM canonical form will contain fewer literals than POM. However, if it has fewer ‘0’ entries then the POM form will have fewer literals than SOM.

Converting to SOM Formn A function that is not in the Sum-of-Minterms form can be

converted to that form by means of a truth tablen Consider F = y’ + x’ z’

x y z F Minterm0 0 0 1 m0 = x y z0 0 1 1 m1 = x y z0 1 0 1 m2 = x y z0 1 1 01 0 0 1 m4 = x y z1 0 1 1 m5 = x y z1 1 0 01 1 1 0

F = ∑(0, 1, 2, 4, 5) =

m0 + m1 + m2 + m4 + m5

x y z + x y z + x y z +

x y z + x y z

Converting to POS Formn A function that is not in the Product-of-Minterms form can

be converted to that form by means of a truth tablen Consider again: F = y + x z

x y z F Minterm0 0 0 10 0 1 10 1 0 10 1 1 0 M3 = (x+y+z)1 0 0 11 0 1 11 1 0 0 M6 = (x+y+z)1 1 1 0 M7 = (x+y+z)

F = ∏(3, 6, 7) =

M3 · M6 · M7 =

(x+y+z) (x+y+z) (x+y+z)

Conversions Between Canonical Forms

F = m1+m2+m3+m5+m7 = ∑(1, 2, 3, 5, 7) = x y z + x y z + x y z + x y z + x y zF = M0 · M4 · M6 = ∏(0, 4, 6) = (x+y+z)(x+y+z)(x+y+z)

x y z F Minterm Maxterm0 0 0 0 M0 = (x + y + z)0 0 1 1 m1 = x’ y’ z0 1 0 1 m2 = x’ y’ z0 1 1 1 m3 = x y z1 0 0 0 M4 = (x’ + y + z)1 0 1 1 m5 = x y’ z1 1 0 0 M6 = (x’ + y’ + z)1 1 1 1 m7 = x y z

Algebraic Conversion to SOM§ Expand all terms first to explicitly list all minterms§ AND any term missing a variable v with (v + v)§ Example 1: f = x + x y (2 variables)

f = x (y + y) + x yf = x y + x y + x yf = m3 + m2 + m0 = ∑(0, 2, 3)

§ Example 2: g = a + b c (3 variables)g = a (b + b)(c + c) + (a + a) b cg = a b c + a b c + a b c + a b c + a b c + a b cg = a b c + a b c + a b c + a b c + a b cg = m1 + m4 + m5 + m6 + m7 = ∑ (1, 4, 5, 6, 7)

Algebraic Conversion to POM§ Expand all terms first to explicitly list all maxterms§ OR any term missing a variable v with v · v§ Example 1: f = x + x y (2 variables)

Apply 2nd distributive law:f = (x + x) (x + y) = 1 · (x + y) = (x + y) = M1

§ Example 2: g = a c + b c + a b (3 variables)g = (a c + b c + a) (a c + b c + b) (distributive)g = (c + b c + a) (a c + c + b) (x + x y = x + y)g = (c + b + a) (a + c + b) (x + x y = x + y)g = (a + b + c) (a + b + c) = M5 . M2 = ∏ (2, 5)

Standard SOP§ A Simplification Example:

§ Writing the minterm expression:F = A B C + A B C + A B C + ABC + ABC

§ Simplifying:F = A B C + A (B C + B C + B C + B C)F = A B C + A (B (C + C) + B (C + C))F = A B C + A (B + B)F = A B C + AF = B C + A

§ Simplified F contains 3 literals compared to 15

)7,6,5,4,1()C,B,A(F S=