electronics 2200-3

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C. E. Stroud Boolean Algebra & SwitchingFunctions (9/07)

1

Boolean Algebra Also known as Switching Algebra

Invented by mathematician George Boole in 1849 Used by Claude Shannon at Bell Labs in 1938

To describe digital circuits built from relays

Digital circuit design is based on

Boolean Algebra

Attributes

Postulates

Theorems

These allow minimization and manipulation of logic

gates for optimizing digital circuits

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2

Boolean Algebra Attributes Binary

A1a: X=0 if X=1 A1b: X=1 if X=0

Complement

aka invert,NOT

A2a: if X=0, X=1

A2b: if X=1, X=0- The tick mark means

complement, invert, orNOT

- Other symbol forcomplement: X=X

XYYX

111

001

010

000

AND operation

A3a: 00=0 A4a: 11=1

A5a: 01=10=0

- The dot means AND

- Other symbol for AND:

XY=XY (no symbol)

OR operation

A3b: 1+1=1 A4b: 0+0=0

A5b: 1+0=0+1=1

- The plus + means OR

X+YYX

111

101

110

000

01

10XX

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3

Boolean Algebra Postulates

Identity Elements P2a: X+0=XP2a: X+0=X

P2b: XP2b: X1=X1=X

Commutativity P3a: X+Y=Y+XP3a: X+Y=Y+X

P3b: XP3b: XY=YY=YXX

Complements

P6a: X+XP6a: X+X=1=1

P6b: XP6b: XXX=0=0

00

00

1111

XX

11

00

0000

XXYY

11

11

0000

XX11

11

00

0000

YYXX XXXXYYXX

001111

000011

001100000000

00

00

11

11XX

11

11

11

00X+YX+Y

11

11

11

00Y+XY+X

11

11

00

00X+0X+0 X+XX+XYYXX

111111

110011

111100

110000

OR operation

AND operation

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4

Boolean Algebra Postulates Associativity

P4a: (X+Y)+Z=X+(Y+Z)P4a: (X+Y)+Z=X+(Y+Z)

P4b: (XP4b: (XY)Y)Z=XZ=X(Y(YZ)Z)

11

11

00

00

00

00

00

00

XXYY

11

00

00

00

00

00

00

00

(X(XY)Y)ZZ

11

11

11

11

11

11

11

00

X+(Y+Z)X+(Y+Z)

0011111111111100

0000001111000011

0000111111110011

0000111111001111

11

11

00

00

YY

11

00

00

00

YYZZ

11

11

11

00

(X+Y)+Z(X+Y)+Z

11

11

11

00

Y+ZY+Z

11

11

00

00

X+YX+Y XX(Y(YZ)Z)ZZXX

111111

000000

001100

000000

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5

Boolean Algebra Postulates Distributivity

P5a: X+(YP5a: X+(YZ) = (X+Y)Z) = (X+Y)(X+Z)(X+Z)

P5b: XP5b: X(Y+Z) = (X(Y+Z) = (XY)+(XY)+(XZ)Z)

11

11

0000

00

00

00

00

XXYY

11

00

1100

00

00

00

00

XXZZ

11

11

1100

00

00

00

00

XXY+Y+

XXZZ

11

11

1100

11

11

11

00

Y+ZY+Z

11

11

1100

00

00

00

00

XX

(Y+Z)(Y+Z)

11

11

1111

11

00

00

00

X+X+

(Y(YZ)Z)

11

00

0000

11

00

00

00

YYZZ

111111111100

111111000011111111110011

111111001111

11

11

00

00

YY

11

00

11

00

X+ZX+Z

11

00

00

00

(X+Y)(X+Y)

(X+Z)(X+Z)

11

11

00

00

X+YX+YZZXX

1111

0000

1100

0000

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6

Boolean Algebra Theorems

Idempotency T1a: X+X=XT1a: X+X=X

T1b: XT1b: XX=XX=X

Null elements T2a: X+1=1T2a: X+1=1

T2b: XT2b: X0=00=0

Involution T3: (XT3: (X))=X=X=X=X

11

00

00

00

XXYY

00

00

00

00

XX00

11

11

11

00

X+YX+Y

00

00

11

11

XX

11

11

00

00

XXXX

11

11

11

11

X+1X+1

11

11

00

00

X+XX+X XXYYXX

111111

110011

001100

000000

OR AND

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Boolean Algebra Theorems Absorption (aka covering)

T4a: X+(XT4a: X+(XY)=XY)=X T4b: XT4b: X(X+Y)=X(X+Y)=X

T5a: X+(XT5a: X+(XY)=X+YY)=X+Y

T5b: XT5b: X(X(X+Y)=X+Y)=XYY

00

00

11

11

XX

00

00

11

00

XXYY

11

11

11

00

X+X+

(X(XY)Y)

11

00

00

00

XXYY

11

00

00

00

XX

(X(X+Y)+Y)

11

11

11

00

X+YX+Y

11

11

00

00

XX

(X+Y)(X+Y)

11

00

11

11

XX+Y+Y

11

11

00

00

X+X+

(X(XY)Y)YYXX

1111

0011

1100

0000

OR AND

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8

Boolean Algebra Theorems

Absorption (aka combining)

T6a: (XT6a: (XY)+(XY)+(XYY)=X)=X

T6b: (X+Y)T6b: (X+Y)(X+Y(X+Y)=X)=X

00

11

00

11

YY

11

00

00

00

XXYY

11

11

00

00

(X+Y)(X+Y)

(X+Y(X+Y))

11

11

11

00

X+YX+Y

11

11

00

00

(X(XY)+Y)+

(X(XYY))

11

11

00

11

X+YX+Y

00

11

00

00

XXYYYYXX

1111

0011

1100

0000

OR AND

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Boolean Algebra Theorems Absorption (aka combining)

T7a: (XT7a: (XY)+(XY)+(XYYZ)=(XZ)=(XY)+(XY)+(XZ)Z)

T7b: (X+Y)T7b: (X+Y)(X+Y(X+Y+Z) = (X+Y)+Z) = (X+Y)(X+Z)(X+Z)

11

11

11

11

11

00

0000

(X+Y)(X+Y)

(X+Y(X+Y+Z)+Z)

11

11

11

11

11

00

1100

X+ZX+Z

11

00

11

00

00

00

0000

XZXZ

11

11

11

00

00

00

0000

(XY)+(XY)+

(XZ)(XZ)

11

11

11

11

11

11

0000

X+YX+Y

11

11

11

11

11

00

1111

X+YX+Y

+Z+Z

11

11

11

11

11

00

0000

(X+Y)(X+Y)

(X+Z)(X+Z)

11

11

11

00

00

00

0000

(XY)+(XY)+

(XY(XYZ)Z)

000000111100

000011000011

110011110011

001100001111

11

11

0000

YY

11

00

0000

XYXY

00

00

0000

XYXYZZ

00

00

1111

YYZZXX

1111

0000

11000000

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10

Boolean Algebra Theorems DeMorgans theorem (very important!) T8a: (X+Y)T8a: (X+Y)= X= XYY

X+Y = XY break (or connect) the bar & change the sign

T8b: (XT8b: (XY)Y)= X= X+Y+Y XY = X+Y break (or connect) the bar & change the sign

Generalized DeMorgans theorem:

GT8a: (X1+X2++Xn-1+Xn)= X1X2Xn-1Xn

GT8b: (X1X2Xn-1Xn)= X1+X2++Xn-1+Xn

00

11

11

11

(X(XY)Y)

00

00

11

11

XX

00

11

00

11

YY

11

00

00

00

XXYY

00

11

11

11

XX+Y+Y

11

11

11

00

X+YX+Y

00

00

00

11

(X+Y)(X+Y)

00

00

00

11

XXYYYYXX

1111

0011

1100

0000

OR AND

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Boolean Algebra Theorems Consensus Theorem

T9a: (XT9a: (XY)+(XY)+(XZ)+(YZ)+(YZ) = (XZ) = (XY)+(XY)+(XZ)Z)

T9b: (X+Y)T9b: (X+Y)(X(X+Z)+Z)(Y+Z) = (X+Y)(Y+Z) = (X+Y)(X(X+Z)+Z)

11

00

11

00

11

11

0000

(X+Y)(X+Y)

(X(X+Z)+Z)

(Y+Z)(Y+Z)

11

00

00

00

11

00

0000

YZYZ

11

11

11

00

11

11

1100

Y+ZY+Z

11

11

00

00

11

00

1100

(XY)+(XY)+

(X(XZ)Z)

11

11

11

11

11

11

0000

X+YX+Y

11

00

11

00

11

11

1111

XX+Z+Z

11

00

11

00

11

11

0000

(X+Y)(X+Y)

(X(X+Z)+Z)

11

11

00

00

11

00

1100

(XY)+(XY)+

(X(XZ)+Z)+

(YZ)(YZ)

110011111100

000000000011

000000110011

001100001111

11

11

0000

YY

11

00

0000

XYXY

00

00

1100

XXZZ

00

11

1111

XXZZXX

1111

0000

11000000

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12

More Theorems? Shannons expansion theorem (also very important!)

T10a:f(X1,X2,,Xn-1,Xn)=(X

1

f(0,X2

,,Xn-1

,Xn

))+(X1

f(1,X2

,,Xn-1

,Xn

)) Can be taken further:

- f(X1,X2,,Xn-1,Xn)= (X1X2f(0,0,,Xn-1,Xn))

+ (X1X2f(1,0,,Xn-1,Xn)) + (X1X2f(0,1,,Xn-1,Xn))

+(X1X2f(1,1,,Xn-1,Xn))

Can be taken even further:- f(X1,X2,,Xn-1,Xn)= (X1X2Xn-1Xnf(0,0,,0,0))

+ (X1X2Xn-1Xnf(1,0,,0,0)) +

+ (X1X2Xn-1Xnf(1,1,,1,1))

T10b:f(X1,X2,,Xn-1,Xn)=(X1+f(0,X2,,Xn-1,Xn))(X1+f(1,X2,,Xn-1,Xn)) Can be taken further as in the case of T10a

Well see significance of Shannons expansion theorem later

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Principle of Duality Any theorem or postulate in Boolean algebra

remains true if:

0 and 1 are swapped, and and + are swapped

- BUT, be careful about operator precedence!!!

Operator precedence order:1) Left-to-right

2) Complement (NOT)

3) AND

4) OR Use parentheses liberally to ensure correct Boolean

logic equation

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Postulates w/ Precedence & Duality

aaaa=0=0a+aa+a=1=16

aa(b+c(b+c) = () = (aab)+(ab)+(acc))a+(ba+(bcc) = () = (a+b)a+b)(a+c(a+c))5

((aab)b)cc==aa(b(bcc))((a+b)+ca+b)+c==a+(b+ca+(b+c))4ab=baa+b=b+a3

aa1=a1=aa+0=aa+0=a2

b. duala. expressionP.

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Theorems w/ Precedence & Duality

f(X)=(x1+f(0,,xn))(x1+f(1,,xn))f(X)=x1f(0,,xn)+x1f(1,,xn)10

(a+b)(a+c)(b+c)= (a+b)(a+c)ab+ac+bc=ab+ac9(ab)=a+b(a+b)=ab8

(a+b)(a+b+c)=(a+b)(a+c)ab+abc=ab+ac7

(a+b)(a+b)=aab+ab=a6a(a+b)=aba+ab=a+b5

a(a+b)=aa+ab=a4

a=a3a0=0a+1=12

aa=aa+a=a1

b. duala. expressionTh.

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16

Switching Functions

For n variables, there are 2npossible

combinations of values From all 0s to all 1s

There are 2 possible values for the output of afunction of a given combination of values ofnvariables

0 and 1

There are 22n

different switching functions forn variables

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Switching Function Examples

n=0 (no inputs) 22n

= 220

= 21 = 2 Output can be either 0 or 1

n=1 (1 input, A) 22n= 22

1= 22 = 4

Output can be 0, 1, A, or A

switch

function

n=0

output

switch

functionn=1

outputAA f0 f1 f2 f3

0 0 1 0 11 0 0 1 1

f0 = 0

f1 = Af2 = A

f3 = 1

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n=2 (2 inputs, A and B) 22n= 22

2= 24 = 16

A B f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f150 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

1 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 11 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

f0 = 0 logic 0

f1 = AB = (A+B) NOT-OR or NOR f2 = AB

f3 = AB+AB = A(B+B) = A invert A

switch

function

n=2

output

A

B

Most frequently used Less frequently used Least frequently used

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2= 24 = 16


0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

1 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 11 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

switch

function

n=2

output

A

B

f4 = AB

f5 = AB+AB = (A+A)B = B invert B

f6 = AB+AB exclusive-OR f7 = AB+AB+AB = A(B+B)+(A+A)B

= A+B = (AB) NOT-AND or NANDMost frequently used Less frequently used Least frequently used

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2= 24 = 16


0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

1 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 11 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

switch

function

n=2

output

A

B

f8 = AB AND

f9 = AB+AB exclusive-NOR

f10 = AB+AB = (A+A)B = B buffer B

f11 = AB+AB+AB = A(B+B)+(A+A)B = A+B


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2= 24 = 16


0 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

1 0 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 11 1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

switch

function

n=2

output

A

B

f12 = AB+AB = A(B+B) = A buffer A

f13 = AB+AB+AB = A(B+B)+AB = A+AB = A+B

f14 = AB+AB+AB = A(B+B)+(A+A)B = A+B OR f15 = AB+AB+AB+AB = A(B+B)+A(B+B)

= A+A = 1 logic 1


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