Boolean Algebra & Switching Functions

C. E. Stroud Boolean Algebra & Switching Functions (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


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, or NOT

- Other symbol for complement: X’=X

X•YYX

111001010000

• AND operation› A3a: 0•0=0› A4a: 1•1=1› A5a: 0•1=1•0=0

- The dot • means AND- Other symbol for AND:

X•Y=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

0110

X’X


3

Boolean Algebra Postulates

• Identity Elements›› P2a: X+0=XP2a: X+0=X›› P2b: XP2b: X••1=X1=X

• Commutativity›› P3a: X+Y=Y+XP3a: X+Y=Y+X›› P3b: XP3b: X••Y=YY=Y••XX

• Complements›› P6a: X+XP6a: X+X’’=1=1›› P6b: XP6b: X••XX’’=0=0

00001111

XX’’

11000000

XX••YY

11110000

XX••11

11000000

YY••XX XX••XX’’YYXX

001111000011001100000000

00001111

XX’’

11111100

X+YX+Y

11111100

Y+XY+X

11110000

X+0X+0 X+XX+X’’YYXX

111111110011111100110000

OR operation

AND operation


4

Boolean Algebra Postulates• Associativity

›› P4a: (X+Y)+Z=X+(Y+Z)P4a: (X+Y)+Z=X+(Y+Z)›› P4b: (XP4b: (X••Y)Y)••Z=XZ=X••(Y(Y••Z)Z)

1111000000000000

XX••YY

1100000000000000

(X(X••Y)Y)••ZZ

1111111111111100

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

0011111111111100

0000001111000011

0000111111110011

0000111111001111

11

110000YY

11

000000

YY••ZZ

11

111100

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

11

111100

Y+ZY+Z

11

110000

X+YX+Y XX••(Y(Y••Z)Z)ZZXX

111111

000000001100000000


5

Boolean Algebra Postulates• Distributivity

›› P5a: X+(YP5a: X+(Y••Z) = (X+Y)Z) = (X+Y)••(X+Z) (X+Z) ›› P5b: XP5b: X••(Y+Z) = (X(Y+Z) = (X••Y)+(XY)+(X••Z)Z)

1111000000000000

XX••YY

1100110000000000

XX••ZZ

1111110000000000

XX••Y+Y+XX••ZZ

1111110011111100

Y+ZY+Z

1111110000000000

XX••(Y+Z)(Y+Z)

1111111111000000

X+ X+ (Y(Y••Z)Z)

1100000011000000

YY••ZZ

111111111100111111000011111111110011111111001111

11

110000YY

11

001100

X+ZX+Z

11

000000

(X+Y)(X+Y)••(X+Z)(X+Z)

11

110000

X+YX+YZZXX

1111

000011000000


6

Boolean Algebra Theorems

• Idempotency›› T1a: X+X=XT1a: X+X=X›› T1b: XT1b: X••X=XX=X

• Null elements›› T2a: X+1=1T2a: X+1=1›› T2b: XT2b: X••0=00=0

• Involution›› T3: (XT3: (X’’))’’=X=X=X=X

11000000

XX••YY

00000000

XX••00

11111100

X+YX+Y

00001111

XX’’

11110000

XX••XX

11111111

X+1X+1

11110000

X+XX+X XX’’’’YYXX

111111110011001100000000

OR AND


7

Boolean Algebra Theorems• Absorption (aka covering)

›› T4a: X+(XT4a: X+(X••Y)=XY)=X›› T4b: XT4b: X••(X+Y)=X(X+Y)=X›› T5a: X+(XT5a: X+(X’’••Y)=X+YY)=X+Y›› T5b: XT5b: X••(X(X’’+Y)=X+Y)=X••YY

0000

1111

XX’’

0000

1100

XX’•’•YY

1111

1100

X+X+(X(X’•’•Y)Y)

1100

0000

XX••YY

1100

0000

XX••(X(X’’+Y)+Y)

1111

1100

X+YX+Y

1111

0000

XX••(X+Y)(X+Y)

1100

1111

XX’’+Y+Y

1111

0000

X+X+(X(X••Y)Y)YYXX

11110011

11000000

OR AND


8

Boolean Algebra Theorems• Absorption (aka combining)

›› T6a: (XT6a: (X••Y)+(XY)+(X••YY’’)=X)=X›› T6b: (X+Y)T6b: (X+Y)••(X+Y(X+Y’’)=X)=X

0011

0011

YY’’

1100

0000

XX••YY

1111

0000

(X+Y)(X+Y)••(X+Y(X+Y’’))

1111

1100

X+YX+Y

1111

0000

(X(X••Y)+Y)+(X(X••YY’’))

1111

0011

X+YX+Y’’

0011

0000

XX••YY’’YYXX

11110011

11000000

OR AND


9

Boolean Algebra Theorems• Absorption (aka combining)

›› T7a: (XT7a: (X••Y)+(XY)+(X••YY’•’•Z)=(XZ)=(X••Y)+(XY)+(X••Z)Z)›› T7b: (X+Y)T7b: (X+Y)••(X+Y(X+Y’’+Z) = (X+Y)+Z) = (X+Y)••(X+Z)(X+Z)

1111111111000000

(X+Y)(X+Y)••(X+Y(X+Y’’+Z)+Z)

1111111111001100

X+ZX+Z

1100110000000000

XZXZ

1111110000000000

(XY)+(XY)+(XZ)(XZ)

1111111111110000

X+YX+Y

1111111111001111

X+YX+Y’’+Z+Z

1111111111000000

(X+Y)(X+Y)••(X+Z)(X+Z)

1111110000000000

(XY)+(XY)+(XY(XY’’Z)Z)

000000111100000011000011110011110011001100001111

11

110000YY

11

000000

XYXY

00

000000

XYXY’’ZZ

00

001111

YY’’ZZXX

1111

000011000000


10

Boolean Algebra Theorems• DeMorgan’s theorem (very important!)

›› T8a: (X+Y)T8a: (X+Y)’’= X= X’’••YY’’• X+Y = X•Y break (or connect) the bar & change the sign

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

› Generalized DeMorgan’s theorem:• GT8a: (X1+X2+…+Xn-1+Xn)’= X1’•X2’•…•Xn-1’•Xn’• GT8b: (X1•X2•…•Xn-1•Xn)’= X1’+X2’+…+Xn-1’+Xn’

0011

1111

(X(X••Y)Y)’’

0000

1111

XX’’

0011

0011

YY’’

1100

0000

XX••YY

0011

1111

XX’’+Y+Y’’

1111

1100

X+YX+Y

0000

0011

(X+Y)(X+Y)’’

0000

0011

XX’•’•YY’’YYXX

11110011

11000000

OR AND


11

Boolean Algebra Theorems• Consensus Theorem

›› T9a: (XT9a: (X••Y)+(XY)+(X’•’•Z)+(YZ)+(Y••Z) = (XZ) = (X••Y)+(XY)+(X’•’•Z)Z)›› T9b: (X+Y)T9b: (X+Y)••(X(X’’+Z)+Z)••(Y+Z) = (X+Y)(Y+Z) = (X+Y)••(X(X’’+Z)+Z)

1100110011110000

(X+Y)(X+Y)••(X(X’’+Z)+Z)••(Y+Z)(Y+Z)

1100000011000000

YZYZ

1111110011111100

Y+ZY+Z

1111000011001100

(XY)+(XY)+(X(X’’Z)Z)

1111111111110000

X+YX+Y

1100110011111111

XX’’+Z+Z

1100110011110000

(X+Y)(X+Y)••(X(X’’+Z)+Z)

1111000011001100

(XY)+(XY)+(X(X’’Z)+Z)+(YZ)(YZ)

110011111100000000000011000000110011001100001111

11

110000YY

11

000000

XYXY

00

001100

XX’’ZZ

00

111111

XX’’ZZXX

1111

000011000000


12

More Theorems?• Shannon’s expansion theorem (also very important!)

› T10a: f(X1,X2,…,Xn-1,Xn)=(X1’•f(0,X2,…,Xn-1,Xn))+(X1•f(1,X2,…,Xn-1,Xn))• Can be taken further:

- f(X1,X2,…,Xn-1,Xn)= (X1’•X2’•f(0,0,…,Xn-1,Xn))+ (X1•X2’•f(1,0,…,Xn-1,Xn)) + (X1’•X2•f(0,1,…,Xn-1,Xn))+(X1•X2•f(1,1,…,Xn-1,Xn))

• Can be taken even further:- f(X1,X2,…,Xn-1,Xn)= (X1’•X2’•…•Xn-1’•Xn’•f(0,0,…,0,0))

+ (X1•X2’•…•Xn-1’•Xn’•f(1,0,…,0,0)) + …+ (X1•X2•…•Xn-1•Xn•f(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

• We’ll see significance of Shannon’s expansion theorem later


13

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-right2) Complement (NOT)3) AND4) OR

• Use parentheses liberally to ensure correct Boolean logic equation


14

Postulates w/ Precedence & Duality

aa••aa’’=0=0a+aa+a’’=1=16aa••(b+c(b+c) = () = (aa••b)+(ab)+(a••cc))a+(ba+(b••cc) = () = (a+b)a+b)••(a+c(a+c) ) 5((aa••b)b)••cc==aa••(b(b••cc))((a+b)+ca+b)+c==a+(b+ca+(b+c))4a••b=b••aa+b=b+a3aa••1=a1=aa+0=aa+0=a2

b. duala. expressionP.


15

Theorems w/ Precedence & Duality

f(X)=(x1+f(0,…,xn))(x1’+ f(1, …,xn))f(X)=x1’f(0,…,xn)+x1f(1, …,xn)10(a+b)(a’+c)(b+c)= (a+b)(a’+c)ab+a’c+bc=ab+a’c9(ab)’=a’+b’(a+b)’=a’b’8(a+b)(a+b’+c)=(a+b)(a+c)ab+ab’c=ab+ac7(a+b)(a+b’)=aab+ab’=a6a(a’+b)=aba+a’b=a+b5a(a+b)=aa+ab=a4

a’’=a3a•0=0a+1=12a•a=aa+a=a1

b. duala. expressionTh.


16

Switching Functions• For n variables, there are 2n possible

combinations of values› From all 0s to all 1s

• There are 2 possible values for the output of a function of a given combination of values of nvariables› 0 and 1

• There are 22n different switching functions for n variables


17

Switching Function Examples

• n=0 (no inputs) 22n = 220 = 21 = 2› Output can be either 0 or 1

• n=1 (1 input, A) 22n = 221 = 22 = 4› Output can be 0, 1, A, or A’

switchfunction

n=0

output

switchfunction

n=1

outputA A f0 f1 f2 f30 0 1 0 11 0 0 1 1

f0 = 0f1 = A’f2 = Af3 = 1


18

Switching Function Examples• n=2 (2 inputs, A and B) 22n = 222 = 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 10 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 11 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 0f1 = A’B’ = (A+B)’ NOT-OR or NORf2 = A’Bf3 = A’B’+A’B = A’(B’+B) = A’ invert A

switchfunction

n=2

outputA

B

Most frequently used Less frequently used Least frequently used


19



switchfunction

n=2

outputA

B

f4 = AB’f5 = A’B’+AB’ = (A’+A)B’ = B’ invert Bf6 = A’B+AB’ exclusive-ORf7 = A’B’+A’B+AB’ = A’(B’+B)+(A’+A)B’

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


20



switchfunction

n=2

outputA

B

f8 = AB ANDf9 = A’B’+AB exclusive-NORf10 = A’B+AB = (A’+A)B = B buffer Bf11 = A’B’+A’B+AB = A’(B’+B)+(A’+A)B = A’+B

Most frequently used Less frequently used Least frequently used


21



switchfunction

n=2

outputA

B

f12 = AB’+AB = A(B’+B) = A buffer A f13 = A’B’+AB’+AB = A(B’+B)+A’B’ = A+A’B’ = A+B’f14 = A’B+AB’+AB = A(B’+B)+(A’+A)B = A+B ORf15 = A’B’+A’B+AB’+AB = A’(B’+B)+A(B’+B)

= A’+A = 1 logic 1Most frequently used Less frequently used Least frequently used

Boolean Algebra & Switching Functions

Documents

Transcript of Boolean Algebra & Switching Functions