Transcript
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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-right2) Complement (NOT)3) AND4) OR
• Use parentheses liberally to ensure correct Boolean logic equation
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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.
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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.
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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 = 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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
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
C. E. Stroud Boolean Algebra & Switching Functions (9/07)
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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
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
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