1 ch. 4 boolean algebra and logic simplification boolean operations and expressions laws and rules...
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Ch. 4 Boolean Algebra and Logic Simplification
Boolean Operations and Expressions Laws and Rules of Boolean Algebra Boolean Analysis of Logic Circuits Simplification Using Boolean Algebra Standard Forms of Boolean Expressions Truth Table and Karnaugh Map Programmable Logic: PALs and GALs Boolean Expressions with VHDL
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2Introduction
Boolean Algebra• George Boole(English mathematician), 1854
“An Investigation of the Laws of Thought, on Which Are Founded the Mathematical Theories of Logic and Probabilities”
Boolean Algebra {(1,0), Var, (NOT, AND, OR), Thms}
• Mathematical tool to expression and analyze digital (logic) circuits
• Claude Shannon, the first to apply Boole’s work, 1938
– “A Symbolic Analysis of Relay and Switching Circuits” at MIT
• This chapter covers Boolean algebra, Boolean expression and its evaluation and simplification, and VHDL program
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Boolean functions : NOT, AND, OR, exclusive OR(XOR) : odd function exclusive NOR(XNOR) : even function(equivalence)
Basic Functions
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• ANDZ=X Y or Z=XY Z=1 if and only if X=1 and Y=1, otherwise
Z=0
• ORZ=X + YZ=1 if X=1 or if Y=1, or both X=1and Y=1.
Z=0 if and only if X=0 and Y=0
• NOTZ=X orZ=1 if X=0, Z=0 if X=1
Basic Functions (계속 )
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5Basic Functions (계속 )
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6Boolean Operations and Expressions
•Boolean Addition– Logical OR operation
Ex 4-1) Determine the values of A, B, C, and D that make the sum term A+B’+C+D’
Sol) all literals must be ‘0’ for the sum term to be ‘0’
A+B’+C+D’=0+1’+0+1’=0 A=0, B=1, C=0, and D=1
•Boolean Multiplication– Logical AND operation
Ex 4-2) Determine the values of A, B, C, and D for AB’CD’=1
Sol) all literals must be ‘1’ for the product term to be ‘1’
AB’CD’=10’10’=1 A=1, B=0, C=1, and D=0
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7Basic Identities of Boolean Algebra
The relationship between a single variable X, its complement X, and the binary constants 0 and 1
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8Laws of Boolean Algebra
•Commutative Lawthe order of literals does not matter–A + B = B + A
–A B = B A
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•Associative Lawthe grouping of literals does not matter–A + (B + C) = (A + B) + C (=A+B+C)
–A(BC) = (AB)C (=ABC)
Laws of Boolean Algebra (계속 )
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•Distributive Law : A(B + C) = AB + AC
A
B
C
X
YX=Y
Laws of Boolean Algebra (계속 )
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(A+B)(C+D) = AC + AD + BC + BD
A
BCD
XY X=Y
Laws of Boolean Algebra (계속 )
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A+0=A
In math if you add 0 you have changed nothing in Boolean Algebra ORing with 0 changes nothing
A
X X=A+0=A
Rules of Boolean Algebra
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A+1=1
ORing with 1 must give a 1 since if any input is 1 an OR gate will give a 1
A
XX=A+1=1
Rules of Boolean Algebra ( 계속 )
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A•0=0
In math if 0 is multiplied with anything you get 0. If you AND anything with 0 you get 0
A
XX=A0 = 0
Rules of Boolean Algebra ( 계속 )
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A•1 =AANDing anything with 1 will yield the anything
A
XX=A1=A
A
Rules of Boolean Algebra ( 계속 )
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A+A = A
ORing with itself will give the same result
A
A
X
A=A+A =A
Rules of Boolean Algebra ( 계속 )
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A+A’=1
Either A or A’ must be 1 so A + A’ =1
A
A’
XX=+A’=1
Rules of Boolean Algebra ( 계속 )
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A•A = A
ANDing with itself will give the same result
A
A
XA=AA=A
Rules of Boolean Algebra ( 계속 )
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A•A’ =0
In digital Logic 1’ =0 and 0’ =1, so AA’=0 since one of the inputs must be 0.
A
A’
XX=AA’=0
Rules of Boolean Algebra ( 계속 )
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A = (A’)’If you not something twice you are back to the
beginning
A
XX=(A’)’=A
Rules of Boolean Algebra ( 계속 )
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A
B
X
A + AB = A
Rules of Boolean Algebra ( 계속 )
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A + A’B = A + B If A is 1 the output is 1 If A is 0 the output is B
AB
XY X=Y
Rules of Boolean Algebra ( 계속 )
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A
B
C
X
Y
(A + B)(A + C) = A + BC
Rules of Boolean Algebra ( 계속 )
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•DeMorgan’s TheoremF(A,A, , + , 1,0) = F(A, A, + , ,0,1)
– (A • B)’ = A’ + B’ and (A + B)’ = A’ • B’– DeMorgan’s theorem will help to simplify digital circuits using NORs and NANDs his theorem states
DeMorgan’s Theorems
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Look at (A +B +C + D)’ = A’ • B’ • C’ • D’
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Ex 4-3) Apply DeMorgan’s theorems to (XYZ)’ and (X+Y+Z)’
Sol) (XYZ)’=X’+Y’+Z’ and (X+Y+Z)’=X’Y’Z’
Ex 4-5) Apply DeMorgan’s theorems to (a) ((A+B+C)D)’ (b) (ABC+DEF)’ (c) (AB’+C’D+EF)’
Sol) (a) ((A+B+C)D)’= (A+B+C)’+D’=A’B’C’+D’(b) (ABC+DEF)’=(ABC)’(DEF)’=(A’+B’+C’)(D’+E’+F’)(c) (AB’+C’D+EF)’=(AB’)’(C’D)’(EF)’=(A’+B)(C+D’)(E’+F’)
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28Boolean Analysis of Logic Circuits
•Boolean Expression for a Logic Circuit
Figure 4-16 A logic circuit showing the development of the Boolean expression for the output.
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•Constructing a Truth Table for a Logic Circuit– Convert the expression into the min-terms containing all the input literals
– Get the numbers from the min-terms – Putting ‘1’s in the rows corresponding to the min-terms and ‘0’s in the remains
Ex) A(B+CD)=AB(C+C’) (D+D’) +A(B+B’)CD =ABC(D+D’) +ABC’(D+D’) +ABCD+AB’CD =ABCD+ABCD’+ABC’D+ABC’D’ +ABCD+AB’CD =ABCD+ABCD’+ABC’D+ABC’D’ +AB’CD =m11+m12+m13+m14+m15=(11,12,13,14,15)
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30Truth Table from Logic Circuit
Input OutputA B C D A(B+C
D)0 0 0 0 00 0 0 1 00 0 1 0 00 0 1 1 00 1 0 0 00 1 0 1 00 1 1 0 00 1 1 1 01 0 0 0 01 0 0 1 01 0 1 0 01 0 1 1 11 1 0 0 11 1 0 1 11 1 1 0 11 1 1 1 1
A(B+CD)=m11+m12+m13+m14+m15 =(11,12,13,14,15)
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Ex 4-8) Using Boolean algebra, simplify this expression
AB+A(B+C)+B(B+C)Sol) AB+AB+AC+BB+BC =B(1+A+A+C)
+AC=B+AC
Simplification Using Boolean Algebra
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Ex 4-9) Simplify the following Boolean expression(AB’(C+BD)+A’B’)C
Sol) (AB’C+AB’BD+A’B’)C=AB’CC+A’B’C=(A+A’)B’C=B’C
Ex 4-10) Simplify the following Boolean expressionA’BC+AB’C’+A’B’C’+AB’C+ABC
Sol) (A+A’)BC+(A+A’)B’C’+AB’C=BC+B’C’+AB’C =BC+B’(C’+AC)=BC+B’(C’+A)=BC+B’C’+AB’
Ex 4-11) Simplify the following Boolean expression(AB +AC)’+A’B’C
Sol) (AB)’(AC)’+A’B’C=(A’+B’)(A’+C’)+A’B’C=A’+A’B’ +A’C’+B’C+A’B’C =A’(1+B’+C’+B’C)+B’C=A’+B’C’
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33Standard Forms of Boolean Expressions
•The Sum-of-Products(SOP) FormEx) AB+ABC, ABC+CDE+B’CD’
•The Product-of-Sums(POS) FormEx) (A+B)(A+B+C), (A+B+C)(C+D+E)(B’+C+D’)
•Principle of Duality : SOP POS
•Domain of a Boolean ExpressionThe set of variables contained in the expressionEx) A’B+AB’C : the domain is {A, B, C}
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• Implementation of a SOP ExpressionAND-OR logic
•Conversion of General Expression to SOP FormA(B+CD)=AB +ACD
Ex 4-12) Convert each of the following expressions to SOP form: (a) AB+B(CD+EF) (b) (A+B)(B+C+D)Sol) (a) AB+B(CD+EF)=AB+BCD+BEF
(b) (A+B)(B+C+D)=AB+AC+AD+ BB+BC+BD =B(1+A+C+D)+ AC+AD=B+AC+AD
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35Standard SOP Form (Canonical SOP Form)
– For all the missing variables, apply (x+x’)=1 to the AND terms of the expression
– List all the min-terms in forms of the complete set of variables in ascending order
Ex 4-13) Convert the following expression into standard SOP form: AB’C+A’B’+ABC’DSol) domain={A,B,C,D}, AB’C(D’+D)+A’B’(C’+C)(D’+D)+ABC’D =AB’CD’+AB’CD+A’B’C’D’+A’B’C’D+A’B’CD’+A’B’CD+ABC’D =1010+1011+0000+0001+0010+0011+1101 =0+1+2+3+10+11+13 = (0,1,2,3,10,11,13)
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36Product-of-Sums Form
• Implementation of a POS ExpressionOR-AND logic
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37Standard POS Form (Canonical POS Form)
– For all the missing variables, apply (x’x)=0 to the OR terms of the expression
– List all the max-terms in forms of the complete set of variables in ascending order
Ex 4-15) Convert the following expression into standard POS form: (A+B’+C)(B’+C+D’)(A+B’+C’+D)Sol) domain={A,B,C,D}, (A+B’+C)(B’+C+D’)(A+B’+C’+D) =(A+B’+C+D’D)(A’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)=(0100) )(0101)(0110)(1101)= (4,5,6,13)
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38Converting Standard SOP to Standard POS
Step 1. Evaluate each product term in the SOP expression. Determine the binary numbers that represent the product terms
Step 2. Determine all of the binary numbers not included in the evaluation in Step 1
Step 3. Write in equivalent sum term for each binary number Step 2 and expression in POS form
Ex 4-17) Convert the following SOP to POSSol) SOP=
A’B’C’+A’BC’+A’BC+AB’C+ABC=0+2+3+5+7 =(0,2,3,5,7)
POS=(1)(4)(6) = (1, 4, 6) (=(A+B+C’)(A’+B+C)(A’+B’+C))
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39Boolean Expressions and Truth Tables
•Converting SOP Expressions to Truth Table FormatEx 4-18) A’B’C+AB’C’+ABC =(1,4,7)
InputsA B C
OutputX
Product Term
0 0 0 00 0 1 1 A’B’C0 1 0 00 1 1 01 0 0 1 AB’C’1 0 1 01 1 0 01 1 1 1 ABC
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• Converting POS Expressions to Truth Table Format
Ex 4-19) (A+B+C)(A+B’+C)(A+B’+C’)(A’+B+C’)(A’+B’+C) = (000)(010)(011)(101)(110) = (0,2,3,5,6)
InputsA B C
OutputX Sum Term
0 0 0 0 A+B+C0 0 1 10 1 0 0 A+B’+C0 1 1 0 A+B’+C’1 0 0 11 0 1 0 A’+B+C’1 1 0 0 A’+B’+C1 1 1 1
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Ex 4-20) Determine standard SOP and POS from the truth tableSol) (a) Standard SOP F=A’BC+AB’C’+ABC’+ABC(b) Standard POS F=(A+B+C)(A+B+C’)(A+B’+C)
(A’+B+C’)
InputsA B C
OutputX
0 0 0 00 0 1 00 1 0 00 1 1 11 0 0 11 0 1 01 1 0 11 1 1 1
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Boolean Expression
Truth Table
Logic Diagram
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43Karnaugh Map
•Simplification methods–Boolean algebra(algebraic method)–Karnaugh map(map method))
XY+XY=X(Y+Y)=X
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Three- and Four-input Kanaugh mapsGray code
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F(X,Y,Z)=m(0,1,2,6) =(XY+YZ)=X’Y’ + YZ’
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Example) F(X,Y,Z)=m(2,3,4,5) =XY+XY
0 1 3 2
4 5 7 6
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Four-Variable Map16 minterms : m0 ~ m15
Rectangle group – 2-squares(minterms) : 3-literals product term
– 4-squares : 2-literals product term– 8-squares : 1-literals product term– 16-squares : logic 1
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F(W, X,Y,Z)=m(0,2,7,8,9,10,11) = WX’ + X’Z’ + W’XYZ
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54Karnaugh Map SOP Minimization
•Mapping a Standard SOP Expression
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Ex 4-21) Ex 4-22)
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•Mapping a Nonstandard SOP Expression– Numerical Expression of a Nonstandard Product Term
Ex 4-23) A’+AB’+ABC’A’ AB’ ABC’000 100 110001 101010011
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Ex 4-24) B’C’+AB’+ABC’+AB’CD’+A’B’C’D+AB’CDB’C’ AB’ ABC’ AB’CD’ A’B’C’D AB’CD0000 1000 1100 1010 0001 10110001 1001 11011000 10101001 1011
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58Karnaugh Map Simplification of SOP Expressions
• Group 2n adjacent cells including the largest possible number of 1s in a rectangle or square shape, 1<=n
• Get the groups containing all 1s on the map for the expression
• Determine the minimum SOP expression form map
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Ex 4-26) F=B+A’C+AC’D
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Ex 4-27) (a) AB+BC+A’B’C’ (b) B’+AC+A’C’ (c) A’C’+A’B+AB’D (d) D’+BC’+AB’C
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Ex 4-28) Minimize the following expressionAB’C+A’BC+A’B’C+A’B’C’+AB’C’
Sol) B’+A’C
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Ex 4-29) Minimize the following expression
B’C’D’+A’BC’D’+ABC’D’+A’B’CD+AB’CD+A’B’CD’+A’BCD’ +ABCD’+AB’CD’
Sol) D’+B’C
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63Mapping Directly from a Truth Table
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64Don’t Care Conditions
•it really does not matter since they will never occur(its output is either ‘0’ or ‘1’)
•The don’t care terms can be used to advantage on the Karnaugh map
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65Karnaugh Map POS Minimization
•Use the Duality PrincipleF(A,A, , + , 1,0) F*(A,A, + , ,0,1)
SOP POS
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Ex 4-30) (A’+B’+C+D)(A’+B+C’+D’)(A+B+C’+D) (A’+B’+C’+D’)(A+B+C’+D’)
Sol)
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Ex 4-31) (A+B+C)(A+B+C’)(A+B’+C)(A+B’+C’)(A’+B’+C)
Sol) (0+0+0)(0+0+1)(0+1+0)(0+1+1)(1+1+0)=A(B’+C) AC+AB’=A(B’+C)
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Ex 4-32) (B+C+D)(A+B+C’+D)(A’+B+C+D’)(A+B’+C+D)(A’+B’+C+D)
Sol) (B+C+D)=(A’A+B+C+D)=(A’+B+C+D)(A+B+C+D)(1+0+0+0)(0+0+0+0)(0+0+1+0)(1+0+0+1)(0+1+0+0)
(1+1+0+0) F=(C+D)(A’+B+C)(A+B+D)
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69Converting Between POS and SOP Using the K-map
Ex 4-33) (A’+B’+C+D)(A+B’+C+D)(A+B+C+D’)(A+B+C’+D’) (A’+B+C+D’)(A+B+C’+D)
Sol)
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71Five/Six –Variable K-Maps
•Five Variable K-Map : {A,B,C,D,E}
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
16 17 19 18
20 21 23 22
28 29 31 30
24 25 27 26
00 01 11 10
00
01
11
10
BCDE A=0
A=1
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•Six Variable K-Map : {A,B,C,D,E,F}
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
16 17 19 18
20 21 23 22
28 29 31 30
24 25 27 26
00 01 11 10
00
01
11
10
CDEF
00
10 01
11
AB
32 33 35 34
36 37 39 38
44 45 47 46
40 41 43 42
48 49 51 50
52 53 55 54
60 61 62 63
56 57 59 58
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Ex 4-34)Sol) A’D’E’+B’C’D’+BCD+ACDE
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74Digital System Application : 7-Segment LED Driver
Seven-Segment LED driver
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A B C D
0 1 3 2
4 5 7 6
12 13 15 14
8 9 11 10
g = m(2,3,4,5,6,8,9) =A+BC’+B’C+CD’CD
AB
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Figure 4-59 Karnaugh map minimization of the segment-a logic expression.
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Figure 4-60 The minimum logic implementation for segment a of the 7-segment display.
End of Ch. 4