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SIMPLIFY THE LOGICAL FUNCTIONFUNCTION
SIMPLIFYING THE LOGIC EQUATIONSIMPLIFYING THE LOGIC EQUATIONSIMPLIFYING THE LOGIC EQUATIONSIMPLIFYING THE LOGIC EQUATION
AA.. DIAGRAMDIAGRAM
Is to illustrate the basic principles of logic orthrough a set / sets..g /DiagramDiagram VennVenn IllustrationIllustration
11.. forfor nn == 11
S
AA’
Cont …..
22.. forfor nn == 22
SA’B’
ABAB’ A’B
Cont …..
22.. forfor nn == 33C’
S
ABC’ABC
AB’C’ A’BC’
AB’C A’BC
A’B’CA’B’C’
Example …..
Specify Boolean function to venn diagram on the shaded areaSpecify Boolean function to venn diagram on the shaded area
S A B
C
B. B. KARNAUGHT MAPKARNAUGHT MAP
Karnaught Map is a method to simplify the logic equations.Karnaught Map with 2 variables f(AB)g p ( )
B’ BA’ A’ B’ A’ BA A B’ A B
Karnaught Map with 3 variables f (ABC)
B’C’ B’C BC BC’B’C’ B’C BC BC’A’ A’B’C’ A’B’C A’BC A’BC’A AB’C’ AB’C ABC ABC’A AB’C’ AB’C ABC ABC’
Cont Cont …..…..
c. Karnaught Map with 4 variabels f(ABCD)
C’D’ C’D CD CD’
A’B’A B
A’B
AB
AB’AB
Cont Cont …..…..
Karnaught Map with 5 variables (ABCDE)
D’E’ D’E DE DE D’E’ D’E DE DE’
A A’
B’C’ B’C’
B’C
BC
B’C
BC
+
BC’ BC’
MAP APPLICATION LOGIC of KARNAUGHT MAP to simplify the equations.
Minimization logic equations in the form kemonisMinimization logic equations in the form kemonisSOP (minterm) with maps karnaughtMinimization of logic equations in the formMinimization of logic equations in the formKemonis POS (MAXTERM) with maps karnaughtForm which can be simplified to eliminate the multi-variable
Simplifying Logic EquationsSimplifying Logic Equations
Canonical SUM REPRESENTATION OF PRODUCT (SOP)Canonical SUM REPRESENTATION OF PRODUCT (SOP)An expression of AND function or method SOP- The series combination of logic
h d f d d b b f- The condition of output is determined by a combination of input - inputExample:pMake the boolean equations and logic circuits from Boolean functions in minterm form as follows:
F (ABC) � (0 3 6 7)F (ABC) = � (0,3,6,7)
AnsAns…..…..
Table expressions for the method SOP / minterm
INPUT OUTPUT
A B C COND AND FUNCTION
0 0 0 1 A’ B’ C’000
001
010
100
A’ B’ C’--0
01
110
010
010
A’ B C-
11
01
10
01
-A B C’
1 1 1 1 A B C
AnsAns ( ( Cont Cont ) ...) ...
Boolean FunctionF = ∑ FiFi
F F F F F F F F= F= F00 + F+ F33 + F+ F6 + F+ F7= = A’B’C’ + A’BC + ABC’ + ABC= A’(B’C’ + BC) + AB(C’ + C)= A (B C + BC) + AB(C + C)= A’(B C) + AB
Logic CircuitLogic Circuitgg
A
BF(ABC) = A(B C) + AB
BC
ExerciseExercise
1. F (ABC) = ∑ ( 0,2,4,5,6 )1. F (ABC) ∑ ( 0,2,4,5,6 )2. F (ABCD) = ∑ ( 0,1,2,4,5,6,9,12,13,14 )
Canonical PRODUCT REPRESENTATION OF SUM (POS)Canonical PRODUCT REPRESENTATION OF SUM (POS)
An expression of the function or method OR POSAn expression of the function or method OR POS- The series combination of logic- The condition of output is determined by a p ycombination of input - inputExample:Make the boolean equations and logic circuits from Boolean functions in the form MAXTERM as follows:
F(ABC) = F(ABC) = ΠΠ ( 0 2 5 7 )( 0 2 5 7 )F(ABC) = F(ABC) = ΠΠ ( 0,2,5,7 )( 0,2,5,7 )
AnsAns …..…..
Expressions Table for method POS / Maxterm
INPUT OUTPUT
A B C COND OR FUNCTION0 0 0 1 A’+B’+C’000
001
010
101
A’+B’+C’- A+B+C’A’+B+C 0
01
110
010
100
A B C - A+B’+C’- A’+B+C
11
01
10
10
A+B’+C- A+B’+C
1 1 1 1 A+B+C
AnsAns ( ( Cont Cont ) ...) ...
Boolean FunctionF = ππ FiFi
F F FF FF FF= F= F00 •• FF2 •• FF5 •• FF7= (= (A’+B’+C’)••(A’+B+C’)••(A+B’+C)••(A+B+C)= (A’+C’) •• (A+C) = (A +C ) •• (A+C) = A’C + AC’= A ⊕ C
Logics GateLogics Gate
F(ABC) A ⊕ CA F(ABC) = A ⊕ CAC
ExerciseExercise
1. F (ABC) = ππ ( 1,3,7 )
2. F (ABCD) = ππ ( 3,8,10,11,15 )
DON’T DON’T CARE CONDITIONCARE CONDITIONDON T DON T CARE CONDITIONCARE CONDITION
Do not Care condition is a condition that can be Do not Care condition is a condition that can be assumed to have state 0 or 1 are also marked with X and to simplify boolean expressions using the map.p y p g pExample:Simplify the Boolean function as follows:
F(A,B,C,D)F(A,B,C,D) == ∑ (( 11,,33,,77,,1111,,1515 ))
WhichWhich hashas don’tdon’t carecare conditioncondition asas followsfollows ::
dd(A,B,C,D)(A,B,C,D) == ∑ ( 0,2,5 )
AnsAns……
C’D’ C’D CD CD’ C’D’ C’D CD CD’
A’B’ X 1 1 X A’B’ X 1 1 X
A’B 0 X 1 0
AB 0 0 1 0
A’B 0 X 1 0
AB 0 0 1 0
OR
AB’ 0 0 1 0
AB 0 0 1 0
AB’ 0 0 1 0
F = CD + A’B’ F = CD + A’B
