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INTRODUCTION TO DIGITAL LOGIC

DeMorgansTheorem

TheoremsofBooleanAlgebra(1)1)A+0=A2)A+1=13)A0=04)A1=A5)A+A=A6)A+A=17)AA=A8)AA=0

TheoremsofBooleanAlgebra(2)9)A=A

10)A+AB=A11)A+AB=A+B

12)(A+B)(A+C)=A+BC13)Commutative:A+B=B+A

AB=BA14)Associative:A+(B+C)=(A+B)+C

A(BC)=(AB)C15)Distributive:A(B+C)=AB+AC

(A+B)(C+D)=AC+AD+BC+BD

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DeMorgansTheorems

16)(X+Y)=X.Y17)(X.Y)=X+Y

Two most important theorems of Boolean Algebra were contributed by De Morgan. Extremely useful in simplifying expression in which product or sum of variables is inverted. The TWO theorems are :

ImplicationsofDeMorgansTheorem

(a) Equivalent circuit implied by theorem (16) (b) Negative- AND (c) Truth table that illustrates DeMorgans Theorem

(a)

(b)

Input Output

X Y X+Y XY

0 0 1 1

0 1 0 0

1 0 0 0

1 1 0 0

(c)

ImplicationsofDeMorgansTheorem

(a) Equivalent circuit implied by theorem (17) (b) Negative-OR (c) Truth table that illustrates DeMorgans Theorem

(a)

(b)

Input Output

X Y XY X+Y

0 0 1 1

0 1 1 1

1 0 1 1

1 1 0 0

(c)

DeMorgansTheoremConversion(1)

Step1:ChangeallORstoANDsandallANDstoOrsStep2:Complementeachindividualvariable(shortoverbar)Step3:Complementtheentirefunction(longoverbars)Step4:EliminateallgroupsofdoubleoverbarsExample:A.B A.B.C =A+B =A+B+C =A+B =A+B+C =A+B =A+B+C =A+B

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DeMorgansTheoremConversion(2)

ABC+ABC (A+B+C)D =(A+B+C).(A+B+C) =(A.B.C)+D=(A+B+C).(A+B+C) =(A.B.C)+D=(A+B+C).(A+B+C) =(A.B.C)+D=(A+B+C).(A+B+C) =(A.B.C)+D

Example:Analyzethecircuitbelow

Y

1. Y=??? 2. Simplify the Boolean expression found in 1

Example:Analyzethecircuitbelow(CONT.)

Followthestepslistbelow(constructingtruthtable) Listalltheinputvariablecombinationsof1and0inbinarysequentially

Placetheoutputlogicforeachcombinationofinput

Baseontheresultfoundwriteoutthebooleanexpression.

Exercises:

Simplify the following Boolean expressions 1. (AB(C + BD) + AB)C 2. ABC + ABC + ABC + ABC + ABC

Write the Boolean expression of the following circuit.

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Standard Forms of Boolean Expressions (1)

Sum of Products (SOP) Products of Sum (POS)

Notes: SOP and POS expression cannot have more than one variable combined in a term with an inversion bar Theres no parentheses in the expression

Standard Forms of Boolean Expressions (2)

Converting SOP to Truth Table Examine each of the products to determine where the product is equal to a 1. Set the remaining row outputs to 0.

Standard Forms of Boolean Expressions (3) Converting POS to Truth Table

Opposite process from the SOP expressions. Each sum term results in a 0. Set the remaining row outputs to 1.

Standard Forms of Boolean Expressions (4)

BCACABCBACBAf ),,(

632),,( mmmCBAf

)6,3,2(),,( mCBAf

The standard SOP Expression All variables appear in each product term. Each of the product term in the expression is called as minterm. Example:

In compact form, f(A,B,C) may be written as

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Standard Forms of Boolean Expressions (5)

)()()(),,( CBACBACBACBAf

The standard POS Expression All variables appear in each product term. Each of the product term in the expression is called as maxterm. Example:

541),,( MMMCBAf In compact form, f(A,B,C) may be written as

)5,4,1(),,( MCBAf

Standard Forms of Boolean Expressions (6)

CBACBACABABCCBAf ),,(

)()()()(),,( CBACBACBACBACBAf

Example:

Convert the following SOP expression to an equivalent POS expression:

Example:

Develop a truth table for the expression:

TheKarnaughMAP(KMap)

K-Map (1)

Karnaugh Mapping is used to minimize the number of logic gates that are required in a digital circuit.

This will replace Boolean reduction when the circuit is large.

Write the Boolean equation in a SOP form first and then place each term on a map.

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The map is made up of a table of every possible SOP using the number of variables that are being used.

If 2 variables are used then a 2X2 map is used If 3 variables are used then a 4X2 map is used If 4 variables are used then a 4X4 map is used If 5 Variables are used then a 8X4 map is used

K-Map (2)

K-Map SOP Minimization

A

A

B B

Notice that the map is going false to true, left to right and top to bottom

The upper right hand cell is A B if X= A B then put an X in that cell

A

A

B B

1

This show the expression true when A = 0 and B = 0

0 1

2 3

2 Variables K-Map (1) If X=AB + AB then put an X in both of these cells

A

A

B B

1

1

From Boolean reduction we know that A B + A B = B

From the Karnaugh map we can circle adjacent cell and find that X = B

A

A

B B

1

1

2 Variables K-Map (2)

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3 Variables K-Map (1)

Gray Code

00 A B

01 A B

11 A B

10 A B

0 1

C C

0 1

2 3

6 7

4 5

X=ABC+ABC+ABC+ABC

Gray Code

00 A B

01 A B

11 A B

10 A B

0 1

C C

Each 3 variable term is one cell on a 4 X 2 Karnaugh map

1 1

1 1

3 Variables K-Map (2)

X=ABC+ABC+ABC+ABC

Gray Code

00 A B

01 A B

11 A B

10 A B

0 1

C C One simplification could be

X = A B + A B

1 1

1 1

3 Variables K-Map (3) X=ABC+ABC+ABC+ABCGray Code

00 A B

01 A B

11 A B

10 A B

0 1

C C Another simplification could be

X = B C + B C

A Karnaugh Map does wrap around

1 1

1 1

3 Variables K-Map (4)

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X=ABC+ABC+ABC+ABC

Gray Code

00 A B

01 A B

11 A B

10 A B

0 1

C C

The Best simplification would be

X = B

1 1

1 1

3 Variables K-Map (4) On a 3 Variables K-Map

One cell requires 3 Variables Two adjacent cells require 2 variables Four adjacent cells require 1 variable Eight adjacent cells is a 1

4 Variables K-Map

Gray Code

00 A B

01 A B

11 A B

10 A B

0 0 0 1 1 1 1 0

C D C D C D C D

0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

Gray Code

00 A B

01 A B

11 A B

10 A B

0 0 0 1 1 1 1 0

C D C D C D C D

1

1

1

1

1

1

X = ABD + ABC + CD

Now try it with Boolean reductions

Simplify: X = A B C D + A B C D + A B C D + A B C D + A B C D + A B C

D

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On a 4 Variables K-Map

One Cell requires 4 variables Two adjacent cells require 3 variables Four adjacent cells require 2 variables Eight adjacent cells require 1 variable Sixteen adjacent cells give a 1 or true

Simplify:

Z=BCD+BCD+CD+BCD+ABC

Gray Code

00 A B

01 A B

11 A B

10 A B

00 01 11 10

C D C D C D C D

1 1 1 1

1 1

1 1

1 1

1

1

Z = C + A B + B D

Simplify using K-Map (1)

Firstly, change the circuit to an SOP expression

Y= A + B + B C + ( A + B ) ( C + D)

Y = A B + B C + A B ( C + D )

Y = A B + B C + A B C + A B D

Y = A B + B C + A B C A B D

Y = A B + B C + (A + B + C ) ( A + B + D)

Y = A B + B C + A + A B + A D + B + B D + AC + C D

Simplify using K-Map (2)

SOP expression

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Gray Code

00 A B

01 A B

11 A B

10 A B

00 01 11 10

C D C D C D C D

1 1

1 1

1 1 1 1

Y = 1

1 1 1 1

1 1

1 1

Simplify using K-Map (3)

K-Map POS Minimization

3 Variables K-Map (1)

Gray Code

0 0

0 1

1 1

1 0

0 1 AB C

0 1

2 3

6 7

4 5

3 Variables K-Map (2)

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0 0

0 1

1 1

1 0

0 0 0 1 1 1 1 0 A B C D

0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

4 Variables K-Map (1) 4 Variables K-Map (2)

4 Variables K-Map (3) Mapping a Standard SOP expression Example: Answer:

Mapping a Standard POS expression Example: Using K-Map, convert the following standard POS expression into a minimum SOP expression Answer: Y = AB + AC or standard SOP:

K-Map - Examples

DCBADCBABCDACDBADCBADCBAY CDADBY

ABCCBACBAY

)( CBAY

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K-Map with Dont Care Conditions (1)

Input Output

Example :

3 variables with output dont care (X)

K-Map with Dont Care Conditions (2)

4 variables with output dont care (X)

Dont Care Conditions Example: Determine the minimal SOP using K-Map:

Answer:

K-Map with Dont Care Conditions (3)

DACBCDDCBAF ),,,(

14,15)D(5,12,13, 9,10)M(0,2,6,8, D)C,B,F(A,

Solution : 14,15)D(5,12,13, 9,10)M(0,2,6,8, D)C,B,F(A,

DACBCDDCBAF ),,,(

AB CD

00

01

11

10

00 01 11 10 0 1 1 0

1 X 1 0

X X X X

0 0 1 0

0 1 3 2

4 5 7 6

12 13 15 14

8 9 11 10

Minimum SOP expression is CD

AD BC

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Exercise

Minimize this expression with a K-Map ABCD + ACD + BCD + ABCD

K-map Product of Sums simplification Example: Simplify the Boolean function F(ABCD)=(0,1,2,5,8,9,10) in (a) S-of-P (b) P-of-S

Using the minterms (1s) F(ABCD)= BD+BC+ACD

Using the maxterms (0s) and complimenting F Grouping as if they were minterms, then using De- Morgens theorem to get F.

F(ABCD)= BD+CD+AB F(ABCD)= (B+D)(C+D)(A+B)

5 variable K-map (1)

5 variables -> 32 minterms, hence 32 squares required

5 variable K-map (2) Adjacent squares. E.g. square 15 is adjacent to

7,14,13,31 and its mirror square 11. The centre line must be considered as the

centre of a book, each half of the K-map being a page

The centre line is like a mirror with each square being adjacent not only to its 4 immediate neighbouring squares, but also to its mirror image.

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5 variable K-Map (3) Example: Simplify the Boolean function F(ABCDE) = (0,2,4,6,11,13,15,17,21,25,27,29,31) Soln: F(ABCDE) = BE+ADE+ABE

6 variable K-map

6 variables -> 64 minterms, hence 64 squares required

End of Chapter 1 - tqvm -