Presentation Ffnal

8/3/2019 Presentation Ffnal

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PRESENTATION BY GROUP # 6


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OUTLINE OF THE

PRESENTATION1. DISCRETE MATHEMATICS2. BOOLEAN ALGEBRA 3. LOGIC GATES4. LOGIC DIAGRAM5. BOOLEAN IDENTITIES6. MINTERMS7. MAXTERMS8. K - MAPS9. ADDERS

10.CONCLUSION


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Contemporary uses of mathematics demand thatstudents learn to deal with uncertainty, to make informed

decisions based on evidence and expectations, toexercise critical judgment about conclusions drawn fromdata, and to apply mathematical models to real-worldphenomena. The technological world in which we live

also depends upon information and communication ofinformation and upon applications of systems withseparate (discrete) entities. Topics of discretemathematics such as counting and permutationproblems, matrix operations, vertex-edge networks, andrelationships among finite sets have significant real-world applications that students will encounter in diversefields of work and study.


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DISCRETE MATHEMATICS

DISCRETE mathematics is the study of mathematical structures that arefundamentally DISCRETE rather than continuous.

The objects studied in discrete mathematics such as integers, graphs, and

statements in logic do not vary smoothly , but have distinct, separated values.

Discrete Mathematics been characterized as branch of mathematics dealing with countable sets.

The set of objects studied in discrete mathematics can be FINITE or INFINITE.

Finite mathematics is sometimes applied to parts of the field of discretemathematics that deals with finite sets, particularly those areas relevant to business.



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Research in discrete mathematics increased in the latter half of thetwentieth century partly due to the development of digitalcomputer which operate in discrete steps and store data in discrete bits.

Concepts and notations from discrete mathematics are useful in

studying and describing objects and problems in branches of computerscience, such as computer algorithms, programming languages, andsoftware developments.

Although the main objects of study in discrete mathematics are discreteobjects, analytic methods from continuous mathematics are oftenemployed as well.


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WHY DISCRETE MATHEMATICS!!!

Discrete math is the mathematics of computing.

Discrete math is very much "real world mathemati

Discrete math teaches mathematical reasoning andproof techniques.

Discrete math is fun.


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Graphs like this are among the objects studied by discrete mathematics, for their interestingmathematical properties, their usefulness as models of real-world problems, and their importance in

developing computer algorithms.


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Boolean Algebra

George Boole (1813-1864)

During development of computer hardware, a decision wasmade to use binary circuits because it greatly simplified theelectronic circuit design.

In order to work with binary circuits, it is helpful to have aconceptual framework to manipulate the circuitsalgebraically, building only the final most simple result .

Boolean logic forms is the basis for computation in modernbinary computer systems.



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Importance of Boolean algebra with

respect to information technology inour day-to-day life :

Thescientists

Thesalespeople

Thedoctors

Thefactories



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Why should managers study

Boolean algebra?

Industriallybased to

technologybased

Expertsystems

Reshapetheir jobs



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Boolean Algebra

When we say that 1 + 1 = 2 or 3 + 4 = 7, we are implying theuse of integer quantities.

Boolean algebra provides the operations and the rules for working with the set {0, 1}.

There is no such thing as "2" or "-1" or "1/2" in the Boolean world

It is a world in which all other possibilities are invalid by fiat.



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In Boolean Algebra numeric operations of multiplication xy , addition x + y, negation x are replaced by the

respective logical operations of conjunction x y , disjunction x y ,negation x.

Operations of Boolean algebra:

Boolean complementation Boolean sum Boolean product



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Boolean OperationsThe complement is denoted by a bar (on the slides, we will

use a minus sign). It is defined by -0 = 1 and -1 = 0.

The Boolean sum , denoted by + or by OR, has thefollowing values:

0 + 0 = 0 , 1 + 0 = 1, 0 + 1 = 1, 1 + 1 = 1,

The Boolean product , denoted by or by AND, has thefollowing values:1 1 = 1, 1 0 = 0, 0 1 = 0, 0 0 = 0



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LOGIC/DIGITAL CIRCUITS

It is a device that accepts one or more input and produces only one output.

Input and output are the digital signals or the electronic signals.

A high signal is represented by 1 and a low signal is represented by 0

Logic gate is the basic building block for a logic circuit.

The relationship between the input and the output can be represented by a TRUTHTABLE

This can also be represented by a BOOLEAN EXPRESSION


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BASIC LOGIC GATES

AND

OR Basic Logic Gates

INVERTER/NOT

BUFFER

NAND

NOR Gates derives from above three gates

XOR

XNOR


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The AND Gate

Z = A.B (Read as A and B)

The AND gate can be represented by the following TRUTHTABLE

A B Z0 0 00 1 01 0 01 1 1

If both the inputs A and B are high then the output will be high. On the other hand if any one input is low the output will be low.

B

A Z


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INPUT HIGH A B

BULB

OUTPUT

Y


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The OR Gate

Z = A+B (Read as A or B)

The OR gate can be represented by the following TRUTHTABLE

A B Z0 0 00 1 11 0 11 1 1

If either of the inputs A or B or both A and B are high then the output will be high. Onthe other hand the output is 0.

Z = AB+AB+AB(Boolean Expression)


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BULB

OUTPUT

Y

A

B


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The INVERTER Gate

Z = A (Read as Z equals not A)

The INVERTER gate can be represented by the following TRUTHTABLE

A Z0 11 0

Inverter circuits inverts logic sense of a binary signal and produces the complement

Z A


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LOW

A

B B

LOW

A

HIGH

HIGH

LOW

A

BULB

OUTPUT Y


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The BUFFER Gate

Z = A

The BUFFER gate can be represented by the following TRUTHTABLE

A Z0 01 1

A BUFFER does not produce any particular logic function since the binary value of theoutput is same as the binary value of input.Circuit is mainly used for amplification.

Z A


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The NAND Gate

Z = (AB)

The NAND gate can be represented by the following TRUTHTABLE

A B AB (AB)=Z 0 0 0 10 1 0 11 0 0 11 1 1 0

The NAND function is the complement of AND function. It is a AND graphic followed by a small circle as shown above.

Z = AB+AB+AB(Boolean Expression)

B

A Z


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The NOR Gate

Z = (A+B)

The NOR gate can be represented by the following TRUTHTABLE

A B A+B (A+B) = Z 0 0 0 10 1 1 01 0 1 01 1 1 0

NOR gate is the complement of the OR gate and uses OR graphic followed by a smallcircle.

(A + B) A B


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The XOR Gate

Z = AB+AB (ODD FUNCTION)

The XOR gate can be represented by the following TRUTHTABLE

A B Z0 0 00 1 11 0 11 1 0

The output of the gate is 1 if any input is 1 but excludes the combination when bothinputs are 1.


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The XNOR Gate

Z = AB+AB

The XOR gate can be represented by the following TRUTHTABLE

A B Z0 0 10 1 01 0 01 1 1

The output of the gate is 1 if both the inputs are 1 or if both the inputs are 0.


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LOGIC DIAGRAMA Boolean function can be transformed from an algebraic expression into a logic diagram.

A logic Diagram is composed of AND , OR , and Inverter gate.

The Boolean Function for example F = x + yzcan be expressed as a logic diagram.


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In the diagram we have inverter for complementing y.

We need a AND gate for anding yz

We also need a OR gate for oring x & yz

z

x

y y

In a logic diagram, the variables of the function are to be taken the inputs of the

circuit, and the variable symbol of the function is taken as the output of the circuit.


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In mathematics , the term identity has several different importantmeanings:

An identity is a relation which is tautologically true. This means that watever the number or value may be the answer stays the same. Forexample, algebraically, this occurs if an equation is satisfied for all values of the involved variables. Definitions are often indicated by the 'triple bar' symbol , such as A2 xx. The symbol can also be

used with other meanings, but these can usually be interpreted insome way as a definition, or something which isotherwise tautologically true (for example, a congruence relation).

BOOLEAN IDENTITIES



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In algebra, an identity or identity element of aset S with a binary operation is an element e that, when combined with any element xof S, produces thatsame x. That is, ex = xe = x for all x in S. An exampleof this is the identity matrix.

The identity function from a set S to itself, oftendenoted id or idS, is the function which maps every

element to itself. In other words,id(x) = x for all x in S.This function serves as the identity element in the setof all functions from S to itself with respectto function composition.


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BOOLEAN IDENTITIES

1. x+0 = x 2. x.1 = x Identity 3. x+1 = 1 4. x.0 = 0 Null5. x+x = x 6. x.x = x Idempotent7. x+x = 1 8. x.x = 0 Inverse9. x+y = y+x 10.xy = yx Commutative11. x + (y + z) = (x + y) + z 12. x(yz) = (xy)z Associative13. x(y+z) = xy + xz 14. x + yz = (x+y)(x+z) Distributive15. (x+y ) = xy 16. (xy ) = x+y De-Morgans17. (x)


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PROVING BOOLEAN IDENTITIES

X 0 Z0 0 01 0 1

X 0 Z0 0 01 0 0

X + 0 = X

X . 0 = 0

IDENTITY

X 1 Z0 1 11 1 1

X + 1 = 1

X 1 Z0 1 01 1 1

X . 1 = X

NULL


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X X Z

0 0 01 1 1

X + X = X

X X Z

0 0 01 1 1

X . X = X

IDEMPOTENT

X X Z1 0 10 1 1

X + X= 1

X X Z1 0 00 1 0

X . X = 0

INVERSE


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X Y X+Y

0 0 00 1 11 0 11 1 1

COMMUTATIVE Y X Y+X0 0 00 1 11 0 1

1 1 1

Y

X

Y+ XX

Y

X + Y


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X Y X.Y

0 0 00 1 01 0 01 1 1

COMMUTATIVE Y X Y.X0 0 00 1 11 0 1

1 1 1

Y

X X.Y X

Y Y.X


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X Y Z Y + Z X + (Y+Z) X + Y (X+Y)+Z0 0 0 0 0 0 00 0 1 1 1 0 1

0 1 0 1 1 1 10 1 1 1 1 1 11 0 0 0 1 1 11 0 1 1 1 1 1

1 1 0 1 1 1 11 1 1 1 1 1 1

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

X Y Z YZ X(YZ) XY (XY)Z

0 0 0 0 0 0 00 0 1 0 0 0 00 1 0 0 0 0 00 1 1 1 0 0 0

1 0 0 0 0 0 01 0 1 0 0 0 01 1 0 0 0 1 01 1 1 1 1 1 1

X(YZ) = (XY)Z

ASSOCIATIVE


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X

Y

X + Y

Z

(X + Y) + Z

Y

Z

Y + Z

X

X + ( Y + Z)


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X.Y X

Y Z

(X.Y)Z

Y.Z Y

Z X

X(Y.Z)

X Y Z Y Z X(Y Z) XY XZ XY XZ


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X Y Z Y + Z X(Y+Z) XY XZ XY + XZ0 0 0 0 0 0 0 00 0 1 1 0 0 0 00 1 0 1 0 0 0 00 1 1 1 0 0 0 01 0 0 0 0 0 0 01 0 1 1 1 0 1 11 1 0 1 1 1 0 1

1 1 1 1 1 1 1 1X Y Z YX X + Y X + Z X + YX (X+Y)(X+0 0 0 0 0 0 0 00 0 1 0 0 1 0 0

0 1 0 0 1 0 0 00 1 1 0 1 1 0 01 0 0 0 1 1 1 11 0 1 0 1 1 1 11 1 0 1 1 1 1 11 1 1 1 1 1 1 1

DISTRIBUTIVE

X (Y+Z) = (XY)+XZ

X + YX = (X+Y)(X+Z)


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Y

ZY+Z

X X(Y+Z)

X

Y

X

Z

XY

XZ

XY+XZ)


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X Y X+Y (X+Y)' X' Y' X'Y'0 0 0 1 1 1 10 1 1 0 1 0 01 0 1 0 0 1 01 1 1 0 0 0 0

(X + Y) = XY

X Y X' Y' XY (XY)' X'+Y'0 0 1 1 0 1 10 1 1 0 0 1 11 0 0 1 0 1 11 1 0 0 1 0 0

(X Y) = X+Y

De- MORGANS


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X

Y

X+Y (X+Y)

X

Y

X

Y XY


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X

Y

XY (XY)

Y

X X

Y X+Y


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MINTERM AND MAXTERM



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We can write expressions in many ways, but some ways are more useful than others

A sum of products (SOP) expressioncontains: Only (sum) operations atthe outermost level Each term that is summed must be a

product of literals f(x,y,z) = y + xyz +xz

A product of sums (POS) expressioncontains: Only (product) operations atthe outermost level

Each term must be a sum of literals f(x,y,z) = y (x + y + z) (x +z)

OR AND

SOP POS



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A minterm is a specialof , in which each input variable appears exactly once with or without bar.

A function with n variables has 2 nminterms.

A three -variable function, such asf(x,y,z), has 2 3 = 8 minterms:x.y.z x.y.z x . y z x.y.z x.y.z x.y.z x.y.zx.y.zHERE X , Y , Z = 1 X , Y , Z = 0

A maxterm is a of , in which each input variable appears exactly once with or without bar.

A function with n variables has 2 nmaxterms

The maxterms for a three-variablefunction f(x,y,z):x + y + z x + y + z x y + z x+ y +zx + y + z x + y + z x + y + z x + y + z HERE X , Y , Z = 0 X , Y , Z = 1

MINTERM AND MAXTERM

productliterals

Sum literalsMAXTERMMINTERM



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MINTERM MAXTERM

Each minterm is true for exactly onecombination of inputs:

Minterm Is true when Shorthand xyz x=0, y=0, z=0 m0 xyz x=0, y=0, z=1 m1

xyz x=0, y=1, z=0 m2 xyz x=0, y=1, z=1 m3xyz x=1, y=0, z=0 m4 xyz x=1, y=0, z=1 m5xyz x=1, y=1, z=0 m6 Xyz x=1, y=1, z=1 m7

Each maxterm is false for exactly onecombination of inputs:

Maxterm Is false when Shorthand x + y + z x=0, y=0, z=0 M0x + y + z x=0, y=0, z=1 M1

x + y + z x=0, y=1, z=0 M2 x + y + z x=0, y=1, z=1 M3 x + y + z x=1, y=0, z=0 M4 x + y + z x=1, y=0, z=1 M5 x + y + z x=1, y=1, z=0 M6 x + y + z x=1, y=1, z=1 M7



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OR GATE AND GATE NOT GATE

A B Z A B Z A Z

0 0 0 0 0 0 1 0

0 1 1 0 1 0 0 1

1 0 1 1 0 0

1 1 1 1 1 1

STANDARD EXPRESSION Z=A+B Z=AB Z=A'

SOP OF MIN-TERM EXPRESSION Z = Z=AB Z=A'

POS OF MAX-TERM EXPRESSION Z=A+B Z= Z=A'

SOP OFMIN-TERM

&

POS OFMAXTERM

A A ZZB Z

A B

A'B + AB' + AB

A+B A+B A+B . . PRESENTATION BY GROUP # 6

CANONICAL FORM


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f = xyz + xyz+ xyz + xyz + xyz = m0 + m1 + m2 + m3 + m6= m(0,1,2,3,6)

f = xyz + xyz+xyz= m4 + m5 + m7= m(4,5,7)

f contains all the minterms not in f

f = (x + y + z)(x + y + z)(x + y + z) = M4 M5 M7= M(4,5,7)

f = (x + y + z)(x + y + z)(x + y + z) (x + y + z)(x + y + z) = M0 M1 M2 M3 M6= M(0,1,2,3,6)

f contains all the maxterms not in f

CANONICAL FORM


CONVERSION OF STANDARD EQUATION INTO


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GATE /EXPRESSION

OR GATE AND GATE

STANDARD EXPRESSION Z=A+B Z=A.B

SOP OF MIN-TERM Z=A(B+B)+B(A+A) =AB+AB+AB+AB =AB+AB+AB

POS OF MAX-TERM Z=A+(BB).B+(AA)

=(A+B).(A+B).(A+B).(A+B)

=(A+B).(A+B).(A+B)

QMINTERM AND MAXTERM

STEPS:-1: Write down the expression. Insert X(or any other

variable) for each missing literal in each term in theexpression.2: Write all combination of literals for each term.

3: Add all terms.4: Drop duplicate terms.

A B

Z Z A B



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Karnaugh mapsLast time we saw applications of Boolean logic to circuit design.

The basic Boolean operations are AND, OR and NOT.These operations can be combined to form complex expressions, which canalso be directly translated into a hardware circuit.Boolean algebra helps us simplify expressions and circuits.

Today well look at a graphical technique for simplifying an expression into a

minimal sum of products (MSP) form:There are a minimal number of product terms in the expression.Each term has a minimal number of literals.

Circuit-wise, this leads to a minimal two-level implementation.


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Re-arranging the truth table A two-variable function has four possible minterms. We can re-arrange theseminterms into a Karnaugh map .

Now we can easily see which minterms contain common literals.Minterms on the left and right sides contain yand y respectively.Minterms in the top and bottom rows contain xand x respectively.

x y minterm0 0 xy0 1 xy1 0 xy1 1 xy

Y

0 10 xy xy

X1 xy xy

Y

0 10 x y x y

X1 x y x y

Y YX x y x yX x y x y


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Karnaugh map simplificationsImagine a two-variable sum of minterms:

xy + xy

Both of these minterms appear in the top row of a Karnaugh map, which meansthat they both contain the literal x.

What happens if you simplify this expression using Boolean algebra?

xy + xy = x(y + y) [ Distributive ]= x 1 [ y + y = 1 ] = x [ x 1 = x ]

Y

xy xyX xy xy


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More two-variable examples Another example expression is xy + xy .

Both minterms appear in the right side, where y is uncomplemented.

Thus, we can reduce xy + xy to just y .

How about xy + xy + xy ? We have xy + xy in the top row, corresponding to x.Theres also xy + xy in the right side, corresponding to y .This whole expression can be reduced to x + y .

Yxy xy

X xy xy

Yxy xy

X xy xy


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A three-variable Karnaugh mapFor a three-variable expression with inputs x, y, z, the arrangement of mintermsis more tricky:

Another way to label the K-map (use whichever you like):

Yxyz xyz xyz xyz

X xyz xyz xyz xyzZ

Ym0 m1 m3 m2

X m4 m5 m7 m6Z

YZ00 01 11 10

0 xyz xyz xyz xyzX

1 xyz xyz xyz xyz

YZ00 01 11 10

0 m0 m1 m3 m2X1 m4 m5 m7 m6

h h f d


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Why the funny ordering? With this ordering, any group of 2, 4 or 8 adjacent squares on the map containscommon literals that can be factored out.

Adjacency includes wrapping around the left and right sides:

Well use this property of adjacent squares to do our simplifications.

xyz + xyz = xz(y + y) = xz 1= xz

xyz + xyz + xyz + xyz = z(xy + xy + xy + xy) = z(y(x + x) + y(x + x)) = z(y+y)

= z

Yxyz xyz xyz xyz

X xyz xyz xyz xyzZ

Yxyz xyz xyz xyz

X xyz xyz xyz xyzZ

l i lifi i


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Example K-map simplificationLets consider simplifying f(x,y,z) = xy + yz + xz.First, you should convert the expression into a sum of minterms form, if its not

already.The easiest way to do this is to make a truth table for the function, and thenread off the minterms. You can either write out the literals or use the minterm shorthand.

Here is the truth table and sum of minterms for our example:

x y z f(x,y,z)0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 01 0 1 11 1 0 11 1 1 1

f(x,y,z) = xyz + xyz + xyz + xyz

= m1 + m5 + m6 + m7

M ki h l K


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Making the example K-map Next up is drawing and filling in the K-map.

Put 1s in the map for each minterm, and 0s in the other squares.

You can use either the minterm products or the shorthand to show you where the 1s and 0s belong.In our example, we can write f(x,y,z) in two equivalent ways.

In either case, the resulting K-map is shown below.

Y0 1 0 0

X 0 1 1 1Z

Yxyz xyz xyz xyzX xyz xyz xyz xyz

Z

f(x,y,z) = xyz+ xyz+ xyz+ xyz

Ym0 m1 m3 m2X m4 m5 m7 m6

Z

f(x,y,z) = m1 + m5 + m6 + m7

K f h bl


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K-maps from truth tables You can also fill in the K-map directly from a truth table.

The output in row i of the table goes into square m i of the K-map.

Remember that the rightmost columns of the K- map are switched. Y

m0 m1 m3 m2X m4 m5 m7 m6

Z

x y z f(x,y,z)0 0 0 00 0 1 10 1 0 00 1 1 0

1 0 0 01 0 1 11 1 0 11 1 1 1

Y0 1 0 0

X 0 1 1 1

Z


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Reading the MSP from the K-map Finally, you can find the MSP.

Each rectangle corresponds to one product term.The product is determined by finding the common literals in thatrectangle.

For our example, we find that xy + yz+ xz = yz + xy . (This is one of theadditional algebraic laws from last time.)

Yxyz x yz xyz xyz

X xyz x yz xyz xyz

Z

Y0 1 0 0

X 0 1 1 1

Z


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ADDERSHALF ADDER

FULL ADDER


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BASIC OPERATION

Digital computer performs various arithmeticoperation. The most basic operation is addition of twobinary digits. Addition consists of four possible elementary operation

0 + 0 = 0

0 + 1 = 11 + 0 = 11 + 1 = 102



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HALF ADDER

Half Adder adds two bits and produce sum and carry.

A B S C0 0 0 00 1 1 01 0 1 01 1 0 1

Expression for SUM = AB + AB

Expression for Carry =

AB


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CIRCUIT FOR HALF ADDER

SUM = AB + AB =

A B

AB + AB

Carry = AB A

B AB


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SUM = AB + AB

Carry = AB

A

B


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LIMITATION OF HALF ADDERIn multi-digit addition we have to add two bits along with the carry of previous digit addition. Each additionrequires addition of 3 bits. This is not possible withHalf Adder.Due to this disadvantage a new adder is introducedcalled Full Adder.


FULL ADDER


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Full Adder takes three bits and produces SUM bit

and a CARRY bit. A B C in SUM CARY 0 0 0 0 0

0 0 1 1 00 1 0 1 00 1 1 0 1

1 0 0 1 01 0 1 0 11 1 0 0 11 1 1 1 1

Expression for SUM = ABC+ABC+ABC+

ABC

Expression for Carry =

ABC+ABC+ABC+ ABC

CIRCUIT DIAGRAM FOR SUM


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A

B

C B xor C

B xnor C

A B C

A(B xor C)

A(B xnor C)

SUM

CIRCUIT DIAGRAM FOR CARRY


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A xor B

AB

C

C(A xor B)

CARRY

A B C


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PRESENTATION BY GROUP #6

L i G t d t Di l


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PRESENTATION BY GROUP #

Logic Gates used to Display ONE WAY Path

Latch T

TT

T

T

T

T

T

T

T

T

F

F

F

F

FF

F

F

F

F
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PRESENTATION BY GROUP #6
http://users.senet.com.au/~dwsmith/_vti_bin/shtml.dll/beginners.htm/map


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PRESENTATION BY GROUP #

The usefulness of Boolean algebra comes from the fact that its rules

can be shown to apply to logical statements.

A logical statement, or proposition, can either be true or false, justas an equation with real numbers can be true or false depending on

the value of the variable.

Boolean algebra, however, variables do not represent the valuesthat make a statement true, instead they represent the truth or falsityof the statement. That is, a Boolean variable can only have one of two values.
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BOOLEAN ALGEBRA
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