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Ms Sandhya Rani Dash UNIT 4: COMBINATIONAL LOGIC MINIMISATION AND DESIGN USING KARNAUGH MAP Structure 4.1 Introduction 4.2 Objectives 4.3 Karnaugh Map 4.3.1 2, 3, 4 variable Karnaugh map 4.3.2 Definition of implicates 4.3.3 Prime implicates 4.3.4 Essential and Non-Essential prime implicates 4.3.5 Truth table representation on Karnaugh Map 4.4 Logic Minimization using Karnaugh Map 4.5Five&Six Variable Maps 4.6 Minimization of Logic Functions using minterm/maxterm 4.7 Don’t care Conditions 4.8 Unit Summary 4.1 Introduction In the previous unit we have discussed about logic operations and Boolean algebra. Boolean algebraic theorems are used for the manipulation of logical expressions. In this unit we will discuss about the minimization of logical functions using karnaugh maps. it i s the simplest and most commonly used method for simplifying Boolean expressions .The n-variable k- map methods for simplification of Boolean or switching functions are described here. The methods for minimization of combinational l ogics are also ex plained in this unit 4.2 Objectives After going through t his unit  you will be able to: i) define a combinational circuit. ii ) explain t he n-variable karnaugh maps. iii) define different types of implicantes iv)convert a standard SOP &POS expression into truth table format iv) si mplify the combinational logic functions usi ng k- map. v) mi nimize the SOP & POS expressions

BADASS Tutorial Karnaugh Maps

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    Ms Sandhya Rani Dash

    UNIT 4: COMBINATIONAL LOGIC MINIMISATION AND DESIGN USING

    KARNAUGH MAP

    Structure

    4.1 Introduction

    4.2 Objectives4.3 Karnaugh Map

    4.3.1 2, 3, 4 variable Karnaugh map4.3.2 Definition of implicates

    4.3.3 Prime implicates4.3.4 Essential and Non-Essential prime implicates

    4.3.5 Truth table representation on Karnaugh Map4.4 Logic Minimization using Karnaugh Map

    4.5Five&Six Variable Maps4.6 Minimization of Logic Functions using minterm/maxterm4.7 Dont care Conditions

    4.8 Unit Summary

    4.1 Introduction

    In the previous unit we have discussed about logic operations and Boolean algebra.

    Boolean algebraic theorems are used for the manipulation of logical expressions. In

    this unit we will discuss about the minimization of logical functions using karnaugh

    maps. it is the simplest and most commonly used method for simplifying Boolean

    expressions .The n-variable k- map methods for simplification of Boolean or

    switching functions are described here. The methods for minimization of

    combinational logics are also explained in this unit

    4.2 Objectives

    After going through this unityou will be able to:

    i) define a combinational circuit.

    ii) explain the n-variable karnaugh maps.

    iii) define different types of implicantes

    iv)convert a standard SOP &POS expression into truth table format

    iv) simplify the combinational logic functions using k- map.

    v) minimize the SOP & POS expressions

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    vi) use the dont care conditions in karnaugh map for further simplification.

    COMBINATIONAL CIRCUITS

    Combinational circuits consist of logic gates whose output depends only on thecurrent inputs at any given time. It does not store the history of the circuit. So memory

    is not required for a combinational circuit.

    A combinational circuit consists of input variables, logic gates and output variables.

    The logic gates accept signals from the inputs and generate signal to the output. The

    information transfer from input to output is in the binary form. A Block diagram of a

    combinational circuit is shown below.

    n input Variables m output variables

    . .

    . .

    . .

    The n input binary variables come from an external source: The m output variables

    go to an external destination. In many applications, the source and the destination are

    storage registers. For n input variables, there are 2npossible combinations of binary

    input values. A combinational circuit can be described by m Boolean functions, one

    for each output variable.

    4.3 Karnaugh Map

    The simplification of logical functions using Boolean laws and theorems become

    complex with the increase in the number of variables and terms. So to avoid this type

    of complexity, the K-Map method was first proposed by Veitch and Gightly. It was

    latter modified by Karnaugh and is now popularly known as the Veitch Diagram or

    Karnaugh Map (or K-map).

    The K-map technique provides a systematic method foe simplifying and manipulating

    Boolean expression. It is a graph composed of an arrangement of an adjacent cells.

    Each cell represent a particular combinations of variables insum or productform. The

    Combinational

    Logic circuit

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    K-Map is actually a modified form of a truth table. Here the combinations are

    conveniently arranged to aid the simplification process by applying the rule

    AB + AB' = A. Although the K-map technique may be used for any number of

    variables, it is generally used up to six variables. Beyond which it would be very

    difficult.

    4.3.1 2, 3, 4 variable Karnaugh map

    In an n-variable K-map, there are 2n

    cells where each cell corresponds to one

    combination of n-variables. Therefore, for a 2-variable K-map, there are 22= 4

    possible combinations of minterms. Similarly, for 3 and 4 variable K-map, there

    should be 23= 8 and 2

    4= 16, number of minterms and so on. The 2, 3, 4 variable K-

    maps are shown in the figure given below.

    The decimal codes corresponding to each combination of variables (minterm) written

    in each cell at the right corner. The variables have been designated as A, B, C, D and

    the binary numbers formed by them are taken as AB, ABC, and ABCD for 2, 3, 4

    variables respectively.

    The decimal codes given in 3 and 4 variable K-maps are cyclic, or unit distance

    codes. This code differs in one variable. This make easier for the simplification of

    Boolean function by grouping of the adjacent cells. The left and right most cells of the

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    three variables K-map are adjacent. For example: The cells 0 & 4 are adjacent and the

    cells 1 & 5 are adjacent. In 4 variables K-map the cells to the extreme left and right as

    well as at the top and bottom portion are adjacent.

    4.3.2 Truth table representation on K-map

    The truth table is a common way of representing the logical expressions of a circuit. It

    consists of a list of the possible combinations of input variable values and the

    corresponding output values (1 or 0). The standard SOP & POS expressions can be

    determined from a truth table. Also the outputs of a truth table of a logic function can

    be represented on the k-map. Let us consider a truth table of 3 variable logical

    functions as follows.

    Input OutputA B C Y

    0 0 0 0

    0 0 1 1

    0 1 0 1

    0 1 1 0

    1 0 0 1

    1 0 1 0

    1 1 0 0

    1 1 1 1

    Corresponding to the output of logic 1, Y can be written in standard SOP form as_ _ _ _ _ _

    Y= ABC + ABC + ABC + ABC

    Corresponding to logic 0, y can be written in POS form as_ _ _ _ _ _

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

    The value of the output variable Y (0 or 1) for each row of truth table is entered in the

    corresponding cells of the K-map The K-Map for the above three variable expressionis shown below.

    Variables

    _ _

    AB

    _

    AB AB

    _

    AB

    00 01 11 10

    _C 0 0 1 0 1

    C 1 1 0 1 0

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    This is the common procedure used to represent a truth table on the K-map. On the

    other hand a truth table can also b e prepared, if a K-map is given by following the

    reverse process.

    Conversion of SOP expression to truth table format

    The steps for converting a SOP expression to truth table format are as follows.

    Step-1: List all possible combinations of binary values of the variables in the

    expression.

    Step-2: Convert the SOP expression to standard form if it is not already.

    Step-3: Finally place a 1 in the output column for each binary value that makes the

    standard SOP expression and place a 0 for all the remaining binary values.

    Example: Develop a truth table for the standard Sop expression

    _ _ _

    ABC + ABC

    SOLUTION:The truth table for the above standard Sop expression is given below

    A B C OUTPUT PRODUCT

    TERM

    0 0 0 0

    0 0 1 0

    0 1 0 1

    _ _

    ABC

    0 1 1 0

    1 0 0 0

    1 0 1 1_

    ABC

    1 1 0 0

    1 1 1 0

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    Conversion of POS expression to truth table format

    A POS expression is equal to 0 only if at least one of the sum term is equal to 0.

    To construct a truth table from a POS expression, the steps are as follows.

    Step-1: List all possible combinations of binary values of the variables in the

    expression just as was done for the SOP expression

    Step-2: Convert the SOP expression to standard form if it is not already.

    Step-3: Finally place a 0 in the output column for each binary value that makes the

    standard SOP expression and place a 1 for all the remaining binary values.

    Example: Develop a truth table for the standard POS expression

    _ _ _ _ _

    (A + B + C) (A + B + C) (A + B + C)

    SOLUTION:The truth table for the above standard POS expression is given below

    A B C OUTPUT PRODUCT

    TERM

    0 0 0 1

    0 0 1 0__

    A + B + C

    0 1 0 0_

    A + B + C

    0 1 1 1

    1 0 0 1

    1 0 1 1

    _ _ _

    A + B + C

    1 1 0 1

    1 1 1 0

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    Self Check Exercise 1

    1)Draw a 2, 3, 4 variable K-map and label each cell according to its binary value.

    2) Write the problem term for each cell in a 2, 3, 4 variable K- map

    3) Develop a truth table for each of the standard SOP expression

    _ _ _i) ABC +ABC

    _ _ _ii) ABC + ABC

    4) Develop a truth table for each of the standard POS expression

    _ _ _ _ _i) (A+B+C) (A+B+C) (A+B+C)

    _ _ _ _ii) (A+B+C) (A+B+C) (A+B+C)

    4.4 Logic Minimization Using Karnaugh Map

    Simplification of logic functions with K-map is based on the principle of combining

    terms in adjacent cells. The adjacent (Horizontal & vertical but not diagonal) cells

    differ in only one variable. So the minimization of logical functions are achieved by

    grouping adjacent 1s or 0s in groups of 2i

    , where i= 1, 2, 3, .,n is the number of

    variables. The process of simplification involves grouping of minterms by drawing a

    loop around the c ells and identifying prime implicants and essential prime implicants.

    Prime Implicant

    A prime implicant is a group of minterms that can not be combined with any other

    minterms or groups.

    Essential Prime Implicant

    An essential prime implicant is a prime implicant in which one or more minterms are

    unique. i.e. it contains at least one minterm which is not contain in any other prime

    implicant.

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    4.5 Five & Six Variable Maps

    A 5-variable K-map contains 25

    =32 number of squares. The no. of square is always

    equal to the no. of minterms. The 32 minterms are designated as m0 , m1, m2 . . . . . m31.

    _ _ _ _ _The input variables are A B C D E ,E , . . . . . . ABCDE in SOP form. In POS form the

    _ _ _ _ _ _input variables are marked as A +B +C + D + E, A + B + C +D +E, A +B +C +D +E

    with max term designations M0, M1, M2. . . . . M31. Boolean functions with five

    variables can be simplified using two 4-variable maps (16 cells each)

    For 6- variable K-map, we need 26 = 64 no. of squares. Maps of more than four

    variables are not as simple to use. Maps with seven or more variables need too many

    squares. So these become more complicated to solve the logical functions. The five

    and six variable K-maps are shown in the figure.

    CDE

    AB 000 001 011 010 110 111 101 100

    00

    01

    11

    10

    0 1 3 2 6 7 5 4

    8 9 11 10 14 15 13 12

    24 25 27 26 30 31 29 28

    16 17 19 18 22 23 21 20

    Five-variable mapDEF

    ABC 000 001 011 010 110 111 101 100

    000

    001

    011

    010

    110

    111

    101

    100

    0 1 3 2 6 7 5 4

    8 9 11 10 14 15 13 12

    24 25 27 26 30 31 29 28

    16 17 19 18 22 23 21 20

    48 49 51 50 54 55 53 52

    56 57 59 58 62 63 61 60

    40 41 43 42 46 47 45 44

    32 33 35 34 38 39 37 36

    six-variable map

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    The decimal code of the rows and columns are given in a reflected code sequence; the

    minterm assigned to each square is read from these numbers. For example the square

    in the second row (01) and fourth column (010), in the five variable map, is number

    01010, the equivalent of decimal 10. Therefore, this square represents minterm m10.

    The simplification of 5 & 6 variable K-maps can be done in the same way as 4-

    variable K-map

    4.6 Minimization of Logic Functions Specified in Minterm/Maxterm

    If an expression is simplified to a stage beyond which it cant be simplified, it will

    require minimum number of gates with minimum number of inputs to the gates. Such

    an expression is referred to as the minimized expression.

    For minimizing a given logic expression specified in minterm/maxterm, we have to

    prepare the k-map first. Then look for combination of 1s for the minterm(SOP form)

    or 0s for the maxterm(POS form). We have to combine the 1s or 0s in such a way

    that the resulting expression is minimum. Toachieve this, following rules can be used

    which will lead to minimized expression.

    Identifying the 1sthat cant be combined with any other 1s and form groupswith such single 1. These are essential prime implicants.

    Identify the 1swhich are adjacent to only one other 1 and form groups anygroup of containing any two cells and which arent part of any group of four

    or eight cells. A group of two cells is called a pair.

    Group the 1s which results in groups of four cells but arent of an eightcells. A group of four cells is called a quad.

    Group the 1s which results in groups of eight cells. A group of eight c ells iscalled a octet.

    After identifying the essential groups of 2 , 4 and 8 1s , if there still remains1s which havent been encircled then these are to be combined

    With each other or with other encircled 1s Discard any redundant group.

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    Form the logical sum of all the terms generated by each group.

    When one or more than one variable appear in both complemented and un

    complemented form with in a group , then that variable is eliminated from the term

    corresponding to that group.

    A larger group of 1s eliminates more variables. A group of two eliminates one

    variable; a group of four eliminates two variables; similarly a group eight eliminatesthree variables.

    The above rules are used to minimize the logical functions in the following example

    given below.

    Example: Simplify the expression Y= (7, 9, 10, 11, 12, 13, 14, 15) , using the k-

    map method.

    SOLUTION:

    The above function is in SOP form. The k-map for the function is as shown in the

    figure.

    In the given k-map there are three pairs and one quad. The output for each of theabove group is given as

    _Y1= CD (AB + AB) = BCD

    _ _ _ _Y2 = AB (CD + CD + CD + CD)

    _ _ _Y3= AB (CD + CD ) + AB (CD + CD)

    _= ABC + ABC

    = AC

    _ _ _Y4= AB (CD + CD) + AB (CD + CD)

    _

    = ABD + ABD

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    " #$

    %&'(')*('+, " ,-. ,/. ,0.,1

    = AB+ AD + AC + BCD

    It can be further simplified by using Boolean algebra.

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

    Example 2: Simplify the expression Y = ( 0, 1 ,4, 5, 6, 8, 9, 12, 13, 14) using K-

    Map method

    SOLUTION:

    The given function is in the POS form. This can also be written as :

    _ _Y = (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) (A + B + C + D) (A + B + C + D)

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

    To simplify a POS expression for each maxterm in the given expression, a 0 has to be

    entered in the corresponding cells and groups must be formed with 0 cells instead of

    one cell to get the minimal expression. Here, a variable corresponding to 0 has to be

    represented in the uncomplemented form and that corresponding to 1 in the

    complemented form

    The K-map for the above function is shown in the figure given below:

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    !" $%& '()*& +,-'./ )"& )0$&$ '"1 )"& 23'1 '4& .4)130&1 (5 0)-(6"6"7 8 0&99:;

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    f ( A, B, C, D) = m (1, 3, 7, 11, 15) + d(0, 2, 5)

    Solution: The above expression is in SOP form. The K- map for this is given below.

    Grouping of 1 gives

    _ _ _ _ _ _Y1= AB (CD + CD + CD + CD)

    _ _ _= AB(C + C)

    _ _= AB

    Grouping of 2 gives_ _ _ _

    Y2= CD (AB + AB + AB + AB)

    _=CD (A + A)

    =CD

    Therefore Y = Y1+ Y2

    _ _= AB +CD

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    Example 2: Using the K-map method, simplify the following Boolean expression.

    f ( A, B, C, D) = M(4, 5, 6, 7, 8, 12) + d(1, 2, 3, 9, 11, 14)

    SOLUTION:

    The above expression is in POS form. The K- map for this is given below.

    Grouping of 1 gives_

    Y1= A + B

    Grouping of 2 gives_

    Y2= A+C +D_ _

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

    Self check exercise 2

    5. Use a K-map to find the minimized SOP for each expression_

    i) A + BC + CD_ _ _ _ _ _ _ _

    ii) ABCD + ABCD + ABCD + ABCD_

    iii) AC (B + C)_ _ _ _ _ _ _ _

    iv) ABC + ABC + ABC + ABC6. Use a K-map to find the minimized POS for each expression

    _ _ _ _ _ _i) (A + B) (A + C) (A + B + C) (A + B + C)

    _ _ _ _ _ _ _ _

    ii) (A + B + C + D) (A + B + C + D) (A + B + C + D)

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    7. Reduce the expression m(1, 5, 6, 12, 13, 14) + d(2, 4)8. Using K-map method, reduce the following expression to the simplest possible

    POS & SOP forms. m (0, 2, 3, 6, 7) + d(8, 10, 11, 15)

    4.8 Unit Summary

    In this unit we have considered the simplification of logical expressions by using the

    n-variable K-map technique. The K-map is a chart or graph composed of an

    arrangement of adjacent cells, each representing a particular combination of variables

    in a sum or product term. The n-variable K-map has 2ncells or squares and its use is

    limited to six variables.

    The truth table representation of the Boolean expressions is also discussed here. The

    truth table representation is a common way of representing the logical expression of a

    circuit in a concise format. The standard SOP & POS expressions can be determined

    from a truth table.

    In K-maps, dont care terms are also used to help in reducing the expression.

    Combinations for which the value of an expression isnt specified, are called dont

    care combinations. A dont care term is a minterm/maxterm in a logic function which

    may or may not included.

    Answers to Check Your Progress

    1) A 3-vaiable K-map has

    i) Eight cells ii) three cells iii) sixteen cells iv) four cells2)In a 4-vaiable K-map, a 2-vaiable product term is produced by

    i) a two cell group of 1sii) an eight cell group of 1siii)a four cell group of 1siv)a four cell group of 0s

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    3) On a K-map, grouping the 0s produces

    i) a product of sum expression

    ii) a sum of product expression

    iii) a dont care condition

    iv) AND-OR logic

    4) A 5-variable K-map has

    i) sixteen cells ii) thirty-two cells iii sixty-four cells iv) eight cells

    5) Dont care terms are represented by which alphabet of the following

    i) Y & Z ii) X & d iii) T & C iv) N & C

    6) The dont care term in k-map can have the value

    i) 0 ii) 1 iii) both 0 or 1

    Answers: 1 (i) 2 (iii) 3 (i) 4 (ii) 5 (ii) 6 (iii)

    Unit End Exercise

    1. Simplify the following expression using K-map

    _ _ _ _ _ _ _ _ _i) ABC + ABC + ABC + ABC

    _ _ _

    ii) ABC + AB + C + BC + DB2.Simplify the following expression using K-map

    _ _ _

    i) A(A + B + C)(A + B + C)(A + B +C)(A + B + C) _ _

    ii) (A + B) (A + B + C) (A + C)3. Draw a truth table for the equations given below:

    _ _i) Y= AC + AB ii) y= A(B + C)