· Web view0 0 0 1 0 0 1 1 f(x,y,z)= xy+ yz ALGORITHM TO FIND ALL PRIME IMPLICANTS A General...

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Example 1: • xy+xy+x y = x+y• LHS x(y+y)+ x y using distributive law• x •1+ x y using y+y = 1 theorem• x(1+ y ) +x y (1+y) =1 theorem• x+ x y + x y using distributive law • x + y(x+ x) • x + y• 1 = x +y RHS•

• Example 2: • (x+y)(x+z) = xz+xy using distributive law • x x + x z + x y + y z using P5 postulate• x z + x y +(x+x) yz using distributive law• xz+ x y + xyz+xyz using distributive law• xz(1+y) +x y(1+z) using T2 theorem• xz•1 +xy • 1 = xz + xy RHS

• PRINCIPLE OF DUALITY• One can transform the given expression by interchanging the operation (+) and (•) as well as the identity elements 0 and 1 . Then the expression will be referred as dual of each other. This is known as the principle of duality.• Example x + x = 1 then the dual expression is • x • x = 0

• BOOLEAN FORMULAS AND FUNCTIONS• Boolean expressions or formulas are constructed by using Boolean constants and variables with the Boolean operations like (+) , (•) and ‘not’ • Example: (x + y) z• f(x,y.z) = (x +y) z or f = (x +y) z

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x y z x y x z x y z f

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

00000011

00001010

00000010

00001011

Truth table for the above Boolean expression is

• Example 2: Write a truth table for following function• f = x y z + x y + x z x y z x y x z x y z f

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

00000011

00001010

00000010

00001011

• NORMAL FORMULAS• Boolean expression can be represented by following structures• 1. Sum of products ( SOP) or disjunctive normal form• 2. Product of sum (POS) or Conjunctive form

• In SOP normal form is a Boolean formula that is written as a single product term or as a sum ( also called disjunctive) of product terms is said to be in the sum of product form or disjunctive normal form.

• Example:• f(w,x,y,z) = x + w y + w y z

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• In the POS normal form is a Boolean formula which is written as a single sum term or as a product of sum ( also called conjunctive) terms is said to be in product of sums form or conjunctive normal form.• Example:• f(w,x,y,z) = z (x + y) (w + y + z) • CANONICAL FORMULAS• A procedure which will be used to write Boolean expressions form truth table is known as canonical formula. The canonical formulas are of two types

• 1. Minterm canonical formulas• 2. Maxterm canonical formulas

• 1. MINTERM CANONICAL FORMULAS• Minterms are product of terms which represents the functional values of the variables appear either in complemented or un complemented form.• Ex: f(x,y,z) = x y z + x y z + x y z• The Boolean expression whichis represented above is also known as SOP or disjunctive formula

• The truth table is x y z f

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

01011000

• m- NOTATION• To simplify the writing of a minterm in canonical formula for a function is performed using the symbol mi. Where i stands for the row number for which the function evaluates to 1.• The m-notation for 3- variable an function Boolean function • f(x,y,z) = x y z + x y z + x y z is written as • f(x,y,z) = m1+ m3 + m4 or • f(x,y,z) = m(1,3,4)

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• A three variable m- notation truth variable x y z Decimal designator of

rowMinterm m-notation

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

01234567

x y zx y zx y zx y zx y zx y zx y zx y z

m0m1m2m3m4m5m6m7

• 2. MAXTERM CANONICAL FORM• Maxterm are sum terms where the variable appear once either in complement or un-complement forms and these terms corresponds to a functional value representing 0.Ex. f(x,y,z) = ( x+ y+ z ) ( x+ y+z ) ( x + y + z ) = M( 0, 2, 5) = M0, M2, M5

M-NOTATION• A maxterm in a canonical form can be represented as Mi. Where i stands for row number for which the the function evaluates to 0. A product of maxterms are represented as M.

• MANIPULATIONS OF BOOLEAN FORMULAS• A Boolean function is described by different formulas. i.e.by applying postulate and theorems of Boolean algebra. Even it is possible to translate the Boolean expressions in different forms.• 1 EQUATION COMPLEMENTATION• For every Boolean function ‘f’ there is associated a complementary function ‘f’ in which • f(x1,x2,….xn) = 1 if f(x1,x2,….xn) = 0 and• f(x1,x2,….xn) = 0 if f(x1,x2,….xn) =1• for all combinations values of x1,x2,….xn.

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• Example 1: f= x y + x y • Example 2: f = w x z+ w ( x + y z)• 2. EXPANSION ABOUT VARIABLE • There are occasions when it is desirable to single out a variable and rewrite a Boolean formula f(x1,x2,…xi,…..xn) so that • xi g1 + xi g2 or• (xi + h1) (xi + h2)• Where g1, g2, h1, and h2 are expression not containing the variable xi. This theorem is known as Shannon expansion theorem.

• Shannon expansion theorem• Theorem 1: f(x1, x2, …..xi, ….xn) can be represented in SOP form as• = xi f(x1,x2,…,1….xn) + xi f(x1,x2,…0,...xn)

• Theorem 2 : f(x1, x2, …..xi, ….xn) can be represented in POS form as• = xi + f(x1,x2,…,0….xn) xi+ f(x1,x2,…1,...xn)• Example : f(w,x,y,z) = w x + (w x + y ) z

• 3. EQUATION SIMPLIFICATION • It is a direct method using postulates and theorem of a Boolean algebra enable the manipulations of a formula into equivalent forms.

• Example: ( x+ x y) ( x + y) + y z

• 4. THE REDUCTION THEOREM• For the purpose of obtaining simple Boolean formula two additional theorems are useful,

• These theorems are known as Shannon’s reduction theorems.• theorem 3:• xi• f( x1, x2, …..xi…..xn) = xi• f( x1, x2, …..1…..xn)• xi+ f( x1, x2, …..xi…..xn) = xi+ f( x1, x2, …..0…..xn)• Where f( x1, x2, …..k,…..xn) for k= 0,1.• Similarly• xi• f( x1, x2, …..xi…..xn) = xi• f( x1, x2, …..0…..xn) • xi+ f( x1, x2, …..xi…..xn) = xi• f( x1, x2, …..1…..xn)• Where f( x1, x2, …..k,…..xn) for k= 0,1.• Example : f(w,x,y,z)= x+ x y + w x ( w+ z) ( y+ w z ) • 5. MINTERM CANONICAL FORMULA• 1. Apply DeMorgan’s law a sufficient no. of times until all the NOT operations appear only with the single variables.• 2. Apply distributive of AND over OR (•) over (+) i.e. x•(y+z) = xy+xz in order to manipulate the formula into disjunctive normal formula.

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• 3. Remove duplicate literals and turns by idempotent law well as any term that are identically zero (x•x = 0)• 4. If any product in the disjunctive normal formula does not contain all the variables of the Boolean function then these missing variables are introduced by ANDing the terms with logic 1 in the form of xi+xi where xi is the missing variable being introduced. This process is repeated for each missing variables in each of the product turns of the disjunctive normal form

• 5. Apply distributive of ( •) over (+) again so that each variable appears exactly once in each term. 6. Remove duplicate term if any. Example : ( x + y) + ( y + x z )( x + y)

• 6. MAXTERM CANONICAL FORMULA• 1. Apply DeMorgan’s law until all the NOT operations appear with single variables.• 2. Apply distributive law of (+) over (•) • i.e. (x+yz) = (x+y)(x+z) and bring the expressions into its conjugate normal form (POS).• 3. The missing variables are introduced into the sum terms by OR ing logic 0’s in the form xi •xi = 0 where xi is missing variable.• 4. Distributive law of (+) over (•) is again applied.• 5. Duplicate literals are deleted.

• Ex: f(x,y,z) = ( x + y ) + ( y+ x z )( x + y )

• 7. COMPEMENTS OF CANONICAL FORMULAS• Even by taking complements minterm expression may result different expressions.• Ex : f(x,y,z) = m(0, 2, 4, 6) its complement expression is • f(x,y,z) = m(1, 3, 5,7)• Similarly • A Maxterm canonical expression may be represented in completed for as follows• Ex : f(x,y,z) = M (1, 2, 4, 7) its complement expression is • f(x,y,z) = M (0 3, 5,6)

• GATES AND COMBINATIONAL NETWORKS

• Gates are electronics circuits whose terminal characteristics correspond to the various Boolean operations. Where as logic networks is a network of

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interconnections of gates and flip – flops. The gates are operated with two possible steady state voltage signal values appear at the terminals of the circuits at any time.• An electronic circuit in which the output signal is a logic-1 if and only if all its inputs are at logic-1 is called AND gate.

• An electronic circuit in which the output signal is a logic-1 if and only if at least one of its input signal is at logic-1 is called OR gate.3. An electronic circuit in which the output signal is always opposite logic with input signal is called as NOT gate or inverter gate.

• COMBINATIONAL NETWORKS

• 1. COMBINATIONAL NETWORK• The inter connections 0f gates result in a gate network. If the network has the property that its outputs at any time are determined strictly by the inputs at that time then the network is said to be a combinational network. Ex. Adders. Multiplexers. etc• Let us consider an set of n signals at any time is called input state or input vector of the network. While a set of resulting signals appearing at the m output terminals is called the output state or output vector. The network can be expressed as

• z1, z2,……zm as Boolean function then • Zi = fi(x1,x2,….xn) for i= 1,2,…m.• 2. SEQUENTIAL NETWORKS• A second type of logic network is the sequential networks. Sequential network have memory property, so that the the outputs from these networks are dependent not only upon the current inputs but upon previous input as well. Feed back path are used in the sequential circuits. Ex. Counters. Shift registers etc.

• ANALYSIS PROCEDURE• Analysis procedure for a combinational network is as follows• 1. Each gate output that is only a function of the input variables is labeled.• 2. Boolean algebraic expression for the outputs of each of these gates are then written.

• 3. Next these gates outputs that are a function of just inputs variables and previously labeled gate outputs.

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• 4. Then each of the previously defined labels is replaced by the already written Boolean expression and this process is continued until the output of the network is labeled and till final expression is obtained.

• f(w,x,y,z) = w•(y+z) + wxy• = w•G1 + G2• f(w,x,y,z) = G2 + G3

• SYNTHESIS PROCEDURE• The synthesis procedure or logic design will be obtained by having specification of the desired behavior of a gate network. The truth table will provide total information of network specifications. If the variables and their compliments are used as inputs then such logic is said to be DOUBLE –RAIL logic. If the variables and their complements are not available, then not-gates are required then it is called SINGLE – RAIL logic.• The gate input signals will pass through input to out is called as levels of logic.

• Example1: f(w,x,y,z) = w x +x (y+z)• In the above logic circuit their exists 3- logic levels G1, G2 & G3 with double -rail logic.

• Example 2: f(w,x,y,z) = w x +x y + x z• In the above logic circuit there exists two level logic levels with double -rail.

• INCOMPLETE BOOLEAN FUNCTION WITH DONOT CARE CONDITIONS• In this type of Boolean functions having n-variables so that it may have 2n combinations of subsets and if all values are not specified such functions are called incompletely specified functions. • Ex. Let us consider a three variable function with following truth table

• We can describe the incomplete function in the SOP form as • f(x,y,z) = m(0,1,7) +dc(3,5)• Also we can represent the function in the pos form as• f(x,y,z) = M(2,4,6) +dc(3,5)• The odd parity generation result output expression • f(w,x,y,z) = m(0,3,5,6,9) +dc(10,11,12,13,14,15)• f(w,x,y,z)=w x y z + x y z + x y z + x y z + w z

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• ADDITIONAL BOOLEAN OPERATIONS AND GATES• The NAND and NOR gates are referred as universal gates or universal building blocks.• 1. NAND FUNCTION• nand(x1, x2, ….xn) = (x1,x2,….. xn) Ay applying DeMorgan’s law we get x1, x2,…..xn = x1+x2+….+xn

• NAND- GATE REALIZATIONS• Any Boolean functions can be realized by nand gates alone• 1. Apply DeMorgan’s law to the expression to obtain in unary operation i.e. appear only with single variables. Draw the logic diagram using and and-or gates.• 2. Replace each and –gate symbol by nand-gate and each or-gate symbol by nand-gates.• 3. Check for bubbles occurring on all lines between two gate symbols. so that two gates will be there.

• 4. Whenever an input variable enters a gate symbol at a bubble then Complement the variable of the gate. If the output has a bubble then insert an output not-gate symbol.• 5. Replace all not-gates by a nand-gate equivalent if desired.

• Example: f(w,x,y,z) = w z + w z ( x + y)

• 2. NOR FUNCTION• nor(x1, x2, ….xn) = (x1+x2+…..+xn) Ay applying DeMorgan’s law we get x1+x2+….+xn = x1, x2,…..xn

NOR-GATE REALIZATIONS To realize an Boolean function we must follow the procedure listed below• 1. Apply DeMorgan’s law to the expression to obtain expression in unary operations i.e. appear only with single variables. Draw the logic diagram using and and-or gates.• 2. Replace each or –gate symbol by nor-gate and each and-gate symbol by nor-gates symbol.• 3. Check for bubbles occurring on all lines between two gate symbols. so that two gates will be there.

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• 4. Whenever an input variable enters a gate symbol at a bubble then Complement the variable of the gate. If the output has a bubble then insert an output not-gate symbol.• 5. Replace all not-gates by a nand-gate equivalent if desired.• Example: f(w,x,y,z) = w z + w z ( x + y)

• EX-OR FUNCTION• A EX-OR gate will have two input and a output. The EX-OR gate will be logic 1 if and only if both inputs are not at same logic state.• The Boolean output expression of EX-OR gate is • f1 = x y +x y• f2 = ( x + y )( x + y )• f3 = x y + x y• f4 = (x + y ) (x + y)

• EX-NOR FUNCTION• A EX-NOR gate will have two input and a output. The EX-NOR gate will be logic 1 if and only if both inputs are at same logic state.• The Boolean output expression of EX-NOR gate is • f1 = x y + x y• f2 = ( x + y )( x + y )• f3 = x y + x y• f4 = ( x + y ) (x + y)

• The truth table of Ex-OR gate

x y f = x y + x y

0 0 0 1 1 0 1 1

0110

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• The truth table of Ex-NOR gate

Example 1:Using Boolean identities prove the following x y + x y + x y = x + yExample 2:Using perfect induction method prove the following identities. f(x, y,z) = x z + x y = ( x + y)(x + z)Example 3: Express the following in min term canonical formulas and construct the truth table.f(x,y,z) = m(0,2,4,5,7)

Example 4: Express the following in maxterm canonical formulas and construct the truth table. f(x,y,z) = M(2,4,7)Example 5:Express the following by a min -term canonical formulas without constructing truth table f(x,y,z) = x y + x z + z Ans f = m(0,1,2,3,4,6)

Example 6: Minimize the following Boolean output expression using Boolean algebra f(A, B, C) = ( A +(A +B))(B B C ) ( C ABC )

CHAPTER 2: SIMPLICATION OF BOOLEAN EXPRESSIONSSIMPLIFICATION OF BOOLEAN ALGEBRA : The given Boolean output expression has to be simplified in terms its equation nor the capacity. The following are the method used

1. Graphical method- Karnaugh method (K-map method)2. Tabular method – Quine McCluskey method.

FORMULATION OF THE SIMPLIFICATION PROBLEM : The merit of the network can be evaluated referring to the following

1. The cost of network.2. The reliability of the network.3. Propagation delay.

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x y f = x y + x y

0 0 0 1 1 0 1 1

1001

11


4. Fabrication density of the network.• CRITERIA OF MINIMALITY : Minimal overall response time- minimum

levels, minimum terms to represent the Boolean function either in SOP or POS, Single output network is to be considered.

• SIMPLIFICATION OF PROBLEM :Using Boolean algebra X+X=1, X.X= 0, X +X=X and X*X= X one can simplify the given Boolean output expression without altering basic function . Only the no. Literal sum or product terms are minimized.

PRIME IMPLICANTS AND IRRENDUNDANT DISJUCTIVE EXPRESSION : IMPLIES: Consider two complete function of n -variables f1 and f2 . The function f1 implies the function f2 if there is no assignment of values to the n-variables that makes f1 equal to 1 and f2 equals to 0.Ex. f1(x,y,z) = xy +yz and f2(x,y,z) = xy + yz+ x z

x y z f1 f2

0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1

00010011

01010011

Similarly in the conjunctive normal formula if f3 implies f4 when if there is no assignment of values to the n-variable makes f4 equal to 0 and f3 equals to 1. Ex.f3(x,y,z) = (x+y)(y+z)(x+z) and f4(x,y,z) = (x+y)(y+z)

SUSUMES: It is a comparison between two product terms or two sum terms is also possible. A term t1 is said to be subsume of term t2 if all the literal of the term t2 are also literal of the term t1.Ex. x y z is subsume of the product x z similarly ( x + y+ z ) is subsume of term (x + z)

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IMPLICANTS : A product term said to be implicant of a complete function if the product term implies the function, which describes the functional values of 1. The term x is equals to 0 for the four 3-tuples of (x,y,z) = (0,0,0), (0,0,1), (0,1,0) and (0,1,1). Hence the x is said to be implicant and other implicant that represent value 1 is y z , where it represent value 1 for two 3-tuples(x,y,z) = ( 0,0,1) and ( 1,0,1).

Truth table

x y z f

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

1 1 1 1 0 1 0 0

PRIME IMPLICANT: If the implicant does not subsumes any other implicant with fewer literals of the same function. In other words if we remove prime implicant term from the expression the remaining product terms no longer implies the function

Ex. x and y z are prime implicants

Theorem 1: When he cost assigned by some criterion, for a minimal Boolean formula is such that decreasing the number of literals in the disjunctive normal formula does not increase the cost of the formula, there is at lest one minimal disjunctive normal formula that corresponds to a sum of the prime implicants.

IRRENDUNDANT DISJUNCTIVE NORMAL FORMULAS :An irredundant disjunctive normal formula describing a complete function is defined as an expression in SOP form such that (1) every product term in the expression is a prime implicant and (2) no product term may be eliminated from the expression without changing the function described by the expression.

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IMPLICATE: A sum term is said to be implicate of a complete Boolean function if the function implies the sum term. Which describes the function value of 0.

Ex. A sum term x+y has the value 0 for two- 3 tuples (x,y,z) = (1,0,0) and (1,1,0) The sum term x +y is implied by the function hence it is referred as implicate of the function.

PRIME IMPLICATE: If the implicate does not subsumes any other implicate with fewer literals of the same function. In other words if we remove prime implicate term from the expression the remaining sum terms no longer implies the function

Ex. x and (y + z) are prime implicates

x y z f

0 0 00 0 10 1 00 1 11 0 01 0 11 1 01 1 1

00001011

IRRENDUNDANT CONJUNCTIVE NORMAL FORMULAS :An irredundant conjunctive normal formula describing a complete function is defined as an expression in POS form such that (1) every sum term in the expression is a prime implicate and (2) no sum term may be eliminated from the expression without changing the function described by the expression.

Note : It should be noted that prime implicates are dual concept of prime implicants and irredundant conjunctive normal formulas are dual concept of irredundant disjunctive normal formulas. Prime implicant f is exactly complement f will be prime implicate and vice versa.

KARNAUGH MAPS ( K- MAP) A method for graphically determining implicants and implicates of a Boolean function was developed by Veitch and modified by Karnaugh . The method involves a diagrammatic representation of a Boolean algebra. This graphic representation is called map.

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It is seen that the truth table can be used to represent complete function of n-variables. Since each variable can have value of 0 or 1. The truth table has 2n rows. Each rows of the truth table consist of two parts (1) an n-tuple which corresponds to an assignment to the n-variables and (2) a functional value.

A Karnaugh map (K-map) is a geometrical configuration of 2n cells such that each of the n-tuples corresponds to a row of a truth table uniquely locates a cell on the map. The functional values assigned to the n-tuples are placed as entries in the cells, i.e. 0 or 1 are placed in the associated cell.

An significant about the construction of K-map is the arrangement of the cells. Two cells are physically adjacent within the configuration if and only if their respective n-tuples differ in exactly by one element. So that the Boolean law x+x=1 cab be applied to adjacent cells. Ex. Two 3- tuples (0,1,1) and (0,1,0) are physically a djacent since these tuples vary by one element.

One variable : One variable needs a map of 21= 2 cells map as shown below

x f(x) 0 f(0) 1 f(1)

TWO VARIABLE : Two variable needs a map of 22 = 4 cells x y f(x,y) 0 0 f(0,0) 0 1 f(0,1) 1 0 f(1,0) 1 1 f(1,1)

THREE VARIABLE : Three variable needs a map of 23 = 8 cells. The arrangement of cells are as follows x y z f(x,y,z) 0 0 0 f(0,0,0) 0 0 1 f(0,0,1) 0 1 0 f(0,1,0)

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0 1 1 f(0,1,1) 1 0 0 f(1,0,0) 1 0 1 f(1,0,1) 1 1 0 f(1,1,0) 1 1 1 f(1,1,1)

FOUR VARIABLE : Four variable needs a map of 24 = 16 cells. The arrangement of cells are as follows w x y z f(w,x,y,z) w x y z f(w,x,y,z) 0 0 0 0 f(0,0,0,0) 1 0 1 0 f(1,0,1,0) 0 0 0 1 f(0,0,0,1) 1 0 1 1 f(1,0,1,1) 0 0 1 0 f(0,0,1,0) 1 1 0 0 f(1,1,0,0) 0 0 1 1 f(0,0,1,1) 1 1 0 1 f(1,1,0,1) 0 1 0 0 f(0,1,0,0) 1 1 1 0 f(1,1,10) 0 1 0 1 f(0,1,0,1) 1 1 1 1 f(1,1,1,1) 0 1 1 0 f(0,1,1,0) 0 1 1 1 f(0,1,1,1) 1 0 0 0 f(1,0,0,0) 1 0 0 1 f(1,0,0,1)

Four variable K-map.

0000 0001 0011 0010

0100 0101 0111 1010

1100 1101 1111 1110

1000 1001 1011 1010

Ex. Obtain the minterm canonical formula of the three variable problem given below f(x, y,z) = x y z+ x y z + x y z + x y z f(x,y,z) = m(0,2,4,5)

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00 01 11 11

Ex. Express the minterm canonical formula of the four variable K-map given below

yz 00 01 11 10

1 1 0 1

1 1 0 0

0 0 0 0

1 0 0 1

f(w,x,y,z) = w x y z + w x y z + w x y z + w x y z + w x y z + w x y zf(w,x,y,z) = m(0, 1, 2, 4, 5,

Ex. Obtain the max term canonical formula (POS) of the three variable problem stated above f(x,y,z) = ( x + y +z)( x + y +z)(x + y +z) (x + y +z) f(x,y,z) = M(1,3,6,7)Ex Obtain the max term canonical formula (POS) of the four variable problem stated abovef(w,x,y,z) = (w + x + y + z) (w + x + y + z) (w + x + y + z) (w + x + y + z) (w + x + y + z) (w + x + y + z) (w + x + y + z) (w + x + y + z) (w + x + y + z)f(w,x,y,z) = M(3,6,7,9,11,12,13,14,15)

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1 0

0 1

1 1 0 0

17


PRODUCT AND SUM TERM REPRESENTATION OFK –MAP 1.The importance of K-map lies in the fact that it is possible to determine the implicants and implicates of a function from the pattern of 0’s and 1’s appearing in the map. The cell of a K-map has entry of 1’s is refereed as 1-cell and that of 0,s is referred as 0-cell. 2. The construction of an n-variable map is such that any set of 1-cells or 0-cells which form a 2ax2b rectangular grouping describing a product or sum term with n-a-b variables , where a and b are non-negative no.s

3. The rectangular grouping of these dimensions referred as Subcubes. The subcubes must be the power of 2 i.e. 2 a+b equals to 1,2,4,8 etc.4. For three variable and four variable K-map it must be remembered that the edges are also adjacent cells or subcubes hence they will be grouped together.5. Given an n-variable map with a pair of adjacent 1-cells or 0-cellscan result n-1 variable.Where as if a group of four adjacent subcubes are formed than it can result n-2 variables. Finally if we have eight adjacent cells are grouped may result n-3 variable product or sum term.

Typical pair of subcubesw x z

1

1 1

1 1 1

1

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Typical group of four adjacent subcubes

1 1

1 1

1 1 1 1

Typical group of four adjacent subcubes.

1 1

1 1

Typical group of eight adjacent subcubes.

1 1 1 1

1 1 1 1

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1

1

1

1

1 1

1 1

1 1

1 1

1 1

1 1

19


1 1 1 1

1 1 1 1

Typical map subcubes describing sum terms

0 0

0 0

0 0

0 0

0 0

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1 1

1 1

1 1

1 1

0 0

0 0

20


USING K-MAP TO OBTAIN MINIMAL EXPRESSION FOR COMPLETE BOOLEAN FUNCTIONS :

How to obtain a minimal expression of SOP or POS of given function is discussed.

PRIME IMPLICANTS and K-MAPS :CONCEPT OF ESSENTIAL PRIME IMPLICANT00 01 11 10

0 0 0 1

0 0 1 1

f(x,y,z)= xy+ yz

ALGORITHM TO FIND ALL PRIME IMPLICANTS

A General procedure is listed below1. For an n-variable map make 2n entries of 1’s. or 0’s.2. Assign I = n , so that find out biggest rectangular group with

dimension 2ax2b = 2 n-1.3. If bigger rectangular group is not possible I = I-1 form the subcubes which consist of all the previously obtained subcube repeat the step till all 1-cell or 0’s are covered.4. Remaining is essential prime implicants

1. Essential prime implicants 2. Minimal sums 3. Minimal products

MINIMAL EXPRESSIONS OF INCOMPLETE BOOLEAN FUNCTIONS 1. Minimal sums 2. Minimal products.

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EXAMPLE TO ILLUSTRATE HOW TO OBTAIN ESSENTIAL PRIMES1. f(x,y,z) = m(0,1,5,7) Ans f(x,y,z) = xz + x y2. f(w,x,y,z) = m(1,2,3,5,6,7,8,13) Ans. f(w,x,y,z) = w z +w y+xyz+w x y z

MINIMAL SUMSf(w,x,y,z)=m(0,1,2,3,5,7,11,15)MINIMAL PRODUCTS F(w,x,y,z)=m(1,3,4,5,6,7,11,14,15)MINIMAL EXPRESSIONS OF INCOMPLETE BOOLEAN FUNCTIONSf(W,X,Y,Z)=m(0,1,3,7,8,12) +dc(5,10,13,14)

QUINE – McCLUSKEY METHOD Using K-maps for simplification of Boolean expressions with more than six variables becomes a tedious and difficult task. Therefore a tabular method illustrate below can be used for the purpose.ALGORITHM FOR GENERATING PRIME IMPLICANTSThe algorithm procedure is listed below1.Express each minterm of the function in its binary representation.2. List the minterms by increasing index.3. Separate the sets of minterms of equal index with lines.4. Let i = 0.5. Compare each term of index I with each term of index I+1. For each pair of terms that can combine which has only one bit position difference.6. Increase I by 1 and repeat step 5 . The increase of I continued until all terms are compared. The new list containing all implicants of the function that have one less variable than those implicants in the generating list.

7. Each section of the new list formed has terms of equal index. Steps 4,5, and 6 are repeated on this list to form another list. Recall that two terms combine only if they have their dashes in the same relative positions and if they differ in exactly one bit position.8. The process terminates when no new list is formed .9. All terms without check marks are prime implicants.

Example: Find all the prime implicants of the function f(w,x,y,z) = m(0,2,3,4,8,10,12,13,14)Step 1: Represent each minter in its 1-0 notation

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no. minterm 1-0 notation index

0234810121314

w x y zw x y zw x y zw x y zw x y zw x y zw x y zw x y zw x y z

0 0 0 00 0 1 00 0 1 10 1 0 01 0 0 01 0 1 01 1 0 01 1 0 11 1 1 0

012112233

Step 2: List the minterm in increasing order of their index.No. w x y z index

0248310121314

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

Index 0

Index 1

Index 2

Index 3

W x y z index

0.20,40,82,32,104,128,108,1210,1412,1312,14

0 0 – 00 – 0 0- 0 0 0

0 0 1 –- 0 1 0- 1 0 0

1 0 – 01 – 0 01 – 1 01 1 0 –1 1 - 0

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w x y z

(0, 2, 8, 10)(0, 4, 8,12 )

__ 0 __ 0__ __ 0 0(index 0)

(8,10,12,14) 1__ __ 0 (index 1)

F(w,x,y,z)=x z + y z +w z+w x y +w x z

PETRICK’S METHOD OF DETERMINING IRREDUNDANT EXPRESSIONSFIND THE PRIME IMPLICANTS AND IRREDUNDANT EXPRESSIONF(W,X,Y,Z)= M(0,1,2,5,7,8,9,10,13,15) A=X Y , B= X Z C= Y Z D= X Z

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

P = (A +C)(BD) = ABD +BCDF1(W,X,Y,Z)= ABD =X Y +X Z +X ZF2(W,X,Y,Z) = BCD = X Z + Y Z +X Z

DECIMAL METHOD FOR OBTAINING PRIME IMPLICANTS The prime implicants can be obtained for decimal number represented minterms.In this procedure binary number are not used to find out prime implicants f(w, x,y,z) =m(0,5,6,7,9,10,13,14,15) fsop= xy +xz+xyz+wyz+w x y z

MAP ENTERED VARIABLE(MEV)It is graphical approach using k-map to have a variable of order n. Where in we are using a K-map of n-1 variable while map is entered with ouput function and variable.f(w,x,y.z) = m(2,3,4,5,13,15) +dc(8,9,10,11)Ans.fsop= w z +x y + w x y

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