Basic Digital Electronics - Unit 4new

8/21/2019 Basic Digital Electronics - Unit 4new

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At the end of the lesson, students should be able to:

1. Explain the INVERTER, AND, OR, NAND and NOR gate using BooleanAlgebra.

2. Describe DeMorganstheorems to Boolean expression and evaluate Booleanexpression.

3. Simplification using Boolean algebra into Sum-Of-Product (SOP) form.4. Explain of dont care condition.

4 BOOLEAN ALGEBRA


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UNIT 4: BOOLEAN ALGEBRA

Fatma Syazana Zaini

Pegawai Latihan Vokasional

IKM Besut, 2010

78

INTRODUCTION

Boolean algebra is the mathematic of digital system. This topic covers laws, rules and

theorem of Boolean algebra and their application to digital circuits. You will also learning

the Boolean operations and expressions in terms of their relationship to NOT, AND, OR,

NAND and NOR gates introduce.

Boolean algebra is the mathematics of digital systems. A basic knowledge of Boolean

algebra is indispensable to the study and analysis of logic circuits.

Variable, complement and literalVariable, complement and literal are the terms used in Boolean algebra.

i. Variable - Symbol (usually an italic uppercase letter to

represent a logical quantity)

- Single variable can have a 1 or 0 value.

ii. Complement - Is the inverse of a variable and is indicated by abar over the variable (over bar)

- Example: AA , if A = 1 then 0A

iii. Literal - Aliteral is a variable or the complement of a

variable.

- For example: B indicates the complement of B.

4.1NOT, AND, OR, NAND AND NOR GATES USING

BOOLEAN ALGEBRA


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Boolean AdditionBoolean addition is equivalent to the OR operation and the basic rules are

illustrated with their relation to the OR gate as follows:

In Boolean algebra, a sum term is a sum of literals. Some examples of sum terms

are:

_ _ _ _

A + B, A + B, A + B + C and A + B + C + D.

A sum term is equal to 1 when one or more of the literals in the term are1. A sum

term is equal to 0 only if each of the literals is 0.

Example 1:

Determine the values of A, B, C and D that make the sum term

DCBA equal to 0.

Solution;

000001010 DCBA

Boolean MultiplicationBoolean Multiplication is the equivalent to the AND operation and the basic rules

are illustrated with their relation to the AND gate as follows:


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In Boolean algebra, a product term is the product of literals. Some examples of

product terms are

andABCDCBAABBA ,,

A product term is equal to 1 only if each of the literals in the term is 1.

A product term is equal to 0 when one or more of the literals are 0.

Example2:

Determine the values of A, B, C and D that make the product term DCBA

equal to 1.

Solution;

11.1.1.10.1.0.1 DCBA

Laws And Rules Of Boolean Algebra Laws of Boolean Algebra

The basic of Boolean algebra: -

1. A + B = B + A

2. AB = BA


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3. A + (B + C) = (A + B) + C

4. A (BC) = (AB) C

5. A (B + C) = AB + AC

Twelve Basic Rules of Boolean AlgebraThere are 12 basic rules that are useful in manipulating and simplifying

Boolean expressions.

Rule 1. A + 0 = A


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Rule 2. A + 1 = 1

Rule 3. A . 0 = 0

Rule 4. A . 1 = A

Rule 5. A + A = A

Rule 6.


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Rule 11. BABAA

BAABABAA - rule 10: A = A + AB

BAABAA

- rule 7: A = AA

BAAAABAA - rule 8: adding AA = 0

BAAA - factoring

BA1 - rule 6: 1 AA

BA - rule 4: drop the 1

Rule 12. (A + B)(A + C) = A + BC

BCABACAACABA

BCABACA - rule 7: AA = A

BCABCA 1 - factoring

BCABA )1( - rule 2: 1 + C=1

BCBA 1 - factoring

BCA 1 - rule 2: 1 + B=1

BCA - rule 4: A .1 =A


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DeMorgans theorem is important because it show us how to find the equivalent of NANDand NOR gates.

State DeMorgans TheoremTwo DeMorgans theorem are: -

a.

b.

Relate DeMorgans Theorem to the equivalency of the: -i. NAND and negative-OR gates

ii. NOR and negative-AND gates

4.2 DEMORGANS THEOREMS

Sign .Change to +, breaks the bar

Sign + Change to ., breaks the bar

Inputs Output 0 0 1 10 1 1 11 0 1 11 1 0 0

Inputs Output 0 0 1 10 1 0 01 0 0 01 1 0 0


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Notice the equality of the two output columns in each truth table. This

shows that the equivalent gates perform the same logic function.

Example 1:

Apply DeMorgans theorems to the expression XYZand ZYX

Solution;

XYZ ZYX

ZYX ZYX

Applying DeMorgans TheoremExample 1:

Applying DeMorgans theorem to the expression below:

Solution;

Assume and From DeMorgan rule (b):

(1)

(2)Insert (2) into (1)


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Example 2:

Simplify the expression

Solution:

Step 1:DeMorgan Rule (a) : Assume and

Step 2: DeMorgan Rule (b) :

Example 3:

Simplify the expression

Solution:

Step 1 : Rule 1: A+(B+C)=(A+B)+CStep 2 : DeMorgan Rule (b) :


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When two or more product terms are summed by Boolean addition, the resulting

expression is a sum-of-products (SOP). Some examples are: -

ABCAB

DCBCDEABC

ACCBABA

Also an SOP expression can contain a single-variable term as in A + A B C + B C D. In an

SOP expression, a single over bar cannot extend over more than one variable; however,

more than one variable in a term can have an over bar. For example, an SOP expression can

have the term CBA but not ABC

Domain of a Boolean expressionThe domain of a general Boolean expression is the set of variables contained in the

expression in either complemented or un-complemented form.

For example: -

i.

CBABA is the set of variables A, B and Cii. DCBEDCCAB is the set of variables A, B, C, D and E

Implementation of an SOP expression.Implementation an SOP expression simply requires from the outputs of two or more

AND gates. A product term is produced by an AND operation, and the sum

(addition) of two or more product terms is produced by an OR operation.

Therefore, an SOP expression can be implemented by AND-OR logic in which the

output of a number of AND gates connect to the inputs of an OR gate. Example AB+ BCD + AC in figure 4.1.

4.3SIMPLIFICATION USING BOOLEAN ALGEBRA INTO SUM OF

PRODUCTS (SOP) AND PRODUCT OF SUM (POS) FORM


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Figure 4.1: Implementation of the SOP expression AB + BCD + AC

Convert a General Expression to SOP FormExample 1:

Convert each of the following Boolean expression to SOP form:

a. A B + B (CD + EF)b. (A + B) (B + C + D)c. CBA

Solution;a. A B + B (CD + EF)

=

b. (A + B) (B + C + D)=AB+AC+AD+BB+BC+BD

= AB+AC+AD+B+BC+BD

c.


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The Standard SOP FormA standard SOP expression is one in which all the variables in the domain appear in

each product term in the expression. For example: -

DCABDCBACDBA is a standard SOP expression.

Convert Product Terms to Standard SOPStep 1: multiply each nonstandard product term by made up of the sum of a

missing variable and its complement.

Step 2: repeat step 1 until all resulting product terms contain all variables in the

domain in either complemented or un-complemented form.

Example 1:

Convert the following Boolean expression into standard SOP form:

DCABBACBA

Solution;

DCABBACBA - A B C is missing variable D or D DCBACDBADDCBA )(

BA - BA is missing variable C or Cand D or D CBACBACCBA )(

DCBACDBADDCBA )(

DCBADCBADDCBA )(

DCABBACBA

DCABDCBADCBA

DCBACDBADCBACDBA


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Binary Representation of a Standard Product TermA standard product term is equal to 1 for only one combination of variable

values.

Example 1:

The product term 1DCBA ; when A = 1, B = 0, C = 1 and D = 0

111110101 DCBA

Example 2:

Determine the binary values for the following standard SOP expression equal to

1.

DCBADCBAABCD

Solution;

The term 1ABCD ; when A = 1, B = 1, C = 1 and D = 1

11111 ABCD

The term 1DCBA ; when A = 1, B = 0, C = 0 and D =1

111111001 DCBA

The term 1DCBA ; when A = 0, B = 0, C = 0 and D = 0

111110000 DCBA

The SOP expression equals 1 when any or all of the three product terms is 1


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Simplification Using Boolean Algebra Into Product Of Sums (Pos) Form

Product-of-Sums (POS).When two or more sum terms are multiplied, the resulting expression is a

product-of-sums (POS). Some examples are: -

))(( CBABA

))()(( DCBEDCCBA

))()(( CACBABA

A POS expression can contain a single-variable term as in

))(( DCBCBAA . In a POS expression, a single over bar cannot extend

over more than one variable; however, more than one variable in a term can have

an over bar. Example, a POS expression can have term CBA but not

CBA .

Implementation of a POS Expression.POS expression simply requires AND in the outputs of two or more OR gates. A

sum term is produced by an OR operation, and the product of two or more sum

terms is produced by AND operation. Therefore, a POS expression can beimplemented by logic in which the outputs of a number of OR gates connect to

the inputs of an AND gate.

Example (A + B)(B + C + D)(A +C) in Figure 4.2.

Figure 4.2: Implementation of the POS expression (A + B)(B + C + D)(A +C)


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The Standard POS Form.A standard POS expression is one in which all the variables in the domain appear

in each sum term in the expression. For example: -

Is a standard POS expression, any nonstandard POS expression can be

converted to the standard form using Boolean algebra.

Convert a Sum Term to Standard POSProcedure: -

1. Add to each nonstandard product term a term made up of the product ofthe missing variable and its complement. This results in two sum terms.

2. Apply rule 12: A + B C =(A + B)(A + C)3. Repeat step 1 until resulting sum terms contain all variables in the domain

in either complemented or un-complemented form.

Example 1:

Convert the following Boolean expression into standard POS form.))()(( DCBADCBCBA

Solution;

)( CBA - Missing DD - Add DD

))(( DCBADCBA

)( DCB - Missing AA - Add AA

))(( DCBADCBA


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))()()()(( DCBADCBADCBADCBADCBA


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Binary representation of a standard sum Term.

A standard sum term is equal to 0, for only one combination of variables

values.

Example 2:

The sum term DCBA = 0 when A = 0, B = 1, C = 0 and D = 1

1010 DCBA

Example 3:

Determine the binary values of the variables for which the following standard

POS expression is equal to 0.

))()(( DCBADCBADCBA

Solution;

The term 0)( DCBA when A=0, B=0, C=0 and D=0

00000)( DCBA


00110)( DCBA


01111)( DCBA

The POS expression equals 0 when any of the three SUM terms equals 0.


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Convert Standard SOP to Standard POS

The binary values of the product terms in a given standard SOP expression arenot present in the equivalent standard POS expression. Also, the binary values

that are not represented in the SOP expression are present in the equivalent

POS expression. Therefore, to convert from standard SOP to standard POS,

the following steps are taken:

Step 1: Evaluate each product term in the SOP expression. That is, determine

the binary numbers that represent the product terms.

Step 2: Determine all of the binary numbers not included in the evaluation in

step 1.

Step 3: Write the equivalent sum term for each binary number from step 2

and express in POS form.

Example 4:

Convert the following SOP expression to an equivalent POS expression

ABCCABCBACBACBA

Solution;

ABCCABCBACBACBA = 000 010 101 110 111Variables = 3

N=2n= 23= 8 possible combinations

Truth table

A B C SOP/POS

0 0 0 SOP0 0 1 POS0 1 0 SOP0 1 1 POS1 0 0 POS1 0 1 SOP1 1 0 SOP1 1 1 SOP


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So, the POS must contain the other three, which are 001, 011 and 100. The Different Between SOP and POS Form

SOP POS

Equation

From Truth Table Output =1Where A = 1

A = 0

Equation

From Truth Table Output = 0Where A = 0

A = 1

Sometimes a situation arises in which some input variable combinations are not allowed. For

example, recall that in the BCD code, there are six invalid combinations: 1010, 1011, 1100,

1101, 1110 and 1111, Since these un-allowed states will never occur in a application involving

BCD code, they can be treated as dont careterms with respect to their effect on the

output. That is, for these dont care terms either a 1 or 0 may be assigned to the output;

it really does not matter since they will never occur.

The dont care terms can be used to advantage on the Karnaughmap. Figure 4.3 shows

that for each dont care term, an X is placed in the cell. When grouping the 1s, the Xs

can be treated as 1s to make larger grouping or as 0s if they cannot be used to advantage.

The larger a group, the simpler the resulting term will be. The truth table in figure 4.3(a)

describes a logic function that has a 1 output only when the BCD code for 7, 8 or 9 ispresent on the inputs. If dont cares are used as 1s, the resulting expression for the

function is A + BCD, as indicated in part (b). If the dont care are not used as 1s, the

resulting expression is:

4.3USE OF DONT CARE CONDITION TO SIMPLIFY LOGIC

FUNCTION


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So you can see the advantage of using dont care terms to get the simplest expression.

Figure 4.3: Example of the use of dont care condition to simplify an expression

In this topic, we have learn about the operation of the basic logic gate like the INVERTER

or NOT, AND, OR, NAND and NOR gate. Student also should be understand about the

operation of the exclusive-OR gate and exclusive-NOR gate.

Student should be able to identify the shape of logic gate symbols according to the

ANSI/IEEE (American National Standard Institute/ International Electrical Electronic

Engineering). To understand more about digital signal input and output, student must beconstruct timing diagrams that showing the proper time relationships of inputs and outputs

for the various logic gates.

Student also learned about the characteristic of IC CMOS(Complimentary Metal Oxide

Semiconductor) and TTL(Transistor-Transistor Logic) families to know the differ from

each other in propagation delay time, power dissipation, speed-power product and fan-out,

SUMMARY


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They also learned how to troubleshoot the logic gates for opens and shorts by using the

oscilloscope.

1. If A = 0, what does A equal?______________________________

2. Determine the values of A, B and C that make the sum term CBA equal to 0._____________________________

3. Determine the values of A, B and C that make the product term CAB equal to 1.______________________________

4. Apply the associative law of addition to the expression A + (B + C + D)

5. Apply the distributive law to the expression A (B + C + D).

6. Which of the following rules states that if one input of an AND gate is always 1, theoutput is equal to the other input?

a. A + 1 = 1b. A + A = Ac. A .A = Ad. A .1 = A

EXERCISE


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7. Apply DeMorgans theorems to each of the following expressions:

a. (A + B + C) Db. ABC + DEF

c. AB + CD + EF

8. Convert A B C + (A + B)(B + C + A B) to SOP form.

9. Convert the expression YXWYZXYWX to standard form.

10.Convert the expression (A + B)(B + C) to standard POS form.

11. Determine the binary values for the POS expression below equal to 0))()()()(( ZYXZYXZYXZYXZYX

1. Digital System Principle And Applications, Tocci, R.J, Prentice Hall international2. Digital Fundamentals, Floyd T.L, Merrill Publishing.3. BPL(K) Module : TFV 2033 Digital Electronics 1.4. Digital Electronics (Teaching Module), KUITHO.

REFERENCE


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Basic Digital Electronics - Unit 4new

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