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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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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
The term 0)( DCBA when A=0, B=1, C=1 and D=0
00110)( DCBA
The term 0)( DCBA when A=1, B=1, C=1 and D=1
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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F tm S n Z ini