SOP to Standard SOP

8/10/2019 SOP to Standard SOP

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Sum-of-Products (SOP)


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The Sum-of-Products (SOP) Form

An SOP expression when two or moreproduct terms are

summed by Booleanaddition.

Examples:

Also: AC C B A B A

DC BCDE ABC

ABC AB

D BC C B A A

In an SOP form, asingle overbar cannotextend over more than

one variable; however,more than one variablein a term can have anoverbar:

example: is OK!

But not :

C B A

ABC


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C BC AC B AC B AC B A

BD BC BB AD AC AB DC B B A

BEF BCD AB EF CD B AB

ACD ABCD B A

)()()(

))((

)(

)(

SOP General Expression

Any logic expression can be changed into SOP formby applying Boolean algebra techniques.ex:


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The Standard SOP Form

A standard SOP expression is one in which all the variables in the domain appear in each product termin the expression.

Example:

Standard SOP expressions are important in:Constructing truth tables

The Karnaugh map simplification method

DC AB DC B ACD B A


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Converting Product Terms toStandard SOP

Step 1: Multiply each nonstandard product term by aterm made up of the sum of a missing variable and itscomplement. This results in two product terms.

As you know, you can multiply anything by 1 withoutchanging its value.

Step 2: Repeat step 1 until all resulting product termcontains all variables in the domain in eithercomplemented or uncomplemented form. Inconverting a product term to standard form, thenumber of product terms is doubled for each missing

variable.


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Converting Product Terms toStandard SOP (example)

Convert the following Boolean expression intostandard SOP form:

DC AB B AC B A

DC AB DC B A DC B A DC B ACD B A DC B ACD B A DC AB B AC B A

DC B A DC B A DC B ACD B A D DC B A D DC B A

C B AC B AC C B A B A

DC B ACD B A D DC B AC B A

)()(

)(

)(

8


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Binary Representation of a StandardProduct Term

A standard product term is equal to 1 for only onecombination of variable values.

Example: is equal to 1 when A=1, B=0, C=1,

and D=0 as shown below

And this term is 0 for all other combinations of values forthe variables.

111110101 DC B A

DC B A


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Product-of-Sums (POS)


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The Product-of-Sums (POS) Form

When two or more sumterms are multiplied, theresult expression is a

product-of-sums (POS):Examples:

Also:

In a POS form, a singleoverbar cannot extendover more than one

variable; however, morethan one variable in aterm can have anoverbar:

example: is OK!

But not :

C B A

C B A

))()((

))()((

))((

C AC B A B A

DC B E DC C B A

C B A B A

))(( DC BC B A A


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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 inthe expression.

Example:

Standard POS expressions are important in:Constructing truth tables

The Karnaugh map simplification method

))()(( DC B A DC B A DC B A


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Converting a Sum Term to StandardPOS

Step 1: Add to each nonstandard product term aterm made up of the product of the missing

variable and its complement. This results in two

sum terms. As you know, you can add 0 to anything withoutchanging its value.

Step 2: Apply rule 12 A+BC=(A+B)(A+C).Step 3: Repeat step 1 until all resulting sumterms contain all variable in the domain in eithercomplemented or uncomplemented form.


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Binary Representation of a StandardSum Term

A standard sum term is equal to 0 for only onecombination of variable values.

Example: is equal to 0 when A=0, B=1, C=0,

and D=1 as shown below

And this term is 1 for all other combinations of values forthe variables.

000001010 DC B A

DC B A


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SOP/POS


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Converting Standard SOP toStandard POS

The Facts: The binary values of the product terms in a givenstandard SOP expression are not present in theequivalent standard POS expression.

The binary values that are not represented in theSOP expression are present in the equivalent POS

expression.


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Converting Standard SOP toStandard POS

What can you use the facts?Convert from standard SOP to standard POS.

How?Step 1: Evaluate each product term in the SOPexpression. That is, determine the binary numbersthat represent the product terms.Step 2: Determine all of the binary numbers notincluded in the evaluation in Step 1.Step 3: Write the equivalent sum term for eachbinary number from Step 2 and express in POSform.


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Converting Standard SOP toStandard POS (example)

Convert the SOP expression to an equivalent POSexpression:

The evaluation is as follows:

There are 8 possible combinations. The SOP expressioncontains five of these, so the POS must contain the other 3

which are: 001, 100, and 110.))()((

111101011010000

C B AC B AC B A

ABC C B A BC AC B AC B A


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Boolean Expressions & Truth Tables

All standard Boolean expression can be easilyconverted into truth table format using binary

values for each term in the expression. Also, standard SOP or POS expression can bedetermined from the truth table.


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Converting SOP Expressions to Truth Table Format

Recall the fact: An SOP expression is equal to 1 only if at least one of theproduct term is equal to 1.

Constructing a truth table: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: Place a 1 in the output column (X) for each binary

value that makes the standard SOP expression a 1 and place 0for all the remaining binary values.


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Converting SOP Expressions to Truth Table Format (example)

Develop a truth table forthe standard SOPexpression

ABC C B AC B A

Inputs Output ProductTermA B C X

0 0 0

0 0 10 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1


0 0 0

0 0 10 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

C B A

C B A

ABC


0 0 0

0 0 1 10 1 0

0 1 1

1 0 0 1

1 0 1

1 1 0

1 1 1 1

C B A

C B A

ABC


0 0 0 0

0 0 1 10 1 0 0

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 1

C B A

C B A

ABC


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Converting POS Expressions to Truth Table Format

Recall the fact: A POS expression is equal to 0 only if at least one of theproduct term is equal to 0.

Constructing a truth table:Step 1: List all possible combinations of binary values of the

variables in the expression.Step 2: Convert the POS expression to standard form if it is

not already.Step 3: Place a 0 in the output column (X) for each binary

value that makes the standard POS expression a 0 and place 1for all the remaining binary values.


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Converting POS Expressions to Truth Table Format (example)

Develop a truth table forthe standard SOPexpression

))((

))()((C B AC B A

C B AC B AC B A


0 0 0

0 0 10 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1


0 0 0

0 0 10 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

)( C B A

)( C B A

)( C B A

)( C B A

)( C B A


0 0 0 0

0 0 10 1 0 0

0 1 1 0

1 0 0

1 0 1 0

1 1 0 0

1 1 1

)( C B A

)( C B A

)( C B A

)( C B A

)( C B A


0 0 0 0

0 0 1 10 1 0 0

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 0

1 1 1 1

)( C B A

)( C B A

)( C B A

)( C B A

)( C B A


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Determining Standard Expressionfrom a Truth Table

To determine the standard SOP expression represented by a truth table.Instructions:

Step 1: List the binary values of the input variables for which the output is 1.Step 2: Convert each binary value to the correspondingproduct term by replacing:

each 1 with the corresponding variable, andeach 0 with the corresponding variable complement.

Example: 1010 DC B A


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Determining Standard Expressionfrom a Truth Table

To determine the standard POS expression represented by a truth table.Instructions:

Step 1: List the binary values of the input variables for which the output is 0.Step 2: Convert each binary value to the correspondingproduct term by replacing:

each 1 with the corresponding variable complement, andeach 0 with the corresponding variable.

Example: 1001 DC B A


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POSSOP

Determining Standard Expressionfrom a Truth Table (example)

I/P O/P

A B C X

0 0 0 0

0 0 1 00 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 1

There are four 1s inthe output and thecorrespondingbinary value are 011,100, 110, and 111.

ABC

C AB

C B A

BC A

111

110

100

011

There are four 0s inthe output and thecorrespondingbinary value are 000,001, 010, and 101.

C B A

C B A

C B A

C B A

101

010

001

000

ABC C ABC B A BC A X

))()()(( C B AC B AC B AC B A X


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The Karnaugh Map


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The Karnaugh Map

Feel a little difficult using Boolean algebra laws,rules, and theorems to simplify logic?

A K-map provides a systematic method forsimplifying Boolean expressions and, if properlyused, will produce the simplest SOP or POSexpression possible, known as the minimumexpression.


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The 3 Variable K-Map There are 8 cells as shown:

C AB

0 1

00

01

11

10

C B A C B A

C B A BC A

C AB ABC

C B A C B A


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The 4-Variable K-Map

CD AB

00 01 11 10

0001

11

10 DC B A

DC AB

DC B A

DC B A

DC B A

DC AB

DC B A

DC B A

CD B A

ABCD

BCD A

CD B A

DC B A

D ABC

D BC A

DC B A


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K-Map SOP Minimization

The K-Map is used for simplifying Booleanexpressions to their minimal form.

A minimized SOP expression contains thefewest possible terms with fewest possible

variables per term.Generally, a minimum SOP expression can beimplemented with fewer logic gates than astandard expression.


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Mapping a Standard SOP Expression

For an SOP expressionin standard form:

A 1 is placed on the K-map for each productterm in the expression.Each 1 is placed in a cellcorresponding to the

value of a product term.

Example: for the productterm , a 1 goes in the101 cell on a 3-variablemap.

C AB

0 1

00

01

11

10

C B A C B A

C B A BC A

C AB ABC

C B A C B A

C B A

1


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C AB

0 1

00

01

11

10

Mapping a Standard SOP Expression(full example)

The expression:

C B AC ABC B AC B A

000 001 110 100

1 1

1

1

DC B A DC B A DC AB ABCD DC AB DC B ACD B A

C B AC B A BC A

ABC C ABC B AC B A

Practice:


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Mapping a Nonstandard SOPExpression

Numerical Expansion of a Nonstandard product term Assume that one of the product terms in a certain 3-variableSOP expression is .

It can be expanded numerically to standard form as follows:Step 1: Write the binary value of the two variables and attach a 0 forthe missing variable : 100.Step 2: Write the binary value of the two variables and attach a 1 forthe missing variable : 100.

The two resulting binary numbers are the values of thestandard SOP terms and .

If the assumption that one of the product term in a 3- variable expression is B. How can we do this?

C

B A

C

C B A C B A


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Mapping a Nonstandard SOPExpression

Map the following SOP expressions on K-maps:

D BC A D AC DC A

CD B A DC B A DC B AC AB B AC B

C A BC

C AB B A A


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K-Map Simplification of SOPExpressions

After an SOP expression has been mapped, wecan do the process of minimization:

Grouping the 1sDetermining the minimum SOP expression from themap


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Grouping the 1s

You can group 1s on the K-map according tothe following rules by enclosing those adjacentcells containing 1s.

The goal is to maximize the size of the groupsand to minimize the number of groups.


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Grouping the 1s (rules)1. A group must contain either 1,2,4,8,or 16 cells

(depending on number of variables in the expression)2. Each cell in a group must be adjacent to one or more

cells in that same group, but all cells in the group donot have to be adjacent to each other.

3. Always include the largest possible number of 1s in agroup in accordance with rule 1.

4. Each 1 on the map must be included in at least one

group. The 1s already in a group can be included inanother group as long as the overlapping groupsinclude noncommon 1s.


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Grouping the 1s (example)

C AB 0 1

00 1

01 1

11 1 110

C AB 0 1

00 1 1

01 1

11 110 1 1


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Grouping the 1s (example)

CD AB 00 01 11 10

00 1 1

01 1 1 1 1

11

10 1 1

CD AB 00 01 11 10

00 1 1

01 1 1 1

11 1 1 1

10 1 1 1


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Determining the Minimum SOPExpression from the Map

The following rules are applied to find theminimum product terms and the minimumSOP expression:

1. Group the cells that have 1s. Each group of cellcontaining 1s creates one product term composedof all variables that occur in only one form (eithercomplemented or complemented) within the

group. Variables that occur both complementedand uncomplemented within the group areeliminated called contradictory variables .


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2. Determine the minimum product term for eachgroup.

For a 3-variable map:1. A 1-cell group yields a 3-variable product term2. A 2-cell group yields a 2-variable product term3. A 4-cell group yields a 1-variable product term4. An 8-cell group yields a value of 1 for the expression.

For a 4-variable map:1. A 1-cell group yields a 4-variable product term2. A 2-cell group yields a 3-variable product term3. A 4-cell group yields a 2-variable product term4. An 8-cell group yields a a 1-variable product term5. A 16-cell group yields a value of 1 for the expression.


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3. When all the minimum product terms are derivedfrom the K-map, they are summed to form theminimum SOP expression.


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Determining the Minimum SOPExpression from the Map (example)

CD AB

00 01 11 10

00 1 1

01 1 1 1 1

11 1 1 1 1

10 1

B

C A

DC A

DC AC A B


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Determining the Minimum SOPExpression from the Map (exercises)

C B A BC AB

C AB 0 1

00 1

01 1

11 1 1

10

C AB 0 1

00 1 1

01 1

11 1

10 1 1

AC C A B


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Determining the Minimum SOPExpression from the Map (exercises)

D B AC A B A C BC B A D

CD AB 00 01 11 10

00 1 1

01 1 1 1 1

11

10 1 1

CD AB 00 01 11 10

00 1 1

01 1 1 1

11 1 1 1

10 1 1 1


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Practicing K-Map (SOP)

DC B A D ABC D BC A DC B A

CD B ACD B A DC AB DC B A DC B

C B AC B AC B A BC AC B A

C A B

C B D


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Mapping Directly from a Truth Table

I/P O/P

A B C X

0 0 0 1

0 0 1 00 1 0 0

0 1 1 0

1 0 0 1

1 0 1 0

1 1 0 1

1 1 1 1

C AB

0 1

00

01

11

10

1

1

1

1


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C AB

0 1

00

01

11

10

Mapping a Standard POS Expression(full example)

The expression:

))()()(( C B AC B AC B AC B A

000 010 110 101

0

0

0

0


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K-map Simplification of POSExpression

))()()()(( C B AC B AC B AC B AC B A

C

AB0 1

00

01

11

10 B A

0 0

0 0

0 AC

C B

A

1

11

)( C B A

AC B A


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Rules of Boolean Algebra

1.6

.5

1.4

00.3

11.2

0.1

A A

A A A

A A

A

A

A A

BC AC A B A

B A B A A

A AB A

A A

A A

A A A

))(.(12.11

.10

.9

0.8

.7

___________________________________________________________A, B, and C can represent a single variable or a combination of variables. 7

SOP to Standard SOP

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