MAX TERMS THEOREMS, MIN TERMS AND BOOLEAN ALGEBRA AND€¦ · chapter 4 boolean algebra and theorems, min terms and max terms

Chapter 4

BOOLEAN ALGEBRA AND THEOREMS, MIN TERMS AND

MAX TERMS

Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006

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Lesson 5

BOOLEAN EXPRESSION, TRUTH TABLE and

product of the sums (POSs) [MAXTERMS]


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Outline

• POS two variables cases• POS for three variable case • POS for four variable case• Conversion of Boolean expression

into POSs {Finding Maxterms]


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Two variable Maxterms

0 0 Mx0= A+B 0 0 0 1

Inputs Maxterms Outputs A B XOR AND OR NAND

1 0 Mx1= A+B 1 0 1 10 1 Mx2= A+B 1 0 1 11 1 Mx3= A+B 0 1 1 0

A B P0 P1 P2 P3


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XOR,AND, OR and NAND Outputs Two XOR,AND, OR and NAND Outputs Two Variable Cases Sum of Product Terms (POSs)Variable Cases Sum of Product Terms (POSs)

XOR: P0 = (A+B) .(A+B ) =

ΠΠΠΠ Mx(0, 3)

AND: P1 = (A+B).(A+B ).(A+B) =

ΠΠΠΠ Mx (0, 1, 2)OR: P2 = (A+B) =

Mx(0)

NAND: P3 = (A+B) = Mx(3)


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Advantage of using POS form is that functions of any two input logic gate functions can be represented by maximum four ORs at the inputs and four ANDs at an output.

POS form advantage


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Outline

• POS two variable cases•• POS for three variable casePOS for three variable case• POS for four variable case• Conversion of Boolean expression



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Three variable Maxterms 4 to 7

0 0 0 Mx0= A+B+C 0 1 1 0

Inputs Maxterms Outputs A B C F6 F7 F8 F9

1 0 0 Mx1= A+B+C 1 1 0 00 1 0 Mx2= A+B+C 1 1 0 01 1 0 Mx3= A+B+C 0 0 0 1

A B C P P’ P’’ P’’’


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Three variable Maxterms 0 to 3

0 0 1 Mx4= A+B+C 0 1 1 0

Inputs Maxterms Outputs A B C F6 F7 F8 F9

1 0 1 Mx5= A+B. C 1 1 0 00 1 1 Mx6= A+B .C 1 1 0 11 1 1 Mx7= A+B .C 0 0 0 1

A B C P P’ P’’ P’’’


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F6, F7, F8 and F9 outputs Three Variable F6, F7, F8 and F9 outputs Three Variable Cases Sum of Product Terms (POSs)Cases Sum of Product Terms (POSs)

F6 = P = (A+B+C). (A+B+C). (A+B+C). (A+B+C) =

ΠΠΠΠ Mx (0, 3, 4, 7)F7 = P’= (A+B+C). (A+B+C) =

ΠΠΠΠ Mx(3, 7)F8 = P’’ = (A+B .C). (A+B +C) . (A+B .C ). (A+B +C) . (A+B +C) . (A+B +C) =

ΠΠΠΠ Mx(1, 2, 3, 5, 6, 7)F9 = P’’’ = (A+B +C). (A+B+C). (A+B +C). (A+B+C) . (A+B + C)=

ΠΠΠΠ Mx(0, 1, 2, 4, 5)


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Advantage of using POS form is that functions of any three input logic gate functions can be represented by maximum eight ORs at the inputs and eight ANDs at an output.

POS form advantage


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Outline

• POS two variable cases• POS for three variable cases • POS for four variable case• Conversion of Boolean expression



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Four variable Maxterms 0 to 3

0 0 0 0 Mx0= A+B+C + D 0

Inputs Maxterms Output A B C D F10

0 1 0 0 Mx1= A+B + C + D 11 0 0 0 Mx2 = A+B + C + D 01 1 0 0 Mx3= A+B + C + D 1

A B C D P


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0 0 1 0 Mx4= A+B+C + D 1


0 1 1 0 Mx5= A+B + C + D 11 0 1 0 Mx6 = A+B + C + D 11 1 1 0 Mx7= A+B + C + D 1

A B C D P


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0 0 0 1 Mx8= A+B+C + D 1


0 1 0 1 Mx9= A+B. C + D 11 0 0 1 Mx10 = A+B+C + D 11 1 0 1 Mx11= A+B + C + D 1

A B C D P


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0 0 1 1 Mx12= A+B+C+D 0


0 1 1 1 Mx13= A+B + C+D 11 0 1 1 Mx14= A+B + C+D 11 1 1 1 Mx15= A+B+C+D 1

A B C D P


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F5 output Four Variable Case Sum of Product Terms (POSs)

F5 = P = (A+B+C+D) . (A+B+C+D) .(A + B+ C + D) . = ΠΠΠΠ Mx(0, 2, 12)


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Advantage of using POS form is that functions of any four input logic gate functions can be represented by maximum sixteen ORs at the inputs and sixteen ANDs at an output.

POS form advantage


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Outline

• POS two variable case• POS for three variable case • POS for four variable case• Conversion of Boolean expression

into POSs [Finding Maxterms]


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Step 1: Perform additional OR operation in the max-term with a term containing that variable ANDed with complement of that and get two POS max-terms using a distributive law X +(Y. Z) = (X+Y). (X+Z) with Y and Z as variable and its complement, respectively.

Finding the max-terms and converting to an n-variable POS standard format


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SStep2: Continuing ORing till all n variable are present in each term of the n-variable POS and 2 Step3: Repeat the process for each term that has a missing variable in the Boolean expression.

Finding the max-terms and converting to an n-variable POS standard format


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• Suppose in a four variable POS, there is a term with three variables, only A+C+D. We perform OR operation with (B.B ) . = (A+B + C + D).(A+B + C +D) = ΠΠΠΠ Mx(0, 2)using distributive law

Example finding the max-terms and converting to an n-variable POS standard

format (A+C+D). (A+ B+C+ D).


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Taking X = A + C + D, Y = B and Z = B.]

(A+B.B + C + D) = (A+B + C + D) . (A+B + C + D) = Mx0 . Mx2Mx2 repeated term is deleted because Mx0 . Mx2. Mx2 = Mx0 . Mx2. [AND law A.A = A]

Using distributive law X +(Y. Z) = (X+Y) . (X+Z)


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Summary


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We learnt:• A Boolean expression output can be

written as an POS expression• POS expression has the Maxterms• Each Maxterm represent that row of

truth table in which output = 0• Each Maxterm is implemented by OR

gate(s)• Maxterms after ANDing gives the

output


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We learnt:• Using AND rules and distribution

law, a Boolean expression with lesser number of variables can be expanded into POS form to get all the Maxterms and POS standard form.


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End of Lesson 4

BOOLEAN EXPRESSION, TRUTH TABLE and

product of the sums (POSs) [MAXTERMS]


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THANK YOU

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MAX TERMS THEOREMS, MIN TERMS AND BOOLEAN ALGEBRA AND€¦ · chapter 4 boolean algebra and theorems, min terms and max terms