max terms theorems, min terms and boolean algebra and€¦ · chapter 4 boolean algebra and...
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Chapter 4
BOOLEAN ALGEBRA AND THEOREMS, MIN TERMS AND
MAX TERMS
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
2
Lesson 5
BOOLEAN EXPRESSION, TRUTH TABLE and
product of the sums (POSs) [MAXTERMS]
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
3
Outline
• POS two variables cases• POS for three variable case • POS for four variable case• Conversion of Boolean expression
into POSs {Finding Maxterms]
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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)
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
7
Outline
• POS two variable cases•• POS for three variable casePOS for three variable case• POS for four variable case• Conversion of Boolean expression
into POSs {Finding Maxterms]
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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’’’
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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’’’
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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)
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
12
Outline
• POS two variable cases• POS for three variable cases • POS for four variable case• Conversion of Boolean expression
into POSs {Finding Maxterms]
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
13
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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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Four variable Maxterms 0 to 3
0 0 1 0 Mx4= A+B+C + D 1
Inputs Maxterms Output A B C D F10
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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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Four variable Maxterms 0 to 3
0 0 0 1 Mx8= A+B+C + D 1
Inputs Maxterms Output A B C D F10
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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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Four variable Maxterms 0 to 3
0 0 1 1 Mx12= A+B+C+D 0
Inputs Maxterms Output A B C D F10
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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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)
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
19
Outline
• POS two variable case• POS for three variable case • POS for four variable case• Conversion of Boolean expression
into POSs [Finding Maxterms]
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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).
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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)
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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Summary
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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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.
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
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End of Lesson 4
BOOLEAN EXPRESSION, TRUTH TABLE and
product of the sums (POSs) [MAXTERMS]
Ch04L4--"Digital Principles and Design", Raj Kamal, Pearson Education, 2006
28
THANK YOU