Chapter 4 BOOLEAN ALGEBRA AND THEOREMS, MIN TERMS AND MAX TERMS

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1 Chapter 4 BOOLEAN ALGEBRA AND THEOREMS, MIN TERMS AND MAX TERMS

2 Lesson 5 BOOLEAN EXPRESSION, TRUTH TABLE and product of the sums (POSs) [MAXTERMS] 2

3 Outline POS two variables cases POS for three variable case POS for four variable case Conversion of Boolean expression into POSs {Finding Maxterms] 3

4 Two variable Maxterms Inputs Maxterms A B Outputs XOR AND OR NAND A B P0 P1 P2 P3 0 0 Mx0= A+B Mx1= A+B Mx2= A+B Mx3= A+B

5 XOR,AND, OR and NAND Outputs Two Variable Cases Sum of Product Terms (POSs) XOR: P0 = (A+B).(A+B ) = Π Mx(0, 3) AND: P1 = (A+B).(A+B ).(A+B) = Π Mx (0, 1, 2) OR: P2 = (A+B) = Mx(0) NAND: P3 = (A+B) = Mx(3) 5

6 POS form advantage 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. 6

7 Outline POS two variable cases POS for three variable case POS for four variable case Conversion of Boolean expression into POSs {Finding Maxterms] 7

8 Three variable Maxterms 4 to 7 Inputs Maxterms A B C Outputs F6 F7 F8 F9 A B C P P P P Mx0= A+B+C Mx1= A+B+C Mx2= A+B+C Mx3= A+B+C

9 Three variable Maxterms 0 to 3 Inputs Maxterms A B C Outputs F6 F7 F8 F9 A B C P P P P Mx4= A+B+C Mx5= A+B. C Mx6= A+B.C Mx7= A+B.C

10 F6, F7, F8 and F9 outputs Three Variable Cases Sum of Product Terms (POSs) F6 = P = (A+B+C). (A+B+C). (A+B+C). (A+B+C) = Π Mx (0, 3, 4, 7) F7 = P = (A+B+C). (A+B+C) = Π Mx(3, 7) F8 = P = (A+B.C). (A+B +C). (A+B.C ). (A+B +C). (A+B +C). (A+B +C) = Π Mx(1, 2, 3, 5, 6, 7) F9 = P = (A+B +C). (A+B+C). (A+B +C). (A+B+C). (A+B + C) = Π Mx(0, 1, 2, 4, 5) 10

11 POS form advantage 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. 11

12 Outline POS two variable cases POS for three variable cases POS for four variable case Conversion of Boolean expression into POSs {Finding Maxterms] 12

13 Four variable Maxterms 0 to 3 Inputs Maxterms A B C D Output F10 A B C D Mx0= A+B+C + D Mx1= A+B + C + D Mx2 = A+B + C + D Mx3= A+B + C + D 1 P 13

14 Four variable Maxterms 0 to 3 Inputs Maxterms A B C D Output F10 A B C D Mx4= A+B+C + D Mx5= A+B + C + D Mx6 = A+B + C + D Mx7= A+B + C + D 1 P 14

15 Four variable Maxterms 0 to 3 Inputs Maxterms A B C D Output F10 A B C D Mx8= A+B+C + D Mx9= A+B. C + D Mx10 = A+B+C + D Mx11= A+B + C + D 1 P 15

16 Four variable Maxterms 0 to 3 Inputs Maxterms A B C D Output F10 A B C D Mx12= A+B+C+D Mx13= A+B + C+D Mx14= A+B + C+D Mx15= A+B+C+D 1 P 16

17 F5 output Four Variable Case Sum of Product Terms (POSs) F5 = P = (A+B+C+D). (A+B+C+D).(A + B+ C + D). = Π Mx(0, 2, 12) 17

18 POS form advantage 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. 18

19 Outline POS two variable case POS for three variable case POS for four variable case Conversion of Boolean expression into POSs [Finding Maxterms] 19

20 Finding the max-terms and converting to an n-variable POS standard format 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. 20

21 Finding the max-terms and converting to an n-variable POS standard format 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. 21

22 Example finding the max-terms and converting to an n-variable POS standard format (A+C+D). (A+ B+C+ D). 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) = Π Mx(0, 2) using distributive law 22

23 Using distributive law X +(Y. Z) = (X+Y). (X+Z) Taking X = A + C + D, Y = B and Z = B.] (A+B.B + C + D) = (A+B + C + D). (A+B + C + D) = Mx0. Mx2 Mx2 repeated term is deleted because Mx0. Mx2. Mx2 = Mx0. Mx2. [AND law A.A = A] 23

24 Summary 24

25 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 25

26 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. 26

27 End of Lesson 4 BOOLEAN EXPRESSION, TRUTH TABLE and product of the sums (POSs) [MAXTERMS] 27

28 THANK YOU 28

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