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

Transcription

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

2 Lesson 4 BOOLEAN EXPRESSION, TRUTH TABLE and SUM OF THE PRODUCTS (SOPs) [MINITERMS] 2

3 Outline SOP two variables cases SOP for three variable case SOP for four variable case Conversion of Boolean expression into SOPs {Finding Miniterms] 3

4 Two variable Miniterms Inputs Miniterms A B Outputs XOR AND OR NAND A B S0 S1 S2 S3 0 0 mn0= A.B mn1= A.B mn2= A.B mn3= A.B

5 XOR,AND, OR and NAND Outputs Two Variable Cases Sum of Product Terms (SOPs) XOR: S0 = A.B + A.B = Σmn(1, 2) AND: S1 = A.B = mn(3) OR: S2 = A.B + A.B + A.B = Σmn(1, 2, 3) NAND: S3 = A.B + A.B + A.B = Σmn(0, 1, 2) 5

6 SOP form advantage Advantage of using SOP form is that functions of any two input logic gate functions can be represented by maximum four ANDs at the inputs and four ORs at an output. 6

7 Outline SOP two variable cases SOP for three variable case SOP for four variable case Conversion of Boolean expression into SOPs {Finding Miniterms] 7

8 Three variable Miniterms 0 to 3 Inputs Miniterms A B C Outputs F1 F2 F3 F4 A B C S S S S mn0= A.B.C mn1= A.B. C mn2= A.B.C mn3= A.B.C

9 Three variable Miniterms 4 to 7 Inputs Miniterms A B C Outputs F1 F2 F3 F4 A B C S S S S mn4= A.B.C mn5= A.B. C mn6= A.B.C mn7= A.B.C

10 F1, F2, F3 and F4 outputs Three Variable Cases Sum of Product Terms (SOPs) F1 = S = A.B.C + A.B.C +A.B.C + A.B.C = Σmn(1, 2, 5, 6) F2 = S = A.B.C + A.B.C = Σmn(3, 7) F3 = S = A.B.C + A.B.C + A.B.C + A.B.C + A.B.C + A.B.C = Σmn(1, 2, 3, 5, 6, 7) F4 = S = A.B.C + A.B.C + A.B.C + A.B.C + A.B.C + A.B.C = Σmn(0, 1, 2, 4, 5, 6) 10

11 SOP form advantage Advantage of using SOP form is that functions of any three input logic gate functions can be represented by maximum eight ANDs at the inputs and eight ORs at an output. 11

12 Outline SOP two variable cases SOP for three variable cases SOP for four variable case Conversion of Boolean expression into SOPs {Finding Miniterms] 12

13 Four variable Miniterms 0 to 3 Inputs Miniterms A B C D Output F5 A B C D mn0= A.B.C.D mn1= A.B. C.D mn2 = A.B.C.D mn3= A.B.C.D 0 S 13

14 Four variable Miniterms 0 to 3 Inputs Miniterms A B C D Output F5 A B C D mn4= A.B.C.D mn5= A.B. C.D mn6 = A.B.C.D mn7= A.B.C.D 1 S 14

15 Four variable Miniterms 0 to 3 Inputs Miniterms A B C D Output F5 A B C D mn8= A.B.C.D mn9= A.B. C.D mn10 = A.B.C.D mn11= A.B.C.D 0 S 15

16 Four variable Miniterms 0 to 3 Inputs Miniterms A B C D Output F5 A B C D mn12= A.B.C.D mn13= A.B. C.D mn14= A.B.C.D mn15= A.B.C.D 0 S 16

17 F5 output Four Variable Case Sum of Product Terms (SOPs) F5 = S = A.B.C.D + A.B.C.D +A.B.C.D + A.B.C.D + A.B.C.D = = Σmn(0, 2, 7, 9, 12) 17

18 SOP form advantage Advantage of using SOP form is that functions of any four input logic gate functions can be represented by maximum sixteen ANDs at the inputs and sixteen ORs at an output. 18

19 Outline SOP two variable case SOP for three variable case SOP for four variable case Conversion of Boolean expression into SOPs [Finding Miniterms] 19

20 Example Suppose in a four variable SOP, there is a term with two variables, only C.D. We perform AND operation with (A + A).(B + B). [Using OR rule Equation (4) that OR of a complement of a variable or term with itself is always 1.] 20

21 Conversion C.D = (A + A).(B+B).C.D = (A + A)(B.C.D + B.C.D) = A.B.C.D + A.B.C.D + A.B.C.D + A.B.C.D = Σ mn (15, 11, 7, 3) = Σ mn (3, 7, 11, 15) and obtain the SOP standard form. 21

22 Summary 22

23 We learnt: A Boolean expression output can be written as an SOP expression SOP expression has the miniterms Each miniterm represent that row of truth table in which output = 1 Each miniterm is implemented by AND gate(s) Miniterms after ORing gives the output 23

24 We learnt: Using OR rules, a Boolean expression with lesser number of variables can be expanded into SOP form to get all the miniterms and obtain SOP standard form 24

25 End of Lesson 4 BOOLEAN EXPRESSION, TRUTH TABLE and SUM OF THE PRODUCT 25

26 THANK YOU 26