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MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 1 Boolean Function Boolean function is an expression form containing binary variable, two-operator binary which.

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Presentation on theme: "MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 1 Boolean Function Boolean function is an expression form containing binary variable, two-operator binary which."— Presentation transcript:

1 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 1 Boolean Function Boolean function is an expression form containing binary variable, two-operator binary which is OR and AND, and operator NOT, sign ‘ and sign = Answer is also in binary We always use sign ‘.’ for AND operator, ‘+’ for OR operator, ‘’’ or ‘  ’ for NOT operator. Sometimes we discard ‘.’ sign if there is no contradiction

2 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 2 Boolean Function Example: From TT we see that F3=F4 Can you prove it using Boolean Algebra?

3 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 3 Complement Function Given function F, complement function for this function is F’, it is obtained by exchanging 1 with 0 on the output function F. Example: F1=xyz’ Complement F1’ = (xyz’)’ = x’+y’+(z’)’ (DeMorgan) = x’+y’+z (Involution)

4 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 4 Complement Function Generally, complement function can be obtained using repeatedly DeMorgan Theorem (A+B+C+…..+Z)’=A’.B’.C’.….Z’ (A.B.C.…..Z)’=A’+B’+C’+.….+Z’

5 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 5 Standard Form There are two standard form: Sum-of-Product (SOP) and Product-of- Sum (POS) Literals: Normal variable or in complement form. Example: x, x’, y, y’ **Product: single literal or several literals with logical product (AND) Example: x, xyz’, A’B, AB

6 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 6 Standard Form **Sum: single literal or several literals with logical sum (OR) Example: x, x+y+z’, A’+B, A+B Sum-of-Product (SOP) expression: single product or several products with logical sum (OR) Example: x, x+yz’,xy’+x’yz, AB+A’B’ Product-of- Sum (POS) expression:single sum or several sum with logical product (AND) Example: x, x.(y+z’),(x+y’)(x’+y+z), (A+B)(A’+B’)

7 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 7 Standard Form Every Boolean expression can be written either inSum-of-Product (SOP) expression or Product-of- Sum (POS)

8 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 8 Minterm & Maxterm Consider two binary variable x,y Every variable can exist as normal literal or in complement form (e.g. x,x’,&y,y’) For two variables, there are four possible combinations with operator AND such as: x’y’,x’y,xy’,xy This product is called minterm Minterm for n variables is the number of “product of n literal from the different variables”

9 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 9 Minterm & Maxterm Generally, n variable will produce 2 n minterm With similar approach, maxterm for n variables is “sum of n literal from the different variables” Example: x’+y’, x’+y, x+y’, x+y Generally, n variable will produce 2 n maxterm

10 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 10 Minterm & Maxterm Minterm and maxterm for 2 variables each is signed with m0 to m3 and M0 to M1. Every minterm is the complement of suitable maxterm Example: m2=xy’ m2’=(xy’)’=x’+(y’)’=x’+y’=M2

11 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 11 Canonical Form What is canonical/normal form? –It is unique form to represent something Minterm is “product term’ –Can state Boolean Function in Sum-of-Minterm

12 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 12 Canonical Form: Sum of Minterm (SOM) Produce TT: Example

13 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 13 Canonical Form: Sum of Minterm (SOM) Produce Sum-of-Minterm by collecting minterm for the function (where the answer is 1)

14 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 14 Canonical Form: Product of Minterm (POM) Maxterm is “sum term” For Boolean function, maxterm for function is term with answer 0 Can state Boolean function in Product-of- Maxterm form

15 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 15 Canonical Form: Product of Minterm (POM) Example:

16 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 16 Canonical Form: Product of Minterm (POM) Why? Take F2 as example Complement function for F2 is

17 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 17 Canonical Form: Product of Minterm (POM) From the previous slide F2’=m0+m1+m2 Therefore: Each Boolean function can be written in Sum-of- Product and Product-of-Sum expression

18 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 18 Canonical Form: Conversion SOP  POS Sum-of-Minterm => Product-of-Maxterm –Change  m to  M –Insert minterm which is not in SOM –E.g. F1(A,B,C)=  m(3,4,5,6,7)=  M(0,1,2) Product-of-Maxterm => Sum-of-Minterm –Change  M to  m –Insert maxterm which is not in POM –E.g. F2(A,B,C)=  M(0,3,5,6)=  m(1,2,4,7)

19 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 19 Canonical Form: Conversion SOP  POS Sum-of-Minterm for F => Sum-of-Minterm for F’ –Minterm list which is not in SOM of F E.g. Product-of-Maxterm for F => Product-of- Maxterm for F’ –Maxterm list which is not in POM of F E.g.

20 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 20 Canonical Form: Conversion SOP  POS Sum-of-Minterm for F => Product-of-Maxterm for F’ –Change  m to  M –E.g. F1(A,B,C)=  m(3,4,5,6,7) F1’(A,B,C)=  M(3,4,5,6,7) Product-of-Maxterm for F=> Sum-of-Minterm for F’ –Change  M to  m –E.g. F2(A,B,C)=  M(0,1,2) F2’(A,B,C)=  m(0,1,2)

21 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 21 Binary Function If n variable, therefore the are 2 n possible minterm Each function can be expressed by Sum-of- Minterm, therefore there are 2 2 different function In two variable case, there is 2 2 =4 possible minterm, and there is 2 4 =16 different binary function The 16 binary function is presented in the next slide n

22 MOHD. YAMANI IDRIS/ NOORZAILY MOHAMED NOOR 22 Binary Function


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