1. ECE- 1551 DIGITAL LOGIC LECTURE 8: BOOLEAN FUNCTIONS Assistant Prof. Fareena Saqib Florida Institute of Technology Spring 2016, 02/11/2016

2. Recap  Chapter 2: Boolean Functions  Operator Precedence  Demorgan’s Theorem  Simplification Theorems  Complement of Function using Demorgan’s Theorem  Canonical and Standard Forms

3. Agenda  Complement of a Function with more than 2 variables  Representation of Boolean Expression in Canonical form  Minterms and Maxterms  Representation of Boolean Expression in Standard Forms  Sum of Products  Product of Sums  Representation of Boolean Expression in Non-standard form  Conversion between different forms.

4. Complement of a Boolean Function  The complement of a function F is F  Is obtained from an interchange of 0’s for 1’s and 1’s for 0’s in the value of F in truth table.  The complement of a function may be derived algebraically through DeMorgan’s theorems as described for two variables and can be extended for three or more variables.  The three ‐ variable form of the first  DeMorgan’s theorem is derived as follows, from postulates and theorems listed in Table 2.1 :  (A + B + C) = (A + x) let B + C = x = Ax by theorem 5(a) (DeMorgan) = A(B + C) substitute B + C = x = A(BC) by theorem 5(a) (DeMorgan) = ABC by theorem 4(b) (associative)  The generalized form of DeMorgan’s theorems states that the complement of a function is obtained by interchanging AND and OR operators and complementing each literal. (NOTE)

5. Complement of Function: Example  F1 = x’yz’ + x’y’z and F2 = x(y’z’ + yz). F1’ =? F1’ =?  F1’ = (x’yz’ + x’y’z )’ = (x’yz’ )’(x’y’z)’ = (x + y’ + z)(x + y + z’ )  F2’ = [x(y’z’ + yz)]’ = x’ + (y’z’ + yz)’ = x’ + (y’z’ ) (yz)’ = x’ + (y + z)(y’ + z’ ) = x’ + yz’ + y’z

6. Complement of Function: Example  Note: A simpler procedure for deriving the complement of a function is to take the dual of the function and complement each literal.  Dual of a function is obtained from the interchange of AND and OR operators and interchanging identity elements1’s and 0’s.  F1 = x’yz’ + x’y’z and F2 = x(y’z’ + yz). F1’ =? F1’ =? F1’ = (x’yz’ + x’y’z)’ = (x’yz’)’ (x’y’z)’ = ((x’)’ + y’ + (z’)’ )(x + y + z’) = (x+y’+z)(x+y+z’) F2’ = x(y’z’ + yz) = x’ + (y’z’+yz)’ = x’ + [(y’z’) (yz) ] = x’ + (y+z)(y’+z’) = x’ + yy’ + zy’ +yz’+zz’ = x’ + zy’ + yz’

7. Complement of Function: Example  A simpler procedure for deriving the complement of a function is to take the dual of the function and complement each literal. Where dual of a function is obtained from the interchange of AND and OR operators and 1’s and 0’s.  F1 = x’yz’ + x’y’z and F2 = x(y’z’ + yz).  F1’ =? F1’ =? F1’ = (x’yz’ + x’y’z) = (x+y’+z) (x+y+z’) F2’ = x(y’z’ + yz) = x’+ (y+z) (y’+z’)

8. Canonical and Standard Forms  In canonical form, boolean function is expressed using min terms and max terms.  In Standard form, boolean function is expressed in the form of sum of products and products of sums.

9. Canonical and Standard Forms  For a given operation “and” with two variables x and y, there are four possible combinations: xy, x’y, xy,’ and x’y.’  Each of these four AND terms is called a minterm, or a standard product.  In a similar fashion, n variables forming an OR term  Each variable being primed or unprimed, provide 2 n possible combinations, called maxterms, or standard sums  Boolean function in terms of minterms from truth table  Any Boolean function can be expressed as a sum of minterms (with “sum” meaning the ORing of terms).  Boolean function in terms of maxterm from truth table:  Any Boolean function can be expressed as a product of maxterms (with “product” meaning the Anding of terms).  Conversion between canonical forms  Boolean function in terms of POS from a given SOP

10. Canonical Form using minterm and maxterms A Boolean function can be expressed algebraically from a given truth table by forming a minterm for each combination of the variables that produces a 1 in the function and then taking the OR of all those terms. Example: f1 = x’y’z + xy’z’ + xyz It may be read from the truth table by forming a minterm for each combination that produces a 1 in the function and then ORing those terms Example: f1 = x’y’z + xy’z’ + xyz = m1 + m4 + m7 = M0. M2. M3. M5. M6

11. Canonical Form using minterm and maxterms  A Boolean function can be expressed algebraically from a given truth table by forming a maxterm for each combination of the variables that produces a 0 in the function.  Each maxterm is obtained from an OR term of the n variables, with each variable being unprimed if the corresponding bit is a 0 and primed if a 1 Example: f1 = x’y’z + xy’z’ + xyz = m1 + m4 + m7 = M0. M2. M3. M5. M6

12. Conversion Between Canonical Forms: From SOP to POS  Example: F1 = x’y’z + xy’z’ + xyz = m1 + m4 + m7  F1’ = m0 + m2+m3+m5+m6  F1’ = x’y’z’ + x’yz’+ x’yz +xy’z+xyz’  Take complement of f1’ = ?  F1 = x’y’z’ + x’yz’+ x’yz +xy’z+xyz’ F1 = x’y’z’ + x’yz’+ x’yz +xy’z+xyz’ F1 = (x+y+z)(x+y’+z)(x+y’+z’)(x’+y+z’)(x’+y’+z)

13. Conversion Between Canonical Forms: … Contd  The complement of a function expressed as the sum of minterms equals the sum of minterms missing from the original function.  Consider an example F(A, B, C) = ∑ (1, 4, 5, 6, 7)  This function has a complement that can be expressed as  F’ (A, B, C) = ∑ (0, 2, 3) = m0 + m2 + m3  Complement of F’ by demorgan’s theorem, we obtain F in form of maxterms  F = (m0 + m2 + m3) = (m’0 m’2 m’3) = M0M2M3 = ∏ (0, 2, 3) where following relation holds: m i ’ =M i  Example F= xy+x’z??

14. Practice Example  Examples  F1 = x’y’z + xy’z’ + xyz  F2 = x’yz + xy’z + xyz’ + xyz  Part 1) Express Boolean function in terms of minterms  Part 2) Express Boolean function in terms of maxterms

15. Standard Form  Another way to express Boolean functions is in standard form.  The two canonical forms of Boolean algebra are basic forms that one obtains from reading a given function from the truth table.  These forms are very seldom the ones with the least number of literals, because each minterm or maxterm must contain, by definition, all the variables, either complemented or uncomplemented  The terms that form the function may contain one, two, or any number of literals. There are two types of standard forms: the sum of products and products of sums.  The sum of products is a Boolean expression containing AND terms, called product terms, with one or more literals each.  The sum denotes the ORing of these terms.

16. Standard Form : Sum of Products - Example  An example of a function expressed as a sum of products is  F1 = y’ + xy + x’yz’  The expression has three product terms, with one, two, and three literals.  Their sum is, in effect, an OR operation.  Logic diagram of a sum-of-products?  The logic sum is formed with an OR gate whose inputs are the outputs of the AND gates and the single literal.  This circuit configuration is referred to as a two ‐ level implementation. (assumed that the input variables are directly available in their complements)

17. Standard Form : Products of Sum - Example  An example of a function expressed as a sum of products is  F2 = x(y’ + z)(x’ + y + z’)  The expression has three sum terms, with one, two, and three literals.  Their product is, in effect, an AND operation.  Logic diagram of a sum-of-products?  The logic sum is formed with an OR gate whose inputs are the outputs of the AND gates and the single literal.  This circuit configuration is referred to as a two ‐ level implementation as well. (assumed that the input variables are directly available in their complements)

18. Example 2.4 is conversion from standard form to canonical form  Express the Boolean function F = A + B’C as a sum of minterms.  The function has three variables: A, B, and C.  The first term A is missing two variables; therefore  A = A(B + B‘ ) = AB + AB‘ AB = AB(C+C‘) AB‘ = AB‘(C+C‘) AB = ABC+ABC‘ AB‘ = AB‘C + AB‘C‘ A = ABC+ABC‘ + AB‘C + AB‘C‘  The second term B‘C is missing one variable; therefore  B‘C = B‘C(A+A‘) B‘C = B‘CA + B‘CA‘  F = A + B‘C = ABC+ABC‘ + AB‘C + AB‘C‘ + B‘CA + B‘CA‘ = ABC + ABC‘ + AB‘C + AB‘C‘ + B‘CA‘  F= ∑(1,4,5,6,7) = m1+m4+m5+m6+m7

19. Example 2.5 is conversion from standard form to canonical form  Express the Boolean function F = xy + x’z as a product of maxterms.  First, convert the function into OR terms by using the distributive law:  F = xy + x’ z = (xy + x’ )(xy + z) = (x + x’ )(y + x’ )(x + z)(y + z) = (x’ + y)(x + z)(y + z)  The function has three variables: x, y, and z. Each OR term is missing one variable  X’ + y = x’ + y + zz’ = (x’ + y + z)(x’ + y + z’ )  x + z = x + z + yy’ = (x + y + z)(x + y’ + z)  y + z = y + z + xx’ = (x + y + z)(x’ + y + z)  Combining all the terms and removing those which appear more than once, we obtain  F = (x + y + z)(x + y’ + z)(x’ + y + z)(x’ + y + z’ ) = M 0 M 2 M 4 M 5

20. Next Class  Karnaugh Maps or K-map  2,3 and 4 input K-Map  Don’t Care conditions