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Boolean Algebra

A. B. C. B × C . A+B × C . 0 0 0 0 1 1 1 1. 0 0 1 1 0 0 1 1. 0 1 0 1 0 1 0 1. 0 0 1 0 0 0 1 0. 0 0 1 0 1 1 1 1. A. B. A + B × C . C. Boolean Algebra. Purpose of BA is to facilitates design and analysis of digital circuits.

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Boolean Algebra

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  1. A B C B×C A+B×C 00001111 00110011 01010101 00100010 00101111 A B A+B×C C Boolean Algebra Purpose of BA is to facilitates design and analysis of digital circuits. For example the value of boolean function F=A + BC’ with the following gate implementation can be shown by this truth table:

  2. Boolean Algebra 1. X + 0 =X 3. X + 1 = 1 5. X + X=X 7. X + X ’ = 1excluded middle 9. (X ’)’ =X involution 10. X + Y=Y + X 12. X+(Y+Z ) = (X+Y )+Z 14. X(Y+Z ) =X×Y + X×Z 16. (X+ Y )=X ×Y  18. X + XY = X 2. X×1 =X identity 4. X×0 = 0 base 6. X×X=X idempotence 8. X×X’ = 0 non contradiction 11. X×Y=Y×X commutative 13. X×(Y×Z ) = (X×Y )×Z associative 15. X+(Y×Z ) =(X+Y )×(X+Z ) distributive 17. (X×Y)’ = X+Y  demorgan 19. X.(X+Y) = x absorbtion • Basic identities of boolean algebra

  3. X+(Y×Z) =(X+Y)×(X+Z) (X+ Y)=X×Y XYZ Y×Z X+(Y×Z) X+Y X+Z (X+Y)×(X+Z) XY (X+ Y) X×Y 1 00 1 0 0 0 0 0 000 0 01 0 0 1 001 0 0 0 0 10 0 0 0 010 0 0 1 0 11 0 1 1 1 011 1 1 100 1 1 1 0 1 1 1 101 1 1 0 0 110 1 1 1 1 1 1 1 1 111 1 Basic identities of B.A. can be proven by truth table: demorgan distributive

  4. Boolean Algebra • For each algebraic expression the dual of of algebraic expression achieved by interchanging AND and OR operators and replacing 0’s and 1’s. • Parallel columns illustrate duality principle. The duality principle states that if E1 and E2 are Boolean expressions then E1= E2 dual (E1)=dual (E2) where dual(E) is the dual of E • Note: 15-17 have no counterpart in ordinary algebra. • Other handy identity. X+X’Y=X+Y (15, 7 and 2)

  5. Boolean Algebra • By using boolean algebra rules, a simpler expression may be obtained • Operator Precedence: when evaluating boolean expression , order of precedence is: • Parentheses • NOT • AND • OR For example :Look at DeMorgan truth table first (x+Y) is computed then complement of (X+Y). But for x’.y’ first the complement of x and complement of y is computed and then the result is ANDed

  6. Boolean Function Simplification

  7. Boolean function simplification • It means by manipulation of B.A. reducing the number of terms and literals in the function. For example: f= x’y’z + x’yz + xy’ = x’(y’z + yz) + xy’ = x’ (z(y’+y)) + xy’ = x’(z.1) +xy’ = x’z + xy’

  8. The Consensus Theorem Theorem. XY +YZ + X Z = XY + X Z Proof. XY +YZ + X Z = XY + (X + X )YZ + X Z 2,7 = XY + XYZ + X YZ + X Z 14 = XY(1 + Z )+ X Z(Y +1) 2,11,14 = XY + X Z 3,2 Dual. (X + Y )(Y + Z )(X  + Z ) = (X + Y )(X  + Z )

  9. Complement of a Function There are two ways for doing that: • Using DeMorgan’s theorem • Taking the dual of the function and complement each literal For example complements of x’yz’ + x’y’z = (x+y’+ z)(x+ y+z’) x(y’z’ + yz) = x’ + (y+z)(y’ + z’)

  10. Canonical and standard Forms • The sum of products is one of two standard forms for Boolean expressions. sum-of-products-expression = p-term+p-term ... +p-term p-term = literal×literal×××× ×literal • example. X Y Z + X Z + XY + XYZ • A minterm is a term that contains every variable, in either complemented or un-complemented form. • example. in expression above, X Y Z is minterm, but X Z is not • A sum of minterms expression is a sum of products expression in which every term is a minterm. • example:X Y Z + X YZ + XYZ  + XYZ is sum of minterms expression that is equivalent to expression above. • shorthand : list minterms numerically, so X Y Z + X YZ +XYZ  + XYZ becomes 001+011+110+111 or Sm (1,3,6,7)

  11. Canonical and standard Forms • The product of sums is the second standard form for Boolean expressions. product-of-sums-expression = s-term×s-term ... ×s-term s-term = literal+literal+××× +literal • example. (X +Y +Z )(X +Z )(X +Y )(X +Y +Z ) • A maxterm is a sum term that contains every variable, in complemented or uncomplemented form. • example. in exp. above, X +Y +Z is a maxterm, but X +Z is not • A product of maxterms expression is a product of sums expression in which every term is a maxterm. • example.(X +Y +Z )(X +Y+Z )(X+Y+Z )(X+Y+Z )is product of maxterms expression that is equivalent to expression above. • shorthand : list maxterms numerically: so, (X +Y +Z )(X +Y+Z) (X+Y+Z )(X+Y+Z ) becomes 110+100+001+000 or • P M(6,4,1,0)

  12. How to build the boolean function from truth table • One way is to find a minterms or standard products by ANDing the terms of the n variable, each being primed if it is 0 and unprimed if it is 1. A boolean function can be formed by forming a minterm for each combination of variables that produce 1 in the function and then taking OR of all those forms. • Another way is by finding maxterms or standard sums by OR term of the n variables, with each variable being unprimed if corresponding bit is 0 and primed if it is 1. A boolean function can be formed as a product of maxterms for each combination of variables that produce 0 in the function and then form And of all those forms

  13. For example: x y z function f1 function f2 minterms maxterms 0 0 0 0 0 m0 M0 0 0 1 1 0 m1 M1 0 1 0 0 0 m2 M2 0 1 1 0 1 m3 M3 1 0 0 1 0 m4 M4 1 0 1 0 1 m5 M5 1 1 0 0 1 m6 M6 1 1 1 1 1 m7 M7 Sum of minterms f1 = x’y’z + xy’z’ + xyz = m1 + m4 + m7 f2= x’yz + xy’z + xyz’ + xyx = m3 + m5 + m6 + m7 Product of maxterms f1= (x+y+ z)(x+y’+z)(x’+ y + z’)(x’ + y + z) = M0M2M3M5M6 f2= (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z) = M0M1M2M4

  14. Conversion between Canonical Forms • By reading from a truth table the two canonical forms ( sum of minterms and product of maxterms) can be obtained. • A boolean function can be converted to the canonical form. For example: F= xy + x’y (in form of sum of the products S.O.P) = (z+ z’)xy + x’z(y+y) = xyz + xyz’ + x’zy + x’zy’ can be converted to sum of minterms that can be shown by m1+m3+m6+m7 or ∑(1,3,6,7)

  15. Conversion between Canonical Forms • To convert it to the product of maxterms F= xy + x’y = (xy + x’)(xy +z) = (x’ + x) (x’+ y)(x+z)(y+z) = (x’+y)(x+z)(y+z) it is in the form of products of sums (P.O.S) but we want the product of maxterms. So = (x’+y+(z.z’)) (x + z + (y.y’))(y +z + (x.x’)) = (x’ + y + z)(x + z + y)(x+ z + y’)(x’ + y +z’) = M0M2M4M5=∏(0,2,4,5)

  16. Conversion between Canonical Forms • To convert from one canonical form to another, interchange the symbol ∑ and ∏ and list those numbers missing from the total number of minterms or maxterms which is 2n,where n is number of variables. • To verify that we convert sum of minterms to the product of maxterms by finding the complement of a function presented as sum of minterms. From the result the product of maxterms can be easily obtained For example: F(x,y,z) = m1+m3+m6+m7= ∑(1,3,6,7) the complement of F (presented in the form of sum of the minterms) are the minterms that makes F to be zero thus F’(x,y,z) = (∑(1,3,6,7))’= (m0 + m2 + m4 + m5) F = (F’(x,y,z))’ = (m0 + m2 + m4 + m5)’ Using Demorgan’s =m0’m2’m4’m5’= since each m’j = Mj it is M0M2M4M5 = ∏(0,2,4,5)

  17. Standard forms • Sometimes boolean functions are shown as standard forms. For example: F1 =y’ + xy + x’y’z’ ( sum of products) F2 = x(y’ + z) (x’ + y + z’) (product of sums) the product and sum can be used to make the gate structure consist of AND and OR gates • Sometimes boolean function can be shown in non standard forms: F3= AB + C(D + E) can be changed to AB + CD + CE • Different forms results different level of implementation of logical gates (see the next slide)

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