A. Abhari CPS2131 Chapter 2: Boolean Algebra and Logic Gates Topics in this Chapter: Boolean Algebra Boolean Functions Boolean Function Simplification.

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A. Abhari CPS2131 Chapter 2: Boolean Algebra and Logic Gates Topics in this Chapter: Boolean Algebra Boolean Functions Boolean Function Simplification Canonical and Standard Forms Minterms and Maxterms Building Boolean Function from the Truth Table Conversion between Canonical Forms Digital Logic Gates

A. Abhari CPS2132 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: A B C A+BCA+BC ABC BCBCA+B  C 0000111100001111 0011001100110011 0101010101010101 0010001000100010 0010111100101111

A. Abhari CPS2133 Boolean Algebra Basic identities of boolean algebra 1. X  0  X 3.X  1  1 5.X  X  X 7.X  X ’  1 excluded 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

A. Abhari CPS2134 Basic identities of B.A. can be proven by truth table: X  Y  Z )   X  Y )  X  Z ) YZYZ 0 0 0 1 0 0 0 1 XYZ 000 001 010 011 100 101 110 111 X  Y  Z ) 0 0 0 1 1 1 1 1 XYXY 0 0 1 1 1 1 1 1 XZXZ 0 1 0 1 1 1 1 1  X  Y )  X  Z ) 0 0 0 1 1 1 1 1  X  Y )  X  Y XY 00 01 10 11 X Y X Y 1 0 0 0  X  Y ) 1 0 0 0 demorgan distributive

A. Abhari CPS2135 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 E 1 and E 2 are Boolean expressions then E 1 = E 2  dual (E 1 )=dual (E 2 ) 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)

A. Abhari CPS2136 Boolean Algebra By using boolean algebra rules, a simpler expression may be obtained Operator Precedence: when evaluating boolean expression, order of precedence is: 1- Parentheses 2-NOT 3-AND 4-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

A. Abhari CPS2137 Boolean Function Simplification

A. Abhari CPS2138 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’

A. Abhari CPS2139 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 )

A. Abhari CPS21310 Complement of a Function There are two ways for doing that: 1)Using DeMorgan’s theorem 2)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’)

A. Abhari CPS21311 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 product 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  m (1,3,6,7)

A. Abhari CPS21312 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 –  M(6,4,1,0)

A. Abhari CPS21313 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

A. Abhari CPS21314 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’)(x’ + y + z) = M0M2M3M5M6 f2= (x+y+z)(x+y+z’)(x+y’+z)(x’+y+z) = M0M1M2M4

A. Abhari CPS21315 Conversion between Canonical Forms To convert it to the product of maxterms F= xy + x’z = (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) Easier than this is by using the truth table

A. Abhari CPS21316 Conversion between Canonical Forms By reading from a truth table the two canonical forms ( sum of minterms and product of maxterms) can be easily obtained. A boolean function can be converted to the canonical form. For example: F= xy + x’z (is in form of the sum of the products S.O.P) by doing = (z+ z’)xy + x’z(y+y) = xyz + xyz’ + x’zy + x’zy’ it can be converted to sum of minterms. Using the truth table the sum of minterms can be shown by m1+m3+m6+m7 or ∑(1,3,6,7)

A. Abhari CPS21317 Conversion between Canonical Forms In general 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 2 n,where n is number of variables. To prove the correctness of the above conversion method lets consider the following example: F(x,y,z) = m1+m3+m6+m7= ∑(1,3,6,7) We know the complement of F (presented in the form of sum of the minterms) is 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 thus M0M2M4M5 = ∏(0,2,4,5)

A. Abhari CPS21318 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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