lecture2 - CHAPTER II BOOLEAN CHAPTER II BOOLEAN ALGEBRA...

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Unformatted text preview: CHAPTER II BOOLEAN CHAPTER II BOOLEAN ALGEBRA Logic circuit 1) combinational circuits Logic circuit 1) combinational circuits 2) sequential circuits a) synchronous b) asynchronous Combinational circuit: output depends only on the inputs now. Sequential circuit: output depends both the inputs now and the inputs past. Synchronous sequential circuit: has a clock to synchronize Asynchronous sequential circuit: has no clock Logic representation There will be a party iff you have time and money, or you have a very good friend come to visit you. T: you have time M: you have money V: friend visit P: party P = T • M +V • AND; + OR You have to know the inputs and output(s), and the relationship between them. Truth table: T 0 1 0 1 0 1 0 1 M 0 0 1 1 0 1 1 1 V 0 0 0 0 1 1 1 1 P 0 0 0 1 1 1 1 1 Definition: A Boolean algebra is a closed algebraic system containing a set of two or more elements and two binary operators + (OR) and ∙ (AND). X +Y ∈K That is, for every X and Y in set K, and X ⋅Y ∈ K Also the following postulates must be satisfied by a Boolean algebra. Postulates P1. Existence of an identity element 0 with respect to the operator + and of an identity element 1 with respect to the operator ∙ . (a) X+0=X (b) X∙1=X P2. Commutativity of the operators + and ∙ . (a) X+Y=Y+X (b) X∙Y=Y∙X P3. Associativity of the operators + and ∙ . (a) X+(Y+Z)=(X+Y)+Z (b) X∙(Y∙Z)=(X∙Y)∙Z P4. dostributivity (a) + over ∙ : X+(Y∙Z)=(X+Y)∙(X+Z) (b) ∙ over + : X∙(Y+Z)=X∙Y+X∙Z X' X ∈K P5. Existence of the complement for every , so that (a) X + X ' = 1 (b) X ⋅ X ' = 0 1. closure: A set is closed with respect to a binary operator is operator unique a, b ∈ S C ∈S → EX. Set of natural numbers is closed with respect to + Set of natural numbers is closed with respect to × but not – and / If S is real number set, then respect to – and / is also closed. Theorems and properties of Boolean algebra 1. Operator precedence a. parentheses high b. NOT c. AND d. OR low 2. Equality of expressions Two expressions are equivalent if they have identical values for all the combinations of the values of their variables. 3. Duality The dual of an expression is obtained by replacing OR by AND and AND by OR, 1 by 0 and 0 by 1, and by preserving the existence of all parentheses. If an equation is valid, then its dual is also valid. Product term: by ANDing two or more variables Sum term: by ORing two or more terms A Boolean function can be represented ad either sum of product (SOP) terms or product of sum (POS) terms. Theorems 1. Idempotency (a) X+X=X (b) X∙X=X 2. properties of 1 and 0 (a) X+1=1 (b) X∙0=0 3. absorption (a) X+XY=X (b) X∙(X+Y)=X 4. absorption (a) (b) X + X 'Y = X + Y X ⋅ (X ' + Y ) = X ⋅Y 5. DeMorgan’s Law ( X ⋅ Y )' = X ' + Y ' ( X + Y )' = X ' ⋅ Y ' (a) (b) After the theorem is proved, it can be used for other proofs Note: the expansion to n variables for some theorems using induction. To change an expression, change each variable to its complement and change each 1 to 0, each 0 to 1, + to ∙, and ∙ to +. Prove the equality of expressions: 1. truth table They have same values for all of the combinations. 2. find SOP or POS forms to see they are the same 3. using the theorems and postulates You will get a good mark iff you understand the material in the class, do exercised and read the problems of the exams carefully. G: good mark M: understand the material E: do exercises R: read questions carefully G=M∙E∙R Write truth table; Write the logic representation for adder; Summation and carry S i , Ci Ai , Bi , Ci −1 Inputs: outputs: Ai Bi Ci −1 Si Ci 000 0 0 S i = Ai Bi' Ci'−1 + Ai' Bi Ci'−1 + Ai' Bi'Ci −1 + Ai Bi Ci −1 100 1 0 010 1 0 110 0 1 001 1 0 101 0 1 011 0 1 Ci = Ai Bi Ci'−1 + Ai Bi'Ci −1 + Ai' Bi Ci −1 + Ai Bi Ci −1 111 1 1 2. From truth table A. canonical SOP form: 1. General a product term corresponding to each row in which the value of the function is 1. 2. In each product term, if the value is 1, then use the variable, if is 0, then use its complement. B. canonical POS form 1. Generate a sum term corresponding to each row in which the value is 0. 2. In each sum term, if the value is 1, then use its complement, if the value is 0, then use the variable itself. EX. '' The SOP form F = A B + AB = ∑ (0,3) ' ' The POS form F = ( A + B)( A + B ) = ∏(1,2) A B F 0 0 0 1 1 0 1 0 2 1 0 0 3 1 1 1 A function in SOP form is said to be in canonical SOP form if each of A function in SOP form is said to be in canonical SOP form if each of the product terms contains each and every variable or its complement. Each term in a canonical SOP form is called a minterm. A function in POS form is said to be in canonical POS form if each of the sum terms contains each and every variable or its complement. Each term in a canonical POS is called a maxterm. How to get canonical form: 1. Algebraic Using distribution law and adding some terms having value o or multiply some terms having value 1. Q=AB+AC=AB(C+C’)+A(B+B’)C=ABC+ABC’+ABC+AB’C =ABC+ABC’+AB’C P=(A’+B)(A+B+C)=(A’+B+CC’)(A+B+C) =(A’+B+C)(A’+B+C’)(A+B+C) EX: F =ABC ' D +A' BCD +ABC ' D ' 1101 0111 1100 =∑7,12,13) ( P = ( A + B ' + C + D ' )( A' + B + C + D ' )( A + B + C ' + D ) 0101 1001 0010 = ∏ ( 2,5,9) EX: Q ( A, B, C , D ) = ∑(0,1,7,8,10,11,12,15) Given Note: Q is four variable function and there will be 24 = 16 Combinations whose decimal values range from 0 to 15. It has 8 minters and should have 16­8=8 maxterms Q ( A, B, C , D ) = ∏ ( 2,3,4,5,6,9,13,14) Prove X+X=X 1. X+X=(X+X)1 postulate 1(b) =(X+X)(X+X’) 5(a) =X+XX’ 4(b) =X+0 5(b) =X Note: give the reason for each step clearly 2. Another way to prove: consider all combinations if X=0 LHS=X+X=0+0=0 1(a) RHS=X=0 if X=1 LHS=1+1 RSH=1=1+0 1(a) =1+1∙0 1(b) =(1+1)(1+0) 4(a) =(1+1)1 1(a) =1+1 1(b) After proving, you can use it for other proves 3. truth table (note different from ordinary algebra: 1+1=1 and + over ∙ distribution) X X+X 0 0 1 1 F=X’YZ’+X’Y’Z F’=(X’YZ’+X’Y’Z)’ =(X’YZ’)’(X’Y’Z)’ =(X+Y’+Z)(X+Y+Z’) Also prove XF+X’G+FG=XF+X’G (F, G are Boolean functions) XF+X’G+FG=XF+X’G+FG(X+X’) =XF+XFG+X’G+X’GF =XF(1+G)+X’G(1+F) =XF+X’G (X+F)(X’+G)(F+G)=(X+F)(X’+G) Prove by duality of the above result ...
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