Podstawy technologii cyfrowej [ang], Notatki z Informatyka w języku angielskim

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[ If A = 1, then Ā = 0 ] [ Ā - "not A" or "A bar" ; sometimes “ A’ “ ; “ ~A “] Literal is a variable or the complement of a variable. Tautology Truth Tables AND ( both must be true ) x y x ∧ y T T T T F F F T F F F F OR ( at least one true) x y x ∨ y T T T T F T F T T F F F NOT ( reverse ) x ~x (NOT x) T F F T IF => THEN ( second one is the result value ) x y x ⇒ y T T T T F F F T T F F T IF AND ONLY IF ( If they re different = False; Both same = True ) x y x ⇔ y T T T T F F F T F F F T T = 1 F = 0 BOOLEAN ALGEBRA TRUE = 1 -> HIGH voltage FALSE = 0 -> LOW voltage [ ^THEY RE THE SAME AS BEFOR^ ] rule (2a) -> The order of taking AND gates is not important BOOLEAN ADDITION A sum term is equal to 1 when one or more of the literals in the term are 1. A sum term is equal to 0 only if each of the literals is 0. BOOLEAN MULTIPLICATION Boolean multiplication == AND operation Product term is equal to 1 only if each of the literals in the term is 1. Product term is equal to 0 when one or more of the literals are 0. The Sum-of-Products (SOP) Form <- When two or more product terms are summed by Boolean addition In an SOP expression a single overbar cannot extend over more than one variable. [ AB + ABC ; ABC + CDE + BCD ] The Standard SOP Form <- is one in which all the variables in the domain appear in each product term in the expression. [ ABCD + ABCD + ABCD ] Converting Product Terms to Standard SOP: A B’ C + A’ B’ + A B C’ D -> 1) A B’ C ( D is missing ) -> A B’ C ( D + D’ ) = A B’ C D + A B’ C D’ 2) A ‘ B’ ( C and D are missing ) -> A’ B’ ( C + C’ ) = A’ B ‘ C + A’ B’ C’ A’ B’ C ( D + D’ ) + A’ B’ C’ ( D + D’ ) = A’ B’ C D + A’ B’ C D’ + A’ B’ C’ D + A’ B’ C’ D’ Result: A B’ C + A’ B’ + A B C’ D -> A B’ C D + A B’ C D’ + A’ B ‘ C + A’ B’ C’ + A’ B’ C D + A’ B’ C D’ + A’ B’ C’ D + A’ B’ C’ D’ + A B C’ D The Product-of-Sums (POS) Form <- Sum of literals; When two or more sum terms are multiplied, the resulting expression is a product-of-sums [ (A + B)(A + B + C) ; (A + B + C)( C + D + E)(B + C + D) ] [ A(A + B + C)(B + C + D) ] In a POS expression, a single overbar cannot extend over more than one variable; however, more than one variable in a term can have an overbar <- A’ + B’ + C’ is okey; but not (A + B + C)’ The Standard POS Form <- ll the variables in the domain appear in each sum term in the expression [ (A + B + C + D)(A + B + C + D)(A + B + C + D) ] Converting a Sum Term to Standard POS (A’ + B + C)(B’ + C + D’)(A + B’ + C’ + D)-> 1) A’ + B + C + DD’ = (A’ + B + C + D)( A’ + B + C + D’) 2) B’ + C + D’ + AA’ = (A + B’ + C + D’ )( A’ + B’ + C + D’ ) Result: (A’ + B + C)(B’ + C + D’)(A + B’ + C’ + D) -> (A’ + B + C + D)( A’ + B + C + D’) (A + B’ + C + D’ )( A’ + B’ + C + D’ ) (A + B’ + C’ + D) 5) F = ABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C Because A + A = A F = ABC + ABC’ + AB’C + AB’C’ + A’B’C F = m7 + m6 + m5 + m4 + m1 In short notation F(A, B, C) = Σ(1, 4, 5, 6, 7) F’(A, B, C) = Σ(0, 2, 3) The complement of a function expressed as the sum of minterms equal to the sum of minterms missing from the original function. A+BC Truth table for F A  POS expression is equal to 0 only if at least one of the sum terms is equal to 0. Simplest SOP or POS Expression <- minimum expression Karnaugh map is an array of cells in which each cell represents a binary value of the input variables. Karnaugh maps can be used for expressions with two, three, four and five variables. The number of cells in a Karnaugh map is equal to the total number of possible input variable combinations. For three variables, the number of cells is 23 = 8. For four variables, the number of cells is 24 = 16. 00 l l A +AB + ABC | s || 000 100 110 ć 001 101 010 10 1 i 011 Example Map the following SOP expression on a Karnaugh map: BC + AB + ABC + ABCD + ABCD + ABCD Solution The SOP expression is obviously not in standard form because each product term does not have four variables. BC AB + ABC + ABCD + ABCD + ABCD 0000 1000 1100 1010 0001 IOH1 0001 1001 1101 1000 1010 1001 1011 Map each of the resulting binary values by placing a 1 in the appropriate cell of the 4l variable Kamaugh map.  Karnaugh Map Simplification of SOP Expressions Grouping the 1s, you can group 1s on the Karnaugh map according to the following rules by enclosing those adjacent cells containing 1s. The goal is to maximize the size of the groups and to minimize the number of groups. A group must contain either 1, 2, 4, 8, or 16 cells, which are all powers of two. In the case of a 3-variable map, 23 = 8 cells is the maximum group. Each cell in a group must be adjacent to one or more cells in that same group. Always include the largest possible number of 1s in a group in accordance with rule 1. Each 1 on the map must be included in at least one group. The 1s already in a group can be included in another group as long as the overlapping groups include noncommon 1s. Example: Group the 1s in each of the Karnaugh maps in Fig.(5-6). © c CD cp AN_ 01 AND 1 AN 0 O 1 IG AN 0 0 1 10 00 1 00 1 1 00 1 1 00 1 m 1 01 1 01 1 1 ! 1 0 ł 1 u 1 1 " U n l 1 ' 10 w 1 1 10 1 1 1a 1 1 Fig.(5-6) Solution: The groupings are shown in Fig.(5-7). In some cases. there may be more than one way to group the 1s to form maximum groupings. © € CD CD AB 1 ABD _1 ANO 0 10 aB [© £ | 0 | -U ee) «Lo Fig.(5-7) Determine the minimum product term for each group. a. For a 3-variable map: (1) Al-cell group yields a 3-variable product term (2) A 2-cell group yields a 2-variable product term (3) A 4-cell group yields a 1-variable term (4) An 8-cell group yields a value of 1 for the expression