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CHAPTER 3: PRINCIPLES OF COMBINATIONAL LOGIC (Sections 3.1 – 3.5)

CHAPTER 3: PRINCIPLES OF COMBINATIONAL LOGIC (Sections 3.1 – 3.5). 3.1 DEFINITION OF COMBINATIONAL LOGIC. Logic circuits without feedback from output to the input, constructed from a functionally complete gate set, are said to be combinational. ·. ·. Combinational Logic Functions.

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CHAPTER 3: PRINCIPLES OF COMBINATIONAL LOGIC (Sections 3.1 – 3.5)

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  1. CHAPTER 3: PRINCIPLES OF COMBINATIONAL LOGIC (Sections 3.1 – 3.5)

  2. 3.1 DEFINITION OF COMBINATIONAL LOGIC Logic circuits without feedback from output to the input, constructed from a functionally complete gate set, are said to be combinational. · · Combinational Logic Functions Inputs · · Outputs · ·

  3. 3.1.1 PROBLEM STATEMENTS TO TRUTH TABLES • E.g : Conveyor system(P85, Ex. 3.1) M=a’bms+ab’ms+abms

  4. 3.1.1 PROBLEM STATEMENTS TO TRUTH TABLES • E.g. A NASA system(P87, Ex. 3.2) C2 C3 C1

  5. 3.1.1 PROBLEM STATEMENTS TO TRUTH TABLES • E.g. Conveyor system(P88, ex. 3.4) S3 m3 S2 m2 m4 S1 m1

  6. 3.1.1 PROBLEM STATEMENTS TO TRUTH TABLES • The process of converting a verbal problem statement into a truth table : • Determine the input variables and output variables that are involved • Assign mnemonic or letter or number symbols to each variable • determine the size of the truth table; how many input combinations exist: 2x=y where x=number of input variables and y=number of combinations. • Construct a truth table containing all of the input variable combinations • By careful reading of the problem statement determine the combinations of inputs that cause a given output to be true

  7. I3 I2 I1 O1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 1 3.1.1 PROBLEM STATEMENTS TO TRUTH TABLES • Exe: Design a truth table to indicate a majority of three inputs is true

  8. 3.1.2 Deriving Switching Equations • Definitions • Product term: literals connected by • • Sum term: literals connected by + • Minterm: a product term in which all the variables appear exactly once, either complemented or uncomplemented • Maxterm: a sum term in which all the variables appear exactly once, either complemented or uncomplemented

  9. Minterm • Represents exactly one combination in the truth table. • Denoted by mj, where j is the decimal equivalent of the minterm’s corresponding binary combination (bj). • A variable in mj is complemented if its value in bj is 0, otherwise is uncomplemented. • M(a,b,m,s)=a’bms+ab’ms+abms (m7) (m11) (m15)

  10. Maxterm • Represents exactly one combination in the truth table. • Denoted by Mj, where j is the decimal equivalent of the maxterm’s corresponding binary combination (bj). • A variable in Mj is complemented if its value in bj is 1, otherwise is uncomplemented. • F(a,b,c)=(a’+b+c’)(a+b+c’)(a+b’+c’) 101 001 011 M5 M1 M3

  11. Truth Table notation for Minterms and Maxterms 3 variables x,y,z (order is fixed)

  12. 3.2 Canonical(规范) Forms • Any Boolean function F( ) can be expressed as a uniquesum of minterms and a unique product of maxterms (under a fixed variable ordering). • every function F() has two canonical forms: • Canonical Sum-Of-Products (sum of minterms) • Canonical Product-Of-Sums (product of maxterms)

  13. Convert switching equations to canonical form • Expand non-canonical terms by inserting equivalent of 1 in each missing variable x: (x + x’) = 1 • Remove duplicate minterms • f1(a,b,c) = a’b’c + bc’ + ac’ = a’b’c + (a+a’)bc’ + a(b+b’)c’ = a’b’c + abc’ + a’bc’ + abc’ + ab’c’ = a’b’c + abc’ + a’bc + ab’c’

  14. Convert switching equations to canonical form • SOP E.g. P=f(a,b,c)=ab’+ac’+bc =ab’(c+c’)+ac’ (b+b’)+bc(a+a’) =ab’c+ab’c’+abc’+ab’c’+abc+a’bc =ab’c+ab’c’+abc’+abc+a’bc

  15. Convert switching equations to canonical form • Expand noncanonical terms by adding 0 in terms of missing variables (e.g., xx’ = 0) and using the distributive law • Remove duplicate maxterms • f1(a,b,c) = (a+b+c)•(b’+c’)•(a’+c’) = (a+b+c)•(aa’+b’+c’)•(a’+bb’+c’) = (a+b+c)•(a+b’+c’)•(a’+b’+c’)• (a’+b+c’)•(a’+b’+c’) = (a+b+c)•(a+b’+c’)•(a’+b’+c’)•(a’+b+c’)

  16. Convert switching equations to canonical form • POS E.g. T=f(a,b,c)=(a+b’)(b’+c) =(a+b’+cc’)(b’+c+aa’) =(a+b’+c) (a+b’+c’)(a+b’+c)(a’+b’+c) =(a+b’+c) (a+b’+c’)(a’+b’+c)

  17. Convert switching equations to canonical form • Exe: P=f(w,x,y,z)=w’x+yz’ T=f(a,b,c,d)=(a+b’+c)(a’+d) • Ans: P=f(w,x,y,z)=wxyz’+wx’yz’+w’xyz+w’xyz’+w’xy’z +w’xy’z’+w’x’yz’ T=f(a,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)

  18. 3.3 GENERATION OF SWITCHING EQUATIONS FROM TRUTH TABLES • Canonical Sum-Of-Products:The minterms included are those mj such that F( ) = 1 in row j of the truth table for F( ). • Canonical Product-Of-Sums:The maxterms included are those Mj such that F( ) = 0 in row j of the truth table for F( ).

  19. Example • Truth table for f1(a,b,c) at right • The canonical sum-of-products form for f1 isf1(a,b,c) = m1 + m2 + m4 + m6 = a’b’c + a’bc’ + ab’c’ + abc’ • The canonical product-of-sums form for f1 isf1(a,b,c) = M0• M3• M5• M7 = (a+b+c)•(a+b’+c’)• (a’+b+c’)•(a’+b’+c’). • Observe that: mj = Mj’ a b c f1 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0

  20. Shorthand: ∑ and ∏ • f1(a,b,c) = ∑m(1,2,4,6), where∑ indicates that this is a sum-of-products form, and m(1,2,4,6) indicates that the minterms to be included are m1, m2, m4, and m6. • f1(a,b,c) = ∏M(0,3,5,7), where ∏ indicates that this is a product-of-sums form, and M(0,3,5,7) indicates that the maxterms to be included are M0, M3, M5, and M7. • ∑m(1,2,4,6) = ∏M(0,3,5,7) = f1(a,b,c)

  21. Conversion Between Canonical Forms • Replace ∑ with ∏ (or vice versa) and replace those j’s that appeared in the original form with those that do not. • Example:f1(a,b,c) = a’b’c + a’bc’ + ab’c’ + abc’ = m1 + m2 + m4 + m6= ∑(1,2,4,6) = ∏(0,3,5,7) = (a+b+c)•(a+b’+c’)•(a’+b+c’)•(a’+b’+c’)

  22. Karnaugh Maps • Why Karnaugh maps？ to simplify boolean equations • Karnaugh maps (K-maps) are graphical representations of boolean functions. • One map cell corresponds to a row in the truth table. • one map cell corresponds to a minterm or a maxterm in the boolean expression

  23. AB A B 0 1 00 01 11 10 0 m0 m2 m0 m2 m6 m4 0 1 m1 m3 1 m1 m3 m7 m5 AB CD 00 01 11 10 00 m0 m4 m12 m8 01 m1 m5 m13 m9 m3 11 m7 m15 m11 10 m2 m6 m14 m10 Karnaugh Maps C • Logic adjacency : Any two adjacent cells in the map differ by ONLY one. • Example:m0 (=x1’x2’) is adjacent to m1 (=x1’x2) and m2 (=x1x2’) but NOT m3 (=x1x2)

  24. 2-Variable Map -- Example • E.g. f(x1,x2) = x1’x2’+ x1’x2 + x1x2’ = m0 + m1 + m2 • What (simpler) function is represented by each dashed rectangle? • x1’x2’+ x1’x2 =x1’(x2’+x2)=x1’ • x1’x2’+ x1x2’=x2’(x1’+x1) =x2’ f(x1,x2) =x1’+x2’ x2 0 1 x1 2 0 0 1 1 1 3 1 1 0

  25. Minimization as SOP using K-map • Enter 1s in the K-map for each product term in the function • Group adjacent K-map cells containing 1s to obtain a product with fewer variables. Groups must be in power of 2 (2, 4, 8, …) • Handle “boundary wrap” for K-maps of 3 or more variables.

  26. Three-Variable Map yz 00 01 11 10 x 0 m0 m1 m3 m2 0 1 3 2 1 m4 m5 m7 m6 4 5 7 6 -Note: variable ordering is (x,y,z); yz specifies column, x specifies row. -Each cell is adjacent to three other cells (left or right or top or bottom or edge wrap)

  27. Three-Variable Map (cont.) minterm The types of structures that are either minterms or are generated by repeated application of the minimization theorem on a three variable map are shown at right. Groups of 1, 2, 4, 8 are possible. group of 2 terms group of 4 terms

  28. Simplification • Example: f(a,b,c) = ac’ + abc + bc’ =a(b+b’)c’+abc+(a+a’)bc’ =abc’+ab’c’+abc+abc’+a’bc’ =abc’+ab’c’+abc+a’bc’ • Result: f(a,b,c) = ac’+ab+bc’ 00 01 11 10 ab c 0 1 1 1 1 1

  29. More Examples xy z 00 01 11 10 • f1(x, y, z) = ∑ m(2,3,5,7) • f2(x, y, z) = ∑ m (0,1,2,3,6) 0 1 1 1 1 1 f1(x, y, z) = x’y + xz 00 01 11 10 xy z 1 1 1 0 f2(x, y, z) = x’+yz’ 1 1 1

  30. Four-Variable Maps YZ 00 01 11 10 WX • Top cells are adjacent to bottom cells. Left-edge cells are adjacent to right-edge cells. • Note variable ordering (WXYZ). m0 m1 m3 m2 00 m4 m5 m7 m6 01 m12 m13 m15 m14 11 m8 m9 m11 m10 10

  31. Four-variable Map Simplification • One square represents a minterm of 4 literals. • A rectangle of 2 adjacent squares represents a product term of 3 literals. • A rectangle of 4 squares represents a product term of 2 literals. • A rectangle of 8 squares represents a product term of 1 literal. • A rectangle of 16 squares produces a function that is equal to logic 1.

  32. Example • Simplify the following Boolean function (A,B,C,D) = ∑m(0,1,2,3,4,5,6,7,8,10,13). • First put the function g( ) into the map, and then group as many 1s as possible. 00 01 11 10 cd ab 1 1 1 1 1 1 00 1 1 1 1 1 1 01 1 1 1 1 11 1 1 1 1 1 1 10 g(A,B,C,D) = a’+b’d’+bc’d

  33. Implicants ,Prime Implicants (PIs) and Essential Prime Implicant(EPI) • Implicant: any single minterm • Prime Implicant (PI):a group of minterms that cannot be combined with any other minterms or groups • Essential Prime Implicant (EPI): a PI in which one or more miniterms are unique • How to find PI and EPI? AB CD 00 01 11 10 1 00 1 01 1 1 11 1 1 1? 10 BCD ABC’ A’BC’D AC’D’ BD

  34. Example • a’b’ is not a prime implicant because it is contained in b’. • acd is not a prime implicant because it is contained in ad. • b’, ad, and a’cd’ are prime implicants. • b’, ad, and a’cd’ are essentialprime implicants. b’ ad cd ab 1 1 1 1 1 1 1 1 a’b’ 1 1 1 acd a’cd’

  35. Another Example • Consider f2(a,b,c,d), whose K-map is shown below. • The only essential PI isb’d. ab cd 1 1 1 1 1 1 1 1 1

  36. Systematic Procedure for Simplifying Boolean Functions • Load the minterms into the K-map by placing a 1 in the appropriate square. • Look for groups of minterms (PI) • The group size must be a power of 2 • Find the largest groups of minterms first, then smaller collections until all groups are found. • Find all essential PIs. • For remaining min-terms not included in the essential PIs, select a set of other PIs to cover them. • The resulting simplified function is the logical OR of the product terms selected above.

  37. Example • f(a,b,c,d) = ∑m(0,1,4,5,8,11,12,13,15). • F(a,b,c,d) = c’d’ + a’c’ + bc’ + acd ab cd 1 1 1 1 1 1 1 1 1

  38. Homework • Simplifying Boolean Functions F(x,y,z)=∑(0,2,3,4,5,7) F(a,b,c,d)=∑(0,3,4,5,7,11,13,15) F(w,x,y,z)=∑(0,1,4,5,9,11,13,15) F(a,b,c,d)=∑(0,1,2,4,5,6,8,9,12,13,14) F(a,b,c,d)=∑(1,3,4,5,7,8,9,11,15) F(w,x,y,z)=∑(1,5,7,8,9,10,11,13,15) • P151: 4-c, 4-e

  39. TO BE CONTINUED

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