数字电子技术(Floyd第十版)Chapter5分析解析课件.ppt

1、 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edDigital FundamentalsTenth EditionFloydChapter 4 2008 Pearson Education 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edIn Boolean algebra,a

2、 variable is a symbol used to represent an action,a condition,or data.A single variable can only have a value of 1 or 0.SummarySummarySummaryBoolean AdditionThe complement represents the inverse of a variable and is indicated with an overbar.Thus,the complement of A is A.A literal is a variable or i

3、ts complement.Addition is equivalent to the OR operation.The sum term is 1 if one or more if the literals are 1.The sum term is zero only if each literal is 0.Determine the values of A,B,and C that make the sum term of the expression A+B+C=0?Each literal must=0;therefore A=1,B=0 and C=1.2009 Pearson

4、 Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edIn Boolean algebra,multiplication is equivalent to the AND operation.The product of literals forms a product term.The product term will be 1 only if all of the literals are 1.SummarySummarySummaryBoolean Mult

5、iplicationWhat are the values of the A,B and C if the product term of A.B.C=1?Each literal must=1;therefore A=1,B=0 and C=0.2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryCommutative LawsIn terms of the result,the order in

6、 which variables are ORed makes no difference.The commutative laws are applied to addition and multiplication.For addition,the commutative law statesA+B=B+AIn terms of the result,the order in which variables are ANDed makes no difference.For multiplication,the commutative law statesAB=BA 2009 Pearso

7、n Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryAssociative LawsWhen ORing more than two variables,the result is the same regardless of the grouping of the variables.The associative laws are also applied to addition and multiplication

8、.For addition,the associative law statesA+(B+C)=(A+B)+CFor multiplication,the associative law statesWhen ANDing more than two variables,the result is the same regardless of the grouping of the variables.A(BC)=(AB)C 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital F

9、undamentals,10th edSummarySummarySummaryDistributive LawThe distributive law is the factoring law.A common variable can be factored from an expression just as in ordinary algebra.That isAB+AC=A(B+C)The distributive law can be illustrated with equivalent circuits:B+CCAXBABBXACAACAB+ACA(B+C)2009 Pears

10、on Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryRules of Boolean Algebra1.A+0=A2.A+1=13.A.0=04.A.1=15.A+A=A7.A.A=A6.A+A=18.A.A=09.A=A=10.A+AB=A12.(A+B)(A+C)=A+BC11.A+AB=A+B 2009 Pearson Education,Upper Saddle River,NJ 07458.All Right

11、s ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryRules of Boolean AlgebraRules of Boolean algebra can be illustrated with Venn diagrams.The variable A is shown as an area.The rule A+AB=A can be illustrated easily with a diagram.Add an overlapping area to represent the variable B.ABAB

12、The overlap region between A and B represents AB.BThe diagram visually shows that A+AB=A.Other rules can be illustrated with the diagrams as well.=2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edASummarySummarySummaryRules of Boolean AlgebraA+A

13、B=A+BThis time,A is represented by the blue area and B again by the red circle.B The intersection represents AB.Notice that A+AB=A+BAABAIllustrate the rule with a Venn diagram.2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummar

14、yRules of Boolean AlgebraRule 12,which states that(A+B)(A+C)=A+BC,can be proven by applying earlier rules as follows:(A+B)(A+C)=AA+AC+AB+BC=A+AC+AB+BC=A(1+C+B)+BC=A.1+BC=A+BCThis rule is a little more complicated,but it can also be shown with a Venn diagram,as given on the following slide 2009 Pears

15、on Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryABCAA+BThe area representing A+B is shown in yellow.The area representing A+C is shown in red.Three areas represent the variables A,B,and C.ACA+CThe overlap of red and yellow is shown i

16、n orange.ABCABCBCORing with A gives the same area as before.ABCBC=ABC(A+B)(A+C)A+BCThe overlapping area between B and C represents BC.2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryDeMorgans TheoremThe complement of a prod

17、uct of variables is equal to the sum of the complemented variables.DeMorgans 1st TheoremAB=A+BApplying DeMorgans first theorem to gates:OutputInputsABABA+B0011010111101110A+BABABABNANDNegative-OR 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th ed

18、SummarySummarySummaryDeMorgans TheoremDeMorgans 2nd TheoremThe complement of a sum of variables is equal to the product of the complemented variables.A+B=A.BApplying DeMorgans second theorem to gates:ABA+BABOutputInputs0011010110001000ABABA+BABNORNegative-AND 2009 Pearson Education,Upper Saddle Rive

19、r,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryApply DeMorgans theorem to remove the overbar covering both terms from the expression X=C+D.DeMorgans TheoremTo apply DeMorgans theorem to the expression,you can break the overbar covering both terms and change the

20、sign between the terms.This results inX=C.D.Deleting the double bar gives X=C.D.=2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edACDBSummarySummarySummaryBoolean Analysis of Logic CircuitsCombinational logic circuits can be analyzed by writing

21、the expression for each gate and combining the expressions according to the rules for Boolean algebra.Apply Boolean algebra to derive the expression for X.Write the expression for each gate:Applying DeMorgans theorem and the distribution law:C(A+B)=C(A+B)+D(A+B)X=C(A B)+D=A B C+DX 2009 Pearson Educa

22、tion,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryBoolean Analysis of Logic CircuitsUse Multisim to generate the truth table for the circuit in the previous example.Set up the circuit using the Logic Converter as shown.(Note that the logic con

23、verter has no“real-world”counterpart.)Double-click the Logic Converter top open it.Then click on the conversion bar on the right side to see the truth table for the circuit(see next slide).2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummary

24、SummarySummaryBoolean Analysis of Logic CircuitsThe simplified logic expression can be viewed by clickingSimplified expression 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummarySOP and POS formsBoolean expressions can be writ

25、ten in the sum-of-products form(SOP)or in the product-of-sums form(POS).These forms can simplify the implementation of combinational logic,particularly with PLDs.In both forms,an overbar cannot extend over more than one variable.An expression is in SOP form when two or more product terms are summed

26、as in the following examples:An expression is in POS form when two or more sum terms are multiplied as in the following examples:A B C+A B A B C+C DC D+E(A+B)(A+C)(A+B+C)(B+D)(A+B)C 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummary

27、SummarySOP Standard formIn SOP standard form,every variable in the domain must appear in each term.This form is useful for constructing truth tables or for implementing logic in PLDs.You can expand a nonstandard term to standard form by multiplying the term by a term consisting of the sum of the mis

28、sing variable and its complement.Convert X=A B+A B C to standard form.The first term does not include the variable C.Therefore,multiply it by the(C+C),which=1:X=A B(C+C)+A B C =A B C+A B C+A B C 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edS

29、ummarySummarySummarySOP Standard formThe Logic Converter in Multisim can convert a circuit into standard SOP form.Click the truth table to logic button on the Logic Converter.Use Multisim to view the logic for the circuit in standard SOP form.See next slide 2009 Pearson Education,Upper Saddle River,

30、NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummarySOP Standard formSOP Standard form 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryPOS Standard formIn POS standard form,every variable in t

31、he domain must appear in each sum term of the expression.You can expand a nonstandard POS expression to standard form by adding the product of the missing variable and its complement and applying rule 12,which states that(A+B)(A+C)=A+BC.Convert X=(A+B)(A+B+C)to standard form.The first sum term does

32、not include the variable C.Therefore,add C C and expand the result by rule 12.X=(A+B+C C)(A+B+C)=(A+B+C)(A+B+C)(A+B+C)2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edThe Karnaugh map(K-map)is a tool for simplifying combinational logic with 3 or

33、 4 variables.For 3 variables,8 cells are required(23).The map shown is for three variables labeled A,B,and C.Each cell represents one possible product term.Each cell differs from an adjacent cell by only one variable.ABCABCABCABCABCABCABCABCSummarySummarySummaryKarnaugh maps 2009 Pearson Education,U

34、pper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th ed0100011110ABCCells are usually labeled using 0s and 1s to represent the variable and its complement.Gray codeSummarySummarySummaryKarnaugh mapsOnes are read as the true variable and zeros are read as the complemented var

35、iable.The numbers are entered in gray code,to force adjacent cells to be different by only one variable.2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryABCABCABCABCABCABCABCABCABABABABC CAlternatively,cells can be labeled w

36、ith the variable letters.This makes it simple to read,but it takes more time preparing the map.Karnaugh mapsCCABABABABCCABABABABABCABCRead the terms for the yellow cells.The cells are ABC and ABC.2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th ed

37、1.Group the 1s into two overlapping groups as indicated.2.Read each group by eliminating any variable that changes across a boundary.3.The vertical group is read AC.SummarySummarySummaryK-maps can simplify combinational logic by grouping cells and eliminating variables that change.Karnaugh maps111AB

38、C00 0111100 1111ABC00 0111100 1Group the 1s on the map and read the minimum logic.B changes across this boundaryC changes across this boundary4.The horizontal group is read AB.X=AC+AB 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edA 4-variable

39、 map has an adjacent cell on each of its four boundaries as shown.ABABABABCDCDCDCDEach cell is different only by one variable from an adjacent cell.Grouping follows the rules given in the text.The following slide shows an example of reading a four variable map using binary numbers for the variablesS

40、ummarySummarySummaryKarnaugh maps 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edXSummarySummarySummaryKarnaugh mapsGroup the 1s on the map and read the minimum logic.1.Group the 1s into two separate groups as indicated.2.Read each group by el

41、iminating any variable that changes across a boundary.3.The upper(yellow)group is read as AD.4.The lower(green)group is read as AD.ABCD00 01111000 01 11 1011111111ABCD00 01111000 01 11 1011111111X=AD+ADB changesC changesB changesC changes across outer boundary 2009 Pearson Education,Upper Saddle Riv

42、er,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryHardware Description Languages(HDLs)A Hardware Description Language(HDL)is a tool for implementing a logic design in a PLD.One important language is called VHDL.In VHDL,there are three approaches to describing logi

43、c:2.Dataflow3.Behavioral1.StructuralDescription is like a schematic(components and block diagrams).Description is equations,such as Boolean operations,and registers.Description is specifications over time(state machines,etc.).2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloy

44、d,Digital Fundamentals,10th edSummarySummarySummaryHardware Description Languages(HDLs)The data flow method for VHDL uses Boolean-type statements.There are two-parts to a basic data flow program:the entity and the architecture.The entity portion describes the I/O.The architecture portion describes t

45、he logic.The following example is a VHDL program showing the two parts.The program is used to detect an invalid BCD code.entity BCDInv isport(B,C,D:in bit;X:out bit);end entity BCDInvarchitecture Invalid of BCDInvbegin X=(B or C)and D;end architecture Invalid;2009 Pearson Education,Upper Saddle Rive

46、r,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSummarySummarySummaryHardware Description Languages(HDLs)Another standard HDL is Verilog.In Verilog,the I/O and the logic is described in one unit called a module.Verilog uses specific symbols to stand for the Boolean logical operators.

47、The following is the same program as in the previous slide,written for Verilog:module BCDInv(X,B,C,D);input B,C,D;output X;assign X=(B|C)&D;endmodule 2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSelected Key TermsSelected Key TermsSelected K

48、ey Terms VariableComplementSum termProduct termA symbol used to represent a logical quantity that can have a value of 1 or 0,usually designated by an italic letter.The inverse or opposite of a number.In Boolean algebra,the inverse function,expressed with a bar over the variable.The Boolean sum of tw

49、o or more literals equivalent to an OR operation.The Boolean product of two or more literals equivalent to an AND operation.2009 Pearson Education,Upper Saddle River,NJ 07458.All Rights ReservedFloyd,Digital Fundamentals,10th edSelected Key TermsSelected Key TermsSelected Key Terms Sum-of-products(S

50、OP)Product of sums(POS)Karnaugh mapVHDLA form of Boolean expression that is basically the ORing of ANDed terms.A form of Boolean expression that is basically the ANDing of ORed terms.An arrangement of cells representing combinations of literals in a Boolean expression and used for systematic simplif