數(shù)字電子技術(shù)chapter

1、DBCACABABDCBACBABAAA,Boolean multiplication is equivalent to the AND operation and the basic rules are illustrated with their relation to the AND gates as follows:ADCBACBABA, ,Example:Determine the value of A, B, C, and D which make the sum term Solution:0DCBA0, 0, 0, 0DCBA1, 0, 1, 0DCBA0DCBAExample

2、:Determine the value of A, B, C, and D which make the product term Solution:CDBACBAAB , ,1DCBA1, 1, 1, 1DCBA0, 1, 0, 1DCBA1DCBABAABABBACABBCACBACBA)()()()(ACABCBA)(AA 011 A00 AAA 1AAA1AAAAA0AAAA AABABABAABCACABA)(This law is similar to absorption in that it can be employed to eliminate extra element

3、s from a Boolean expressionBABAAThe dual of distributive lawBCACABA)(BABABAAABAABABAABABAA1)()()(PROOFPROOFBCABCCBAABCACABABCACABAACBAABACABA)()()()(Consensus theoremCAABCBACAABCABBCAABCCAABBCAACAABBCCAAB)()()(PROOFCAABBCCAAB The key to using this theorem is to find a variable and its complement, no

4、te the associated terms, and eliminate the included term (the consensus term), which is composed of the associated terms.Duality theorem（對偶定理）: if an expression is valid in Boolean algebra, the dual of the expression is also valid. The dual expression is found by replacing all + operators with , and

5、 operators with +, all 1s with 0s, and all 0s with 1s.The duality theorem will be used extensively in proving Boolean algebra theorems. Example. Find the dual of the expression)(CABABCA Solution. Changing all + operators to , and vice versa, the dual expression isACABCBA)(YXXY(equivalency of the NAN

6、D and negative-OR gates)XYYX YXYX(equivalency of the NOR and negative-AND gates.)YX YXThe gate equivalencies and truth tables for DeMorgans theorems are shown below:The proof is given below:YXXYYXYX,YXXYYXYXYXABBAYXYXBAAAAAXYYXYXYXBAYXYYXXYXYXABYXBYXA1011000)(,ZYXZYXYZXXYZ )(ZYXZYXZYXZYXXYZ,Solution

7、:Example: Apply DeMorgans theorems to each of the following expressions:(a) (b)DCBA)(DEFABC (a) Let A+B+C=X,D=Y:DCBA)(DCBAYXXYCBACBACBADCBA)(DCBA Solution:)( )( )(FEDCBADEFABCDEFABCbExample: Apply DeMorgans theorem to the following expression:)(FEDCBASolution:)( FEDCBAYXYXYXXY) )()( FEDCBA) )()( FED

8、CBA)( FEDCBAA(B+CD)B+CDBDCACDExample: Logic functionstruth tableCBACBAABCEFCDABACDABCDBA)(Example: Convert each of the following Boolean expression to SOP form:CBADCBBAEFCDBAB)c( )()(b( )()a(SolutionBEFBCDABEFCDBAB)()a (BDBCBADACABBDBCBBADACABDCBBA )()(b(CBCACBACBACBA)()c(DCBACDBAABCDCBABCAABCExampl

9、eThe standard SOP form is important in constructing truth table or for Karnaugh map simplification which we will discuss later.Examples: Convert the following Boolean expression into standard SOP form.DCABBACBADCABDDCCBADDCBA)()(DCABDCBADCBADCBACDBADCBACDBASolution)()()()(DCBADCABACBACBACBA)()()()(D

10、CBADCBADCBACBACBACBAC)B)(A(ABCAExample: Convert following Boolean algebra expression into standard POS form.)()(DCBADCBCBASolution)(. 2DCBADCBADCBAADCB)()(. 1DCBADCBA DDCBACBA)()(DCBADCBCBA12)()()()()(DCBADCBA DCBADCBADCBADCBADCBCBADCBA1010DCBAMintermMaxterm10101010Example:m10M6Minterms and Maxterms

11、 for Three Binary Variables iimfiiMfNote: Minterms and maxterms for n variables can converted to each other by applying DeMorgans Theorem. iiMm Example: Determine the truth table for following expression: )()()()(CBACBACBACBACBASolution: InputOutputABCx00000011010001101001101011001111)()()()(CBACBAC

12、BACBACBA000010011101110InputOutputABCx00000010010001111001101011011111Example: From the truth table, determine the standard SOP expression and the equivalent standard POS expression.Sum of minterms expression:Solution:)7 , 6 , 4 , 3(7643 mmmm ABCCABCBABCAXProduct of maxterms expression:),()()(5210 5

13、210MMMMCBACBACBACBAX00000010010001111001101011011111)5 , 2 , 1 , 0()7 , 6 , 4 , 3( XikkiiMmfikkiiiiMMmmmfTheoremProof:Exercise: From the truth table, determine the standard SOP expression and the equivalent standard POS expression.InputOutputABCx00000011010001101001101011001111CCBACBAABCExample: Con

14、vert the following function to a logic diagram, use AND, OR, and NOT gates.ABCCBACBACBAfABABCCBAfExample:ABCBACA )(CABAFBABA) 3(CBABA CBABA)( )4(CBABACBABA)()6(CBABA)(5(BA) 1 (BA)2(CBA)( Example 4-8 Using Boolean algebra, simplify this expression:)()(CBBCBAABXSolution:BACBACABBCBACABBCBACABABBCBBACA

15、BABCBBCBAABX)()(BACABA(B+C)B(B+C)B+CX3 AND,2 OR1 AND,1 ORCABXCBCBCBAACBACBACBACCBACBACBACBADACBACBABDBACBACBABDCBA1)()()0()()(Example:4 AND,2 OR1 ANDABAABAA1. 1AABA. 2AAAAAA. 4BABAA. 3ABAABAA1. 1CDBACDBAFAExampleABAABAABA. 2BCDCBABCAAF)()(DCBABCABCAYXYXAABAExample:BCADCAAC)(BABAA. 3DCDAACFDACAC YXYX

16、BABAAExample:DAC ABCBCABCACBAterms repeated usingAAA AA, 1. 4ABCBCACBAFBCAACCBA)()()()(ABCBCABCACBA1 AAExample:BCBAAAACBAABCCCBABA)()(CBBCBABAF)()()(CBACBABCABCCBABA CBACBABCCBABCABAExample:BCBACBABCBCBACBABCBCAACBACBABCBA)(1 AAExamples:EDCAEDBDECADCBAYCEADBBCBADCACYBACBADCDABAYBACBAYCBABCAYADECADCB

17、AABDYBCACBAY7654321)()()(Solutions:11)()(1) 1(4321BACBABACBAYCCCBABACBABCACBABCAYAADECDCBBDAADECADCBAABDYBCBAABBCCAABBCBCACBAYSolutions:EDCEEBEABEAEDEECBABEAEDCDEECDCBADEAEBEECDCBADADEAECBBEECDCBCCEDABAEAEDEBDBECAEDBECEADBCADBECEAEDBECADCBAEDBEDDCADCBAEDCAEDBDECADCBAYEABCDEBCADECBCADCEBCADCEBBCCEADB

18、BCBAACDCACCEADBBCBADCACYBACBADCDABABACBADCDABAY)()()()()()()()()(0)()(7650mABBABABA3m2m1m0110m1 m2 m3m0 BA011001 10 1100 BABCA1000110110m1 m0 m3 m2 m5 m4 m7 m6 Note: The combination of BC is 00011110 instead of 00011011.CDAB0011011000110110m1 m0 m3 m2 m5 m4 m7 m6 m13 m12 m15 m14 m9 m8 m11 m10 CDAB00

19、11011000110110m1 m0 m3 m2 m5 m4 m7 m6 m13 m12 m15 m14 m9 m8 m11 m10 A=0A=1AB0011011000110110m1 m0 m3 m2 m5 m4 m7 m6 m13 m12 m15 m14 m9 m8 m11 m10 B=0B=0B=100110110CDAB0011011000110110m1 m0 m3 m2 m5 m4 m7 m6 m13 m12 m15 m14 m9 m8 m11 m10 D=1AB00110110m1 m0 m3 m2 m5 m4 m7 m6 m13 m12 m15 m14 m9 m8 m11

20、m10 C=1CDEAB00110110m1 m0 m3 m2 m9 m8 m11 m10 m25 m24 m27 m26 m17 m16 m19 m18 000011001010100111101110m7 m6 m5 m4 m15 m14 m13 m12 m31 m30 m29 m28 m23 m22 m21 m20 CDAB0011011000110110m1 m0 m3 m2 m5 m4 m7 m6 m13 m12 m15 m14 m9 m8 m11 m10 m5 m0 CBACABCBACBAXBCA1000110110m1 m0 m3 m2 m5 m4 m7 m6 0001 001

21、1 1101001 1 ExampleDCBADCBADCAB ABCDDCABDCBACDBAF0011010111011111110000011010CDAB00110110001101101 1 1 1 1 1 1 CABBAAX000001010011100101010 110BCA1000110110m1 m0 m3 m2 m5 m4 m7 m6 1 1 1 1 1 1 1 ExampleExercise: Plot the following expression on a Karnaugh map.)(DCBACDAB00110110001101101 1 1 1 1 1 1 1

22、 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4-variable Karnaugh map1 1 1 1 11 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 1 11 1 1 1 11 1 1 1 1 1 1 11 1 1 111 1 11 1 11 1 1 1 1 1 11Example1 1 1 1 1 1 11 1 1 CDAB00110110001101101 Two 1s:DCAFour 1s:CA Eight 1s:BBCADCAF 1 ABBCFBCACABABCCBAF

23、ABBCBCA0011011011 110100111110011 1 ABBCF)7 , 6 , 4 , 3(),(7643mmmmFABBCBCA0011011011 110BCACABABCCBAF100111110011CDAB00110110001101101 0 3 2 5 4 7 6 13 12 15 149 8 11 10 Example)14,12,11,10, 8 , 6 , 4 , 3 , 2 , 0(F1 1 1 1 1 1 1 1 1 1 CDAB0011011000110110DCBCBDFExampleInputsOutputABCDY00000000100010

24、000110010000101001100011111000110011101010111100110111101111BCDcode0123456789CDCDABAB00001111010110100000111101011010 1 1 1 1 1 1 CBABCDAABCD Without “dont care” terms With “dont care” termsBCDACBAYBCDAY1110000000000000111100001100110010101010FDCBA1111111111111100001100110010101010FDCBA1) Truth table2) Karnaugh mapCDCDABAB00001111010110100000111101011010 1 1 1 11 1 1 1 1 1 ABDBCBDBCAF)()()(CBACBACBACBAYBCA100