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* Property of STI Page 1 of 12

Basic Computer Concepts Number Systems Concepts. “Deka” – Greek word denoting a value equal. Ø. to ten Table 1: Decimal Example 2,048 TABLE 2: The value denotation of a number “place” or position. v. Given a base-n number system, the value denotation of a position m is: m

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* Property of STI Page 1 of 12

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  1. BasicComputerConcepts NumberSystemsConcepts “Deka”–Greekworddenotingavalueequal Ø toten Table1:DecimalExample2,048 TABLE2:Thevaluedenotationofanumber “place”orposition v Givenabase-nnumbersystem,thevalue denotationofapositionmis: m Valuedenotation 1 1 2 1xn 3 1xnxn 4 1xnxnxn M 1xnxn...xn(nmultipliedM-1times)0 ComputerOrganization *PropertyofSTI Page1of12

  2. BasicComputerConcepts NumberSystemsConcepts ExampleA: Findingthevalueofanumberinanynumber Ø system Givenabase-nsystemandanumberPQRST,the valueisgivenby: Tx1 +Sxn +Rxnxn +Qxnxnxn +Pxnxnxnxn ExampleB: Givenabase-8systemandanumber56472, thevalueisgivenby: Ø Step 2x1 +7x8 +4x8x8 +6x8x8x8 +5x8x8x8x8 Total:23,866 ComputerOrganization Value 2x1=2 7x8=56 4x64=256 6x512=3,072 5x4,096=20,480

  3. BasicComputerConcepts NumberSystemsConcepts TABLE3:Generalmethodforobtainingthevalue denotationofabase-nnumeral ValuedenotationVm Thevaluedenotationofapositionminabase-n numbersystemis: valuedenotationVm=n(m-1) DecimalvalueNofabase-nnumeral ThedecimalvalueNofanumber pkpk-1pk-2,..,p2p1 inabase-nnumbersystemisgivenby: numericalvalueN=i=1?i=kpiVi =i=1?i=kpin(i-1) ComputerOrganization *PropertyofSTI Page3of12

  4. BasicComputerConcepts NumberSystemsConcept Examples: 1.Derivethedecimalvalueof6409 Solution: From Table 3: 6409=0x90+4x91+6x92 =0x1+4x9+6x9x9 =0+36+486 =52210 2.Derivethedecimalvalueof64012 Solution: From Table 3: 64012=0x120+4x121+6x122 =0x1+4x12+6x12x12 =0+48+864 =91210 ComputerOrganization *PropertyofSTI Page4of12

  5. BasicComputerConcepts NumberSystemsConcept UsingthenotationofTable3: 3.Derivethedecimalvalueof64016 Solution: From Table 3: 64016=i=1?i=3pi16(i-1) =0x160+4X161+6X162 Value=0+64+1536=1600 ComputerOrganization *PropertyofSTI Page5of12

  6. BasicComputerConcepts Hexadecimal,Octal,& BinarySystems 1.Derivethedecimalvalueof110111012 Solution: 110111012=1x20+0x21+1x22 +1x23+1x24+0x25 +1x26+1x27 Value=1+0+4+8+16+0+64+128 =22110 2.Derivethedecimalvalueof541760238 Solution: 541760238=3x80+2x81+0x82 +6x83+7x84+1x85 +4x86+5x87 Value=3+16+0+3072+28672 +32768+1048576+10485760 =1159886710 ComputerOrganization *PropertyofSTI Page6of12

  7. BasicComputerConcepts Hexadecimal,Octal,& BinarySystems 3.Derivethedecimalvalueof24CF16 Solution: 24CF16=Fx160+Cx161+4x162+2x163 Value=15+192+1024+8192 =94231010 ComputerOrganization *PropertyofSTI Page7of12

  8. BasicComputerConcepts Converting:Decimalto AnotherNumberSystem GivenadecimalnumberM,toconverttoa numbersystemwithbase-n: Asanexample,letM=2,654andn=8 Ø Ø ComputerOrganization *PropertyofSTI Page8of12

  9. BasicComputerConcepts Converting:Decimalto AnotherNumberSystem 2,65410=51368 ComputerOrganization *PropertyofSTI Page9of12

  10. BasicComputerConcepts Converting:Decimalto AnotherNumberSystem Derivetheoctalvalueof10112 Solution: a.Convert10112todecimal: 10112=1x20+1x21+0x22+1x23 decimalvalue=1+2+0+8=1110 b.Convert1110tooctal: :11/8=1remainder3 :3/1=3remainder0 octalvalue=138 ComputerOrganization *PropertyofSTI Page10of12

  11. BasicComputerConcepts Converting:Decimalto AnotherNumberSystem ShortcutMethod Derivetheoctalvalueof101110112 Solution: Groupthebinarydigitsintothree: 101110112=101110112=0101110112 421 Now 010=2 421 111=7 421 011=3 Therefore 0101110112=2738 ComputerOrganization *PropertyofSTI Page11of12

  12. BasicComputerConcepts Converting:Decimalto AnotherNumberSystem ShortcutMethod Derivethehexadecimalvalueof101110112 Solution: Groupthebinarydigitsintofour: 101110112=101110112 8421 Now1011=11 8421 1011=11 But1116=B;therefore 101110112=BB16 ComputerOrganization *PropertyofSTI Page12of12

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