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The couteut of PC in the basic computer is 9AA (all aumbers are in hexadecimal). The coutent of AC is 7EA7. The content of memory at address 9AA is 933F. The couteut of memory at address 33F is 07AB. The conteut of memory at address 7AB is 8885. Table 1 Basic Computer Instructions Hesadesimal sode 1-1 Symbol 1-0 Deseription AND memory words to AC Add memory word to AC Load memory word to AC Store content of AC in memory Branch unconditionaally AND ADD LDA Axx STA Bxxx BUN 4xxx CXx BSA Dxx Branch and save retum address ISZ Exxx Increment and skip if zero What is the instruction that will be fenched and executed next? Oa 9AB Ob.9AA +1 Oc 93F O da and b are correct Refer to Table 1 for the binary operation that will be performed in the AC when the instruction is executed and give its value. O a OAAS b.0ASC c. 7EAS d. none of the above is correct Give the contents of registers PC, AR, DR, AC, and IR in hexadecimal at the end of the instruction cyele. Oa PC - 9AB, AR-7AB, DR = 8BB5, AC -0ASC, IR = 933F Ob PC - 9AA, AR = 33F, DR - 07AB, AC = OA5C, IR = 933F OCPC-9AB, AR = 7AB, DR = 8BBS, AC 0AAS, IR = 7EA7 Od PC-9AA , AR-33F, DR-07AB, AC = 0AA5, IR = 933F %3D %3D %3D %3! %3D !!

The couteut of PC in the basic computer is 9AA (all aumbers are in hexadecimal). The coutent of AC is 7EA7. The content of memory at address 9AA is 933F. The couteut of memory at address 33F is 07AB. The conteut of memory at address 7AB is 8885. Table 1 Basic Computer Instructions Hesadesimal sode 1-1 Symbol 1-0 Deseription AND memory words to AC Add memory word to AC Load memory word to AC Store content of AC in memory Branch unconditionaally AND ADD LDA Axx STA Bxxx BUN 4xxx CXx BSA Dxx Branch and save retum address ISZ Exxx Increment and skip if zero What is the instruction that will be fenched and executed next? Oa 9AB Ob.9AA +1 Oc 93F O da and b are correct Refer to Table 1 for the binary operation that will be performed in the AC when the instruction is executed and give its value. O a OAAS b.0ASC c. 7EAS d. none of the above is correct Give the contents of registers PC, AR, DR, AC, and IR in hexadecimal at the end of the instruction cyele. Oa PC - 9AB, AR-7AB, DR = 8BB5, AC -0ASC, IR = 933F Ob PC - 9AA, AR = 33F, DR - 07AB, AC = OA5C, IR = 933F OCPC-9AB, AR = 7AB, DR = 8BBS, AC 0AAS, IR = 7EA7 Od PC-9AA , AR-33F, DR-07AB, AC = 0AA5, IR = 933F %3D %3D %3D %3! %3D !!

Systems Architecture
Systems Architecture
7th Edition
ISBN:9781305080195
Author:Stephen D. Burd
Publisher:Stephen D. Burd
Chapter2: Introduction To Systems Architecture
Section: Chapter Questions
Problem 2VE: A(n) __________ is a storage location implemented in the CPU.
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The content of PC in the basic computer is 9AA (all mumbers are in hexadecimal). The coutent of AC is 7EA7. The content of memory at address 9AA is 933F. The content of memory at address 33F is 07AB. The content of memory at address 7AB is 8BBS. Table 1 Basic Computer Instructions Hexadesimal code Symbol Description 1-0 1-1 AND memory words to AC Add memory word to AC AND ADD Ixxx LDA Load memory word to AC Store content of AC in memory 2xxx Axxx STA BUN Branch unconditionally BSA Sxxx Dx Branch and save retum address ISZ Exxx Increment and skip if zero What is the instruction that will be fetched and executed next? O a. 9AB Ob.9AA +1 Oc 933F O d.a and b are correct Refer to Table 1 for the binary operation that will be performed in the AC when the instruction is executed and give its value. O a. OAAS b. 0ASC O. 7EAS O d. none of the above is correct Give the contents of registers PC, AR, DR, AC, and IR in hexadecimal at the end of the instruction cycle. Oa. PC -9AB, AR = 7AB, DR = 8BB5, AC = 0ASC, IR = 933F Ob. PC = 9AA, AR = 33F, DR = 07AB, AC = 0A5C, IR = 933F OC PC = 9AB , AR = 7AB, DR = 8BB5, AC = 0AA5, IR = 7EA7 Od. PC = 9AA , AR = 33F, DR = 07AB, AC = 0AAS, IR = 933F %3D %3D %3! %3D
Transcribed Image Text:The content of PC in the basic computer is 9AA (all mumbers are in hexadecimal). The coutent of AC is 7EA7. The content of memory at address 9AA is 933F. The content of memory at address 33F is 07AB. The content of memory at address 7AB is 8BBS. Table 1 Basic Computer Instructions Hexadesimal code Symbol Description 1-0 1-1 AND memory words to AC Add memory word to AC AND ADD Ixxx LDA Load memory word to AC Store content of AC in memory 2xxx Axxx STA BUN Branch unconditionally BSA Sxxx Dx Branch and save retum address ISZ Exxx Increment and skip if zero What is the instruction that will be fetched and executed next? O a. 9AB Ob.9AA +1 Oc 933F O d.a and b are correct Refer to Table 1 for the binary operation that will be performed in the AC when the instruction is executed and give its value. O a. OAAS b. 0ASC O. 7EAS O d. none of the above is correct Give the contents of registers PC, AR, DR, AC, and IR in hexadecimal at the end of the instruction cycle. Oa. PC -9AB, AR = 7AB, DR = 8BB5, AC = 0ASC, IR = 933F Ob. PC = 9AA, AR = 33F, DR = 07AB, AC = 0A5C, IR = 933F OC PC = 9AB , AR = 7AB, DR = 8BB5, AC = 0AA5, IR = 7EA7 Od. PC = 9AA , AR = 33F, DR = 07AB, AC = 0AAS, IR = 933F %3D %3D %3! %3D
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