Question: The machine code for instruction LDAB $3088 is (7B, 30, 88). (Assuming data bus is 16 bit.) (1) In which computer component is this instruction

The machine code for instruction LDAB $3088 is (7B, 30, 88). (Assuming data bus is 16 bit.)

(1) In which computer component is this instruction stored?

(2) How many memory accesses are needed for fetching this instruction to CPU?

(3) How many memory accesses are required to execute this instruction? (4) Explain in detail what happens in the computer during each memory access.

The machine code for instruction LDAB $3088 is (7B, 30, 88). (Assuming data bus is 8 bit.

(1) How many memory accesses are needed for fetching this instruction to CPU?

(2) How many memory accesses are required to execute this instruction?

(3) Explain in detail what happens in the computer during each memory access.

The machine code for instruction STD $3088 is (7C, 30, 88). (Assuming data bus is 16 bit.)

(1) How many memory accesses are needed for fetching this instruction to CPU?

(2) How many memory accesses are required to execute this instruction?

(3) Explain in detail what happens in the computer during each memory access.

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We ought to isolate the requests for each direction and expect a typical von Neumann design where rules and data are taken care of in memory Direction LDAB 3088 16digit data bus 1 The direction LDAB 3... View full answer

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