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Convert the following from hex to binary and draw it on the memory map.
RAM = 0000 -> 00FF EPROM = FF00 -> FFFF
Answer:
0000 0000 0000 0000 (0) RAM start
0000 0000 1111 1111 (255 byte) RAM stop
1111 1111 0000 0000 (63.75K) Eprom start
1111 1111 1111 1111 (64K) Eprom stop The memory map is shown below:-
Due to the tri-state nature of the I.C's (Ram, Rom etc.) they may be placed in parallel and 'turned on ' when they are required. This process is termed decoding. There are two types of decoding:- 1. Partial decoding 2. Absolute decoding
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Q. Regard as a logical address space of eight pages of 1024 words every mapped onto a physical memory of 32 frames. a. How several bits are there in the logical address? b. H
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What makes a Just-in-Time inventory system differ from a traditional inventory system?
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What are scheduling queues? Since the processes enter the system they are put into job queue. This queue contains all processes in the system. The processes that are exist in m
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How can the deadlock be prevented Bankers algorithm : The Resource allocation graph algorithm isn't applicable to a resource allocation system with multiple instances of every
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