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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
Consider a computer system with a 32-bit logical address and 4KB page size. The system supports up to 512MB of physical memory. How many entries are there in a conventional single-
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Q. How could a system be designed to allow a choice of operating systems to boot from? What would the bootstrap program need to do? Answer: Delieve a system that would like to
combined approaches (ULT/KLT)
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