18. We wish to construct a byte-addressable memory system with a capacity of 2,0

Question

18. We wish to construct a byte-addressable memory system with a capacity of 2,048 bytes using 64x8 RAM chips. How many address lines must go to each RAM chip? 5 4 6 7

19. We wish to construct a byte-addressable memory system with a capacity of 2,048 bytes using 64x8 RAM chips.

How many address lines must be decoded to generate the chip select inputs for the RAM chips?

7

20. Which of the following are advantages for Big endian over Little endian?

21. Using the MARIE instruction set on page 252 of your book, convert the following assembly code to binary:

LoadI 101

22. Using the MARIE instruction set on page 252 of your book, convert the following assembly code to binary:

Halt

Explanation / Answer

SOLUTION:-

(18) 2048 = 211 = 11 address lines will be needed for addressing 2048 bytes and 64 =26 = 6 lines will be needed for each RAM chip. So the correct answer is 6 .

(19) 11 - 6 = 5 . so 5 lines will be decoded to generate the chip select inputs for the RAM chips. So the correct answer is 5 .

(20) Big endian is more natural for people most people.

Explanation:- This feature made it convenient for reading the hex dumps.

=============================================================================

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.