Question

Q2. (4 pts) A certain microprocessor (uP) has a 37-bit address bus and a 32-bit wide data bus. Here, similar to Q1, we are using byte packing, that is, we should be able to access each byte in the memory. Assume that you are using a memory chips organized as 128K by 8 bits.

Q2-1.Divide the 37-bit address lines into page number bits, offset bits and byte address bits.

Q2-2.How many 128K by 8 memory devices would you need to completely fill the memory space of this processor?

Q2-3.Complete the following table for the corresponding pages of memory. You should write an absolute address in hexadecimal.

Page (decimal)	Starting Address (hex)	Ending Address (hex)
0	0000000	7FFFF
50
1,000

  

Answer 1

Q2-1) 37 bit address line can access 237 bytes of memory.

each page has a memory of 512KB derived from Q2-3

so offset bits would be 512KB/128KB = 4 = 22 --- SO 2 bits will be offset bits.

Byte Address Bits will be 128K = 27  * 210 = 219 -- so 17 bits will be for address bits

37-17-2 = 18 bits will be used for page number bits

Q2-2) memory space will be 237

number of 128K by 8 bits required are = 237/ 217  = 220

convert into decimal and then convert them into hexadecimal for easeQ2-3)

7FFFF - 524287

so next starting will be 524288 so for 50 page will be = 524288*50 = 26214400 conevrting into hex we get 1900000

similarly for 1000 it will be 524288 *1000 = 524288000

similarly for 1048575 it will be 524288*1048575 = 549755289600

ending in decimal is 524287 so for 50 page it will be = 524287 * 51 = 26738637 converting into hex we get

similarly for 1000 we get 524287*1001 = 524811287 into hex 1F47FC17

similarly for 1048575 we get 524287*1048576 = 549754765312 into hex 7FFFE80001

Page (decimal)	Starting Address(hex)	Ending Address(hex)
0	0000000	7FFFF
50	1900000	197FFCD
1000	1F400000	1F47FC17

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