Addressing is indicated by a numeric address a byte of memory cells, rather than indicative of this storage unit, so do not multiplied by 8.
1、
A CPU addressing capability of 8KB, then its address bus width is how much?
8 * 1024 = 8192 a total of 8192 rooms trying to find how much room to find the biggest root address bus 2 ^ n = 8192 n = 13
Addressing a CPU capacity is 16KB, then its address bus width is how much?
solution:
16*1024 =16384
A total of 16,384 rooms with an address bus can address two rooms so that 2 ^ n = 16384 n = 14
2、
___2048__ 2KB memory has memory cells. Number from the storage unit to ____2047__ __0__.
A: a storage unit Byte 1KB = 1024B 2KB = 2048B
3、
2KB of memory may store __1024 * 2 * 8 ___ bit, ____ 1024 * 2 ____ byte.
2KB =1024B*2*8
4、
1GB, 1MB, 1KB are __________________byte.
1GB=1024*1024*1024
1MB=1024*1024
1KB=1024
5、
8080,8088,8086,80286,80386 data line widths are respectively 8, 8, 16, 16, 32, a data which can be transmitted as is: _1 __ (B), __ 1_ (B), __ 2_ (B), __ 2_ (B), _ 4 __ (B)
6、
8080,8088,80286,80386 address bus width were 16, 20, 24, 32, they are addressing capability __2 ^ 16/1024 __ (64 KB), __ 2 ^ 20/1024 / 1024 __ (1 MB), __ 2 ^ 24/1024/1024 ___ (MB), _ 2 ^ 32/1024/1024/1024 ____ (GB)
7、
Read 1024 bytes (B) data from memory, at least 8086 times ___512_ need to read, read at least 80386 times __256_
A data transfer bus 16 B-2B 1024 times required 1024/2
8、
Memory to store data and program __2 ___ hexadecimal form.
9、
How to drive the CPU work? Please give examples?
The input device inputs an instruction via the control bus and the data found through the address bus data transfer command transmitted via the data bus