1.1
-
The addressing capability of 1 CPU is 8KB, then the width of its address bus is 13
-
1KB memory has 1024 storage units, and the number of storage units is from 0 to 1023
-
1KB memory can store 1024*8=2^13=8192 bits, 1024 Bytes
-
1GB is 1024^3 Byte , 1MB is 1024^2 Byte , 1KB is 1024 Byte
-
The address bus widths of 8080, 8088, 80296, and 80386 are 16, 20, 24, and 32 respectively, so their addressing capabilities are: 2^6=64 (KB), 2^0=1 (MB ), 2^4=16 (MB), 2^2=4 (GB)
A line can have two states, and 16 lines can be converted into 2^16 states
- The data bus widths of 8080, 8088, 8086, 80286, and 80386 are 8, 8, 16, 16, and 32 respectively. Then the data they can transmit at one time are: 1 (B), 1 (B), 2 (B), 2 (B), 4 (B)
8 roots can transmit 1Byte at a time, so the first two spaces are 1, 16/8=2, so the latter is 2, and the last 32/8=4
- To read 1024 bytes of data from memory, 8086 must read at least 512 times, and 80386 must read at least 256 times
- In the memory, data and programs are stored in the form of 01 (binary)