Group planning chapter I

Chapter one

1.2.1 The basic composition of hardware  

 

• Main memory ： Run a memory

• Auxiliary deposit ： Fuselage storage

 

1.2.2 Know each hardware component

 

• MAR Store all addresses in , therefore MAR The number of digits of the address is the number of all addresses , This corresponds to the number of storage units

• The data removed from each storage unit is stored in MDR Medium , therefore MDR The number of bits of the code in this storage unit , That is, the storage word length

• MDR The number of digits of represents a word = How many? bit（ That is, the storage word length ）

 

The first order ：

 

First, the entire table is stored in the memory

Then you have to go through the first step of completing an instruction ： Take command .PC You need to store the address of the first instruction , Namely 0. When you have an address, you take instructions from memory , We need to PC Give the address stored in to MAR,MAR The address of the first instruction is stored in ：0; Then look for the address from the memory 0 And put MDR In the store . here MDR The data in is 000001 000000101. Then you have to store instructions , will MDR The instruction fetcher in IR Storage in . At this point, the first step is completed . here PC Self adding 1.

Then the second step , Analysis instructions . take IR The opcode stored in is sent to CU Middle analysis , The analyzed instruction is " Count ".

After the analysis, proceed to the third step ： Execution instruction . Start fetching data from memory , And store the operands in ACC. Send the address code of the data to be fetched into MAR,MAR Address in 000000101, That is to say 5. Then from the memory address to 5 Take out the stored data and store it in MDR, here MDR The data in is 0000000000000010. And then from MDR Take out the data and store it in ACC, here ACC The data in is 0000000000000010. The first instruction is executed .

1.2.3 The hierarchical structure of a computer system

 

Compile the language only once , More efficient

Each execution of the interpreted language requires recompilation , Less efficient

 

1.3 Computer performance index

Memory performance index

 

CPU Performance indicators

GHz：CPU Main frequency

Main frequency 10hz: The digital pulse signal oscillates every second 10 Time

 

1KIPS-- A computer can execute 1000 instructions per second

1GFLOPS-- It means that the computer performs billions of floating-point operations per second

Overall system performance index （ static state ）

Data path bandwidth :

• data bus 1 The number of bits of information that can be transmitted in parallel ( Each hardware component transmits data through the data bus )

 

throughput :

• It refers to the number of requests processed by the system in unit time .

response time :

Sending a request from a user to a computer , The waiting time until the system responds to the request and obtains the results it needs .

 

Overall system performance index （ dynamic ）

 

The running score software has designed a program containing various instructions to let the computer run in and get the performance of the computer

 

storage capacity ：k=2^10 M=2^20 G=2^30 T=2^40

frequency 、 rate ：k=10^3 M=10^6 G=10^9 T=10^12