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1、如有侵权，请联系网站删除，仅供学习与交流第5章-计算机组成【精品文档】第 9 页Chapter 5 Computer OrganizationBy 黄永平，权勇，孙永雄，王智，丰小月，孙铭会Knowledge point:5.1. three subsystems that make up a computer5.2. functionality of each subsystem5.3. memory addressing and calculating the number of bytes5.4. addressing system for input/output devices.5.
2、5. the systems used to connect different components together.Multiple-choice Questions39. The _ is a computer subsystem that performs operations on data.( Knowledge point 5.1)a. CPUb. memoryc. I/O hardwared. none of the above40. _ is a stand-alone storage location that holds data temporarily. ( Know
3、ledge point 5.1)a. An ALUb. A registerc. A control unitd. A tape drive41. _ is a unit that can add two inputs. ( Knowledge point 5.2)a. An ALUb. A registerc. A control unitd. A tape drive42. A register in a CPU can hold _.( Knowledge point 5.2)a. datab. instructionsc. program counter valuesd.all of
4、the above43. A control unit with five wires can define up to _ operations. ( Knowledge point 5.2)a. 5b. 10c. 16d. 32 44. A word is _ bits. ( Knowledge point 5.3)a. 8b. 16c. 32d. any of the above45. If the memory address space is 16 MB and the word size is 8 bits, then _ bits are needed to access eac
5、h word. ( Knowledge point 5.3)a. 8b. 16c. 24d.3246. The data in _ are erased if the computer is powered down. ( Knowledge point 5.5.3)a. RAMb. ROMc. a tape drived. a CD-ROM47. _ is a memory type with capacitors that need to be refreshed periodically. ( Knowledge point 5.3)a. SRAMb. DRAMc. ROMd. all
6、of the above48. _ is a memory type with traditional flip-flop gates to hold data. ( Knowledge point 5.3)a. SRAMb. DRAMc. ROMd. all of the above 49. There are _ bytes in 16 terabytes. ( Knowledge point 5.3)a.b. c.d. 50. _ can be programmed and erased using electronic impulses but can remain in a comp
7、uter during erasure. ( Knowledge point 5.3)a. ROMb. PROMc. EPROMd. EEPROM51. _ is a type of memory in which the user, not the manufacturer, stores programs that cannot be overwritten. ( Knowledge point 5.3)a. ROMb. PROMc. EPROMd. EEPROM52. CPU registers should have _ speed memory. ( Knowledge point
8、5.3)a. highb. mediumc. lowd. any of the above53. Main memory in a computer usually consists of large amounts of _ speed memory. ( Knowledge point 5.3)a. highb. mediumc. lowd. any of the above54. The _ memory contains a copy of a portion of main memory. ( Knowledge point 5.3)a. CPUb. cachec. maind. R
9、OM55. The _ is nonstorage I/O device. ( Knowledge point 5.4)a. keywordb. monitorc. printerd. all of the above56.A _ is an optical storage device. ( Knowledge point 5.4)a. CD-ROMb. CD-Rc. CD-RMd. all of the above57. The _ is a storage device in which the manufacturer writes information to the disc. (
10、 Knowledge point 5.4)a. CD-ROMb. CD-Rc. CD-RMd. all of the above58. The _ is a storage device in which the user can writes information only once to the disc. ( Knowledge point 5.4)a. CD-ROMb. CD-Rc. CD-RMd. all of the above59. The _ is a storage device that can undergo multiple writings and erasings
11、. ( Knowledge point 5.4)a. CD-ROMb. CD-Rc. CD-RMd. all of the above60. The smallest storage area on a magnetic disk that can be accessed at one time is a _.( Knowledge point 5.4)a. trackb. sectorc. framed. head61. For a magnetic disk, the _ time is the time it takes for the read/write head to move t
12、o the desired track where the data are stored. ( Knowledge point 5.4)a. rotationb. seekc. transferd. location62. Polycarbonate resin is used in _ . ( Knowledge point 5.4)a. CD-ROMsb. CD-Rsc. CD-RWsd. all of the above63. In a _, a high-power laser beam simulates pits in an alloy of silver, indium, an
13、timony, and tellurium. ( Knowledge point 5.4)a. CD-ROMb. CD-Rc. CD-RMd. all of the above64. In a _, a high-power laser beam simulates pits in the dye layer. ( Knowledge point 5.4)a. CD-ROMb. CD-Rc. CD-RMd. all of the above65. Which optical storage device has the highest capacity ? ( Knowledge point
14、5.4)a. CD-ROMb. CD-Rc. CD-RMd. DVD66. In a DVD, a _ beam reads the disk. ( Knowledge point 5.4)a. high-power laserb. infraredc. red laserd. blue laser67. A _ bus connects the CPU and memory. ( Knowledge point 5.5)a. datab. addressc. controld. all of the above68. If the word size is 2 bytes, a data b
15、us with _ wires is needed. ( Knowledge point 5.5)a. 2b. 4c. 8d. 1669. If the memory haswords, the address bus needs to have _ wires. ( Knowledge point 5.5)a. 8b. 16c. 32d. 6470. A control bus with eight wires can define _ operations. ( Knowledge point 5.5)a. 8b. 16c. 256d. 51271. The _ controller fe
16、atures a parallel interface and daisy-chained connection for I/O devices. ( Knowledge point 5.5)a. SCSIb. Fire Wirec. USBd. IDE72. The _ controller is a serial device that connects slow devices such as the keyboard and mouse to the computer. ( Knowledge point 5.5)a. SCSIb. Fire Wirec. USBd. IDE73. T
17、he _ controller is a high-speed serial interface that transfers data in packets. ( Knowledge point 5.5)a. SCSIb. Fire Wirec. USBd. IDE74. The three steps in the running of a program on a computer are performed _ in this specific order. ( Knowledge point 5.5)a. fetch, execute, and decode b. fetch, ex
18、ecute, and fetchc. fetch, decode, and executed. decode, fetch, and execute75. In the _ method to synchronize the operation of the CPU with the I/O device ,the I/O device informs the CPU when it is ready for data transfer. ( Knowledge point 5.5)a. programmed I/Ob. interrupt-driven I/Oc. DMA d. isolat
19、ed I/O76. In the _ method to synchronize the operation of the CPU with the I/O device, the CPU is idle until the I/O operation is finished. ( Knowledge point 5.5)a. programmed I/Ob. interrupt-driven I/Oc. DMA d. isolated I/O77. In the _ method to synchronize the operation of the CPU with the I/O dev
20、ice, a large block of data can be passed from an I/O device to memory directly. ( Knowledge point 5.5)a. programmed I/Ob. interrupt-driven I/Oc. DMA d. isolated I/OReview questions:1. What are the three subsystems that make up a computer?(Knowledge point 5.1)Answer: the CPU, main memory, and the inp
21、ut/output (I/O) subsystem.2. What are the parts of a CPU? (Knowledge point 5.1)Answer: The CPU performs operations on data and has a ALU, a control unit, and a set of registers.3. Whats the function of the ALU? (Knowledge point 5.2)Answer: The ALU performs arithmetic and logical operations.Exercises
22、:78. A computer has 64MB of memory. Each word is 4 bytes. How many bits are needed to address each single word in memory? (Knowledge point 5.3)Solution:The memory address space is 64 MB, that is 2 raised to the power 26. The size of each word in bytes is 2 raised to the power 2. So we need 24(subtra
23、ct 2 from 26) bits to address each single word in memory.79. How many bytes of memory are needed to store a full screen of data if the screen is made of 24 lines with 80 characters in each line? The system uses ASCII code, with each ASCII character store as a byte. (Knowledge point 5.3)Solution:The
24、quantity of bytes in a full screen is 1920 (24*80) while the system uses ASCII code with each ASCII character store as a byte. So we need 1920 bytes of memory to store the full screen of data.80. An imaginary computer has four data registers(R0 to R3), 1024 words in memory, and 16 different instruct
25、ions(add, subtract, etc.). What is the minimum size of an instruction in bits if a typical instruction uses the following format: add 565 R2. (Knowledge point 5.5)Solution:The number of data registers in this computer is 4, that is 2 raised to the power 2. The number of words in this computer is 102
26、4, that is 2 raised to the power 10. The number of instructions in this computer is 16, that is 2 raised to the power 4. So the minimum size of an instruction in bits is 16 (2+10+4) bits.81. If the computer in Exercise 80 uses the same size of word for data and instructions. What is the size of each
27、 data register? (Knowledge point 5.5)Solution:The size of an instruction in bits is 16. Data size is the same with that of an instruction. So 16 is the size of each data register.82. What is the size of the instruction register of the computer in Exercise 80? (Knowledge point 5.5)Solution:The size o
28、f an instruction in bits is 16. So the size of the instruction register of the computer is also 16.83. What is the size of the program counter of the computer in Exercise 80? (Knowledge point 5.5)Solution:The number of words in this computer is 1024, that is 2 raised to the power 10. So the size of
29、the program counter of the computer is 10.84. What is the size of the data bus in Exercise 80? (Knowledge point 5.5)Solution:Data size of this computer in bits is 16. So the size of the data bus is 16.85. What is the size of the address bus in Exercise 80? (Knowledge point 5.5)Solution:The number of
30、 words in this computer is 1024, that is 2 raised to the power 10. So the the size of the address bus is 10.86. What is the minimum size of the control bus in Exercise 80? (Knowledge point 5.5)Solution:There are 2 control actions(read and write to memory) at most. So the minimum size of the control
31、bus is 1. 87. A computer uses isolated I/O addressing. Memory has 1024 words. If each controller has 16 registers, how many controllers can be accessed by this computer? (Knowledge point 5.4)Solution:Memory has 1024 words. So the address space is 1024. Each controller has 16 registers. Then we get 6
32、4 (divide 16 by 1024)controllers which can be accessed by this computer. 88. A computer uses memory-mapped I/O addressing. The address bus uses 10 lines. If memory is made of 1000 words, how many four-register controllers can be accessed by this computer? (Knowledge point 5.4)Solution:The address bus uses 10 lines. So, the address space is 1024(2 raised to the power 10). The memory is made of 1000 words and each controller has four registers. Then we get (1024-1000)/4 = 6 four-register controllers which can be accessed by this computer.