A Level Computer Science AQA - Fundamentals of Computer Organisation and Architecture - Paper 2 Flashcards

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1
Q

What is meant by the processor (sometimes referred to as the CPU)?

A

The part of the computer system that processes data by executing program instructions

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2
Q

What is meant by the processor instruction set?

A
  • The list of low level operations that can be executed by a processor
  • An instruction set is processor specific
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3
Q

For the processor to be able to execute a program, what must first happen?

A

The program instructions must first be transferred from secondary storage to main memory, from where they are fetched, decoded and executed

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4
Q

Describe main memory

A
  • Main memory is memory that can be accessed directly by the processor
  • Each memory location has a unique physical address, which is a number used to locate that memory location and access its contents
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5
Q

What are the two types of main memory?

A
  • RAM (Random Access Memory)
  • ROM (Read Only Memory)
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6
Q

What is meant by a type of memory being volatile?

A

When power is switched off, all data and instructions stored in the memory are lost

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7
Q

Describe RAM (Random Access Memory)

A
  • RAM is used to hold all of the data and instructions that are currently being processed
  • Each location can be accessed directly
  • The processor can read and write to it
  • It is usually volatile
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8
Q

Describe ROM (Read Only Memory)

A
  • It is non-volatile
  • It is written by the computer manufacturer
  • Usually stores the BIOS (Basic Input Output System)
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9
Q

What is meant by a bus?

A

A set of parallel lines that are used to connect together components internal components in order to pass signals between them

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10
Q

Why do buses consist of multiple lines?

A

So that multiple bits can travel at the same time allowing data to be transferred quickly between components

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11
Q

Name the three types of bus

A
  • Address bus
  • Data bus
  • Control bus
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12
Q

Describe the address bus

A
  • Carries address locations of stored data from the processor to main memory and input/output controllers
  • Unidirectional (one way connection from processor to address bus and one-way connection from address bus to main memory and to the I/O controllers)
  • In general, if the width of the address bus is expressed as n bits, then there are 2ⁿ numbers that can be used to address memory locations
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13
Q

Finish the sentence:
The width of the address bus refers to its number of parallel lines, which determines …

A

determines the number of bits that can be used to form an address of a memory location

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14
Q

Describe the data bus

A
  • Carries data to and from the processor, main memory, and input/output controllers
  • The data bus is bidirectional (allows a two-way connection between internal components of the system)
  • In general, if the width of the data bus is expressed as n bits, then n bits can be transferred at one time
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15
Q

Finish the sentence:
The width of the data bus refers to its number of parallel lines, which determines …

A

determines the number of bits that can be transferred in one operation

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16
Q

Describe the control bus

A
  • Carries control signals that coordinate the operation of the components
  • The control bus is bidirectional (there are two way connections between the components that the control bus connects)
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17
Q

The control bus carries control signals that coordinate operations that take place inside a computer system.
Give 2 things that a control signal can be used to do.

A
  • Synchronise communication between the components using clock pulses
  • Specify the type of data that is being transferred via the other buses (i.e, data values, instructions or addresses)
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18
Q

Give 2 examples of control signals that a control bus might carry.

A
  • Bus request: Signifies that a component needs to access a bus
  • Interrupt request: Signifies that an error or exception has occurred that requires the attention of the processor (CPU)
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19
Q

Name the two common system architectures

A
  • Von Neumann architecture
  • Harvard architecture
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20
Q

What are the key properties of the Von Neumann architecture?

A
  • Both program instructions and data are stored in main memory and moved between memory and processor when needed by the processor
  • The same address and data buses are used in the process of transferring instructions and data between main memory and the processor
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21
Q

What does the Von Neumann architecture consist of?

A

Consists of:
- a processor
- a memory unit that can communicate directly with the processor
- connections for input and output devices
- secondary storage for saving/backing up data

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22
Q

What are the key properties of the Harvard architecture?

A
  • The Harvard architecture keeps instructions and data in separate memories
  • The processor accesses these memories using separate data and address buses
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23
Q

Where is Von Neumann architecture more common?

A

It is used extensively in general purpose computer systems

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24
Q

Where is Harvard architecture more common?

A

Embedded systems such as digital signal processing (DSP) systems use Harvard architecture processors extensively

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25
Q

Name 4 advantages of Harvard architecture over Von Neumann

A
  • Instructions and data can be accessed simultaneously
  • Instruction and data memories can have different word lengths
  • Avoids possibility of data being executed as coded, which can be exploited by hackers
  • Different technologies can be used to implement instruction and data memory
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26
Q

Give one advantage of Von Neumann architecture over Harvard architecture

A

Allows the processor to run a variety of programs that are not known in advance

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27
Q

Describe the stored program concept

A

Machine code instructions stored in main memory are fetched and executed serially by a processor that performs arithmetic and logical operations

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28
Q

What is the role of the control unit?

A
  • To control operation of fetch-decode-execute cycle
  • To synchronise operations of processor
  • To control the transfer of data between registers
  • To handle interrupts
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29
Q

Give 2 examples of tasks that the control unit is responsible for

A
  • Decodes instructions to determine what needs to be done
  • Sends and receives control signals to and from other components
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30
Q

What is the role of the arithmetic logic unit?

A

Performs calculations and makes logical decisions

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31
Q

What is meant by the term register?

A

A very fast memory location within the processor

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32
Q

What is the purpose of the Program Counter (PC)?

A

Holds the address of the next instruction to be executed by the processor

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33
Q

What is the purpose of the Current Instruction Register (CIR)?

A

Holds the instruction that the processor is currently executing

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34
Q

What is the purpose of the Memory Address Register (MAR)?

A

Holds the address of the memory location (in main memory) that the processor needs to access, either to read from or write to

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35
Q

What is the purpose of the Memory Buffer Register (MBR)?
(sometimes also known as Memory Data Register)

A

Holds the data (data values or instructions) that are read from or written to the main memory

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36
Q

What is the purpose of the Accumulator?

A

Stores the result of any calculation processed by the ALU

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37
Q

What is the purpose of the Status Register (SR)?

A
  • Stores information about the result of the last instruction that the ALU executed
  • Each bit within the status register acts as a flag to indicate if an error or exception has occurred within the process
  • The ST can also be set to enable or disable interrupts
38
Q

What is the role of the clock?

A

Generates a signal used to synchronise the operations of the processor

39
Q

The speed of the clock is measured by the number of clock cycles in one second. One clock cycle per second is 1Hz.
What is the speed of the clock there are one billion clock cycles per second?

A

1 gigahertz (GHz)

40
Q

Explain the steps involved in the Fetch-Decode-Execute cycle and how the registers are used.

A

FETCH
- contents of PC transferred to MAR
- address bus used to transfer this address to main memory
- contents of addressed memory location moved into the MBR using data bus
- PC incremented
- transfer content of MBR to CIR
DECODE
- the control unit decodes instruction held by the CIR
- instruction split into opcode and operand
EXECUTE
- if necessary, data is fetched
- opcode identifies operation to perform
- instruction executed by relevant part of processor
- result stored in accumulator

41
Q

What do machine language instructions consist of?

A

An opcode and one or more operand (value, memory address or register)

42
Q

What is meant if by an instruction that uses immediate addressing?

A

The operand is the data

43
Q

What is meant if by an instruction that uses direct addressing?

A
  • The operand is the address of the data
  • Address to be interpreted as meaning either main memory or register
44
Q

What does the following AQA assembly language instruction do?
LDR Rd,

A

Loads the value stored in the memory location specified by into register d

45
Q

What does the following AQA assembly language instruction do?
STR Rd,

A

Store the value that is in register d into the memory location specified by

46
Q

What does the following AQA assembly language instruction do?
ADD Rd, Rn,

A

Add the value specified in to the value stored in register n and store the result in register d

47
Q

What does the following AQA assembly language instruction do?
SUB Rd, Rn,

A

Subtract the value specified by from the value in register n and store the result in register d

48
Q

What does the following AQA assembly language instruction do?
MOV Rd,

A

Copy the value specified by into register d

49
Q

What does the following AQA assembly language instruction do?
CMP Rn,

A

Compare the value stored in register n with the value specified by

50
Q

What does the following AQA assembly language instruction do?
B

A

Always branch to the instruction at position in the program

51
Q

What does the following AQA assembly language instruction do?
B

A

Branch to the instruction at position if the last comparison met the criterion specified by .
Possible values for and their meanings are:
EQ: equal to
NE: not equal to
GT: greater than
LT: less than

52
Q

What does the following AQA assembly language instruction do?
AND Rd, Rn,

A

Perform a bitwise logical AND operation between the value in register n and the value specified by and store the result in register d

53
Q

What does the following AQA assembly language instruction do?
ORR Rd, Rn,

A

Perform a bitwise logical OR operation between the value in register n and the value specified by and store the result in register d

54
Q

What does the following AQA assembly language instruction do?
EOR Rd, Rn,

A

Perform a bitwise logical XOR (exclusive or) operation between the value in register n and the value specified by and store the result in register d

55
Q

What does the following AQA assembly language instruction do?
MVN Rd,

A

Perform a bitwise logical NOT operation on the value specified by and store the result in register d

56
Q

What does the following AQA assembly language instruction do?
LSL Rd, Rn,

A

Logically shift left the value stored in register n by the number of bits specified by and store the result in register d

57
Q

What does the following AQA assembly language instruction do?
LSR Rd, Rn,

A

Logically shift right the value stored in register n by the number of bits specified by and store the result in register d

58
Q

What does the following AQA assembly language instruction do?
HALT

A

Stops the execution of the program

59
Q

Figure 5 shows an incomplete assembly language program. The intended purpose of the code is to count from 1 to 10 inclusive, writing the values to memory location 17, which is used to control a motor.
Complete the code in Figure 5. You may not need to use all four lines for your solution and you should not write more than one instruction per line. (4)
Figure 5
——————-MOV R0, #1
startloop:
——————-STR R0, 17
——————-……………………………………………
——————-……………………………………………
——————-……………………………………………
——————-……………………………………………
endloop:
——————-HALT

A

ADD R0, R0, #1
CMP R0, #11
BEQ endloop
B startloop

60
Q

The greatest common divisor of two common integers A and B is the largest positive integer that divides both of the numbers without leaving a remainder.
For example, if A = 4 and B = 6 then:
- 4 has the divisors 1, 2 and 4
- 6 has the divisors 1, 2, 3, and 6
Therefore, the greatest common divisor of 4 and 6 is 2, since this is the biggest number which appears both in the list of divisors of both 4 and 6.
Figure 1
WHILE A =/= B
———IF A > B THEN
————-A = A - B
———ELSE
————-B = B - A
———ENDIF
ENDWHILE
When the procedure described in this algorithm terminates, the value in A (and also B) is the greatest common divisor of A and B.
The available general purpose registers that the programmer can use are numbered 0 to 12.
Write a program (using the AQA assembly language instruction set) that uses the method described in a Figure 1 to calculate the greatest common divisor of two positive integers.
- At the start, the positive integer A will be stored in memory location 102 and the positive integer B in memory location 103. Your program should use these values to find their greatest common divisor
- When your program terminates, it should store the greatest common divisor of these two numbers in memory location 104
(8)

A

——————-LDR R1, 102
——————-LDR R2, 103
loop:
——————-CMP R1, R2
——————-BEQ endwhile
——————-BGT AminusB
——————-B BminusA
AminusB:
——————-SUB R1, R1, R2
——————-B loop
BminusA:
——————-SUB R2, R2, R1
——————-B loop
endwhile:
——————-STR R1, 104
——————-HALT

61
Q

Describe the role of interrupts

A
  • Allows the currently executing process to be suspended
  • So that a device that needs the immediate attention of the processor can be dealt with/serviced
62
Q

Explain why the volatile environment (the contents of registers) must be saved before an interrupt is serviced.

A
  • So that the currently running process can be returned to
  • As the interrupt will change/overwrite register values
63
Q

An interrupt can occur during the fetch-execute cycle.
Explain what happens when an interrupt occurs.

A
  • Volatile environment/current processor state saved on stack
  • Source of interrupt identified
  • Appropriate interrupt service routine (ISR) called
  • Volatile environment/processor state restored
64
Q

What is meant by cache memory?

A
  • A small amount of very quickly accessible memory located on the processor
  • It stores the most frequently used data and instructions
65
Q

How does cache memory affect processor performance?

A
  • If more data and instructions are stored on the cache, fewer fetches from main memory will be required
  • As cache is located on the processor itself, data can also be fetched from here more quickly than RAM so processor performance would improve
66
Q

How does increasing the clock speed affect processor performance?

A

By increasing the clock speed, the processor will execute more instructions per unit time, resulting in higher processor performance

67
Q

How would increasing the word length affect processor performance?

A
  • By increasing the word length, the processor can handle more data per instruction
  • The processor could have a larger instruction set
  • The processor could access a larger amount of memory through direct addressing
68
Q

How does address bus width affect processor performance?

A
  • By increasing the address bus width we can address more unique memory locations
  • Increasing the address bus width increases the maximum potential memory size
69
Q

How does data bus width affect processor performance?

A

By increasing the data bus width, we can more more data around per unit time so processor performance would improve

70
Q

How would increasing the number of cores affect processor performance?

A
  • Two or more independent processing units means that tasks can potentially be run in parallel
  • Tasks split up to make use of extra cores so speed of execution is increased
71
Q

Name 6 factors that affect processor performance.

A
  • Multiple cores
  • Cache memory
  • Clock Speed
  • Word length
  • Address bus width
  • Data bus width
72
Q

What does a barcode reader consist of?

A
  • light source (low-powered laser diode)
  • lens
  • photoelectric detectors (photodiodes)
  • decoder circuitry to analyse the barcode’s image data, and generate character codes
73
Q

Describe the principles of operation of a barcode reader

A
  • Laser shone at barcode
  • black/white bands reflect different amounts of light
    (black bands reflect less light, white areas reflect more light)
  • light sensor measures amount of reflected light
  • light reflected converted into an electrical signal
  • electrical form of reflection analysed to decode barcode into character form
  • scanner outputs the character codes (e.g. ASCII) as binary for processing by a computer
74
Q

Give 3 advantages of barcodes

A
  • barcodes are very cheap
  • often smaller and lighter than an RFID tag
  • barcodes not susceptible to radio interference unlike RFID
75
Q

Give 4 disadvantages of barcode systems

A
  • Barcodes could be obscured (e.g. by clothing)
  • Difficult to scan a code that is being moved around and that is not on a flat surface
  • Scanner needs to be quite close to barcode
  • Barcode more likely to be damaged than RFID tag (when damaged it is unreadable)
76
Q

Describe the principals of operation of a digital camera

A
  • Shutter opens for a fraction of a second to admit light
  • Light reflected from object focused by lens onto a 2D array of light sensitive cells (photosensors) to form analogue signal
  • Photoelectric effect produced a voltage in the pixel
  • To process analogue image digitally, the magnitude of charge in each photosensor is converted into a digital format by an analogue to digital converter (ADC)
  • For colour images, red, green and blue filters cover photosensors. Each row of photodiodes shares an ADC located to the right of each row. The accumulated charges in a row are shifted rightwards 1 cell at a time, converted into an analogue voltage, then into digitally equivalent voltage by ADC
77
Q

Give 1 advantage of digital cameras

A

Not tag/code that can be lost/damaged

78
Q

Give 2 disadvantages of digital cameras

A
  • Difficulties taking a clear picture e.g caused by blocked line of sight due to other competitors
  • High data storage requirements for the number of photos that will be needed in a large event
79
Q

Describe the principles of operation of a laser printer

A
  • To print a page, a bitmap image of the page is created
  • Negative charge applied to print drum and laser used to reverse charge on the drum, in line with bitmap image
  • Charge applied to toner and toner sticks to drum where the laser struck
  • Paper passed over drum and toner transfers to it
  • Positively charged transfer roller assists transfer of toner from drum to paper
  • Heat fuses toner onto paper
  • For colour printing, four different colour toners are required
80
Q

Give 3 advantages of laser printers

A
  • Prints fast so can handle large volumes
  • Laser printers can print text accurately
  • Durable and reliable
81
Q

Give 2 disadvantages of laser printers

A
  • Can’t print images as accurately as text
  • Can be expensive to buy
82
Q

Describe how RFID operates.

A
  • RFID scanner transmits signal to interrogate tag
  • Tag receives signal and current induced
  • RFID transponder/tag transmits its data (unique identifier and any other stored data) by radio waves to the reader
83
Q

Give 4 advantages of RFID

A
  • No line-of-sights issues (unlike with barcodes)
  • Readers can process multiple tags at the same time as each has a unique identifier
  • RFID tags less easily damaged than barcodes
  • Tags potentially reusable
84
Q

Where might RFID systems be used?

A
  • For passports
  • Also good for tracing because of unique identity of tag
85
Q

Give 2 problems of RFID

A
  • Data flooding (extra data produced when items moved a small distance in stocking systems)
  • Security (unauthorised readers can potentially use RFID to identify or track packages)
86
Q

Explain the need for secondary storage within a computer system

A
  • To store data/programs whilst the computer is turned off (secondary storage is non- volatile)
  • As the contents of RAM are lost when the computer is turned off
87
Q

Describe the principles of operation of a hard disk drive

A
  • Binary 0s and 1s represented by magnetising spots on disk
  • Disk divided into tracks and sectors
  • Disk continuously spinning at high speed
  • Data read/written as correct sector passes under read/write head
  • May be multiple platters, one read/write head per platter
  • Data transferred in sectors/blocks
88
Q

Describe the principles of operation of an optical disk drive

A
  • Low power laser beam shone at disk
  • Light is focussed on spot on track
  • Light sensor detects amount of light reflected back
  • Disc spins at constant linear velocity
  • Data is stored on one/spiral track
  • Continuation of land/pit reflects light whereas transition between land and pit scatters light
  • Transition between land and pit indicates a 1 and continuation of land/pit represents 0
89
Q

Describe the principles of operation of an Solid State Disk (SSD)

A
  • Data is stored electronically (no moving parts)
  • Data is stored in floating gate transistors that do not lose their charge when power is no longer applied
  • Presence of trapped electrons or absence indicates 0/1
  • Data is organised into pages/blocks
  • A block must be erased before it can be overwritten
  • Controller manages the organisation of the data
90
Q

Give 2 advantages of Hard Disk over Solid State Drives

A
  • Greater capacity for same price
  • Higher capacity drives available
91
Q

Give 5 advantages of a Solid State Drives over Hard Disk

A
  • Lower power consumption
  • Faster read/write times
  • Lower latency
  • Less vulnerable to damage from physical impact
  • Less heat generates
92
Q

Give 1 development in the design of hard disk drives that has enabled an increase in storage capacity

A

More tracks on a platter