A-Level Computer Science · ⁨מדעי המחשב A-Level⁩

The three number systems 数制 you must use:

• denary 十进制 (decimal, base 10) — uses digits 0–9. Place values are powers of ten.
• binary 二进制 (base 2) — uses 0 and 1. Place values are powers of two. Every byte 字节 is 8 bits 位.
• hexadecimal 十六进制 (base 16) — uses 0–9 then A–F for 10–15. Each hex digit 数位 stands for exactly 4 bits.

Conversions

Denary → binary: keep dividing by 2 and record the remainders, read bottom-up. Or subtract the largest place value 位值 (power of 2) that fits.

Example: $558_{10}$: $558 = 512 + 32 + 8 + 4 + 2 = 2^{9} + 2^{5} + 2^{3} + 2^{2} + 2^{1}$. In 12 bits: 0010 0010 1110.

Binary → hex: group the bits into nibbles 半字节 (4 bits) from the right and convert each. 0010 0010 1110 → 2 2 E → 22E.

Hex → binary: replace each hex digit with its 4-bit pattern. Hex → denary: multiply each digit by its place value. 22E $= 2 \times 256 + 2 \times 16 + 14 = 558$.

Worked example. Convert denary 200 to 8-bit binary, then to hexadecimal.

$200 = 128 + 64 + 8$, so the binary is 11001000. In nibbles, 1100 1000 $= 12$ and $8$, i.e. $\text{C}$ and $8$, so the hexadecimal is C8.

How many bits?

Exam questions fix the register width 寄存器宽度 (8, 12 or 16 bits). Pad with leading zeros to that width: $558$ in 12 bits is 0010 0010 1110, never 10 0010 1110.

To find the minimum number of bits that can store a value, ask which place values you need:

• an unsigned integer from $0$ to $2^{n} - 1$ needs $n$ bits: $200$ needs 8 bits (the top is $255$), $1000$ needs 10 bits (the top is $1023$), $16$ needs 5 bits (4 bits stop at $15$).
• a signed two's-complement integer from $-2^{n-1}$ to $2^{n-1} - 1$ needs $n$ bits: $-200$ needs 9 bits, because 8 bits stop at $-128$.
• one hexadecimal digit needs 4 bits, one BCD digit needs 4 bits, and one ASCII character needs 7 bits (8 for extended ASCII).

Binary vs decimal prefixes

Two prefix families look similar but differ — decimal (powers of 10) and binary (powers of 2):

So a tebibyte (TiB) is slightly more than a terabyte (TB). A "1 TB" drive holds $10^{12}$ bytes, but an operating system that reports in TiB shows a smaller number.

Binary addition

Add column by column from the right, carrying as in denary:

Overflow 溢出 happens when the result needs more bits than the register 寄存器 can hold — the carry-out of the leftmost column is the overflow bit.

Worked example. Add the 8-bit unsigned integers $10110101$ and $01101100$, and comment on the result.

$10110101 + 01101100 = 1\,00100001$. The answer needs 9 bits, so it does not fit in an 8-bit register: overflow has occurred. A full answer names the error and says why, using the word size the question gave: "Overflow: the true result ($289$) is larger than the largest value an 8-bit register can hold ($255$), so the carry out of the most significant bit is lost and the stored result ($00100001 = 33$) is wrong."

Binary subtraction

The usual way is two's complement 补码 addition: to do $A - B$, form the two's complement of $B$ (invert every bit and add 1), then add, and discard any final carry-out.

To subtract $00011110$ from $01100100$ (unsigned 8-bit):

• two's complement of $00011110$: invert → $11100001$, add 1 → $11100010$.
• add to $01100100$: result $1\,01000110$ (9 bits) — discard the leading 1 → $01000110 = 70_{10}$. Check: $100 - 30 = 70$. ✓

Two's complement signed integers

In an $n$-bit two's-complement number:

• the most significant bit 最高有效位 (MSB) is the sign bit 符号位: 0 = positive, 1 = negative.
• to read a negative number: invert every bit, add 1, then negate.

So $11100010$ is negative; invert → $00011101$, add 1 → $00011110 = 30$, so it is $-30$. This is a signed integer 有符号整数 (unlike an unsigned 无符号 one). The range for $n$ bits is $-2^{n-1}$ to $+2^{n-1} - 1$; for 8 bits, $-128$ ($10000000$) to $+127$ ($01111111$).

The same bits mean different numbers depending on the agreed reading. As an unsigned integer every bit is a place value, so 8 bits run from $0$ to $255$; as a signed two's-complement integer the top bit is the sign, so the same 8 bits run from $-128$ to $+127$. The pattern $11111111$ is $255$ read one way and $-1$ read the other — nothing in the bits themselves says which.

The same byte read as unsigned and as signed: only the agreed interpretation tells them apart 8-bit two's complement: the sign bit splits the range into negative ($-128$ to $-1$) and positive ($0$ to $127$)

Worked example. What denary value does the 8-bit two's-complement number $10110100$ represent?

The MSB is 1, so it is negative. Invert → $01001011$, add 1 → $01001100 = 76$, so the value is $-76$. Check with place values: $-128 + 32 + 16 + 4 = -76$.

Worked example. Write $-108$ as a 12-bit two's-complement integer.

Start from $+108$ in 12 bits: $108 = 64 + 32 + 8 + 4$, so 0000 0110 1100. Invert every bit: 1111 1001 0011. Add 1: 1111 1001 0100. Check with place values, where the top bit is worth $-2^{11} = -2048$: $-2048 + 1024 + 512 + 256 + 128 + 16 + 4 = -108$. ✓

For 12 bits the range is $-2048$ (1000 0000 0000) to $+2047$ (0111 1111 1111). Questions that ask for the smallest and largest values want these two patterns, so learn the rule: the most negative number is a 1 followed by zeros; the most positive is a 0 followed by ones.

An arithmetic shift 算术移位 moves every bit left or right but keeps the sign: a shift right by one place halves the value and copies the sign bit into the empty space on the left, so a negative number stays negative (1111 1001 0100 shifted right three places is 1111 1111 0010, which is $-14$: $-108 / 8 = -13.5$, and a shift right rounds down). A shift left doubles the value. Shifts belong to the assembly instruction set in topic 4, but this question is asked with the number work here.

Overflow in signed arithmetic happens when the true result falls outside this range — spotted when the sign bit flips wrongly (two positives giving a negative, or two negatives giving a positive).

One's complement

Before two's complement, an older scheme called one's complement 反码 represented a negative number by simply inverting every bit of the positive — there is no "add 1" step.

• $+30 = 00011110$, so in one's complement $-30 = 11100001$ (just the inverse).
• Drawback: it has two zeros — $00000000$ ($+0$) and $11111111$ ($-0$) — which wastes a bit pattern and makes arithmetic awkward.

Two's complement (invert and add 1) removes the negative zero: it has a single zero and lets addition and subtraction use the same circuit. That is why modern computers store signed integers in two's complement, not one's complement.

In BCD 二进码十进数, each denary digit is written as its own 4-bit pattern. The number $93$ is 1001 0011 in BCD — not binary 93 ($01011101$). Each nibble uses only 0–9; patterns $1010$–$1111$ are invalid.

BCD reading: 0010 0111 0101 → 2, 7, 5 → 275.

Use: calculators, digital clocks, and devices that show denary digits — each digit drives a 7-segment display 七段显示器. Currency code often uses BCD to avoid the rounding errors of converting fractions like 0.1 to binary.

A "justify" answer must link the use to a property of BCD: each denary digit has its own 4 bits, so a digit can be sent straight to its display, or added digit by digit, with no conversion of the whole number; and a decimal fraction such as $0.10$ is stored exactly, which a binary fraction cannot do.

Hex is a compact way to write binary (1 hex digit = 4 bits):

• memory addresses 内存地址 in low-level programming — 0x7FFE.
• colour values in HTML/CSS — #FF8800.
• MAC addresses — AC:DE:48:00:11:22.

Hex does not change the stored data — it just makes binary easier for humans.

Computers store text as numbers; each character has a numeric code point 码点 set by a character set 字符集.

ASCII

• ASCII uses 7 bits — 128 code points. Basic Latin letters, digits, punctuation, and control codes.
• Extended ASCII uses 8 bits — 256 code points; the lower 128 match ASCII, the upper 128 vary by region.

Unicode

• Unicode is a universal character set covering almost every script, plus symbols and emoji.
• common encodings 编码: UTF-8 (1–4 bytes, ASCII-compatible), UTF-16 (2 or 4 bytes), UTF-32 (fixed 4 bytes).

Why Unicode beats ASCII

• it represents far more characters (every script, emoji); ASCII covers only basic English.
• files are portable with no code-page confusion, and allow multilingual text in one document.
• trade-off: Unicode files are usually larger for English-only text.

When a question asks for differences, give them in pairs with numbers: ASCII uses 7 bits (extended ASCII 8), so 128 (256) characters; Unicode uses up to 32 bits (UTF-8 uses 1 to 4 bytes), so more than a million code points. ASCII covers basic English only; Unicode covers every script, and its first 128 code points are the ASCII ones. In UTF-8 an English letter still takes 1 byte, so a 40-letter English file name is 40 bytes in ASCII and in UTF-8 alike, while a Chinese character takes 3 bytes.

A bitmap 位图 image (also called a bitmapped image) stores the colour of every pixel 像素 in a grid. At the start of the file a file header 文件头 records the image's metadata — its width, height and colour depth — so software knows how to read the pixel data that follows.

• image resolution 图像分辨率: the bitmap's own size, width × height in pixels (e.g. 1920 × 1080).
• screen resolution 屏幕分辨率: the width × height the display can show. If an image's resolution is larger than the screen it is scaled down to fit; a low-resolution image looks blocky when stretched onto a higher-resolution screen.
• colour depth 颜色深度 (bit depth 位深度): bits per pixel. 1 bit → black/white; 8 bits → 256 colours; 24 bits → 16.7 million ("true colour").

File size

Divide by 8 for bytes, by 1024 for KiB, etc. Example: a $3000 \times 2000$ image at 24 bpp is $3000 \times 2000 \times 24 = 1.44 \times 10^{8}$ bits $\approx 17.2\ \text{MiB}$.

State the units you used. The mark scheme accepts $1\ \text{MB} = 10^{6}$ bytes (the SI prefix) or $1\ \text{MiB} = 1024 \times 1024$ bytes (the binary prefix), as long as your working shows which one; the same image is $18.0\ \text{MB}$ or $17.2\ \text{MiB}$. Add the size of the file header if the question gives one.

A video is a sequence of bitmap images, each one a frame 帧. Before compression its size is the size of one frame $\times$ the frame rate 帧率 (frames per second) $\times$ the duration in seconds: 30 frames per second of $1920 \times 1080$ pixels at 24 bits is $30 \times 1920 \times 1080 \times 24 \approx 1.5 \times 10^{9}$ bits, about $187\ \text{MB}$, for every second. That is why video is always compressed.

Changing settings

• lower resolution → smaller file, less detail (looks blocky when enlarged).
• lower colour depth → smaller file, but smooth shades show banding.
• higher of either → larger file, better quality.

A vector graphic 矢量图形 stores the instructions to draw the image as a drawing list 绘图列表 — an ordered list of drawing objects 绘图对象 (geometric primitives 图元: lines, curves, polygons, circles). Each drawing object has properties 属性 such as colour, fill, line width and position (coordinates). To show it, the program renders 渲染 the drawing list at any resolution needed.

Bitmap vs vector

Vector advantage: it scales without losing quality — a vector logo stays sharp at any size, while a bitmap blurs when enlarged. Vector disadvantage: it cannot describe arbitrary pixel detail (photographs).

A "justify" answer links the choice to the task. "The logo must appear on a business card and on a billboard, so it should be a vector graphic: it is stored as drawing objects and is re-rendered sharply at any size, whereas a bitmap would show its pixels when enlarged." For a photograph the argument runs the other way: there are no shapes to describe, so every pixel's colour must be stored.

A continuous wave of analogue data 模拟数据 (the sound) is converted into digital data 数字数据 by sampling 采样:

• sampling rate 采样率 — samples per second (Hz). CD quality is $44.1\ \text{kHz}$.
• sampling resolution 采样分辨率 (bit depth) — bits per sample's amplitude 振幅. CD quality is 16 bits.

File size

A 10-second stereo CD clip: $44100 \times 16 \times 10 \times 2 = 14\,112\,000$ bits $\approx 1.68\ \text{MiB}$.

Changing settings

• higher sampling rate → captures higher pitches, larger file.
• higher sample resolution → finer amplitude steps, less quantisation 量化 noise, larger file.
• lower of either → smaller file, clear quality loss.

(The sampling rate must be at least twice the highest frequency you want to keep.)

Compression 压缩 reduces file size, saving storage and transmission bandwidth 带宽. Two kinds:

• lossless 无损 — the original data is recovered exactly (text, programs, ZIP/PNG).
• lossy 有损 — some detail is dropped for much smaller files (JPEG, MP3, video).

When to use which

• lossless for documents, source code, medical images — anything needing exact data.
• lossy for streaming media. Real-time video streaming uses lossy compression because it must send huge amounts of data in real time over limited bandwidth; lossless would not shrink it enough. Raw HD video is gigabytes per minute, so without compression the picture would keep freezing.

A "justify" answer names the method, then the reason from the situation: "Lossless, because the spreadsheet must be restored exactly; a single changed value would make the accounts wrong." Or: "Lossy, because the photographs are viewed on a phone screen where the dropped detail is not visible, and the smaller files upload faster and use less storage."

Lossless methods

• run-length encoding 行程编码 (RLE): store "the next $n$ values are $x$" instead of repeating $x$. Great for flat areas; useless for noisy data.
• dictionary methods 字典编码 (ZIP, PNG): replace repeated byte sequences with a short reference. Good for text and code.
• Huffman coding 霍夫曼编码: give short codes to common symbols and long codes to rare ones, bringing the average code length near the data's entropy 熵.

How each kind of file is compressed:

• text file: dictionary methods and Huffman coding turn repeated words and common characters into short codes. Text must stay lossless, because one changed character changes the meaning.
• bitmap image: RLE for runs of identical pixels (icons, diagrams, black-and-white scans); lossy JPEG for photographs, or a lower colour depth or resolution.
• vector graphic: the drawing list is already small; remove drawing objects that are not needed, store coordinates to fewer decimal places, or apply a lossless method such as ZIP to the file.
• sound file: lossy MP3 or AAC removes what the ear cannot hear; a lower sampling rate or resolution is also lossy; lossless formats keep every sample and shrink the file much less.

Lossy methods

• images (JPEG): drop fine detail and colour differences the eye barely sees.
• sound (MP3, AAC): drop pitches we hear less well, and quiet sounds hidden by louder ones.
• video combines spatial 空间 compression (within each frame, like JPEG) with temporal 时间 compression (most frames store only the differences from the previous frame).

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• Show working for base conversions: denary → binary by place values, binary → hexadecimal in nibbles (groups of 4 bits).
• For two's complement the MSB is negative; to negate, invert and add 1; watch for overflow when the sign bit flips wrongly.
• Distinguish bitmap (pixels; file size $=$ width $\times$ height $\times$ colour depth) from vector (drawing commands; scales without loss).
• Sound file size depends on sample rate $\times$ bit depth $\times$ time — more of each means better quality but a bigger file.
• Compare lossless vs lossy compression and give a use for each.

Common mistakes

• Explaining an overflow with "the answer was greater than 255" or "it has 9 bits". State the word size the question gave, then say the result cannot be represented in it.
• Making a negative number by setting the top bit to 1 and leaving the rest (sign and magnitude). Two's complement means invert every bit of the positive value, then add 1.
• Forgetting to pad a converted number to the register width the question asks for.
• Mixing bits and bytes in a file-size calculation. Work in bits, divide by 8 once, and say whether you used 1000 or 1024.
• Answering "describe" in everyday words ("the picture gets worse"). Use the syllabus terms: fewer colours, banding, lower image resolution, larger pixels.

A general-purpose computer has four building blocks:

• input devices 输入设备 — get data in (keyboard, mouse, microphone, scanner, sensors).
• output devices 输出设备 — give results out (monitor, speakers, printer, actuators).
• primary memory 主存储器 — fast memory the processor 处理器 (CPU) reaches directly (RAM and ROM). Holds the running program and its data.
• secondary storage 辅助存储器 — slower, larger, keeps programs and data when not in use (hard disk, SSD, optical disc, USB stick).

The syllabus asks why each is needed. Input devices are needed because the computer can only work on data and instructions that have been entered. Output devices are needed to present the results in a form people can use. Primary memory is needed because the processor can only execute instructions and use data that are held in memory it can address directly, and it must reach them fast. Secondary storage is needed because primary memory is volatile and small: programs and data must survive the power being switched off, in a larger and cheaper store, and removable storage lets data be moved between computers or kept as a backup.

An embedded system 嵌入式系统 is a computer built into another device to do one fixed job (washing machine, microwave, car engine unit, thermostat).

• benefits: optimised for one task, so it is small, uses little power and is cheap to make in volume; reliable, because it runs one fixed program with few chances to go wrong; starts quickly and needs no user set-up; easy to use through a simple interface.
• drawbacks: limited to its one task, so it cannot be upgraded to do more; hard to update (its firmware 固件 may need special tools or cannot be changed at all); difficult to troubleshoot, and usually the whole device must be replaced when it fails; if it is connected to a network it can be a security weakness, because its software is rarely patched.

A "describe the drawbacks" question wants each drawback as a full point: what the limitation is and what it means for the user, for example "the firmware cannot be updated, so a security fault found later cannot be fixed".

Laser printer

A laser printer 激光打印机 scans the page image onto a charged photosensitive drum 感光鼓. Toner 墨粉 sticks to the charged areas, transfers to the paper, and is melted on by a fuser. Fast, sharp, high-volume.

How it works, in the steps the mark scheme lists:

1. The data for the page is sent to the printer's buffer.
2. The drum is given a uniform electrostatic charge.
3. A laser, reflected off a rotating mirror, scans the page image onto the drum, removing the charge where it strikes, so the charge left on the drum matches the image.
4. Toner, a charged powder, is attracted to the charged parts of the drum only.
5. The paper is given the opposite charge and rolled against the drum, so the toner transfers onto it.
6. The fuser 定影器, a pair of heated rollers, melts the toner into the paper. The drum is then discharged and cleaned for the next page.

3D printer

A 3D printer 3D打印机 builds an object layer by layer: FDM melts plastic filament through a nozzle; stereolithography cures liquid resin with a UV laser. Used for prototypes and custom medical parts.

How it works:

1. A model of the object is designed in CAD software (or scanned).
2. Slicing software divides the model into thin horizontal layers and produces the instructions for each one.
3. The printer builds the object one layer at a time: an FDM printer melts plastic filament 塑料丝 and lays it down through a moving nozzle; a resin printer cures liquid resin with a laser or UV light; a powder printer fuses powder with a laser.
4. Each layer bonds to the layer below, and the platform (or nozzle) moves by one layer's thickness.
5. When the last layer is done, any support material is removed. Uses include prototypes, custom medical parts such as prosthetics, and spare parts printed on demand.

Microphone and speakers

A microphone 麦克风 turns sound into an electrical signal (a diaphragm vibrates, changing capacitor 电容器 charge or coil position); the signal is digitised by an analogue-to-digital converter 模数转换器 (ADC). A speaker does the reverse — a varying signal drives a coil in a magnetic field, moving a cone to make sound.

How a microphone works: sound waves make a diaphragm 膜片 vibrate; in a dynamic microphone a coil attached to the diaphragm moves in a magnetic field, so a varying current is induced in it, and in a condenser microphone the diaphragm is one plate of a capacitor whose capacitance changes as it moves; the varying analogue signal is then sampled by an ADC and stored as digital data. A speaker runs the chain backwards: a digital-to-analogue converter 数模转换器 (DAC) produces a varying current, the current in the coil creates a changing magnetic field that pushes against the permanent magnet, the coil and cone move in and out, and the cone's movement makes pressure waves in the air.

Magnetic hard disk (HDD)

A hard disk 硬盘 stores data on spinning platters coated with magnetic material. Each platter has tracks 磁道 divided into sectors 扇区. A read/write head 读写头 floats just above and magnetises tiny regions (write) or senses them (read). Cheap per gigabyte, but slower than SSDs and has moving parts.

How it works: the platters spin at high speed (thousands of revolutions per minute); each surface is divided into concentric tracks and each track into sectors; read/write heads on actuator arms 磁头臂 move across the platters to the right track; to write, the head magnetises a tiny region with one of two polarities, representing 0 or 1; to read, it detects the polarity as the region passes beneath it. The delays, waiting for the arm to reach the track and for the sector to spin round, are why a hard disk is slower than an SSD.

Solid-state (flash) memory

A solid-state drive 固态硬盘 stores data as charge in transistors 晶体管, with no moving parts. Faster random access than HDDs, tougher, lower power, but dearer per gigabyte; each cell wears out after many writes.

How it works: each cell is a floating-gate transistor 浮栅晶体管; a charge trapped on the floating gate represents a bit and stays there when the power is off; a controller chip maps each address to a cell and spreads writes across the cells, because a cell survives only a limited number of writes.

A "why a server uses hard disks rather than SSDs" question wants the left column: cheaper per gigabyte for very large capacities, a long life under constant rewriting, and easier data recovery.

Optical disc

A laser detects reflections from tiny pits on an optical disc 光盘 (CD, DVD, Blu-ray). The drive is an optical disc reader/writer: writing uses a stronger laser to change the surface's reflectivity.

How it works: the disc carries one long spiral track of pits 凹坑 and lands 平台 (the flat areas between them); the disc spins and a laser is focused on the track; light reflected from a land differs from light reflected at the edge of a pit, and a light sensor reads each change as a 1 and no change as a 0. Writing uses a stronger laser to change the reflectivity of a dye or alloy layer. A Blu-ray uses a blue laser with a shorter wavelength, so its pits are smaller and closer together, which is why it holds more data than a DVD.

Touchscreen

A touchscreen 触摸屏 senses contact. Resistive 电阻式: two conductive layers pressed together; works with anything but is less accurate. Capacitive 电容式: a finger disturbs a charge field; accurate, multi-touch, used in phones.

How it works: a resistive screen has two thin conductive layers separated by spacers; pressing pushes the top layer onto the bottom one, closing a circuit at that point, and the controller reads the voltage to find the coordinates. A capacitive screen has a glass layer coated with a transparent conductor that holds a charge; a finger touching it draws a tiny current, the current is measured at each corner, and the controller works out the touch position from the differences. Capacitive screens respond to a light touch and to several fingers at once, but not to a gloved finger or an ordinary stylus.

Virtual reality headset

A virtual reality 虚拟现实 (VR) headset has two small displays (one per eye) and motion sensors (accelerometer 加速度计, gyroscope 陀螺仪) that track head movement so the scene shifts as you look around.

How it works: each eye sees its own display through a lens, and the two images differ slightly, so the brain sees depth; sensors (accelerometer, gyroscope, sometimes cameras) report where the head is and which way it points; the computer re-renders the scene from that viewpoint many times a second, so turning the head turns the view; headphones give sound that matches the direction. Used for games, for training such as flight or surgery simulators, and for viewing designs before they are built.

A buffer 缓冲 is memory that holds data temporarily while it moves between devices of different speeds. Example: the CPU writes a document to a printer buffer quickly, then is free to do other work while the printer prints from the buffer at its own pace. Buffers stop the fast device waiting for the slow one (also used in streaming, the keyboard, and disk access).

"State why a 3D printer needs a buffer": the computer sends the print data much faster than the printer can build the layers, so the data is held in the buffer until the printer is ready for it, and the processor is freed to do other work. When the buffer runs low the printer sends an interrupt 中断 to ask for more (topic 4). A video stream works the same way: the buffer fills ahead of playback so a short drop in the network speed does not stop the picture.

• RAM 随机存取存储器 (Random Access Memory) — volatile 易失性 (loses data without power). Holds the OS, running programs and their data; read and written constantly.
• ROM 只读存储器 (Read-Only Memory) — non-volatile 非易失性 (keeps data without power). Usually written once; holds firmware needed at start-up (the BIOS / boot loader).

ROM starts the system; RAM then holds the active work.

More RAM lets a computer hold more programs and data at once, so it swaps less between memory and disk and runs faster; that is the answer to "explain why the computer with more RAM performs better".

The same memory split matters in a wearable device: its fixed program must remain available after power off, while its live readings change during use.

• SRAM 静态RAM (Static RAM) stores each bit in a flip-flop 触发器 of several transistors. Fast, but expensive and not dense. Used for CPU cache 高速缓存.
• DRAM 动态RAM (Dynamic RAM) stores each bit as charge on a tiny capacitor. Cheaper and denser but slower, and must be refreshed 刷新 (rewritten) thousands of times a second. Used for main memory.

Use SRAM for small fast memory (cache); DRAM for large main memory.

"Explain why the embedded system uses DRAM": it needs a large amount of memory at low cost in a small space, and its speed requirement is modest, so the cheaper, denser DRAM is the right choice; SRAM is kept for the small cache where speed matters most.

ROM variants you can program after manufacture:

• PROM (Programmable ROM) — written once (fuses burned by a programmer); cannot be changed.
• EPROM (Erasable Programmable ROM) — erased by strong UV light through a window, then rewritten (whole chip at once).
• EEPROM (Electrically Erasable Programmable ROM) — erased and rewritten electrically, a byte at a time, in circuit. Flash memory is a derivative optimised for block erase.

"Give two differences between EPROM and EEPROM" wants two rows of this table, each stated for both types.

Both read sensors; the difference is what they do next.

• monitoring 监控 — collects and reports data but takes no action (a weather station logging readings).
• control system 控制系统 — uses sensor data to decide and act through actuators, usually in a feedback loop (a thermostat turning a boiler on/off).

The three-mark "describe the differences" answer: a monitoring system only measures, records or displays the readings, and at most raises a warning; a control system compares each reading with a preset value 预设值 and, if it is outside the range, sends signals to actuators that change the physical process; the change is then measured again, so a control system contains feedback and a monitoring system does not. Whether a given system is one or the other is decided by that test: a bridge system that measures a vehicle's height and switches on a warning sign is monitoring, because nothing it does changes the vehicle; a system that lowers a barrier is control.

Worked example. Describe how an automated system opens a door when a person is within 2 metres and closes it when nobody is.

An infra-red or ultrasonic sensor measures the distance to anything in front of the door; the analogue reading is converted to digital by an ADC and sent to the processor; the processor compares the distance with the preset 2 metres; if it is less, the processor sends a signal to the actuator (a motor) to open the door; the sensor keeps measuring, and when no reading below 2 metres is received the processor signals the motor to close the door. The repeated measuring after each action is the feedback that stops the door opening and closing at the wrong times.

Sensors and actuators

A sensor 传感器 turns a physical quantity into a signal: temperature (a thermistor 热敏电阻 or thermocouple), pressure (strain gauge), infra-red, sound. Analogue signals need an ADC first. An actuator 执行器 does the reverse — turns a signal into an action (a motor, valve, heater, buzzer).

Feedback

In a control system the actuator changes the environment, which the sensors then re-measure — a feedback 反馈 loop. Without feedback the system cannot correct itself or know when to stop (a thermostat with no temperature feedback would heat forever).

A logic gate 逻辑门 is a small circuit that does one Boolean 布尔 operation. Inputs and outputs are 0 (false, low) or 1 (true, high). Know the symbol, function and truth table 真值表 for each gate.

NOT (inverter)

AND — output 1 only if all inputs are 1

OR — output 1 if at least one input is 1

NAND (NOT AND) — output 0 only when all inputs are 1

NOR (NOT OR) — output 1 only when all inputs are 0

XOR (Exclusive OR, also called EOR) — output 1 if the inputs are different

A logic circuit 逻辑电路 is a network of gates that carries out a Boolean expression. You should be able to move between a problem statement, a logic expression, a truth table, and a circuit diagram.

The paper writes expressions in words, X = (A AND NOT B) OR (B AND C), and accepts the algebraic form $X = A\overline{B} + BC$ where a dot (or nothing) is AND, a plus is OR, and a bar is NOT. Use whichever the question uses.

From expression to circuit

Draw one gate per operator and wire them up. For $X = (A \text{ AND } B) \text{ OR } (\text{NOT } C)$: a NOT gate on $C$, an AND gate on $A$ and $B$, then an OR gate on the two results.

From circuit to expression

Work forwards from the inputs, labelling each gate's output, until you reach the final output.

Worked example. Write the expression for the circuit below, then complete its truth table.

Label the gate outputs: $P = A \text{ AND NOT } B$, $Q = B \text{ AND } C$, so $X = P \text{ OR } Q = (A \text{ AND NOT } B) \text{ OR } (B \text{ AND } C)$. Then give the truth table a column for each intermediate output, so every row can be checked one gate at a time:

Drawing a circuit from an expression is the same walk in reverse: start from the innermost brackets, draw one gate per operator, draw a NOT gate on the wire of any input that appears with NOT, keep the inputs on the left and the single output on the right, and label the output with its letter. Every line must end at a gate input or the output; a line that goes nowhere loses the mark.

From circuit to truth table

For $n$ inputs there are $2^{n}$ rows. List every input combination; for each, work out the internal gates then the output.

From truth table to expression (sum of products)

For each row that outputs 1, write an AND of the inputs (with NOT on any input that is 0 in that row); OR these together. Example: a table that is 1 only on $(A=0,B=1)$ and $(A=1,B=0)$ gives $\overline{A}B + A\overline{B}$, which is $A \text{ XOR } B$.

From a problem statement

Turn the English into a Boolean expression first: "A and B" → A AND B; "A or B or both" → A OR B; "exactly one of A and B" → A XOR B; "neither A nor B" → A NOR B; "not both" → A NAND B.

Worked example. A machine's alarm $X$ sounds when the guard is open ($A=1$) and either the motor is running ($B=1$) or the temperature is high ($C=1$). Write the Boolean expression, and give the rows where $X=1$. Turn the English into logic one clause at a time: "either B or C" is $B + C$, and "A and that" is $X = A\cdot(B + C)$. For the rows, $X=1$ needs $A=1$ and at least one of $B$, $C$ equal to 1 - so $(A,B,C) = (1,0,1)$, $(1,1,0)$ and $(1,1,1)$, three rows out of eight. Notice $A=0$ can never sound the alarm, whatever $B$ and $C$ do. Bracket the OR before ANDing it: $X = A\cdot B + C$ is a different circuit altogether, one that would sound the alarm on a high temperature even with the guard closed.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• Distinguish RAM (volatile, read/write) from ROM (non-volatile, holds the bootstrap); SRAM (cache, faster) from DRAM (main memory, needs refreshing).
• For a logic circuit, build the Boolean expression gate by gate, then a truth table covering every input combination.
• Learn the symbol, expression and truth table for each gate (AND, OR, NOT, NAND, NOR, XOR).
• Explain a buffer (a temporary store bridging two different speeds) and the role of an interrupt.

Common mistakes

• Naming the device instead of describing its operation. "It uses a laser" earns nothing; the steps (charge the drum, laser removes charge, toner attracted, transferred, fused) earn the marks.
• Saying a monitoring system "controls" something. If nothing changes the physical process, it is monitoring; add the actuator and the feedback and it becomes control.
• Writing that RAM "stores files permanently" or that ROM "stores the user's data". RAM is volatile working memory; ROM holds the fixed start-up instructions.
• A truth table with fewer than $2^{n}$ rows, or rows in a random order. Count in binary from 000 to 111 so no combination is missed.
• Drawing two lines from one output of a gate to be safe, or leaving a wire that ends nowhere. Draw exactly the connections the expression needs.

The Von Neumann architecture 冯·诺依曼体系结构 underlies almost every general-purpose computer:

• a single memory — the Immediate Access Store 立即存取存储器 (IAS) — holds both program instructions and data (the stored program 存储程序 concept).
• a processor 处理器 (CPU) fetches instructions from memory and runs them one at a time.
• instructions run in order unless a branch changes the flow.

The stored-program idea is what makes a computer flexible: change the program and you change what it does, with no rewiring.

All of these parts sit inside one small chip. The diagram later in this section shows how they connect; the photo below shows the real thing.

Arithmetic and Logic Unit (ALU)

The ALU 算术逻辑单元 does the arithmetic (add, subtract, …) and logic (AND, OR, comparisons). It takes operands from registers 寄存器 and puts results back in a register.

Control Unit (CU)

The control unit 控制单元 decodes each instruction and sends the control signals to carry it out — opening data paths, telling the ALU what to do, and controlling memory reads and writes.

System clock

The clock sends a steady stream of pulses that keep the CPU in step. Each instruction takes a fixed number of cycles, and the clock speed 时钟频率 (e.g. 3.8 GHz) is one factor in performance.

"Explain how the CU and the system clock work together": the clock emits pulses at a fixed frequency; the control unit uses each pulse to move the fetch-execute cycle on by one step, sending its control signals in time with the pulses, so every part of the processor changes state together. A faster clock means more steps per second, up to the point where the circuits cannot settle between pulses.

Registers

Registers are tiny, very fast stores inside the CPU. The special purpose registers 专用寄存器 each have a fixed job in the cycle:

• Program Counter 程序计数器 (PC) — the address of the next instruction.
• Memory Address Register 内存地址寄存器 (MAR) — the address being read or written.
• Memory Data Register 内存数据寄存器 (MDR) — the data going to or from memory.
• Current Instruction Register 当前指令寄存器 (CIR) — the instruction being decoded.
• Accumulator 累加器 (ACC) — the value the ALU is working on.
• Status Register 状态寄存器 — holds flags 标志 (carry, zero, negative, overflow) used by branches. Each flag is one bit, set or cleared by the ALU after an operation: the zero flag after a comparison that matched, the carry flag when an addition overflowed the register, the negative flag when a result is negative. A conditional jump reads the flags to decide whether to branch, and an overflow flag can raise an interrupt.
• Index Register 变址寄存器 — an offset added to an address in indexed addressing; incrementing it steps through an array one element at a time.

The "complete the table describing the role of each register" question wants one precise sentence per register in these terms: the PC holds the address of the next instruction to be fetched; the MAR holds the address of the location being read from or written to; the MDR holds the data or instruction just read from, or about to be written to, that location; the CIR holds the instruction currently being decoded and executed; the ACC holds the result of the last arithmetic or logic operation.

General-purpose registers 通用寄存器 are used by the programmer for temporary values during a calculation. Movements of data between registers and memory are written in register transfer 寄存器传送 notation — e.g. MAR ← [PC] ("copy the contents of PC into MAR").

Three internal buses 总线 (sets of parallel wires) connect the parts:

• address bus 地址总线 — carries the memory address. One-way (CPU → memory).
• data bus 数据总线 — carries the data. Two-way.
• control bus 控制总线 — carries control signals (read, write, interrupt). Two-way.

An $n$-bit address bus can reach $2^{n}$ memory locations. The data-bus width sets how many bits move per access (often the word size).

• clock speed — more cycles per second.
• number of cores 核心 — a multi-core CPU runs several threads at once.
• word size 字长 — a 64-bit CPU handles 64-bit chunks per cycle and can address far more memory than a 32-bit one.
• amount of RAM 随机存取存储器 — more RAM holds more of the working set; too little forces the OS to page 页 to disk.
• cache memory 高速缓存 size — more cache cuts average memory access time.
• secondary storage 辅助存储器 type — an SSD loads programs far faster than an HDD.
• bus width and speed — wider/faster buses move data more quickly.

Match the specs to the workload: a quad-core beats a dual-core on parallel work, but higher per-core speed wins on single-threaded work.

Each factor is a two-mark answer with a reason attached:

• More cores: each core can fetch and execute its own instruction at the same time, so several programs, or the threads of one program, run in parallel. But a program must be written to use more than one core, so doubling the cores does not double the speed.
• Higher clock speed: more fetch-execute cycles per second, so more instructions per second; the limit is the heat produced.
• Wider bus: a wider data bus moves more bits in each transfer, so fewer transfers are needed for the same data; a wider address bus can address more memory locations.
• Cache memory: a small, fast memory inside or next to the processor that keeps the instructions and data used most recently or most often. Reading them from cache is much faster than from RAM, so the processor spends less time waiting.

"Explain why the new computer performs better" is answered by comparing the two specifications line by line: a higher clock speed executes more instructions per second, more cores run more tasks at once, more cache means fewer slow accesses to RAM, and more RAM means fewer transfers to disk.

A port 端口 is a physical socket for connecting a peripheral 外围设备:

• USB (Universal Serial Bus) — general-purpose (keyboards, drives, phones).
• HDMI (High Definition Multimedia Interface) — digital video and audio to a screen.
• VGA (Video Graphics Array) — older analogue video output to a monitor.
• Ethernet (RJ-45) — wired LAN. Audio jacks — headphones/microphone.

Different ports use different signals, so an HDMI cable will not fit a USB socket. USB-C is unusual in carrying video, data and power.

"Explain how the computer connects to the monitor through HDMI": the HDMI port sends the video and the audio as one digital signal down a single cable, so no conversion to analogue is needed and the picture is not degraded; the cable carries high-definition resolutions and the monitor's own port decodes the signal. A USB device is plug-and-play: when it is connected the computer detects it, identifies it, loads or installs the driver it needs, and can supply it with power, all without a restart.

The CPU repeats the fetch-execute cycle 取指-执行周期, one run per machine instruction.

Fetch

1. the PC's address is copied to the MAR.
2. the PC is incremented to point to the next instruction.
3. a read signal goes over the control bus.
4. memory puts the instruction on the data bus.
5. it is copied into the MDR, then into the CIR.

The exam asks for these steps in register transfer notation 寄存器传送记法, where [X] means the contents of register X and [[MAR]] means the contents of the memory location whose address is in the MAR:

MAR ← [PC]          the address of the next instruction goes to the MAR
PC  ← [PC] + 1      the PC now points to the following instruction
MDR ← [[MAR]]       the instruction at that address is read into the MDR
CIR ← [MDR]         the instruction is copied into the CIR for decoding

The order matters: the PC is incremented straight after its address has been copied, so that a jump executed later can still overwrite it. During execution the same notation describes each instruction; for LDD 200, for example, MAR ← 200, MDR ← [[MAR]], ACC ← [MDR].

Decode

The CU decodes the instruction in the CIR — what operation, and which operands or addresses.

Execute

The CU carries it out: arithmetic/logic goes to the ALU (result to the ACC); a load/store moves data between memory and a register; a branch changes the PC. Then the cycle repeats.

An interrupt 中断 is a signal that pauses the normal cycle so the CPU can handle an urgent event (a key press, a packet arriving, a hardware fault, division by zero, the OS timer).

Handling one:

1. finish the current instruction.
2. save the state (PC and registers).
3. load the address of the interrupt service routine 中断服务程序 (ISR) into the PC and run it.
4. the ISR handles the event.
5. restore the saved state and carry on.

Interrupts let the system respond promptly without the CPU constantly checking devices, and are how the OS multitasks.

"Explain how an interrupt from an input device is detected and handled in the F-E cycle" is a four-mark answer with these points: the device sends an interrupt signal that sets the interrupt flag in the interrupt register 中断寄存器; the processor checks that register at the end of every fetch-execute cycle, after the current instruction has finished executing; if a flag is set and the interrupt has a higher priority than the current task, the contents of the PC and the other registers are saved onto the stack 栈; the address of the interrupt service routine is loaded into the PC and the routine runs; when it finishes, the saved values are restored from the stack and the interrupted program continues from where it stopped.

Causes worth naming: a hardware interrupt from a device (a key pressed, a printer buffer empty, a network packet arriving), a software interrupt from a fault (division by zero, an illegal instruction, arithmetic overflow), a timer interrupt from the operating system marking the end of a time slice, and a power failure warning.

The CPU actually runs machine code 机器码 — bit patterns, specific to one architecture. Assembly language 汇编语言 is a readable form, with one instruction per machine instruction, written using mnemonics 助记符 like LDD, ADD, JMP. An assembler 汇编器 translates it to machine code.

Two-pass assembler

A two-pass assembler reads the source twice:

• pass 1 builds a symbol table 符号表: each time a label 标签 (like LOOP:) appears, record its address; no code yet.
• pass 2 generates code: translate each instruction, and when one refers to a label (like JMP LOOP), look up its address in the symbol table.

Two passes handle forward references 前向引用 (a jump to a label defined later).

Worked example. Apply the two-pass process to this program, whose first instruction is stored at address 100.

        LDD  COUNT
LOOP:   DEC  ACC
        CMP  #0
        JPN  LOOP
        END
COUNT:  5

Pass 1 reads each line, counts the address it will occupy, and records every label in the symbol table: LOOP = 101 (the DEC line) and COUNT = 105 (the data line). No code is produced. Pass 2 reads the program again and translates each line into machine code, replacing each mnemonic by its opcode 操作码 and each symbolic address by the number from the symbol table: LDD COUNT becomes the opcode for LDD with operand 操作数 105, and JPN LOOP becomes the opcode for JPN with operand 101. The jump back to LOOP could have been resolved in one pass, but a jump forward to a label not yet seen could not, which is why the assembler makes two.

Example instruction set

Cambridge uses a small generic set, printed in the paper's reference table, with one general-purpose register, the accumulator (ACC), and an index register (IX). An operand written #n is a denary number, Bn a binary number and &n a hexadecimal number; <address> is a location number or a label.

The "assembly language instructions are grouped" question wants the group names, and an instruction from each: data movement, input and output, arithmetic, unconditional and conditional jumps, compare, and bit manipulation.

The addressing mode 寻址方式 (the modes of addressing) says how the CPU finds the operand:

• immediate addressing 立即寻址 — the operand is the value in the instruction. LDM #10 loads 10.
• direct addressing 直接寻址 — the instruction holds an address; the operand is the value there. LDD 200.
• indirect addressing 间接寻址 — the instruction holds an address that holds another address, which is the data. LDI 200.
• indexed addressing 变址寻址 — effective address is address + index register; used for arrays. LDX 100 with IR = 5 reads address 105.

(Relative addressing 相对寻址 gives the address as an offset from the PC — used for jumps.)

Worked example. Memory holds: location 200 = 250, location 250 = 99, location 105 = 7. The index register holds 5. What is in the accumulator after each of LDM #200, LDD 200, LDI 200 and LDX 100? Follow how far each mode has to look. LDM #200 is immediate - the operand is the number written in the instruction, so the accumulator holds 200. LDD 200 is direct - go to location 200 and take what is there: 250. LDI 200 is indirect - location 200 holds 250, which is another address, so go on to location 250: 99. LDX 100 is indexed - add the index register to the address, $100 + 5 = 105$, and read location 105: 7. Count the hops to keep them apart: immediate 0, direct 1, indirect 2, indexed 1 (once the index has been added).

To trace it: make a table with columns for the PC, ACC, index register, each variable and any flags. Step through the instructions, updating the table after each; follow branches when they change the PC; stop at END. A common pattern is a loop over an array using indexed addressing.

Worked example. Trace this program. Address 200 holds 5 and address 201 holds 0.

100   LDD  200
101   CMP  #0
102   JPE  108
103   OUT
104   DEC  ACC
105   STO  200
106   LDD  201
107   JMP  100
108   END

Write one row for each instruction executed, filling in only the columns that change:

and so on, until LDD 200 loads 0, the compare sets the equal flag, JPE 108 is taken and the program ends. Three things the examiner checks: a CMP changes no register, only a flag; a jump not taken still counts as executed; and OUT outputs a character, so it goes in the output column, not the ACC column. "State the effect of changing LDD 10 to LDM #10": the ACC would hold the number 10 instead of the contents of address 10.

A logical shift 逻辑移位 moves all the bits left or right by some places, filling new positions with 0.

• left shift by 1 (LSL #1) — bits move left, a 0 enters on the right; for an unsigned number this is × 2.
• right shift by 1 (LSR #1) — bits move right, a 0 enters on the left; for an unsigned number this is integer ÷ 2.

Shifting by $n$ places multiplies or divides by $2^{n}$. Example: 00001011 (11) LSL #1 → 00010110 (22).

Bits shifted off the end are lost, so the multiplication is only correct while they were zeros. LSL #2 on the two's-complement integer 11001010 gives 00101000: the two 1s that fell off the left are gone, the sign bit has changed, and the result is no longer four times the original.

An arithmetic right shift keeps the sign bit so a negative signed number stays negative. A cyclic shift 循环移位 (rotate) feeds the bit that drops off one end back in at the other end, so no bits are lost.

"Show the result of an arithmetic right shift of 3 places on 10011110": copy the sign bit into each vacated place, 11110011. The same shift on 01011100 gives 00001011. A cyclic left shift of 1 on 10000110 gives 00001101: the leading 1 reappears on the right.

The difference between the two right shifts is a single bit. Take 11110000, which is 240 read as unsigned and $-16$ read as signed. LSR #1 brings in a 0 and gives 01111000 $= 120$, which is the correct half of 240. ASR #1 copies the sign bit instead and gives 11111000 $= -8$, which is the correct half of $-16$. Neither is wrong — each halves the value under one reading.

Bit manipulation for monitoring/control

Embedded devices often use one bit 位 of a register per signal (e.g. bit $n$ = LED $n$). Using a mask 掩码 — bit masking — you can:

• set bit $n$: R = R OR a mask with bit $n$ set.
• clear bit $n$: R = R AND a mask with bit $n$ clear and the rest set.
• toggle bit $n$: R = R XOR a mask with bit $n$ set.
• test bit $n$: R AND the mask, then check if the result is non-zero.

Bit manipulation is fast, uses little memory, and lets one byte hold up to 8 on/off states.

In the exam's instruction set these are AND, OR and XOR with a mask written as a denary, binary or hexadecimal operand. With the ACC holding 10101100:

"Write the instruction that sets the least significant bit to 1 and leaves the others unchanged": OR #1, or OR B00000001. To clear a bit use AND with a mask that has a 0 in that place and 1s elsewhere; to test a bit, AND with a mask that has a 1 only in that place, then compare the result with zero. In a monitoring device, one bit of a register per sensor lets a single AND check whether a particular sensor is on, and one OR switches an actuator's control bit on without disturbing the others.

A definition question is marked against fixed wording. Learn these exactly, and give one answer only.

• Learn the fetch-execute cycle in register-transfer terms (PC, MAR, MDR, CIR, ACC) and what increments the PC.
• Name each register's job; the address bus is one-way, the data bus is two-way.
• Distinguish the addressing modes (immediate, direct, indirect, indexed) — a frequent question.
• Explain how clock speed, number of cores, cache size and word length affect performance.
• For a binary shift, state whether it is logical or arithmetic; a left shift multiplies by 2, a right shift divides by 2.

Common mistakes

• Saying the PC holds the current instruction, or the MDR holds an address. The PC holds the address of the next instruction; the MDR holds data or an instruction, never an address.
• Leaving the increment of the PC out of the fetch, or putting it after the execute. It happens as soon as the address has been copied to the MAR.
• Reading LDD 10 as "load 10". LDD 10 loads the contents of address 10; LDM #10 loads the number 10.
• Putting a value in the ACC column for CMP or OUT. A compare sets a flag only; an output goes to the output column.
• Saying an interrupt is handled "immediately". The processor finishes the current instruction and checks for interrupts at the end of the cycle.
• Using a logical right shift on a negative two's-complement number. Only an arithmetic shift keeps the sign bit.