AQA GCSE 数据表示 二进制 十六进制 考点
数据表示 Data Representation 是 AQA GCSE Computer Science 的必考基础模块，覆盖存储单位、进制转换、字符编码、图像与声音的数字存储方式。本文用中英双语把核心概念和典型计算题一次性讲透，适合考前快速复习。
Data Representation is a compulsory foundation module in AQA GCSE Computer Science. It covers units of storage, number base conversion, character encoding, and the digital storage of images and sound. This article explains every core idea with worked calculations in both Chinese and English, ideal for last-minute revision.
1. 存储单位 Units of Storage
计算机中最小的数据单位是比特 bit，一个 bit 只能表示 0 或 1。4 个 bit 组成一个半字节 nibble，8 个 bit 组成一个字节 byte，字节是日常计量的基本单位。更大的单位按 1024 进位：1 KB = 1024 B，1 MB = 1024 KB，1 GB = 1024 MB，1 TB = 1024 GB。这里的 1024 等于 2 的 10 次方，因为计算机采用二进制。
The smallest unit of data is the bit, which can only store 0 or 1. Four bits make one nibble and eight bits make one byte, the unit used for everyday measurement. Larger units scale by 1024: 1 KB = 1024 B, 1 MB = 1024 KB, 1 GB = 1024 MB, 1 TB = 1024 GB. The number 1024 equals 2 to the power of 10, because computers work in binary.
典型计算：一块 U 盘容量 2 GB，问等于多少字节？2 × 1024 × 1024 × 1024 = 2147483648 字节。注意考试中单位一律按 1024 换算，不要写成 1000，这是最常见的失分点。
Worked example: how many bytes are in a 2 GB flash drive? 2 × 1024 × 1024 × 1024 = 2147483648 bytes. In exams always multiply by 1024, never 1000; this is the most common mark-losing mistake.
另一类常见题型是按大小排序：把 1 MB、500 KB、1 GB、2048 B 从小到大排列，答案是 2048 B < 500 KB < 1 MB < 1 GB。做这种题先把所有单位换算成字节再比较，就不会出错。
Another frequent question type is ordering by size: arrange 1 MB, 500 KB, 1 GB and 2048 B from smallest to largest. The answer is 2048 B < 500 KB < 1 MB < 1 GB. Convert everything to bytes first and the comparison becomes trivial.
2. 二进制与十进制互转 Binary and Denary Conversion
十进制 denary 是我们日常使用的逢十进一系统，二进制 binary 是逢二进一，只含 0 和 1。把二进制转十进制，只需要把每一位上的 1 乘上它对应的位权再相加。8 位二进制的位权从右到左依次是 1、2、4、8、16、32、64、128。
Denary is the base-10 system we use every day, while binary is base-2 and contains only the digits 0 and 1. To convert binary to denary, multiply each 1 by its place value and add the results. For an 8-bit number the place values from right to left are 1, 2, 4, 8, 16, 32, 64 and 128.
例题：二进制 00001101 转十进制。从右往左看，8 和 1 两个位上是 1，所以 8 + 4 + 1 = 13。反向操作，把十进制 178 转二进制：178 = 128 + 32 + 16 + 2，对应的二进制是 10110010。推荐用减位权法：先找不超过该数的最大的位权，减掉后继续，直到结果为 0。
Example: convert binary 00001101 to denary. The 1s sit at place values 8, 4 and 1, so 8 + 4 + 1 = 13. To go backwards, convert denary 178: 178 = 128 + 32 + 16 + 2, giving binary 10110010. The reliable method is repeated subtraction: take the largest place value not exceeding the number, subtract it, and repeat until zero.
再练一组找手感：把二进制 11110000 转十进制，结果是 128 + 64 + 32 + 16 = 240；把十进制 37 转二进制，37 = 32 + 4 + 1，得到 00100101。换算完成后用直觉检查一遍：8 位二进制最大是 255，任何超过 255 的结果都说明算错了。
Practise another pair: binary 11110000 becomes 128 + 64 + 32 + 16 = 240 in denary; denary 37 = 32 + 4 + 1, giving binary 00100101. Always sanity-check your answer: the maximum 8-bit value is 255, so any result above 255 means you made an error.
3. 十六进制 Hexadecimal
十六进制是逢十六进一，用 0 到 9 再加 A 到 F 表示 0 到 15。十六进制和二进制换算非常方便：1 个十六进制位恰好等于 4 个 bit，因此 2 个十六进制位正好表示 1 个字节。
Hexadecimal is base-16 and uses the digits 0 to 9 followed by the letters A to F to represent the values 0 to 15. It maps neatly onto binary: one hexadecimal digit equals exactly four bits, so two hex digits represent exactly one byte.
例题：把十六进制 1E 转十进制。E = 14，所以 1E = 1 × 16 + 14 = 30。转二进制时把每个十六进制位拆成 4 位即可：1E = 0001 1110。十六进制常用于显示内存地址、MAC 地址以及网页颜色代码，例如红色可以写成 #FF0000，比写 24 位二进制简短得多。
Example: convert hex 1E to denary. Since E = 14, 1E = 1 × 16 + 14 = 30. To convert to binary, expand each hex digit into four bits: 1E = 0001 1110. Hexadecimal is commonly used for memory addresses, MAC addresses and web colours; for example red is written #FF0000, far shorter than the equivalent 24-bit binary string.
记忆技巧：十进制 10 到 15 对应 A 到 F，白色是 #FFFFFF，黑色是 #000000。看到形如 # 开头加 6 个十六进制位的写法，就是网页中的 RGB 颜色代码，前两位控制红色，中间两位控制绿色，最后两位控制蓝色。
Memory aid: the denary values 10 to 15 map to the letters A to F; white is #FFFFFF and black is #000000. When you see a hash followed by six hexadecimal digits, that is an RGB colour code in a web page: the first pair controls red, the middle pair green and the last pair blue.
4. 二进制加法与移位 Binary Addition and Shifts
二进制加法规则：0 + 0 = 0，1 + 0 = 1，1 + 1 = 0 并向左进 1，1 + 1 + 1 = 1 并进 1。例题：0110 + 0101。最低位 0 + 1 = 1；第二位 1 + 0 = 1；第三位 1 + 1 = 0 进 1；第四位 0 + 0 再加进位 1 得 1，结果是 1011，也就是 6 + 5 = 11。
Binary addition rules: 0 + 0 = 0, 1 + 0 = 1, 1 + 1 = 0 with a carry of 1, and 1 + 1 + 1 = 1 with a carry of 1. Example: 0110 + 0101. The units column gives 0 + 1 = 1; the twos column gives 1 + 0 = 1; the fours column gives 1 + 1 = 0 carry 1; the eights column is 0 + 0 plus the carry, giving 1. The answer is 1011, which matches 6 + 5 = 11.
溢出 overflow 发生在结果超出可用位数时，例如 8 位最大数是 255，计算 200 + 100 需要 9 位，最高位的 1 无处存放就会溢出。二进制左移一位相当于乘以 2，右移一位相当于除以 2。例题：0011 左移一位变成 0110，从 3 变成 6。如果左移后最高位被挤掉，同样会发生溢出。
Overflow happens when a result needs more bits than are available. An 8-bit number has a maximum of 255, so 200 + 100 needs a ninth bit and the leading 1 is lost. Shifting binary left by one place multiplies by 2, and shifting right by one place divides by 2. Example: 0011 shifted left becomes 0110, turning 3 into 6. If a shift pushes the leading bit off the end, overflow also occurs.
5. 字符编码 Character Sets
字符集 character set 是计算机给每个字符分配的唯一二进制代码。ASCII 是最早的通用字符集，用 7 位表示 128 个字符，包括英文字母、数字和常用符号。扩展 ASCII 用满 8 位，最多 256 个字符。
A character set gives every character its own unique binary code. ASCII, the first widely used set, uses 7 bits to represent 128 characters including letters, digits and common symbols. Extended ASCII uses all 8 bits and supports up to 256 characters.
记一个具体例子帮助理解：大写字母 A 的 ASCII 码是十进制 65，对应二进制 01000001。因为编码是连续的，B 就是 66，小写 a 是 97。考试若给出一部分 ASCII 码表，剩下的可以按顺序推出来。
One concrete example helps: the ASCII code for capital A is denary 65, which is binary 01000001. Codes are consecutive, so B is 66 and lowercase a is 97. If an exam gives part of an ASCII table, the remaining codes can be derived in order.
ASCII 的局限是字符数量太少，放不下中文、日文、阿拉伯文等文字。Unicode 用更多位给全球几乎所有语言的字符以及 emoji 分配唯一编码，因此现代计算机普遍使用 Unicode。考试常问：为什么需要 Unicode？标准答案是 ASCII 只能表示 128 或 256 个字符，无法覆盖全球文字。
The limitation of ASCII is that it cannot hold Chinese, Japanese, Arabic or other scripts. Unicode uses more bits to give a unique code to characters from almost every language in the world, plus emoji, which is why modern computers use it. A common exam question asks why Unicode is needed; the standard answer is that ASCII can represent only 128 or 256 characters and cannot cover the world's writing systems.
6. 图像存储 Images
位图图像由像素 pixel 组成，每个像素是图像中的一个点。图像分辨率 resolution 用宽乘高表示，例如 800 × 600。颜色深度 colour depth 表示每个像素用多少 bit 存储颜色，颜色深度越大，可显示颜色越多，画质越好，文件也越大。
A bitmap image is made of pixels, and each pixel is a single dot of colour. Resolution is written as width times height, for example 800 × 600. Colour depth is the number of bits used to store each pixel; a higher depth means more possible colours, better quality, and a larger file.
文件大小计算公式：图像大小（字节）= 宽 × 高 × 颜色深度 ÷ 8。例题：一张 100 × 80 的图像，颜色深度 4 bit，则文件大小 = 100 × 80 × 4 = 32000 bit = 4000 字节。把颜色深度提高到 8 bit，文件大小会翻倍到 8000 字节。
File size formula: size in bytes = width × height × colour depth ÷ 8. Example: a 100 × 80 image with 4-bit colour has a file size of 100 × 80 × 4 = 32000 bits = 4000 bytes. Raising the colour depth to 8 bits doubles the file to 8000 bytes.
对照数据：4 位颜色深度只能显示 16 种颜色，8 位可显示 256 种，24 位真彩色可显示约 1670 万种颜色。分辨率不变时，颜色深度每增加一位，文件大小就按比例增加，所以高画质照片的文件通常很大。
Comparison data: 4-bit colour shows only 16 colours, 8-bit shows 256, and 24-bit true colour shows about 16.7 million colours. With a fixed resolution, every extra bit of depth enlarges the file proportionally, which is why high-quality photos take so much space.
7. 声音存储 Sound
声音是连续的模拟信号，计算机要先把模拟信号转成数字信号，这个过程叫采样 sampling。采样率 sample rate 是每秒采样的次数，单位是赫兹 Hz；位深度 bit depth 是每次采样用多少 bit 记录音量。
Sound is a continuous analogue signal, so a computer must convert it to digital form by sampling. The sample rate is the number of samples taken per second, measured in hertz (Hz), and the bit depth is the number of bits used to record each sample.
声音文件大小计算公式：文件大小 = 采样率 × 位深度 × 时长 × 声道数。例题：一段 2 秒的单声道录音，采样率 8000 Hz，位深度 16 bit，文件大小 = 8000 × 16 × 2 = 256000 bit = 32000 字节，约 31 KB。采样率和位深度越高，声音越接近原声，文件也越大；降低它们能压缩文件但音质变差。
Sound file size formula: size = sample rate × bit depth × duration × number of channels. Example: a 2-second mono recording at 8000 Hz with 16-bit depth gives 8000 × 16 × 2 = 256000 bits = 32000 bytes, about 31 KB. Higher sample rates and bit depths give sound closer to the original but create larger files; lowering them shrinks the file at the cost of quality.
实际对照：CD 音质标准是采样率 44100 Hz、位深度 16 bit，而语音通话常用 8000 Hz。采样率按用途选择，说明采样率不是越高越好，而要平衡文件大小与音质需求。
Real-world comparison: CD-quality audio uses a 44100 Hz sample rate with 16-bit depth, while telephone calls typically use 8000 Hz. The rate is chosen for the purpose, showing that higher is not always better; you must balance file size against the quality actually needed.
8. 数据压缩 Compression
压缩用来减小文件体积，节省存储空间并加快传输。压缩分两类：有损压缩 lossy 会永久删除部分数据，文件明显变小，但无法还原，适合图片、音乐和视频，例如 JPEG 和 MP3；无损压缩 lossless 不丢任何数据，可以完全还原，但压缩比例有限，例如 PNG 和 ZIP。
Compression reduces file size to save storage and speed up transfer. There are two types: lossy compression permanently removes some data, giving much smaller files that cannot be restored, and suits images, music and video such as JPEG and MP3; lossless compression keeps every piece of data and allows perfect restoration, but achieves less reduction, as in PNG and ZIP files.
选型判断：证件照和合同扫描件必须用无损，因为任何细节丢失都不可接受；在线视频流媒体用有损，因为观看体验优先、原文件在服务器另有备份。
How to choose: scanned contracts and ID photos must use lossless compression because losing detail is unacceptable; online video streaming uses lossy because playback experience matters more, and the original file remains safe on the server.
9. 考点提醒 Exam Tips
最后总结 AQA 考试中数据表示部分最容易丢分的几个地方：第一，单位换算一律按 1024，KB、MB、GB 之间相差 2 的 10 次方；第二，进制转换必须写出计算过程，直接写答案容易丢方法分；第三，区分 bit 和 byte，文件大小公式最后要除以 8 得到字节；第四，颜色深度和位深度单位都是 bit；第五，回答 Unicode 与 ASCII 的区别时，要点是字符数量和覆盖语言范围，而不是速度或占用。
Final checklist of the most common lost marks in this topic: first, always convert units by 1024; second, show your working in base conversions because answers alone rarely earn full marks; third, distinguish bits from bytes and divide by 8 at the end of file size calculations; fourth, remember colour depth and bit depth are measured in bits; fifth, when comparing ASCII and Unicode, focus on character count and language coverage, not speed or file size.
最后留一道自测题：一张 200 × 150 的图像，颜色深度 8 bit，文件大小是多少？按公式 200 × 150 × 8 = 240000 bit，除以 8 得 30000 字节。先自己算一遍再看答案，比直接读效果更好。
A final self-test: what is the file size of a 200 × 150 image with 8-bit colour? Using the formula, 200 × 150 × 8 = 240000 bits, which divided by 8 is 30000 bytes. Attempt it yourself before reading the answer; active recall beats passive reading.
把这九个考点逐个练熟，Data Representation 的选择题和计算题基本可以拿满。建议把例题重新算一遍，并计时做一套真题检验速度。
Master these nine points one by one and you can expect full marks on the Data Representation questions. Recalculate every worked example yourself, then time yourself on a past paper to check your speed.
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