📚 Year 10 Edexcel Computer Science: Formula & Theorem Quick Reference Guide | Year 10 Edexcel 计算机科学：公式定理速查手册
This guide compiles the essential formulas and theorems you need to master for your Year 10 Edexcel Computer Science exams. From data unit conversions and image file sizes to logic gate identities and algorithm efficiencies, these quick references will help you solve calculation questions accurately and efficiently.
本指南汇集了 Year 10 Edexcel 计算机科学考试中必须掌握的基本公式和定理。从数据单位换算、图像文件大小到逻辑门恒等式和算法效率，这些速查参考将帮助你准确、高效地解决计算题。
1. Data Units and Conversions | 数据单位与换算
All digital data is stored in bits. One byte is equal to eight bits. Larger units are derived using decimal prefixes (1000) or binary prefixes (1024).
所有数字数据以比特（bit）存储。1 字节（byte）等于 8 比特。更大的单位使用十进制前缀（1000）或二进制前缀（1024）派生。
1 byte = 8 bits
1 KB = 1000 bytes (1 KiB = 1024 bytes)
1 MB = 1000 KB (1 MiB = 1024 KiB)
1 GB = 1000 MB (1 GiB = 1024 MiB)
1 TB = 1000 GB (1 TiB = 1024 GiB)
To convert from a smaller unit to a larger unit, divide; to convert from a larger unit to a smaller unit, multiply.
从小单位转换为大单位用除法；从大单位转换为小单位用乘法。
Example: Convert 5000 KB to MB. 5000 ÷ 1000 = 5 MB.
示例：将 5000 KB 转换为 MB：5000 ÷ 1000 = 5 MB。
2. Image File Size | 图像文件大小
The size of an uncompressed raster image depends on its resolution (width × height in pixels) and the colour depth (bits per pixel).
未压缩位图文件的大小取决于其分辨率（以像素为单位的宽度 × 高度）和颜色深度（每像素位数）。
Image file size (bits) = width (pixels) × height (pixels) × colour depth (bits per pixel)
Including a fourth channel for transparency (alpha) adds extra bits per pixel. Typical colour depths: 1 bit for monochrome, 8 bits for 256 colours, 24 bits for true colour.
若包含透明度的第四通道（alpha 通道），每像素会额外增加位数。典型的颜色深度：单色 1 位、256 色 8 位、真彩色 24 位。
Example: An image of 640 × 480 pixels with a 24-bit colour depth: 640 × 480 × 24 = 7,372,800 bits = 921,600 bytes ≈ 921.6 KB.
示例：640 × 480 像素、24 位颜色深度的图像：640 × 480 × 24 = 7,372,800 bits = 921,600 bytes ≈ 921.6 KB。
3. Sound File Size | 声音文件大小
Uncompressed sound file size is determined by sample rate, bit depth, duration, and the number of channels.
未压缩音频文件大小由采样率、位深度、时长和声道数决定。
Sound file size (bits) = sample rate (Hz) × bit depth × duration (seconds) × number of channels
For mono sound, the number of channels is 1; for stereo, it is typically 2. Common sample rates are 44.1 kHz (CD quality) and 48 kHz.
单声道声道数为 1，立体声通常为 2。常见的采样率有 44.1 kHz（CD 质量）和 48 kHz。
Example: 60 seconds of stereo audio at 44.1 kHz with 16-bit depth: 44100 × 16 × 60 × 2 = 84,672,000 bits = 10,584,000 bytes ≈ 10.1 MB.
示例：60 秒 44.1 kHz、16 位深度的立体声音频：44100 × 16 × 60 × 2 = 84,672,000 bits = 10,584,000 bytes ≈ 10.1 MB。
4. Text File Size | 文本文件大小
Plain text file size is calculated by multiplying the number of characters by the bits used per character, plus any control characters or metadata.
纯文本文件大小通过字符数乘以每字符所用位数计算，再加上控制字符或元数据。
Text file size (bits) ≈ number of characters × bits per character
Standard ASCII uses 7 bits per character, but extended ASCII and many simple text files use 8 bits. Unicode encodings (UTF-8) can use 8 to 32 bits per character, but for basic Latin characters UTF-8 uses 8 bits.
标准 ASCII 每字符 7 位，但扩展 ASCII 和许多简单文本文件使用 8 位。Unicode 编码（UTF-8）每字符可使用 8 至 32 位，但对于基本拉丁字符 UTF-8 使用 8 位。
Example: A file with 2000 characters stored in ASCII (7-bit) requires 2000 × 7 = 14,000 bits = 1750 bytes. If stored as 8-bit extended ASCII, 2000 × 8 = 16,000 bits = 2000 bytes.
示例：一个包含 2000 个字符、以 ASCII（7 位）存储的文件需要 2000 × 7 = 14,000 bits = 1750 bytes。若以 8 位扩展 ASCII 存储，则需 2000 × 8 = 16,000 bits = 2000 bytes。
5. Binary & Hexadecimal Conversions | 二进制与十六进制转换
Binary (base‑2) and hexadecimal (base‑16) are essential number systems in computing. Conversion methods rely on weighted place values.
二进制（基数为 2）和十六进制（基数为 16）是计算中基本的数字系统，转换方法依赖于加权位值。
Binary to decimal: Each digit represents a power of 2. Example: 1011₂ = (1×2³) + (0×2²) + (1×2¹) + (1×2⁰) = 8 + 0 + 2 + 1 = 11.
二进制转十进制：每位数字代表 2 的幂。示例：1011₂ = (1×2³) + (0×2²) + (1×2¹) + (1×2⁰) = 8 + 0 + 2 + 1 = 11。
Hexadecimal to decimal: Each digit is multiplied by a power of 16. Letters A–F correspond to 10–15. Example: (A2)₁₆ = (10×16¹) + (2×16⁰) = 160 + 2 = 162.
十六进制转十进制：每位数字乘以 16 的幂。字母 A–F 对应 10–15。示例：(A2)₁₆ = (10×16¹) + (2×16⁰) = 160 + 2 = 162。
Decimal to binary: Repeatedly divide the number by 2 and record the remainder. Read the remainders backwards (from last to first).
十进制转二进制：将数字反复除以 2 并记录余数，从下往上读取余数。
Binary and hexadecimal are easily interconverted by grouping binary digits into groups of four (nibbles) and replacing each group with the corresponding hex digit.
二进制与十六进制可快速互转：将二进制数字每四位一组（半字节），用对应的十六进制数字替换。
Binary arithmetic follows simple rules: 0+0=0, 0+1=1, 1+0=1, 1+1=0 carry 1, 1+1+1=1 carry 1. When an addition exceeds the maximum value that can be stored, overflow occurs.
二进制算术遵循简单规则：0+0=0, 0+1=1, 1+0=1, 1+1=0 进位 1, 1+1+1=1 进位 1。当加法结果超出可存储的最大值时会发生溢出。
Left shift by n places multiplies a binary number by 2ⁿ; right shift by n places divides by 2ⁿ (integer division, truncating any remainder).
左移 n 位相当于乘以 2ⁿ；右移 n 位相当于除以 2ⁿ（整数除法，丢弃余数）。
6. Logic Gates & Boolean Algebra | 逻辑门与布尔代数
Logic gates implement Boolean functions. The basic gates and their algebraic expressions are summarised below.
逻辑门实现布尔函数。基本门及其代数表达式总结如下。
AND gate: Output Q = A AND B. Boolean expression: A · B or A ∧ B. True only when both inputs are true.
与门：输出 Q = A AND B。布尔表达式：A · B 或 A ∧ B。仅当所有输入为真时输出才为真。
OR gate: Output Q = A OR B. Boolean expression: A + B or A ∨ B. True when at least one input is true.
或门：输出 Q = A OR B。布尔表达式：A + B 或 A ∨ B。至少一个输入为真时输出为真。
NOT gate: Output Q = NOT A. Boolean expression: ¬A or A’. Inverts the input.
非门：输出 Q = NOT A。布尔表达式：¬A 或 A’。将输入取反。
NAND gate: Output Q = NOT (A AND B). Boolean expression: ¬(A · B). Inverse of AND.
与非门：输出 Q = NOT (A AND B)。布尔表达式：¬(A · B)。与门的取反。
NOR gate: Output Q = NOT (A OR B). Boolean expression: ¬(A + B). Inverse of OR.
或非门：输出 Q = NOT (A OR B)。布尔表达式：¬(A + B)。或门的取反。
XOR gate: Output Q = A XOR B. Boolean expression: A ⊕ B or (A · ¬B) + (¬A · B). True when inputs differ.
异或门：输出 Q = A XOR B。布尔表达式：A ⊕ B 或 (A · ¬B) + (¬A · B)。输入不同时为真。
Key Boolean identities (laws) used for simplification include: A · A = A, A + A = A, A + 0 = A, A · 1 = A, A · 0 = 0, A + 1 = 1, A + ¬A = 1, A · ¬A = 0, and De Morgan’s Laws: ¬(A · B) = ¬A + ¬B, ¬(A + B) = ¬A · ¬B.
用于化简的关键布尔恒等式包括：A · A = A, A + A = A, A + 0 = A, A · 1 = A, A · 0 = 0, A + 1 = 1, A + ¬A = 1, A · ¬A = 0，以及德摩根定律：¬(A · B) = ¬A + ¬B, ¬(A + B) = ¬A · ¬B。
7. Network Transmission Time | 网络传输时间
The time needed to send a file over a network is calculated by dividing the file size by the available bandwidth. Both must be in the same units (bits or bytes).
通过网络发送文件所需的时间通过文件大小除以可用带宽来计算。两者必须使用相同单位（同为比特或同为字节）。
Transmission time (seconds) = File size (bits) / Bandwidth (bits per second, bps)
If file size is given in bytes, multiply by 8 to obtain bits. Network speeds are often quoted in kilobits per second (kbps), megabits per second (Mbps), or gigabits per second (Gbps). 1 Mbps = 1,000,000 bps.
如果文件大小以字节给出，需乘以 8 得到比特。网络速度通常以千比特每秒（kbps）、兆比特每秒（Mbps）或吉比特每秒（Gbps）表示。1 Mbps = 1,000,000 bps。
Example: A 100 Megabit file transmitted over a 50 Mbps connection takes 100 / 50 = 2 seconds. A 20 Megabyte file over a 10 Mbps connection: (20 × 8) / 10 = 160 / 10 = 16 seconds.
示例：在 50 Mbps 连接上传输 100 Mb 文件需要 100 / 50 = 2 秒。在 10 Mbps 连接上传输 20 MB 文件：(20 × 8) / 10 = 160 / 10 = 16 秒。
This formula ignores additional overhead (protocol headers, retransmissions, latency), which Edexcel questions may ask you to consider separately.
该公式忽略了额外开销（协议头、重传、延迟），Edexcel 试题可能会要求单独考虑这些因素。
8. Storage Device Capacity & Transfer Speed | 存储设备容量与传输速度
Storage capacity is frequently expressed in decimal prefix units (e.g., 1 GB = 10⁹ bytes) by manufacturers
Published by TutorHao | Year 10 Computer Science Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply