📚 Year 10 AQA Computer Science: Quick Reference to Formulas and Theorems | Year 10 AQA 计算机：公式定理速查手册
This handbook collects the essential formulas and theorems you will meet in the Year 10 AQA GCSE Computer Science course. From calculating file sizes and data transfer speeds to applying the laws of Boolean logic, these are the building blocks for both Paper 1 and Paper 2. Keep it handy for rapid revision.
本手册汇集了 Year 10 AQA GCSE 计算机科学课程中必须掌握的核心公式与定理。从文件大小和数据传输速度的计算，到布尔逻辑定律的应用，这些内容构成了试卷一和试卷二的基础。随身携带本手册，便于快速复习。
1. Data Storage Units & File Size Calculations | 数据存储单位与文件大小计算
A bit (b) is the smallest unit of data. A byte (B) is 8 bits. The metric prefixes kilo, mega, giga, tera are used in computing, but careful: file sizes often use binary (1024) scaling, while storage manufacturers may use decimal (1000). For AQA, the expected conversion is typically 1 kilobyte = 1000 bytes unless stated otherwise. Use the constant conversion factors when calculating size.
位（bit）是数据的最小单位。字节（Byte）等于 8 位。计算机领域使用千、兆、吉、太等词头，但需注意：文件大小常采用二进制换算（1024 倍数），而存储厂商可能使用十进制（1000）。AQA 考试通常遵循 1 千字节 = 1000 字节，除非题目另有说明。计算时请使用固定的换算系数。
- 1 bit = 1 b
- 1 byte = 8 bits
- 1 kB (kilobyte) ≈ 1000 B
- 1 MB (megabyte) ≈ 1000 kB
- 1 GB (gigabyte) ≈ 1000 MB
- 1 TB (terabyte) ≈ 1000 GB
File size (in bits) = Total number of elements × bits per element
文件大小（位） = 元素总数 × 每个元素的位数
2. Image File Size Formula | 图像文件大小公式
A bitmap image is a grid of pixels. The colour depth (bit depth) determines how many bits are used to encode the colour of each pixel. The file size of an uncompressed image is simply the total number of pixels multiplied by the colour depth, then converted to bytes if needed.
位图图像是由像素组成的网格。颜色深度（位深度）决定了用多少位来编码每个像素的颜色。未压缩图像文件的大小就是总像素数乘以颜色深度，如果需要，再转换为字节单位。
Image file size (bits) = Width (px) × Height (px) × Colour depth (bpp)
图像文件大小（位）= 宽度（像素）× 高度（像素）× 颜色深度（位/像素）
To obtain the result in bytes, divide by 8. For example, a 1920×1080 image with 24-bit colour depth requires 1920 × 1080 × 24 = 49,766,400 bits ≈ 6.22 MB (using 1 MB = 1,000,000 bytes).
要得到以字节为单位的结果，需除以 8。例如，一幅 1920×1080 的 24 位颜色深度图像需要 1920 × 1080 × 24 = 49,766,400 位 ≈ 6.22 MB（按 1 MB = 1,000,000 字节计算）。
3. Sound File Size Formula | 声音文件大小公式
A sound is represented digitally by taking samples of the analogue signal at regular intervals. The sample rate, sample resolution (bit depth), and number of channels all affect file size. The formula multiplies these three factors by the duration.
声音通过以固定间隔对模拟信号采样来数字化。采样率、采样分辨率（位深度）和声道数都会影响文件大小。公式是将这三个因素与时长相乘。
Sound file size (bits) = Sample rate (Hz) × Sample resolution (bits) × Channels × Duration (s)
声音文件大小（位）= 采样率（赫兹）× 采样分辨率（位）× 声道数 × 时长（秒）
For CD-quality stereo sound, a common set of values is 44,100 Hz, 16 bits, 2 channels. A 10-second recording would need 44,100 × 16 × 2 × 10 = 14,112,000 bits ≈ 1.76 MB. Remember that compression can significantly reduce this size.
对于 CD 质量的立体声，常用的参数是 44,100 Hz、16 位、2 声道。一段 10 秒的录音需要 44,100 × 16 × 2 × 10 = 14,112,000 位 ≈ 1.76 MB。请记住，压缩可以大幅减小文件大小。
4. Data Transfer Time Formula | 网络传输时间公式
When sending a file across a network, the transfer time depends on the file size and the transfer speed (bit rate). This formula is fundamental for understanding network performance and for answering practical questions.
通过网络传输文件时，传输时间取决于文件大小和传输速度（比特率）。该公式是理解网络性能以及解答实际题目的基础。
Transfer time (s) = File size (bits) ÷ Transfer speed (bps)
传输时间（秒） = 文件大小（位） ÷ 传输速度（位/秒）
Make sure both quantities are in the same unit before dividing. If a 100 MB file is downloaded at 10 Mbps, first convert everything to bits: 100 MB = 800 Mb, then time = 800 ÷ 10 = 80 seconds. Always state your final answer with appropriate units.
计算前请确保两个量使用相同单位。如果要下载一个 100 MB 的文件，网速是 10 Mbps，首先将所有量转换为位：100 MB = 800 Mb，然后时间 = 800 ÷ 10 = 80 秒。最终答案务必注明合适的单位。
5. Binary and Hexadecimal Conversions | 二进制与十六进制转换
The exam expects you to fluently convert between denary (base 10), binary (base 2), and hexadecimal (base 16). The key relationships are: one hex digit represents exactly 4 bits (a nibble). Two hex digits represent a full byte. Use these groupings for rapid conversions.
考试要求熟练进行十进制、二进制和十六进制之间的转换。关键关系是：一个十六进制数字正好代表 4 个二进制位（一个半字节）。两个十六进制数字代表一个完整的字节。利用这些分组可以快速转换。
| Denary | Binary (4-bit) | Hexadecimal |
|---|---|---|
| 0 | 0000 | 0 |
| 10 | 1010 | A |
| 15 | 1111 | F |
To convert binary to hex, split the binary number into groups of 4 starting from the right, then replace each group with its hex equivalent. For denary to hex, you can use repeated division by 16 and read remainders in reverse, or convert to binary first. Practice ensures speed.
将二进制转换为十六进制时，从右开始每 4 位分为一组，然后将每组替换为对应的十六进制。十进制转十六进制可以使用反复除以 16 并倒序取余的方法，也可以先转换为二进制。多加练习才能快速完成。
6. Basic Laws of Boolean Algebra | 布尔代数的基本定律
Boolean algebra simplifies logic circuits and conditions in programming. The following laws apply to Boolean variables (A, B, C) and the operations AND (·), OR (+), NOT (ˉ). Memorising these helps in reducing expressions to their simplest forms.
布尔代数用于化简逻辑电路和程序中的条件。以下定律适用于布尔变量（A, B, C）以及运算 AND（·）、OR（+）、NOT（ˉ）。熟记这些定律有助于将表达式化简为最简形式。
- Identity Law: A + 0 = A, A · 1 = A / 恒等律：A + 0 = A，A · 1 = A
- Null Law: A + 1 = 1, A · 0 = 0 / 零一律：A + 1 = 1，A · 0 = 0
- Idempotent Law: A + A = A, A · A = A / 幂等律：A + A = A，A · A = A
- Inverse Law: A + Aˉ = 1, A · Aˉ = 0 / 互补律：A + Aˉ = 1，A · Aˉ = 0
- Double Negation: Aˉˉ = A / 双重否定律：Aˉˉ = A
- Commutative Law: A + B = B + A, A · B = B · A / 交换律
- Associative Law: (A + B) + C = A + (B + C), (A · B) · C = A · (B · C) / 结合律
- Distributive Law: A · (B + C) = A · B + A · C / 分配律
- Absorption Law: A + (A · B) = A, A · (A + B) = A / 吸收律
These laws are not just theoretical; they are directly tested in logic circuit simplification and truth table problems.
这些定律不仅具有理论意义，还会在逻辑电路化简和真值表问题中直接考查。
7. De Morgan’s Laws | 德摩根定律
De Morgan’s Laws are powerful tools for transforming logical expressions. They describe how the negation of an AND or OR can be rewritten. You must be able to apply them fluently for both hardware and software contexts.
德摩根定律是变换逻辑表达式的强大工具。它们描述了如何改写 AND 或 OR 的取反形式。无论是在硬件还是软件情境中，都必须能够熟练运用。
(A · B)ˉ = Aˉ + Bˉ
(A + B)ˉ = Aˉ · Bˉ
In words: the NOT of an AND is the OR of the individual NOTs. The NOT of an OR is the AND of the individual NOTs. These allow a circuit built only from NAND gates (or NOR gates) to implement any Boolean function.
用语言表述：AND 的取反等于各自取反的 OR。OR 的取反等于各自取反的 AND。这些定律使得仅用与非门（或仅用或非门）构成的电路能够实现任何布尔函数。
8. Logic Gate Truth Tables | 逻辑门真值表
A truth table captures every possible input combination and the resulting output for a logic gate or circuit. The basic gates tested are NOT, AND, OR, XOR, NAND, and NOR. You need to draw and interpret them accurately.
真值表记录了逻辑门或电路每一种可能的输入组合及其对应的输出。考试涉及的基本门电路包括 NOT、AND、OR、XOR、NAND 和 NOR。需要能够准确地绘制和解读它们。
| Gate | Symbol | Boolean Expression | Key Truth Table Rows (A, B → Output) |
|---|---|---|---|
| NOT | Triangle + bubble | Q = Aˉ | 0 → 1, 1 → 0 |
| AND | D-shape | Q = A · B | 0,0→0; 0,1→0; 1,0→0; 1,1→1 |
| OR | Curved shape | Q = A + B | 0,0→0; 0,1→1; 1,0→1; 1,1→1 |
| XOR | OR with extra curve | Q = A ⊕ B | 0,0→0; 0,1→1; 1,0→1; 1,1→0 |
| NAND | AND + bubble | Q = (A·B)ˉ | Opposite of AND |
| NOR | OR + bubble | Q = (A+B)ˉ | Opposite of OR |
For combined circuits, work layer by layer, applying the truth table of each gate to the intermediate signals. This methodical approach minimises errors.
对于组合电路，应逐层分析，将每个门的真值表应用于中间信号。这种有序的方法可以最大限度地减少错误。
9. Operator Precedence & Logical Expressions | 运算优先级与逻辑表达式
When evaluating or simplifying a Boolean expression, it is essential to apply operators in the correct order. The standard precedence (from highest to lowest) is: brackets, NOT, AND, OR. XOR has the same precedence as OR. This is analogous to BODMAS in arithmetic.
在计算或化简布尔表达式时，按照正确的顺序应用运算符至关重要。标准优先级（由高到低）为：括号、NOT、AND、OR。XOR 的优先级与 OR 相同。这与算术中的 BODMAS 规则类似。
For example, the expression A + B · Cˉ must be evaluated as A + (B · (Cˉ)) – perform NOT first, then AND, then OR. Adding parentheses clarifies the intended evaluation and is often required in exam answers. Being consistent with precedence ensures your truth tables and simplifications are correct.
例如，表达式 A + B · Cˉ 必须按 A + (B · (Cˉ)) 计算——先执行 NOT，然后 AND，最后 OR。添加括号可以明确计算顺序，考试中通常要求这么做。保持优先级的一致性可以确保真值表和化简结果正确无误。
10. Compression Ratio | 压缩比计算
Compression reduces file size for storage or transmission. The compression ratio tells you how much smaller the compressed file is compared to the original. It is a simple but examinable numeric measure.
压缩可以减小文件大小，以便存储或传输。压缩比表示压缩后的文件相对于原始文件缩小了多少。这是一个简单但会考查的数值指标。
Compression ratio = Original file size ÷ Compressed file size
压缩比 = 原始文件大小 ÷ 压缩文件大小
A compression ratio of 4:1 means the compressed file is one quarter of the original size. The space saving can also be expressed as a percentage: saved space (%) = (1 – (compressed/original)) × 100. When a question asks how much storage is saved, remember to subtract the compressed size from the original.
4:1 的压缩比意味着压缩文件只有原始文件大小的四分之一。节省的空间也可以用百分比表示：节省百分比 = (1 – （压缩文件/原始文件）) × 100。当题目问及节省了多少存储空间时，请记得用原始文件大小减去压缩文件大小。
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