📚 Formula & Theorem Quick Reference Handbook | 公式定理速查手册
This handbook brings together all the essential formulas and theorems you need for Year 7 AQA Computer Science. From binary conversions to logic gate truth tables, these quick-reference notes will help you solve problems confidently and build a strong foundation in computing.
这本手册汇集了 Year 7 AQA 计算机课程中所有必不可少的公式和定理。从二进制转换到逻辑门真值表，这些速查笔记将帮助你自信地解决问题，并为计算思维打下坚实基础。
1. Binary and Denary Conversion | 二进制与十进制转换
ere a 1 appears. Each place value is a power of 2, starting from 2⁰ on the right. For example, the binary number 1101₂ equals (1×8) + (1×4) + (0×2) + (1×1) = 13 in denary.
计算机使用二进制（基数为2，数字为0和1），而人类使用十进制（基数为10）。要将二进制数转换为十进制，只需将出现1的位值相加。每个位值都是2的幂次，最右边为2⁰。例如，二进制数1101₂等于(1×8) + (1×4) + (0×2) + (1×1) = 13（十进制）。
Denary = Sum of (binary digit × 2^position from right, starting at 0)
十进制值 = (每个二进制位 × 2^从右起的位置) 之和
he remainders from bottom to top. For 13: 13÷2=6 r1, 6÷2=3 r0, 3÷2=1 r1, 1÷2=0 r1 → 1101₂.
将十进制转换为二进制，反复除以2并记录余数（0或1），直到商为0。然后从下往上读取余数。例如13：13÷2=6余1，6÷2=3余0，3÷2=1余1，1÷2=0余1 → 1101₂。
2. Binary Addition Rules | 二进制加法规则
Binary addition follows four simple rules. These are the basis for all arithmetic inside a computer's ALU (Arithmetic Logic Unit).
二进制加法遵循四条简单规则。这些是计算机算术逻辑单元（ALU）所有运算的基础。
| Rule | Sum | Carry |
|---|---|---|
| 0 + 0 | 0 | 0 |
| 0 + 1 | 1 | 0 |
| 1 + 0 | 1 | 0 |
| 1 + 1 | 0 | 1 |
| 1 + 1 + 1 | 1 | 1 |
When adding two binary numbers, align them to the right and apply the rules column by column. If a column produces a carry, add it to the next left column. Overflow occurs if the result needs more bits than available.
当两个二进制数相加时，将它们右对齐并逐列应用规则。如果某一列产生进位，就加到左侧下一列中。如果结果需要比可用比特更多的位数，就会发生溢出。
3. Hexadecimal Basics | 十六进制基础
ely used to represent binary more compactly, as one hex digit equals four bits (a nibble).
十六进制（基数为16）使用数字0-9和字母A-F（A=10, B=11, C=12, D=13, E=14, F=15）。它被广泛用于更紧凑地表示二进制，因为一个十六进制数字等于四位二进制（一个半字节）。
binary to hex, split the binary into groups of four from the right, then write each group's denary equivalent as a hex digit. For example, 11011010₂ → 1101 1010 → D A → DA₁₆.
要将二进制转换为十六进制，从右侧将二进制数分成四位一组，然后将每组的十进制等值写成十六进制数字。例如，11011010₂ → 1101 1010 → D A → DA₁₆。
Hexadecimal digit = denary value of 4-bit nibble, using A-F for 10-15
十六进制数字 = 4位半字节的十进制值，10-15用A-F表示
4. Storage Unit Conversions | 存储单位换算
All data in a computer is measured in bits. Larger units are built using powers of 2 (though sometimes metric prefixes are used for simplicity). The key relationships are:
计算机中的所有数据都以比特为单位进行度量。更大的单位使用2的幂次构建（尽管为简单起见有时使用公制前缀）。关键关系如下：
- 1 nibble = 4 bits
- 1 byte (B) = 8 bits
- 1 kilobyte (kB) = 1000 bytes (often used, but 1 KiB = 1024 B)
- 1 megabyte (MB) = 1000 kB or 1 MiB = 1024 KiB
- 1 gigabyte (GB) = 1000 MB or 1 GiB = 1024 MiB
- 1 terabyte (TB) = 1000 GB or 1 TiB = 1024 GiB
在考试中，如果没有特别说明，通常使用1000的倍数。但是，在处理内存或精确计算文件大小时，1 KiB = 1024 B 更准确。请记住：1字节 = 8比特，向上每级通常是 ×1000（或 ×1024）。
File size (bytes) = number of pixels × colour depth (bits) / 8
文件大小（字节）= 像素数 × 色深（比特） / 8
5. Image File Size Formula | 图像文件大小公式
Bitmap images are made of pixels, each with a colour. The file size of an uncompressed image depends on resolution and colour depth.
位图图像由像素组成，每个像素有一种颜色。未压缩图像的文件大小取决于分辨率和色深。
mage file size (bits) = Width in pixels × Height in pixels × Colour depth (bits per pixel)
图像文件大小（比特） = 宽像素数 × 高像素数 × 色深（每像素比特数）
Colour depth is the number of bits used for each pixel. For example, 8-bit colour allows 256 colours; 24-bit colour allows over 16 million colours. To get the size in bytes, divide the total bits by 8. To get it in kilobytes, divide by 8000 (or 8192 if required).
色深是每个像素使用的比特数。例如，8位色允许256种颜色；24位色允许超过1600万种颜色。要获得以字节为单位的大小，将总比特数除以8。要获得以千字节为单位的大小，再除以8000（如果需要，除以8192）。
6. Sound File Size Formula | 声音文件大小公式
Digital sound is captured by taking samples of the sound wave at regular intervals. The quality and size of the file increase with a higher sample rate and bit depth.
数字声音是通过定期对声波进行采样来捕获的。采样率越高、位深度越高，文件的质量和大小就越大。
file size (bits) = Sample rate (Hz) × Bit depth × Number of channels × Duration (seconds)
声音文件大小（比特） = 采样率（Hz）× 位深度 × 声道数 × 时长（秒）
ne minute of such audio equals: 44,100 × 16 × 2 × 60 = 84,672,000 bits, or about 10.58 MB.
例如，CD 音质的采样率为 44,100 Hz，位深度为 16 位，声道数为 2（立体声）。一分钟这样的音频等于：44,100 × 16 × 2 × 60 = 84,672,000 比特，约 10.58 MB。
7. ASCII and Character Encoding | ASCII与字符编码
Computers store text by mapping each character to a binary code. The most common standard is ASCII (American Standard Code for Information Interchange), which uses 7 bits to represent 128 characters, including letters, digits, punctuation and control codes. Extended ASCII uses 8 bits to add another 128 characters.
计算机通过将每个字符映射到一个二进制代码来存储文本。最常见的标准是 ASCII（美国信息交换标准代码），它使用 7 个比特表示 128 个字符，包括字母、数字、标点符号和控制代码。扩展 ASCII 使用 8 个比特，增加了另外 128 个字符。
Number of characters = 2n, where n = number of bits
字符数 = 2n，其中 n 为比特数
56 characters. Modern systems often use Unicode, which can use up to 32 bits per character and includes symbols from almost all writing systems.
例如，7 位 ASCII：2⁷ = 128 个字符。8 位扩展 ASCII：2⁸ = 256 个字符。现代系统通常使用 Unicode，它每个字符最多可使用 32 位，并包含几乎所有书写系统的符号。
8. Logic Gate Truth Tables | 逻辑门真值表
Logic gates are the building blocks of digital circuits. The three basic gates are AND, OR and NOT. Their behaviour is fully described by truth tables. A NOT gate has only one input; the output is the opposite of the input.
逻辑门是数字电路的构建块。三种基本门是 AND（与门）、OR（或门）和 NOT（非门）。它们的特性完全由真值表描述。NOT 门只有一个输入；输出是输入的反值。
| A | B | A AND B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
AND: Output is 1 only if both inputs are 1. OR: Output is 1 if at least one input is 1. NOT: Input A, output NOT A. If A=0, output=1; if A=1, output=0.
AND： 只有当两个输入都为 1 时，输出才为 1。OR： 只要至少一个输入为 1，输出就为 1。NOT： 输入 A，输出 NOT A。若 A=0，输出为 1；若 A=1，输出为 0。
9. Boolean Algebra Laws | 布尔代数定律
Boolean algebra simplifies logic expressions. The main laws include identity, commutative, associative, distributive, and De Morgan's laws. These help reduce the number of logic gates needed in a circuit.
布尔代数用于简化逻辑表达式。主要定律包括同一律、交换律、结合律、分配律和德摩根律。这些有助于减少电路中所需的逻辑门数量。
Commutative law: A AND B = B AND A; A OR B = B OR A.
Associative law: (A AND B) AND C = A AND (B AND C); (A OR B) OR C = A OR (B OR C).
Distributive law: A AND (B OR C) = (A AND B) OR (A AND C); A OR (B AND C) = (A OR B) AND (A OR C).
交换律： A AND B = B AND A；A OR B = B OR A。
结合律： (A AND B) AND C = A AND (B AND C)；(A OR B) OR C = A OR (B OR C)。
分配律： A AND (B OR C) = (A AND B) OR (A AND C)；A OR (B AND C) = (A OR B) AND (A OR C)。
De Morgan's laws: NOT A) AND (NOT B). These are especially useful for converting between NAND and NOR gates.
德摩根律： NOT (A AND B) = (NOT A) OR (NOT B)；NOT (A OR B) = (NOT A) AND (NOT B)。这在 NAND 门和 NOR 门之间转换时特别有用。
10. Network Data Transfer Time | 网络数据传输时间
When sending data across a network, the transfer time depends on the size of the data and the bandwidth (bit rate) of the connection.
当通过网络发送数据时，传输时间取决于数据的大小和连接的带宽（比特率）。
Time (seconds) = Data size (bits) / Bandwidth (bits per second, bps)
时间（秒） = 数据大小（比特） / 带宽（比特每秒，bps）
ion, transfer time = 40,000,000 / 100,000,000 = 0.4 seconds (ignoring protocol overheads).
计算前请务必将所有单位转换为比特。例如，一个 5 MB 的文件为 5 × 8 × 1,000,000 = 40,000,000 比特。通过 100 Mbps 的连接，传输时间 = 40,000,000 / 100,000,000 = 0.4 秒（忽略协议开销）。
This formula also applies to estimating download times for files from the internet. Always check if the file size is given in bytes or bits, and whether prefixes use powers of 10 or 2.
此公式也适用于估计从互联网下载文件的时间。请始终检查文件大小是以字节还是比特给出的，以及前缀使用的是 10 的幂次还是 2 的幂次。
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