📚 Year 8 Edexcel Computer Science: Quick Reference Guide of Formulas and Theorems | Year 8 Edexcel 计算机：公式定理速查手册
This quick reference guide brings together the essential formulas, theorems, and conversion methods required for Year 8 Edexcel Computer Science. Use it to revise key concepts in data representation, logic, and calculations involving files and networks.
本速查手册汇总了 Year 8 Edexcel 计算机课程必备的重要公式、定理与转换方法。涵盖数据表示、逻辑运算以及文件和网络相关的计算，帮助你高效复习核心概念。
1. Binary Weight Values (Powers of 2) | 二进制位权值（2 的幂）
In the binary number system, each digit (bit) holds a weight that is a power of 2. The rightmost bit has the lowest weight, 20 = 1, and weights double as you move left.
在二进制数制中，每一位（比特）的权值是 2 的幂次。最右侧位权值最小，为 20 = 1，向左移动权值翻倍。
| Power of 2 (2n) | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 |
|---|---|---|---|---|---|---|---|---|---|
| Decimal Value | 1 | 2 | 4 | 8 | 16 | 32 | 64 | 128 | 256 |
2. Binary to Decimal Conversion | 二进制转十进制
To convert a binary number to decimal, multiply each bit by its corresponding power of 2, then add all the results together.
将二进制数转换为十进制时，用每一位上的数字乘以其对应的 2 的幂，然后将所有结果相加。
Decimal = dn-1 × 2n-1 + dn-2 × 2n-2 + … + d0 × 20
For example, the binary number 10112 is calculated as 1×23 + 0×22 + 1×21 + 1×20 = 8 + 0 + 2 + 1 = 11 in decimal.
例如，二进制数 10112 的计算过程为 1×23 + 0×22 + 1×21 + 1×20 = 8 + 0 + 2 + 1 = 11（十进制）。
3. Decimal to Binary Conversion | 十进制转二进制
The standard method to convert a decimal number into binary is repeated division by 2. You record the remainder at each step, and the binary result is the sequence of remainders read from bottom to top.
把十进制数转换成二进制的标准方法是反复除以 2。记录每一步的余数，从下往上读取余数序列即得二进制结果。
- Divide the decimal number by 2. 将十进制数除以 2。
- Write down the remainder (0 or 1). 记下余数（0 或 1）。
- Use the integer quotient for the next division. 将整数商用于下一次除法。
- Repeat until the quotient becomes 0. 重复直到商为 0。
- The binary number is the remainders read from the last remainder upwards. 从最后一个余数向上读取，即为二进制数。
4. Binary Addition and Overflow | 二进制加法与溢出
Binary addition follows simple rules. When a result exceeds the maximum value that can be stored in a fixed number of bits, an overflow error occurs.
二进制加法遵循简单规则。当结果超过固定位数所能存储的最大值时，会发生溢出错误。
0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, 1 + 1 = 0 carry 1, 1 + 1 + 1 = 1 carry 1
When adding two 8-bit numbers, if the result requires a 9th bit, the leftmost carry is lost and the answer becomes incorrect. For example, 11111111 + 00000001 gives 00000000 with an overflow carry.
当两个 8 位数相加时，如果结果需要第 9 位，最左侧的进位会丢失，答案将出错。例如 11111111 + 00000001 得到 00000000 并产生溢出进位。
5. Logic Gates and Boolean Expressions | 逻辑门与布尔表达式
Logic gates are the building blocks of digital circuits. Each gate corresponds to a Boolean operator that defines its output for given inputs.
逻辑门是数字电路的基本构件。每种门对应一个布尔运算符，定义了在给定输入下的输出。
AND gate: Output is true only when all inputs are true. 与门： 仅当所有输入都为真时输出才为真。
Q = A AND B (also written as A ∧ B or A · B)
OR gate: Output is true when at least one input is true. 或门： 只要有一个输入为真，输出即为真。
Q = A OR B (also written as A ∨ B or A + B)
NOT gate: Output is the inverse of the single input. 非门： 输出是单一输入的反相。
Q = NOT A (also written as ¬A or A’)
6. Truth Tables | 真值表
A truth table lists all possible combinations of inputs for a logic circuit and shows the resulting output. Below are the truth tables for basic gates.
真值表列出逻辑电路所有可能的输入组合，并显示对应的输出。以下是基本门的真值表。
| A | B | A AND B | A OR B | NOT A |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 0 |
7. De Morgan’s Laws | 德摩根定律
De Morgan’s Laws describe how an AND operation with inverted inputs can be transformed into an OR operation with inverted outputs, and vice versa. These theorems are essential for simplifying logic circuits.
德摩根定律描述了如何将带有取反输入的 AND 运算转换为带有取反输出的 OR 运算，反之亦然。这些定理对简化逻辑电路至关重要。
¬(A ∧ B) = ¬A ∨ ¬B
¬(A ∨ B) = ¬A ∧ ¬B
In words: the negation of (A AND B) is equivalent to (NOT A) OR (NOT B); the negation of (A OR B) is equivalent to (NOT A) AND (NOT B).
换言之：(A AND B) 的否等价于 (NOT A) OR (NOT B)；(A OR B) 的否等价于 (NOT A) AND (NOT B)。
8. Data Storage Unit Conversions | 数据存储单位换算
Digital data quantities are measured using standard binary prefixes. Knowing the exact conversions is vital for calculating file sizes and storage requirements.
数字数据量使用标准二进制前缀度量。了解准确的换算关系对计算文件大小和存储需求至关重要。
| Unit | Equivalent |
|---|---|
| 1 bit (b) | smallest unit of data |
| 1 Byte (B) | 8 bits |
| 1 Kilobyte (KB) | 1024 Bytes |
| 1 Megabyte (MB) | 1024 KB |
| 1 Gigabyte (GB) | 1024 MB |
| 1 Terabyte (TB) | 1024 GB |
9. Image File Size Calculation | 图像文件大小计算
The size of an uncompressed bitmap image depends on its dimensions and colour depth. The formula uses the number of pixels and the bits per pixel.
未压缩的位图图像大小取决于其尺寸和颜色深度。公式利用像素数量以及每像素的比特数来计算。
Image size (bits) = width × height × colour depth
Image size (bytes) = (width × height × colour depth) / 8
Colour depth is the number of bits used to represent the colour of a single pixel. For example, an image 100×100 pixels with a 24-bit colour depth requires 100 × 100 × 24 = 240,000 bits, which is 30,000 bytes (about 29.3 KB).
颜色深度是用来表示单个像素颜色的比特数。例如，一张 100×100 像素、24 位颜色深度的图像需要 100 × 100 × 24 = 240,000 比特，即 30,000 字节（约 29.3 KB）。
10. Sound File Size Calculation | 声音文件大小计算
Uncompressed sound file size is determined by the sampling rate, bit depth, number of channels, and the duration of the audio.
未压缩的声音文件大小由采样率、位深度、声道数和音频时长共同决定。
File size (bits) = sampling rate (Hz) × bit depth × number of channels × duration (s)
File size (bytes) = (sampling rate × bit depth × channels × duration) / 8
For stereo audio (2 channels) recorded at 44.1 kHz with 16-bit depth for 10 seconds, the raw file size is 44,100 × 16 × 2 × 10 = 14,112,000 bits, which equals 1,764,000 bytes or about 1.68 MB.
对于以 44.1 kHz 采样率、16 位位深度录制 10 秒的立体声（2 声道），原始文件大小为 44,100 × 16 × 2 × 10 = 14,112,000 比特，即 1,764,000 字节或约 1.68 MB。
11. Data Transmission Time | 数据传输时间
When sending data across a network, the transfer time can be calculated if you know the total data size and the transmission rate. Ensure both quantities use the same unit prefix.
通过网络发送数据时，如果知道数据总量和传输速率，就可以计算传输时间。确保两者使用相同的单位前缀。
Time (seconds) = Data size (bits) / Transfer rate (bps)
For example, downloading a 5 Megabyte file over a 10 Megabits per second (Mbps) connection: 5 MB = 5 × 8 = 40 Megabits; time = 40 Mb / 10 Mbps = 4 seconds.
例如，通过 10 Mbps（兆比特每秒）的连接下载一个 5 MB（兆字节）的文件：5 MB = 5 × 8 = 40 兆比特；时间 = 40 Mb / 10 Mbps = 4 秒。
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