Year 10 SQA Computing: Formula & Theorem Quick Reference Handbook | Year 10 SQA 计算机：公式定理速查手册

📚 Year 10 SQA Computing: Formula & Theorem Quick Reference Handbook | Year 10 SQA 计算机：公式定理速查手册

This quick reference guide collects the essential formulas, laws, and calculation methods you will encounter in the Year 10 SQA Computing course. It covers data representation, digital logic, file sizes, network performance, processor metrics, and basic algorithmic principles. Use it as a portable revision companion to reinforce your understanding and solve calculation questions with confidence.

这份速查手册汇总了你在 SQA 计算机 Year 10 课程中会遇到的核心公式、定律和计算方法。内容涵盖数据表示、数字逻辑、文件大小、网络性能、处理器指标以及基础算法原理。将它作为随身复习伙伴，帮助你巩固理解，自信应对各类计算题。

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1. Data Units and Storage Capacity | 数据单位与存储容量

The smallest unit of data is a bit (b), which can be 0 or 1. A group of 8 bits forms a byte (B). Larger units are based on powers of 2, although storage manufacturers sometimes use decimal approximations. Always read question contexts carefully to decide whether you should use binary (1024) or decimal (1000) multipliers.

数据的最小单位是比特（bit），只能取 0 或 1。8 个比特组成一个字节（Byte）。更大的单位以 2 的幂次为基础，但存储设备厂商有时采用十进制近似值。答题时务必仔细阅读题目背景，判断应使用二进制乘数（1024）还是十进制乘数（1000）。

1 kilobyte (KB) = 2¹⁰ bytes = 1024 bytes (binary) or 1000 bytes (disk marketing)

1 megabyte (MB) = 2²⁰ bytes = 1 048 576 bytes (binary) or 10⁶ bytes

1 gigabyte (GB) = 2³⁰ bytes (binary) or 10⁹ bytes

1 terabyte (TB) = 2⁴⁰ bytes (binary) or 10¹² bytes

When calculating storage capacity, multiply the number of storage units (e.g. sectors, blocks) by the size of each unit. For main memory, the total addressable space is given by 2ⁿ, where n is the number of address lines.

计算存储容量时，将存储单元的数量（如扇区数、块数）乘以每个单元的大小。主存的总可寻址空间可由 2ⁿ 得出，其中 n 为地址线的条数。

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2. Binary, Denary and Hexadecimal Conversions | 二进制、十进制与十六进制转换

Computers store data in binary (base‑2). Humans often need to read numbers in denary (base‑10) or hexadecimal (base‑16). Conversions follow place‑value weighting and division/remainder algorithms.

计算机以二进制（base‑2）存储数据。人类通常需要以十进制（base‑10）或十六进制（base‑16）阅读数字。转换遵循位值加权和除余算法。

Denary to binary: repeatedly divide by 2, record remainders from bottom to top.

十进制转二进制：反复除以 2，从下往上记录余数。

Binary to denary: sum (bit × 2^position), position starts at 0 from right.

二进制转十进制：求和 (位值 × 2^位置)，位置从右端 0 开始。

Hex to binary: replace each hex digit with its 4‑bit binary equivalent.

十六进制转二进制：将每个十六进制数字替换为对应的 4 位二进制。

A = 0010 1010. Leading zeros are often dropped unless the context specifies a fixed width.

例如，十六进制数字 A（十进制 10）的二进制为 1010，因此 2A = 0010 1010。除非上下文要求固定宽度，否则前导零通常会省略。

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3. Boolean Algebra Laws | 布尔代数定律

Boolean algebra is used to simplify logic circuits. The laws operate on variables that take values TRUE (1) or FALSE (0). The basic operations are AND (conjunction), OR (disjunction), and NOT (negation). Simplification reduces the number of gates and cost.

布尔代数用于简化逻辑电路。这些定律作用于取值为 TRUE（1）或 FALSE（0）的变量。基本运算包括与（AND）、或（OR）、非（NOT）。简化可减少门数量和成本。

Commutative Law: A + B = B + A, A · B = B · A

交换律：A + B = B + A，A · B = B · A

Associative Law: (A + B) + C = A + (B + C)

结合律：(A + B) + C = A + (B + C)

Distributive Law: A · (B + C) = (A · B) + (A · C)

分配律：A · (B + C) = (A · B) + (A · C)

Identity Law: A + 0 = A, A · 1 = A

同一律：A + 0 = A，A · 1 = A

Complement Law: A + Ā = 1, A · Ā = 0

互补律：A + Ā = 1，A · Ā = 0

Using these laws, complex expressions can be reduced. For instance, A + A·B simplifies to A (absorption law), which is derived from the distributive and identity laws.

运用这些定律可以简化复杂表达式。例如，A + A·B 可简化为 A（吸收律），这是结合分配律和同一律推导出来的。

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4. De Morgan's Theorems | 德摩根定律

De Morgan's Theorems are crucial for transforming logic expressions and implementing circuits using only NAND or NOR gates. They describe how negation distributes over AND and OR operations.

德摩根定律对于逻辑表达式变换以及仅用 NAND 或 NOR 门实现电路至关重要。它们描述了否定如何在 AND 和 OR 运算上分配。

First theorem: ¬(A · B) = ¬A + ¬B

第一定律：¬(A · B) = ¬A + ¬B

Second theorem: ¬(A + B) = ¬A · ¬B

第二定律：¬(A + B) = ¬A · ¬B

In words: the negation of a conjunction is the disjunction of the negations; the negation of a disjunction is the conjunction of the negations. These theorems are heavily tested when creating NAND‑only or NOR‑only circuits.

简言之：与运算的非等于非运算的或；或运算的非等于非运算的与。在构建纯 NAND 或纯 NOR 电路时，这两个定律是重点考查内容。

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5. Image File Size Calculation | 图像文件大小计算

ght in pixels) and the colour depth (bits per pixel). No compression is assumed unless stated otherwise. The basic formula is:

位图图像的文件大小取决于图像尺寸（宽×高，单位像素）和色深（每像素位数）。除非特别说明，否则假定无压缩。基本公式为：

File size (bits) = Width (pixels) × Height (pixels) × Colour Depth (bits per pixel)

文件大小（位） = 宽度（像素） × 高度（像素） × 色深（每像素位数）

dividing by 8, and then to larger units if required. For example, a 1024×768 image with 24‑bit colour requires 1024 × 768 × 24 = 18 874 368 bits = 2 359 296 bytes ≈ 2.25 MB.

将结果除以 8 得到字节数，然后根据需要换算为更大单位。例如，一幅 1024×768、24 位色深的图像需要 1024 × 768 × 24 = 18 874 368 位 = 2 359 296 字节 ≈ 2.25 MB。

e, the colour depth is the ceil(log₂(number of colours)). For 256 colours, depth = 8 bits.

若图像使用调色板，色深为 ceil(log₂(颜色数))。256 色时，深度 = 8 位。

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6. Sound File Size Calculation | 声音文件大小计算

Digital sound is sampled at regular intervals. The file size is determined by the sample rate, bit depth, number of channels, and duration. The underlying formula is:

数字声音按固定间隔采样。文件大小由采样率、位深度、声道数和时长决定。基本公式为：

File size (bits) = Sample Rate (Hz) × Bit Depth × Channels × Duration (seconds)

文件大小（位） = 采样率（Hz） × 位深度 × 声道数 × 时长（秒）

For CD‑ f such audio occupies 44 100 × 16 × 2 × 60 = 84 672 000 bits = 10 584 000 bytes ≈ 10.1 MB.

CD 品质音频的采样率通常为 44 100 Hz，位深度 16 位，声道数 2（立体声）。一分钟此类音频占用 44 100 × 16 × 2 × 60 = 84 672 000 位 = 10 584 000 字节 ≈ 10.1 MB。

Higher quality or longer durations increase the size proportionally. Remember to divide by 8 to get bytes, then by 1024 for kibibytes if binary units are expected.

更高的质量或更长的时长会成比例地增加文件大小。记得除以 8 得到字节数；若要求二进制单位，再除以 1024 得到 KiB。

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7. Network Data Transfer Rate | 网络数据传输速率

Network speed is usually measured in bits per second (bps) while file sizes are in bytes. To calculate transfer time, ensure both are in the same unit. The fundamental relationship is:

网络速度通常以比特每秒（bps）为单位，而文件大小以字节为单位。计算传输时间时，需确保两者单位一致。基本关系为：

Transfer Time (seconds) = File Size (bits) / Network Speed (bps)

传输时间（秒） = 文件大小（位） / 网络速度（bps）

8 000 000 bits (decimal). Many exam questions assume 1 byte = 8 bits and ignore the binary‑decimal distinction unless specified.

若以兆字节和兆比特给出，需转换：1 MB = 8 × 1 048 576 位（二进制）或 8 000 000 位（十进制）。许多试题假定 1 字节 = 8 位，除非特别说明，否则忽略二进制与十进制的区别。

Network throughput may also be affected by latency and protocol overheads, but the basic formula ignores these.

网络吞吐量还可能受延迟和协议开销影响，但基本公式忽略这些因素。

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8. Processor Performance Formulae | 处理器性能公式

CPU performance is often described by its clock speed, number of cores, and the average cycles per instruction (CPI). The execution time of a program can be estimated using these parameters.

CPU 性能通常通过时钟速度、核心数量和平均每指令周期数（CPI）来描述。程序的执行时间可由这些参数估算：

CPU Time = Instruction Count × CPI / Clock Rate

CPU 时间 = 指令数 × CPI / 时钟频率

, and the average CPI is 1.5, the execution time is (1×10⁹ × 1.5) / (3×10⁹) = 0.5 seconds.

时钟频率以赫兹（Hz）为单位。3 GHz 的处理器每秒有 3×10⁹ 个周期。若某程序包含 1×10⁹ 条指令，平均 CPI 为 1.5，则执行时间为 (1×10⁹ × 1.5) / (3×10⁹) = 0.5 秒。

For multi‑core processors, ideal speedup follows Amdahl's Law, which is beyond Year 10 but introduces the idea that parallelisation does not scale perfectly.

对于多核处理器，理想加速遵循阿姆达尔定律，这超出 Year 10 范围，但引入了并行化并不能完美扩展的思想。

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9. Binary Addition and Overflow | 二进制加法与溢出

Binary addition follows the same place‑value rules as denary addition. When the sum of two bits exceeds 1, a carry is generated to the next column. The rules for single‑bit addition are:

二进制加法遵循与十进制加法相同的位值规则。当两个比特之和超过 1 时，会产生向下一列的进位。单比特加法规则为：

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 0, carry 1 to the next bit

1 + 1 + carry‑in = 1, carry 1

Overflow occurs when the result exceeds the maximum value representable in the given number of bits. In signed binary (two's complement), overflow is detected when the carry into the most significant bit differs from the carry out, or when adding two positive numbers yields a negative result and vice versa.

当结果超出给定位数所能表示的最大值时，溢出就会发生。在有符号二进制（补码）中，如果进位到最高位与从最高位的进位不同，或者两个正数相加得到负数（反之亦然），即检测到溢出。

For 4‑ (14) + 0011 (3) = 1 0001, but only 4 bits stored → 0001 with overflow (carry out of MSB).

4 位无符号：1110 (14) + 0011 (3) = 1 0001，但仅存储 4 位 → 0001 并产生溢出（最高有效位进位）。

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10. Memory Addressing Range | 内存寻址范围

The number of unique memory locations directly addressable by a CPU depends on the width of the address bus. Each address line carries one bit, so n address lines can address 2ⁿ distinct locations.

CPU 可直接寻址的唯一内存单元数取决于地址总线的宽度。每条地址线传输一个比特，因此 n 条地址线可以寻址 2ⁿ 个不同的存储位置。

Addressable locations = 2ⁿ, where n = number of address lines

可寻址单元数 = 2ⁿ，其中 n = 地址线数量

For example, a 16‑ es 1 byte, the total addressable memory is 65 536 bytes = 64 KiB. Modern processors use 32‑bit or 64‑bit address buses, allowing terabytes of addressable space.

例如，16 位地址总线可以寻址 2¹⁶ = 65 536 个存储位置。如果每个位置存储 1 字节，可寻址总容量为 65 536 字节 = 64 KiB。现代处理器使用 32 位或 64 位地址总线，可寻址空间达到 TB 级别。

ddress bus width and data bus width, you can calculate total capacity: Capacity (bits) = 2^{address lines} × data bus width (bits).

当题目结合地址总线和数据总线宽度时，可以计算总容量：容量（位） = 2^地址线数 × 数据总线宽度（位）。

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11. Simple Algorithm Time Complexity | 简单算法时间复杂度

Time complexity describes how the runtime of an algorithm grows relative to the input size, n. In Year 10, you focus on identifying whether an algorithm runs in constant, linear, or quadratic time by counting basic operations.

时间复杂度描述了算法运行时间随输入规模 n 的增长情况。在 Year 10，你需要通过统计基本操作次数来识别算法是常数时间、线性时间还是二次时间。

• Constant time O(1): the number of steps does not depend on n. Accessing an array element by index is O(1).

  常数时间 O(1)：步骤数与 n 无关。通过索引访问数组元素为 O(1)。

• Linear time O(n): a single loop that iterates n times. Summing all elements of an array is O(n).

  线性时间 O(n)：单层循环执行 n 次。对数组所有元素求和为 O(n)。

• Quadratic time O(n²): a nested loop where both loops run n times. Bubble sort in its worst case is O(n²).

  二次时间 O(n²)：嵌套循环，两层循环各运行 n 次。冒泡排序的最坏情况为 O(n²)。

The exact number of operations can be counted for a given algorithm, but Big O notation captures the dominant term and ignores constants. This helps compare the efficiency of different approaches.

对于给定算法，可以统计精确的操作次数，但大 O 符号只捕捉主导项并忽略常数。这有助于比较不同方案的效率。

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12. Text Encoding and Checksums | 文本编码与校验和

Characters are stored as numeric codes. ASCII uses 7 bits (0–127) to represent English letters, digits and symbols. Extended ASCII uses 8 bits (0–255). Unicode (UTF‑8, UTF‑16) can represent characters from almost all writing systems using variable‑length encoding.

字符以数字编码储存。ASCII 使用 7 位（0–127）表示英文字母、数字和符号。扩展 ASCII 使用 8 位（0–255）。Unicode（UTF‑8、UTF‑16）可使用变长编码表示几乎所有书写系统的字符。

ASCII 'A' = 65 decimal = 0100 0001 binary

ASCII 字符 'A' = 65 十进制 = 0100 0001 二进制

A checksum is a simple error‑detection value appended to data. One common method is the 8‑bit sum complement: add all data bytes ignoring overflow, take the one's complement of the sum, and append it. The receiver performs the same addition including the checksum; a result of all 1s (or zero in some schemes) indicates no error.

校验和是一种附加到数据后的简单检错值。常见方法之一是 8 位求和补码：将所有数据字节相加（忽略溢出），对结果取反码并附加。接收方对包括校验和在内的数据进行同样的加法；结果为全 1（或某些方案中为零）则表明无错。

Checksum = ¬(sum of data bytes) truncated to 8 bits

校验和 = 数据字节和的 8 位截断反码

More robust methods like CRC are used in modern networks, but the basic checksum principle remains the same.

现代网络使用 CRC 等更可靠的方法，但基本的校验和原理依然相同。

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