📚 Z Transforms and Difference Equations | Z 变换与差分方程
Z transforms provide a powerful tool for analysing discrete-time signals and systems. They convert difference equations – which describe sequences – into algebraic equations that are much easier to manipulate, especially when initial conditions are present. In the IB Mathematics curriculum, Z transforms appear in the further mathematics or discrete mathematics options, where students learn to handle linear difference equations with constant coefficients using this transform method. Understanding Z transforms also bridges the gap between discrete mathematics and real-world applications such as digital signal processing.
Z 变换为分析离散时间信号与系统提供了强大的工具。它将描述序列的差分方程转化为更容易处理的代数方程，尤其是在给定初始条件时。在 IB 数学课程中，Z 变换出现在进阶数学或离散数学选修部分，学生将学习如何用这种变换方法求解常系数线性差分方程。掌握 Z 变换还能在离散数学和数字信号处理等实际应用之间搭建桥梁。
1. Introduction to Z Transforms | Z 变换简介
Just as the Laplace transform is used for continuous-time systems, the Z transform is the primary mathematical tool for discrete-time systems. It maps a sequence x[n] (often defined for n ≥ 0) to a complex function X(z). This transformation turns convolution into multiplication and shifting into algebraic operations, making it invaluable for solving systems described by difference equations.
正如拉普拉斯变换用于连续时间系统，Z 变换是离散时间系统的主要数学工具。它将序列 x[n]（通常定义在 n ≥ 0）映射为复函数 X(z)。这种变换将卷积转化为乘法，将移位转化为代数运算，因此在求解由差分方程描述的系统时极为有用。
2. Definition of the Z Transform | Z 变换的定义
The one-sided (or unilateral) Z transform of a sequence x[n] is defined by the infinite sum
X(z) = Σn=0∞ x[n] z-n
序列 x[n] 的单边 Z 变换由以下无穷级数定义：
X(z) = Σn=0∞ x[n] z-n
sual sequences where x[n] = 0 for n < 0, and we often assume that the transform exists for |z| > R, where R is some radius.
其中 z 是复变量，收敛域（ROC）是使该级数绝对收敛的 z 的集合。在大多数 IB 考题中，我们处理的因果序列满足当 n < 0 时 x[n] = 0，并且通常假设变换在 |z| > R 时存在，R 为某个半径。
n]} = X(z). The transform is linear, which follows directly from the linearity of the sum.
我们用 Z{x[n]} = X(z) 表示 Z 变换。该变换是线性的，这直接来自求和的线性性质。
3. Linearity and Basic Properties | 线性性质与基本性质
If Z{x[n]} = X(z) and Z{y[n]} = Y(z), then for any constants a and b:
Z{a x[n] + b y[n]} = a X(z) + b Y(z)
若 Z{x[n]} = X(z) 且 Z{y[n]} = Y(z)，则对于任意常数 a 与 b：
Z{a x[n] + b y[n]} = a X(z) + b Y(z)
Another fundamental property is the multiplication by an exponential: Z{an ndy when we know the transform of a basic sequence and need to find the transform of an exponentially weighted version.
另一个基本性质是指数相乘：Z{an x[n]} = X(z/a)。当我们已经知道某个基本序列的变换，而需要求其指数加权版本的变换时，这一性质特别方便。
4. Z Transforms of Common Sequences | 常见序列的 Z 变换
The table below summarises the Z transforms of fundamental sequences that appear frequently in IB problems. All sequences are casual (zero for n < 0).
下表总结了 IB 题目中经常出现的基本序列的 Z 变换。所有序列均为因果序列（n < 0 时值为 0）。
| Sequence x[n] | Z Transform X(z) | ROC |
|---|---|---|
| δ[n] (unit impulse) | 1 | all z |
| u[n] (unit step, u[n] = 1 for n ≥ 0) | 1 / (1 − z-1) = z / (z − 1) | |z| > 1 |
| an u[n] | 1 / (1 − a z-1) = z / (z − a) | |z| > |a| |
| n an u[n] | a z / (z − a)2 | |z| > |a| |
| cos(ω n) u[n] | z(z − cos ω) / (z2 − 2z cos ω + 1) | |z| > 1 |
The transform of δ[n] being 1 is particularly important because it means the impulse response of a system plays a central role in transfer function analysis.
δ[n] 的变换为 1 尤为重要，因为它意味着系统的冲激响应在传递函数分析中起着核心作用。
5. Time-Shifting Property | 时移性质
The time-shifting property is the key to solving difference equations. For a casual sequence x[n], the Z transform of the delayed version x[n−k] (with k a positive integer) is:
Z{x[n−k] u[n−k]} = z-k X(z)
时移性质是求解差分方程的关键。对于因果序列 x[n]，其延迟版本 x[n−k]（k 为正整数）的 Z 变换为：
Z{x[n−k] u[n−k]} = z-k X(z)
However, when we work with the one-sided transform and non-zero initial conditions, the property must account for those initial values. For example, the transform of x[n+1] is:
Z{x[n+1]} = z X(z) − z x[0]
然而，当使用单边变换且初始条件非零时，该性质必须反映这些初始值。例如，x[n+1] 的变换为：
Z{x[n+1]} = z X(z) − z x[0]
Similar formulas exist for higher shifts. For x[n+2] we have Z{x[n+2]} = z2 X(z) − z2 x[0] − z x[1]. These expressions allow us to convert a difference equation into a polynomial equation in z and X(z), where the initial conditions appear as coefficients.
对于更高阶的移位也存在类似的公式。对于 x[n+2]，有 Z{x[n+2]} = z2 X(z) − z2 x[0] − z x[1]。这些表达式使我们能够将差分方程转化为关于 z 和 X(z) 的多项式方程，其中初始条件体现为系数。
6. Convolution and Multiplication Properties | 卷积与乘法定理
onvolution of two causal sequences x[n] and y[n] is defined as (x ∗ y)[n] = Σk=0n x[k] y[n−k]. Its Z transform is simply the product:
Z{(x ∗ y)[n]} = X(z) Y(z)
两个因果序列 x[n] 和 y[n] 的卷积定义为 (x ∗ y)[n] = Σk=0n x[k] y[n−k]。其 Z 变换就是简单的乘积：
Z{(x ∗ y)[n]} = X(z) Y(z)
This property is the discrete-time equivalent of the convolution theorem for Laplace and Fourier transforms. It tells us that a system's output is the inverse transform of the product of the input transform and the system's transfer function H(z).
这一性质是拉普拉斯变换和傅里叶变换中卷积定理的离散时间对应。它告诉我们，系统的输出是输入变换与系统传递函数 H(z) 乘积的逆变换。
(z)/dz. This is useful when inverting transforms involving denominators with squared terms.
时域中乘以 n 对应于 z 域中的微分：Z{n x[n]} = −z dX(z)/dz。当分母含有平方项时，这一性质有助于逆变换。
7. Inverse Z Transform: Inspection and Tables | 逆 Z 变换：观察法与查表
into one. For instance, if X(z) = z/(z − 0.5), we immediately identify x[n] = (0.5)n u[n] using the table.
逆 Z 变换从 X(z) 恢复 x[n]。在许多 IB 考题中，最简单的途径是识别出 X(z) 是标准形式，或者将它变形为标准形式。例如，若 X(z) = z/(z − 0.5)，我们可以立即通过查表识别出 x[n] = (0.5)n u[n]。
When the expression is rational, we often rewrite it in terms of z-1 before matching. The term 1/(1 − a z-1) is the transform of an u[n]. Splitting a rational function into partial fractions naturally leads to such terms.
当表达式为有理分式时，我们通常会先将其写成关于 z-1 的形式再进行匹配。项 1/(1 − a z-1) 就是 an u[n] 的变换。将有理函数分解为部分分式会自然产生此类项。
8. Inverse Z Transform: Partial Fraction Expansion | 逆 Z 变换：部分分式展开
Suppose X(z) = (z + 1) / [(z − 2)(z − 3)] with ROC |z| er if the degree of the numerator is less than the denominator. Then we perform expansion:
X(z)/z = (z + 1)/[z(z − 2)(z − 3)] = A/z + B/(z − 2) + C/(z − 3)
设 X(z) = (z + 1) / [(z − 2)(z − 3)]，收敛域为 |z| > 3。我们先将其写为关于 z 的真分式（若分子次数低于分母，已是真分式）。然后进行展开：
X(z)/z = (z + 1)/[z(z − 2)(z − 3)] = A/z + B/(z − 2) + C/(z − 3)
− 3). Using the table and the ROC, we obtain x[n] = (1/6)δ[n] + [(−3/2)2n + (4/3)3n] u[n].
解出 A、B、C 得 A = 1/6，B = −3/2，C = 4/3。乘以 z 得到 X(z) = (1/6) + (−3/2)z/(z − 2) + (4/3)z/(z − 3)。利用表格及收敛域，我们得到 x[n] = (1/6)δ[n] + [(−3/2)2n + (4/3)3n] u[n]。
This method works whenever X(z) is a rational function with distinct poles. For repeated poles, the expansion includes terms like A/(z − p)k.
当 X(z) 是有理函数且具有互异极点时，该方法总是有效
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