📚 Exploring Z-Transforms and Difference Equations in IB Mathematics | IB数学：z变换与差分方程的解法探究
The Z-transform is a powerful mathematical tool that converts discrete-time signals and difference equations into algebraic forms. In IB Mathematics HL, students encounter this technique as a bridge between recurrence relations and real-world modelling, especially in topics such as sequences, series, and discrete dynamical systems.
Z变换是一种强大的数学工具，能够将离散时间信号和差分方程转化为代数形式。在IB数学高级课程中，学生将其视为连接递推关系与实际建模的桥梁，特别是在数列、级数和离散动力系统相关主题中尤为重要。
1. What Is a Z-Transform? | 什么是Z变换？
The Z-transform maps a discrete sequence \(a_n\) (for n = 0, 1, 2, …) into a function of a complex variable z. Formally, the one-sided Z-transform is defined as:
Z变换将离散序列 aₙ（其中 n = 0, 1, 2, …）映射为复变量 z 的函数。形式上，单边Z变换定义为：
Z{aₙ} = A(z) = Σₙ₌₀^∞ aₙ z⁻ⁿ
Here z is a complex number, and the series converges within a certain region of the z-plane. In IB problems, we typically work with simple sequences where the transform can be written as a rational function of z.
这里 z 是复数，该级数在 z 平面的一定区域内收敛。在IB问题中，我们通常处理简单的序列，其变换可以写成 z 的有理函数。
2. Basic Z-Transform Pairs | 基本Z变换对
Just as with Laplace transforms, a small table of transform pairs is essential for solving problems. The following table includes the most common sequences encountered in IB:
与拉普拉斯变换类似，掌握一张基本的变换对表格是解题的关键。下表列出了IB中常见的基本序列：
| Sequence aₙ | Z-Transform A(z) |
| 1 | z/(z-1) for |z|>1 |
| n | z/(z-1)² for |z|>1 |
| aⁿ | z/(z-a) for |z|>|a| |
| naⁿ | az/(z-a)² for |z|>|a| |
| n² | z(z+1)/(z-1)³ for |z|>1 |
Notice that all transforms are expressed in z rather than z⁻¹. This form is often easier to manipulate when solving difference equations.
注意所有变换都用 z 而不是 z⁻¹ 表示。这种形式在解差分方程时通常更便于运算。
3. Linearity and Shifting Properties | 线性性质与移位性质
The Z-transform is linear. If xₙ and yₙ have transforms X(z) and Y(z), then for constants α and β:
Z变换是线性的。若 xₙ 和 yₙ 的变换分别为 X(z) 和 Y(z)，则对常数 α 和 β 有：
Z{αxₙ + βyₙ} = αX(z) + βY(z)
More importantly, the right-shift property handles initial conditions. Suppose xₙ is defined for n ≥ 0 and xₙ = 0 for n < 0. Then:
更重要的是，右移性质用于处理初始条件。设 xₙ 对 n ≥ 0 有定义，且 n < 0 时 xₙ = 0，则：
Z{xₙ₋₁} = z⁻¹ X(z) + x₋₁
In most IB problems, x₋₁ is zero, so the shift simply multiplies by z⁻¹. This property transforms a recurrence relation into an algebraic equation.
在大多数IB题目中，x₋₁ 为零，因此移位只是乘以 z⁻¹。这一性质将递推关系转化为代数方程。
4. Inverse Z-Transform Using Partial Fractions | 利用部分分式求逆Z变换
To recover a sequence from its Z-transform, we rewrite A(z) as a sum of simpler fractions whose inverses are known. A useful trick is to first divide by z, find the partial fractions for A(z)/z, and then multiply back by z.
要从Z变换还原序列，我们将 A(z) 写成若干个简单分式之和，这些分式的逆变换是已知的。一个常用技巧是先除以 z，对 A(z)/z 做部分分式分解，然后再乘以 z。
If A(z) = z·B(z), then decompose B(z) = p/(z-a) + q/(z-b) + …
Then each term of the form z/(z-a) corresponds to aⁿ. This method is especially effective for rational transforms with distinct poles.
这样，形如 z/(z-a) 的每一项对应 aⁿ。此方法对具有不同极点的有理变换尤其有效。
5. Solving First-Order Difference Equations | 一阶差分方程的求解
Consider a first-order linear difference equation with constant coefficients:
考虑一个常系数一阶线性差分方程：
yₙ₊₁ + a yₙ = fₙ, y₀ given
Taking the Z-transform of both sides and using Z{yₙ₊₁} = zY(z) – z y₀, we obtain an algebraic equation for Y(z). Solving for Y(z) and then using partial fractions gives the explicit form of yₙ.
对两边取Z变换，利用 Z{yₙ₊₁} = zY(z) – z y₀，得到一个关于 Y(z) 的代数方程。解出 Y(z) 后，利用部分分式即可得到 yₙ 的显式表达式。
Example: yₙ₊₁ – 3yₙ = 0, y₀ = 2. Then zY(z) – 2z – 3Y(z) = 0, so Y(z) = 2z/(z-3). Hence yₙ = 2·3ⁿ.
例如：yₙ₊₁ – 3yₙ = 0，y₀ = 2。则 zY(z) – 2z – 3Y(z) = 0，所以 Y(z) = 2z/(z-3)。因此 yₙ = 2·3ⁿ。
6. Solving Second-Order Difference Equations | 二阶差分方程的求解
A second-order equation such as yₙ₊₂ + p yₙ₊₁ + q yₙ = rₙ can be solved by taking the Z-transform and using initial values y₀ and y₁. We use:
形如 yₙ₊₂ + p yₙ₊₁ + q yₙ = rₙ 的二阶方程，可以通过取Z变换并利用初值 y₀ 和 y₁ 来求解。我们使用：
Z{yₙ₊₂} = z²Y(z) – z²y₀ – z y₁
Substituting the initial conditions transforms the recurrence into a rational function of z. Partial fraction decomposition then yields the solution in terms of known sequences.
代入初始条件后，递推关系转换为 z 的有理函数。接着用部分分式分解，即可用已知序列表示解。
For example, the Fibonacci recurrence Fₙ₊₂ = Fₙ₊₁ + Fₙ, with F₀=0, F₁=1, leads to F(z) = z/(z² – z – 1). Decomposing gives the Binet formula.
例如，斐波那契递推 Fₙ₊₂ = Fₙ₊₁ + Fₙ，且 F₀=0，F₁=1，得到 F(z) = z/(z² – z – 1)。分解后即可得到比奈公式。
7. Applying Z-Transforms to Recurrence Relations | 应用Z变换解递推关系
In IB, recurrence relations often appear as arithmetic or geometric sequences with additional terms. For example, the relation aₙ₊₁ = 2aₙ + 5, a₀ = 1, can be solved by Z-transforms:
在IB中，递推关系常以等差或等比数列并附加项的形式出现。例如，关系 aₙ₊₁ = 2aₙ + 5，a₀ = 1，可通过Z变换求解：
zA(z) – z·1 = 2A(z) + 5z/(z-1)
Solving gives A(z) = (z² + 4z)/((z-1)(z-2)). Partial fractions yield aₙ = 6·2ⁿ – 5.
解出 A(z) = (z² + 4z)/((z-1)(z-2))。部分分式得到 aₙ = 6·2ⁿ – 5。
This approach works for any linear recurrence with constant coefficients, including those with exponential or constant forcing terms.
这种方法适用于任何常系数线性递推，包括带有指数型或常数强迫项的递推。
8. Stability and the Unit Circle | 稳定性与单位圆
The Z-transform also helps analyse the long-term behaviour of discrete systems. A system described by a difference equation is stable if all poles of its Z-transform lie inside the unit circle on the complex plane (|z| < 1).
Z变换还可以帮助分析离散系统的长期行为。如果一个差分方程所描述的系统的Z变换的所有极点都位于复数平面上的单位圆内（|z| < 1），则该系统是稳定的。
For example, the sequence yₙ = 2ⁿ grows without bound because its pole z = 2 is outside the unit circle. In contrast, yₙ = (1/2)ⁿ tends to zero since |z| = 1/2 < 1.
例如，序列 yₙ = 2ⁿ 因为其极点 z = 2 在单位圆外而无界增长。相比之下，yₙ = (1/2)ⁿ 因为 |z| = 1/2 < 1 而趋于零。
When the sequence oscillates, a pole on the negative real axis inside the unit circle produces alternating decay, such as (-0.5)ⁿ.
当序列振荡时，单位圆内负实轴上的极点会产生交替衰减，例如 (-0.5)ⁿ。
9. Z-Transforms vs. Differential Equations | Z变换与微分方程的比较
Z-transforms are the discrete-time analogue of Laplace transforms. While Laplace transforms are used for continuous differential equations, Z-transforms handle discrete difference equations. The shift property in Z-transforms corresponds to differentiation in Laplace transforms.
Z变换是拉普拉斯变换在离散时间领域的类比。拉普拉斯变换用于连续微分方程，而Z变换用于离散差分方程。Z变换中的移位性质对应拉普拉斯变换中的微分性质。
This correspondence is useful because a differential equation can be approximated by a difference equation, and Z-transforms allow us to solve such discretised models exactly.
这种对应关系很有用，因为微分方程可以用差分方程近似，而Z变换可以精确求解这类离散化模型。
10. Common Pitfalls in IB Exams | IB考试常见陷阱
Students often forget to include initial conditions when applying the shift property. Always check whether the sequence is defined for n ≥ 0 and whether any shifted term like x₋₁ is nonzero.
学生在应用移位性质时常常忘记包含初始条件。务必检查序列是否仅在 n ≥ 0 时有定义，以及 x₋₁ 这类移位项是否非零。
Another mistake is applying partial fractions to A(z) directly instead of first considering A(z)/z. This can lead to incorrect inverses.
另一个错误是直接对 A(z) 做部分分式，而不是先考虑 A(z)/z。这会导致错误的逆变换。
Finally, be careful with the signs of exponents: Z{ yₙ₊₁ } = zY(z) – zy₀, not zY(z) + y₀.
最后，注意指数符号：Z{ yₙ₊₁ } = zY(z) – zy₀，不是 zY(z) + y₀。
11. Worked Example: Population Model | 实例：种群模型
A population pₙ evolves according to pₙ₊₁ = 1.5 pₙ + 100, with p₀ = 200. Find pₙ explicitly.
某种群 pₙ 按照 pₙ₊₁ = 1.5 pₙ + 100 演化，且 p₀ = 200。求 pₙ 的显式表达式。
Taking the Z-transform gives zP(z) – 200z = 1.5P(z) + 100z/(z-1). Thus P(z) [z – 1.5] = 200z + 100z/(z-1).
取Z变换得 zP(z) – 200z = 1.5P(z) + 100z/(z-1)。因此 P(z)[z – 1.5] = 200z + 100z/(z-1)。
So P(z) = z(200z – 100)/((z-1)(z-1.5)). Decomposing P(z)/z = 200/(z-1.5) – 100/(z-1). Hence pₙ = 200(1.5)ⁿ – 100.
所以 P(z) = z(200z – 100)/((z-1)(z-1.5))。分解 P(z)/z = 200/(z-1.5) – 100/(z-1)。因此 pₙ = 200(1.5)ⁿ – 100。
12. Practice Problems | 练习问题
1. Find the Z-transform of the sequence aₙ = 3ⁿ + 2n.
1. 求序列 aₙ = 3ⁿ + 2n 的Z变换。
2. Solve the difference equation yₙ₊₂ – 5yₙ₊₁ + 6yₙ = 0, with y₀ = 1, y₁ = 5.
2. 求解差分方程 yₙ₊₂ – 5yₙ₊₁ + 6yₙ = 0，其中 y₀ = 1，y₁ = 5。
3. Determine whether the system described by yₙ₊₁ = 0.75yₙ + 2 is stable. Justify your answer using the pole location.
3. 判断系统 yₙ₊₁ = 0.75yₙ + 2 是否稳定，并利用极点位置说明理由。
4. Use the Z-transform to solve aₙ₊₂ = aₙ₊₁ + aₙ with a₀ = 0, a₁ = 1, and hence write the first six terms.
4. 用Z变换求解 aₙ₊₂ = aₙ₊₁ + aₙ，其中 a₀ = 0，a₁ = 1，并写出前六项。
Check your answers by substituting back into the original recurrence relations.
将答案代回原递推关系进行验证。
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