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数学代写|线性代数代写Linear algebra代考|MATH2022 Linear Difference Equations

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数学代写|线性代数代写Linear algebra代考|Linear Difference Equations

Given scalars $a_0, \ldots, a_n$, with $a_0$ and $a_n$ nonzero, and given a signal $\left{z_k\right}$, the equation
$$a_0 y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}+a_n y_k=z_k \quad \text { for all } k$$
is called a linear difference equation (or linear recurrence relation) of order $\boldsymbol{n}$. For simplicity, $a_0$ is often taken equal to 1 . If $\left{z_k\right}$ is the zero sequence, the equation is homogeneous; otherwise, the equation is nonhomogeneous.

EXAMPLE 3 In digital signal processing, a difference equation such as (3) describes a linear filter, and $a_0, \ldots, a_n$ are called the filter coefficients. If $\left{y_k\right}$ is treated as the input and $\left{z_k\right}$ as the output, then the solutions of the associated homogeneous equation are the signals that are filtered out and transformed into the zero signal. Let us feed two different signals into the filter
$$.35 y_{k+2}+.5 y_{k+1}+.35 y_k=z_k$$
Here $.35$ is an abbreviation for $\sqrt{2} / 4$. The first signal is created by sampling the continuous signal $y=\cos (\pi t / 4)$ at integer values of $t$, as in Figure 3(a). The discrete signal is
$$\left{y_k\right}={\ldots, \cos (0), \cos (\pi / 4), \cos (2 \pi / 4), \cos (3 \pi / 4), \ldots}$$
For simplicity, write $\pm .7$ in place of $\pm \sqrt{2} / 2$, so that
$$\left{y_k\right}={\ldots, 1, .7,0,-.7,-1,-.7,0, .7,1, .7,0, \ldots}$$
Table 1 shows a calculation of the output sequence $\left{z_k\right}$, where $.35(.7)$ is an abbreviation for $(\sqrt{2} / 4)(\sqrt{2} / 2)=.25$. The output is $\left{y_k\right}$, shifted by one term.

数学代写|线性代数代写Linear algebra代考|Solution Sets of Linear Difference Equat

Given $a_1, \ldots, a_n$, consider the mapping $T: \mathbb{S} \rightarrow \mathbb{S}$ that transforms a signal $\left{y_k\right}$ into a signal $\left{w_k\right}$ given by
$$w_k=y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}+a_n y_k$$
It is readily checked that $T$ is a linear transformation. This implies that the solution set of the homogeneous equation
$$y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}+a_n y_k=0 \quad \text { for all } k$$
is the kernel of $T$ (the set of signals that $T$ maps into the zero signal), and hence the solution set is a subspace of $\mathrm{S}$. Any linear combination of solutions is again a solution.
If $a_n \neq 0$ and if $\left{z_k\right}$ is given, the equation
$$y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}+a_n y_k=z_k \quad \text { for all } k$$
has a unique solution whenever $y_0, \ldots, y_{n-1}$ are specified.
PROOF If $y_0, \ldots, y_{n-1}$ are specified, use (7) to define
$$y_n=z_0-\left[a_1 y_{n-1}+\cdots+a_{n-1} y_1+a_n y_0\right]$$
And now that $y_1, \ldots, y_n$ are specified, use (7) to define $y_{n+1}$. In general, use the recurrence relation
$$y_{n+k}=z_k-\left[a_1 y_{k+n-1}+\cdots+a_n y_k\right]$$
to define $y_{n+k}$ for $k \geq 0$. To define $y_k$ for $k<0$, use the recurrence relation
$$y_k=\frac{1}{a_n} z_k-\frac{1}{a_n}\left[y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}\right]$$
This produces a signal that satisfies (7). Conversely, any signal that satisfies (7) for all $k$ certainly satisfies (8) and (9), so the solution of (7) is unique.

数学代写|线性代数代写线性代数代考|线性差分方程

$$a_0 y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}+a_n y_k=z_k \quad \text { for all } k$$

$$.35 y_{k+2}+.5 y_{k+1}+.35 y_k=z_k$$

$$\left{y_k\right}={\ldots, \cos (0), \cos (\pi / 4), \cos (2 \pi / 4), \cos (3 \pi / 4), \ldots}$$

$$\left{y_k\right}={\ldots, 1, .7,0,-.7,-1,-.7,0, .7,1, .7,0, \ldots}$$

数学代写|线性代数代写线性代数代考|线性差分方程的解集

$$w_k=y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}+a_n y_k$$

$$y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}+a_n y_k=0 \quad \text { for all } k$$

$$y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}+a_n y_k=z_k \quad \text { for all } k$$

PROOF如果指定了$y_0, \ldots, y_{n-1}$，则使用(7)定义
$$y_n=z_0-\left[a_1 y_{n-1}+\cdots+a_{n-1} y_1+a_n y_0\right]$$
。现在已经指定了$y_1, \ldots, y_n$，则使用(7)定义$y_{n+1}$。通常，使用递归关系
$$y_{n+k}=z_k-\left[a_1 y_{k+n-1}+\cdots+a_n y_k\right]$$

$$y_k=\frac{1}{a_n} z_k-\frac{1}{a_n}\left[y_{k+n}+a_1 y_{k+n-1}+\cdots+a_{n-1} y_{k+1}\right]$$

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。