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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

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