第五章作业

5.1

1. $x_1(n)=n\cdot u(n+2)$$x_1(n) = n \cdot u(n+2)$.

   

2. $x_2(n)=(2^{-n}+1)u(n+1)$$x_2(n) = (2^{-n} + 1) u(n+1)$.

   

3. $\begin{array}{c}{x}_3(n)=\{\begin{array}{ll}n+2,& n\ge 0,\\ 3\cdot 2^n,& n<0.\end{array}\end{array}$$x_3(n) = \begin{cases} n+2, & n \ge 0, \\ 3 \cdot 2^n, & n < 0. \end{cases}$

   

4. $x_4(n)=x_2(n)+x_3(n)$$x_4(n) = x_2(n) + x_3(n)$.

   

附 绘图 mathematica 代码如下：

​x
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x1 = n UnitStep[n + 2];
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x2 = (2^-n + 1) UnitStep[n + 1];
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x3 = Which[n < 0, 3*2^n, n >= 0, n + 2];
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x4 = x2 + x3;
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DiscretePlot[x1, {n, -3, 3}, PlotRange -> All, GridLines -> Automatic]
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DiscretePlot[x2, {n, -2, 5}, PlotRange -> All, GridLines -> Automatic]
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DiscretePlot[x3, {n, -5, 3}, PlotRange -> All, GridLines -> Automatic]
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DiscretePlot[x4, {n, -3, 3}, PlotRange -> All, GridLines -> Automatic]

5.2

1. $(n-1)[u(n-1)-u(n-5)]$$(n-1) \bqty{u(n-1) - u(n-5)}$.

2. $u(n-3)-u(n-6)$$u(n-3) - u(n-6)$.

3. $(-1{)}^{n}u(n-1)$$(-1)^n u(n-1)$.

4. $-u(n)+2u(n-3)-u(n-6)$$-u(n) + 2u(n-3) - u(n-6)$.

5.3

1. 周期性，$N=14$$N = 14$.

2. 非周期.

备注 对于 $\sin (\omega n+\phi )$$\sin(\omega n + \varphi)$ 或 ${\text{e}}^{\mathrm{j}\omega n}$$\e^{\j\omega n}$，若 ${\displaystyle \frac{2\pi }{\omega }}\in \mathbb{Q}$$\dfrac{2\pi}{\omega} \in \Q$，则为周期序列，否则为非周期序列.

5.4

1. 迭代得 $y(0)=1$$y(0) = 1$，齐次通解为 $y(n)={\displaystyle \frac{C}{3^n}}u(n)$$y(n) = \dfrac{C}{3^n} u(n)$，代入有 $y_1(n)=h(n)={\displaystyle \frac{1}{3^n}}u(n)$$y_1(n) = h(n) = \dfrac{1}{3^n} u(n)$.

2. $y_2(n)=x(n)\ast h(n)={\displaystyle \frac{3-3^{-n}}{2}}u(n)$$y_2(n) = x(n) * h(n) = \dfrac{3 - 3^{-n}}{2} u(n)$.

3. 输出信号为

   $$\begin{array}{rl}{y}_3(n)& ={\displaystyle \frac{3-3^{-n}}{2}}u(n)-{\displaystyle \frac{3-3^{5-n}}{2}}u(n-5)\\ & ={\displaystyle \frac{3-3^{-n}}{2}}[u(n)-u(n-5)]+{\displaystyle \frac{121}{3^n}}u(n-5).\end{array}$$

波形图略.

5.5

1. $y(n)-ay(n-1)+by(n-2)=x(n)$$y(n) - a y(n-1) + b y(n-2) = x(n)$.

2. 二阶差分方程（二阶系统）.

5.6

1. $y(n)-b_{1}y(n-2)-b_{2}y(n-2)=a_{0}x(n)+a_{1}x(n-1)$$y(n) - b_1 y(n-2) - b_2 y(n-2) = a_0 x(n) + a_1 x(n-1)$.

2. 二阶差分方程（二阶系统）.

5.7

备注 不知道有什么便捷的工具可以绘制系统框图；上图直接截的标答.

5.8

1. $y(n)=2^{-n}$$y(n) = 2^{-n}$.

2. $y(n)=2^{n+1}$$y(n) = 2^{n+1}$.

3. $y(n)=(-3{)}^{n-1}$$y(n) = (-3)^{n-1}$.

4. $y(n)={\displaystyle \frac{(-3{)}^{-n}}{3}}=-(-3{)}^{-n-1}$$y(n) = \dfrac{(-3)^{-n}}{3} = -(-3)^{-n-1}$.

5.9

1. $y(n)=4(-1{)}^n-12(-2{)}^n$$y(n) = 4 (-1)^n - 12 (-2)^n$.

2. $y(n)=(2n+1)(-1{)}^n$$y(n) = (2n + 1) (-1)^n$.

3. $\cos {\displaystyle \frac{n\pi }{2}}+2\sin {\displaystyle \frac{n\pi }{2}}$$\cos\dfrac{n\pi}{2} + 2 \sin\dfrac{n\pi}{2}$。

5.10

1. 特征根为 $2,2,3$$2, 2, 3$，设 $y(n)=(an+b)2^n+c\cdot 3^n$$y(n) = (a n + b) 2^n + c \cdot 3^n$，代入得

于是 $y(n)=3^n-(n+1)2^n,n\ge 0$$y(n) = 3^n - (n + 1) 2^n, n \ge 0$.

2. 大概是题目打错了吧...这个计算量挺离谱的：

xxxxxxxxxx
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RSolveValue[{
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    y[n] - 2 y[n - 1] + 26 y[n - 2] - 2 y[n - 3] + y[n - 4] == 0,
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    y[0] == 0, y[1] == 1, y[2] == 2, y[3] == -3
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}, y[n], n]

5.11

1. 齐次通解为 $y(n)=C(-2{)}^n$$y(n) = C (-2)^n$.

2. 设特解为 $y_0(n)=an+b$$y_0(n) = a n +b$，代入得 $a={\displaystyle \frac{1}{3}},b=-{\displaystyle \frac{4}{9}}$$a = \dfrac{1}{3}, b = -\dfrac{4}{9}$.

3. 于是全解为 $y(n)=C(-2{)}^n+{\displaystyle \frac{n}{3}}-{\displaystyle \frac{4}{9}}$$y(n) = C (-2)^n + \dfrac{n}{3} - \dfrac{4}{9}$.

4. 代入初值，得 $y(n)={\displaystyle \frac{13}{9}}(-2{)}^n+{\displaystyle \frac{n}{3}}-{\displaystyle \frac{4}{9}}$$y(n) = \dfrac{13}{9} (-2)^n + \dfrac{n}{3} - \dfrac{4}{9}$.

5.12

1. 齐次通解为 $y(n)=(an+b)(-1{)}^n$$y(n) = (an + b) (-1)^n$.

2. 设特解为 $y_0(n)=A\cdot 3^n$$y_0(n) = A \cdot 3^n$，代入得 $A={\displaystyle \frac{9}{16}}$$A = \dfrac{9}{16}$.

3. 于是全解为 $y(n)=(an+b)(-1{)}^n+{\displaystyle \frac{9}{16}}\cdot 3^n$$y(n) = (an + b) (-1)^n + \dfrac{9}{16} \cdot 3^n$.

4. 代入初值，得 $y(n)=(-{\displaystyle \frac{3}{4}}n-{\displaystyle \frac{9}{16}})(-1{)}^n+{\displaystyle \frac{9}{16}}\cdot 3^n$$y(n) = \pqty{-\dfrac{3}{4} n - \dfrac{9}{16}} (-1)^n + \dfrac{9}{16} \cdot 3^n$.

5.13

1. 全解为 $y(n)=a\cos {\displaystyle \frac{n\pi }{2}}+b\sin {\displaystyle \frac{n\pi }{2}}+A\sin n+B\cos n$$y(n) = a \cos\dfrac{n\pi}{2} + b\sin\dfrac{n\pi}{2} + A\sin n + B \cos n$.

   代入即可解得.

2. $\frac{1}{8}(2\lfloor n{\rfloor }^2-2((-1{)}^{\lfloor n\rfloor +n}+1)\lfloor n\rfloor +(-1{)}^{\lfloor n\rfloor +n}-5(-1{)}^n-12)$$\frac{1}{8} \left(2 \lfloor n\rfloor ^2-2 \left((-1)^{\lfloor n\rfloor +n}+1\right) \lfloor n\rfloor +(-1)^{\lfloor n\rfloor +n}-5 (-1)^n-12\right)$.

   标答有误.

3.