第六章

6.1

由备注中的结论：

1. $2^{-n}u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{2z}{2z-1}},\left|z\right|>{\displaystyle \frac{1}{2}}$$2^{-n} u(n) \zt \dfrac{2z}{2z - 1}, \vqty{z} > \dfrac{1}{2}$.

2. $-2^{-n}u(-n-1)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{2z}{2z-1}},\left|z\right|<{\displaystyle \frac{1}{2}}$$-2^{-n} u(-n-1) \zt \dfrac{2z}{2z - 1}, \vqty{z} < \dfrac{1}{2}$.

3. $(-3{)}^{n}u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z}{z+3}},\left|z\right|>3$$(-3)^n u(n) \zt \dfrac{z}{z + 3}, \vqty{z} > 3$.

4. 这里求双边 z 变换：$\delta (n+1)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}\delta (n+1)z^{-n}=z,\left|z\right|<\mathrm{\infty }}$$\delta(n + 1) \zt \nsuminf \delta(n + 1) z^{-n} = z, \vqty{z} < \infty$.

5. $\delta (n)-{\displaystyle \frac{1}{8}}\delta (n-3)\stackrel{\mathcal{Z}}{\leftrightarrow }1-{\displaystyle \frac{1}{8z^3}},\left|z\right|>0$$\delta(n) - \dfrac{1}{8} \delta(n-3) \zt 1 - \dfrac{1}{8 z^3}, \vqty{z} > 0$.

6. 法一：直接由定义与等比数列求和公式

   $2^{-n}u(n)-2^{-n}u(n-10)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{1-(2z{)}^{-10}}{1-(2z{)}^{-1}}},\left|z\right|>0$$2^{-n} u(n) - 2^{-n} u(n-10) \zt \dfrac{1-(2z)^{-10}}{1-(2z)\inv}, \vqty{z} > 0$.

   法二：利用微分性质

   $F(z)={\displaystyle \frac{z}{z-\frac{1}{2}}}-{\displaystyle \frac{z\cdot z^{-10}}{z-\frac{1}{2}}}={\displaystyle \frac{1-(2z{)}^{-10}}{1-(2z{)}^{-1}}},\left|z\right|>0$$F(z) = \dfrac{z}{z - \cfrac{1}{2}} - \dfrac{z \cdot z^{-10}}{z - \cfrac{1}{2}} = \dfrac{1-(2z)^{-10}}{1-(2z)\inv}, \vqty{z} > 0$.

7. $(2^{-n}+3^n)u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z}{z-\frac{1}{2}}}+{\displaystyle \frac{z}{z-3}},\left|z\right|>3$$\pqty{2^{-n} + 3^{n}} u(n) \zt \dfrac{z}{z - \cfrac{1}{2}} + \dfrac{z}{z - 3}, \vqty{z} > 3$.

8. 由备注中的结论有，

   $$\begin{array}{rl}\sin ({\displaystyle \frac{n\pi }{2}}+{\displaystyle \frac{\pi }{4}})u(n)& ={\displaystyle \frac{\sqrt{2}}{2}}(\sin {\displaystyle \frac{n\pi }{2}}+\cos {\displaystyle \frac{n\pi }{2}})u(n)\\ & \stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{\sqrt{2}}{2}}{\displaystyle \frac{z^2+z}{z^2+1}},\left|z\right|>1.\end{array}$$

9. $\cos \left({\displaystyle \frac{n\pi }{4}}\right)u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z(z-\frac{\sqrt{2}}{2}z)}{z^2-\sqrt{z}+1}},\left|z\right|>1$$\cos\pqty{\dfrac{n\pi}{4}} u(n) \zt \dfrac{z (z - \cfrac{\sqrt 2}{2} z)}{z^2 - \sqrt z + 1}, \vqty{z} > 1$.

备注 该题使用的结论及其证明现罗列如下：

• 由等比数列求和公式，有：

  • $a^{n}u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\left({\displaystyle \frac{a}{z}}\right)}^n={\displaystyle \frac{z}{z-a}},\mathrm{ROC}:\left|z\right|>\left|a\right|}$$a^n u(n) \zt \nosum \pqty{\dfrac{a}{z}}^n = \dfrac{z}{z - a}, \rm{ROC} \colon \vqty{z} > \vqty{a}$.

  • $-a^{n}u(-n-1)\stackrel{\mathcal{Z}}{\leftrightarrow }-{\displaystyle \sum _{n=-1}^{-\mathrm{\infty }}{\left({\displaystyle \frac{z}{a}}\right)}^{-n}={\displaystyle \frac{z}{z-a}},\mathrm{ROC}:\left|z\right|<\left|a\right|}$$-a^{n} u(-n - 1) \zt -\dsum_{n=-1}^{-\infty} \pqty{\dfrac{z}{a}}^{-n} = \dfrac{z}{z - a}, \rm{ROC} \colon \vqty{z} < \vqty{a}$.

• 利用线性性质与上述结论 (序列指数加权)，有

  $$\begin{array}{rl}\cos \left({\omega }_{0}n\right)u(n)& ={\displaystyle \frac{{\text{e}}^{\mathrm{j}{\omega }_{0}n}+{\text{e}}^{-\mathrm{j}{\omega }_{0}n}}{2}}u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{1}{2}}({\displaystyle \frac{z}{z-{\text{e}}^{\mathrm{j}{\omega }_0}}}+{\displaystyle \frac{z}{z-{\text{e}}^{-\mathrm{j}{\omega }_0}}})\\ & ={\displaystyle \frac{z(z-\cos {\omega }_0)}{z^2-2z\cos {\omega }_0+1}},\left|z\right|>1,\\ \sin \left({\omega }_{0}n\right)u(n)& ={\displaystyle \frac{{\text{e}}^{\mathrm{j}{\omega }_{0}n}-{\text{e}}^{-\mathrm{j}{\omega }_{0}n}}{2\mathrm{j}}}u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{1}{2\mathrm{j}}}({\displaystyle \frac{z}{z-{\text{e}}^{\mathrm{j}{\omega }_0}}}-{\displaystyle \frac{z}{z-{\text{e}}^{-\mathrm{j}{\omega }_0}}})\\ & ={\displaystyle \frac{z\sin {\omega }_0}{z^2-2z\cos {\omega }_0+1}},\left|z\right|>1.\end{array}$$

6.2

1. 其 z 变换及其收敛域为

2. 零极点图的代码及其图像

附

• 调用代码：

1
a = [2, -5, 2];
2
b = [-3, 0];
3
plotpzd(a, b, true)

• 函数代码：

​x
1
function plotpzd(a, b, showCircle)
2
% Plot the zeros-poles distribution map
3
% Variable a is a denominator coefficient vector,
4
% and b is a nominator coefficient vector.
5

6
ps = roots(a);          % roots of denominator polynomial
7
zs = roots(b);          % roots of nominator polynomial
8
legStr = [];            % legend string
9

10
if (~isempty(zs))
11
    plot(real(zs), imag(zs), 'o');  hold on;
12
    legStr = [legStr; 'Zeros'];
13
end
14
if (~isempty(ps))
15
    plot(real(ps), imag(ps), 'rx', 'markersize', 12);
16
    legStr = [legStr; 'Poles'];
17
end
18
if(~exist('showCircle','var'))
19
    showCircle = false;
20
end
21
if (showCircle)
22
    rectangle('position', [-1, -1, 2, 2], 'curvature', [1,1], 'LineStyle', ':')
23
end
24

25
xmin = floor(min([real(ps); real(zs)]));    xmin = min(xmin, -1);
26
xmax = ceil(max([real(ps); real(zs)]));     xmax = max(xmax, 1);
27
ymin = floor(min([imag(ps); imag(zs)]));    ymin = min(ymin, -1);
28
ymax = ceil(max([imag(ps); imag(zs)]));     ymax = max(ymax, 1);
29
axis([xmin xmax ymin ymax]), axis equal;
30
legend(legStr), grid on;
31

32
% set(gca,'YAxisLocation','origin');
33
% set(gca,'XAxisLocation','origin');
34

35
end

6.3

1. $x(n)=\delta (n)$$x(n) = \delta(n)$.

2. $x(n)=\delta (n+3)$$x(n) = \delta(n + 3)$.

3. $x(n)=\delta (n-1)$$x(n) = \delta(n-1)$.

4. $x(n)=\delta (n)+2\delta (n+1)-2\delta (n-2)$$x(n) = \delta(n) + 2\delta(n+1) - 2 \delta(n - 2)$.

5. $x(n)=a^{n}u(n)$$x(n) = a^n u(n)$.

6. $x(n)=-a^{n}u(-n-1)$$x(n) = -a^n u(-n-1)$.

备注

• $\delta (n-m)\stackrel{\mathcal{Z}}{\leftrightarrow }{z}^{-m}$$\delta(n-m) \zt z^{-m}$.

• $\begin{array}{c}{\displaystyle \frac{z}{z-a}}\stackrel{{\mathcal{Z}}_B}{\leftrightarrow }\{\begin{array}{ll}{a}^{n}u(n),& \left|z\right|>\left|a\right|,\\ -a^{n}u(-n-1),& \left|z\right|<\left|a\right|.\end{array}\end{array}$$\dfrac{z}{z - a} \bzt \begin{cases} a^n u(n), & \vqty{z} > \vqty{a}, \\ -a^n u(-n - 1), & \vqty{z} < \vqty{a}. \end{cases}$

6.4

1. 展开 ${\displaystyle \frac{X(z)}{z}}={\displaystyle \frac{2}{z}}+{\displaystyle \frac{1}{z+0.5}}$$\dfrac{X(z)}{z} = \dfrac{2}{z} + \dfrac{1}{z + 0.5}$，

   于是 $X(z)=2+{\displaystyle \frac{z}{z+0.5}}$$X(z) = 2 + \dfrac{z}{z + 0.5}$，

   从而 $x(n)=2\delta (n)+{(-{\displaystyle \frac{1}{2}})}^{n}u(n)$$x(n) = 2\delta(n) + \pqty{-\dfrac{1}{2}}^n u(n)$.

2. ${\displaystyle \frac{X(z)}{z}}={\displaystyle \frac{z-\frac{1}{2}}{(z+\frac{1}{2})(z+\frac{1}{4})}}={\displaystyle \frac{4}{z+\frac{1}{2}}}-{\displaystyle \frac{3}{z+\frac{1}{4}}}$$\dfrac{X(z)}{z} = \dfrac{z - \cfrac{1}{2}}{\pqty{z + \cfrac{1}{2}} \pqty{z + \cfrac{1}{4}}} = \dfrac{4}{z + \cfrac{1}{2}} - \dfrac{3}{z + \cfrac{1}{4}}$,

   $x(n)=[4{(-{\displaystyle \frac{1}{2}})}^n-3{(-{\displaystyle \frac{1}{4}})}^n]u(n)$$x(n) = \bqty{4 \pqty{-\dfrac{1}{2}}^n - 3\pqty{-\dfrac{1}{4}}^n} u(n)$.

3. $x(n)=(2^{1-n}-4^{-n})u(n)$$x(n) = \pqty{2^{1-n} - 4^{-n}} u(n)$.

4. ${\displaystyle \frac{X(z)}{z}}={\displaystyle \frac{-a}{z}}+(a-{\displaystyle \frac{1}{a}}){\displaystyle \frac{1}{z-\frac{1}{a}}}$$\dfrac{X(z)}{z} = \dfrac{-a}{z} + \pqty{a - \dfrac{1}{a}} \dfrac{1}{z - \cfrac{1}{a}}$.

   $x(n)=-a\delta (n)+(a-{\displaystyle \frac{1}{a}})a^{-n}u(n)$$x(n) = -a \delta(n) + \pqty{a - \dfrac{1}{a}} a^{-n} u(n)$.

备注 当 $\left|z\right|>\left|a\right|$$\vqty{z} > \vqty{a}$ 时，有

6.5

由长除法：

1. $x(n)=\{1,3,7,\cdots \}$$x(n) = \Bqty{1, 3, 7, \cdots}$.

2. $x(n)=\{1,{\displaystyle \frac{3}{2}},{\displaystyle \frac{9}{4}},\cdots \}$$x(n) = \Bqty{1, \dfrac{3}{2}, \dfrac{9}{4}, \cdots}$.

3. $x(n)=\{0,1,2,\cdots \}$$x(n) = \Bqty{0, 1, 2, \cdots}$.

备注

• 有如下思路

  1. 直接求出 z 逆变化，从而得到序列的前若干项.

  2. 幂级数展开（求导法）

  3. 幂级数展开（长除法）

  如果无需 z 逆变换的闭合表达式，则第三种方法一般最为简便.

• 这里采用前两种算法的 mathematica 代码如下：

  0. 题述函数的定义：

  xxxxxxxxxx
  3
  1
  X1[z_] := z^2/((z - 2) (z - 1))
  2
  X2[z_] := (z^2 + z + 1)/((z - 1) (z + 0.5))
  3
  X3[z_] := (z^2 - z)/(z - 1)^3
  

  1. 求逆变换的代码：

  xxxxxxxxxx
  3
  1
  Table[InverseZTransform[X1[z], z, n], {n, 0, 2}]
  2
  Table[InverseZTransform[X2[z], z, n], {n, 0, 2}]
  3
  Table[InverseZTransform[X3[z], z, n], {n, 0, 2}]
  

  2. 幂级数展开的代码：

  xxxxxxxxxx
  3
  1
  Series[X1[1/z], {z, 0, 2}]
  2
  Series[X2[1/z], {z, 0, 2}]
  3
  Series[X3[1/z], {z, 0, 2}]
  

  经检验，结果是一致的.

6.6

0. $X(z)={\displaystyle \frac{8z}{z-1}}-{\displaystyle \frac{z}{{(z-\frac{1}{2})}^2}}-{\displaystyle \frac{6z}{z-\frac{1}{2}}}$$X(z) = \dfrac{8z}{z - 1} - \dfrac{z}{\pqty{z - \cfrac{1}{2}}^2} - \dfrac{6z}{z - \cfrac{1}{2}}$.

1. $x(n)=[8-(2n+6)2^{-n}]u(n)$$x(n) = \bqty{8 - (2n + 6) 2^{-n}} u(n)$.

2. $x(n)=[(2n+6)2^{-n}-8]u(-n-1)$$x(n) = \bqty{(2n + 6) 2^{-n} - 8} u(-n-1)$.

3. $x(n)=-8u(-n-1)-(2n+6)2^{-n}u(n)$$x(n) = -8 u(-n-1) - (2n + 6) 2^{-n} u(n)$.

备注 当 $\left|z\right|<\left|a\right|$$\vqty{z} < \vqty{a}$ 时，有

6.7

1. ${\displaystyle \frac{X(z)}{z}}={\displaystyle \frac{1}{(z-1{)}^2(z+1)}}={\displaystyle \frac{1}{4(z+1)}}+{\displaystyle \frac{1}{2(z-1{)}^2}}-{\displaystyle \frac{1}{4(z-1)}}$$\dfrac{X(z)}{z} = \dfrac{1}{(z-1)^2 (z + 1)} = \dfrac{1}{4 (z + 1)} + \dfrac{1}{2 (z - 1)^2} - \dfrac{1}{4 (z - 1)}$.

   $x(n)={\displaystyle \frac{(-1{)}^n+2n-1}{4}}u(n)$$x(n) = \dfrac{(-1)^n + 2n - 1}{4} u(n)$.

2. ${\displaystyle \frac{X(z)}{z}}={\displaystyle \frac{1}{(z-6{)}^2}}$$\dfrac{X(z)}{z} = \dfrac{1}{(z - 6)^2}$.

   $x(n)=6^{n-1}n\cdot u(n)$$x(n) = 6^{n-1} n \cdot u(n)$.

3. $X(z)={\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{z^n}{n!}}={\displaystyle \sum _{n=0}^{-\mathrm{\infty }}{\displaystyle \frac{z^{-n}}{(-n)!}}}}$$X(z) = \nosum \dfrac{z^n}{n!} = \dsum_{n=0}^{-\infty} \dfrac{z^{-n}}{(-n)!}$.

   $x(n)={\displaystyle \frac{u(-n)}{(-n)!}}$$x(n) = \dfrac{u(-n)}{(-n)!}$.

4. 由 6.1 备注中的结论：

   $$\begin{array}{rl}x(n)& =[\cos \left(\omega n\right)+{\displaystyle \frac{1+\cos \omega }{\sin \omega }}\sin \left(\omega n\right)]u(n)\\ & ={\displaystyle \frac{\sin (n+1)\omega +\sin \left(n\omega \right)}{\sin \omega }}u(n)\end{array}$$

备注 ${\text{e}}^{az}\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{a^{-n}}{(-n)!}}u(-n)$$\e^{az} \zt \dfrac{a^{-n}}{(-n)!} u(-n)$.

6.8

1. 右边序列：$x(n)=[2^{-n}-2^n]u(n)$$x(n) = \bqty{2^{-n} - 2^n} u(n)$.

2. 左边序列：$x(n)=[2^n-2^{-n}]u(-n-1)$$x(n) = \bqty{2^n - 2^{-n}} u(-n-1)$.

3. 双边序列：$x(n)=2^{-n}u(n)+2^{n}u(-n-1)$$x(n) = 2^{-n} u(n) + 2^n u(-n-1)$.

附 零极点图如下图所示：

6.9

1. 思路一：求 z 逆变换.

   • ${\displaystyle \frac{X(z)}{z}}={\displaystyle \frac{z^2+z+1}{z(z-1)(z-2)}}={\displaystyle \frac{1}{2z}}-{\displaystyle \frac{3}{z-1}}+{\displaystyle \frac{7}{2(z-2)}}$$\dfrac{X(z)}{z} = \dfrac{z^2 + z + 1}{z (z - 1) (z - 2)} = \dfrac{1}{2z} - \dfrac{3}{z-1} + \dfrac{7}{2 (z - 2)}$.

   • $x(n)={\displaystyle \frac{1}{2}}\delta (n)-3u(n)+{\displaystyle \frac{7}{2}}{2}^{n}u(n)$$x(n) = \dfrac{1}{2} \delta(n) - 3 u(n) + \dfrac{7}{2} 2^n u(n)$，

   • 因此 $x(0)=1$$x(0) = 1$，$x(\mathrm{\infty })$$x(\infty)$ 不存在.

   思路二：特值定理

   • $x(0)=\underset{z\to \mathrm{\infty }}{lim}X(z)=1$$x(0) = \clim z X(z) = 1$.

   • 由于极点绝对值大于 1，$x(\mathrm{\infty })$$x(\infty)$ 不存在.

2. $x(0)=\underset{z\to \mathrm{\infty }}{lim}X(z)=1$$x(0) = \clim z X(z) = 1$.

   $x(\mathrm{\infty })=\underset{z\to 1}{lim}(z-1)X(z)=0$$x(\infty) = \liml_{z \to 1} (z-1) X(z) = 0$.

3. $x(0)=\underset{z\to \mathrm{\infty }}{lim}X(z)=0$$x(0) = \clim z X(z) = 0$.

   $x(\mathrm{\infty })=\underset{z\to 1}{lim}{\displaystyle \frac{z}{z-0.5}}=2$$x(\infty) = \liml_{z\to1} \dfrac{z}{z-0.5} = 2$.

6.10

1. $X(z)=\ln (1+{\displaystyle \frac{a}{z}})=a{z}^{-1}-{\displaystyle \frac{a^2{z}^{-2}}{2}}+{\displaystyle \frac{a^3{z}^{-3}}{3}}-\cdots$$X(z) = \ln\pqty{1 + \dfrac{a}{z}} = az\inv - \dfrac{a^2 z^{-2}}{2} + \dfrac{a^3 z^{-3}}{3} - \cdots$.

2. $x(n)=(-1{)}^{n+1}{\displaystyle \frac{a^n}{n}}u(n-1)$$x(n) = (-1)^{n+1} \dfrac{a^n}{n} u(n-1)$.

备注 类似的，有以下结论：

• ${\text{e}}^{az}\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{a^{-n}}{(-n)!}}u(-n)$$\e^{az} \zt \dfrac{a^{-n}}{(-n)!} u(-n)$.

• $\ln (1-{\displaystyle \frac{a}{z}})\stackrel{\mathcal{Z}}{\leftrightarrow }-{\displaystyle \frac{a^n}{n}}u(n-1)$$\ln\pqty{1 - \dfrac{a}{z}} \zt -\dfrac{a^n}{n} u(n-1)$.

• $\ln {\displaystyle \frac{z-b}{z-a}}\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{a^n-b^n}{n}}u(n-1)$$\ln\dfrac{z-b}{z-a} \zt \dfrac{a^n - b^n}{n} u(n-1)$.

6.11

1. 思路一：利用定义，$x(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{1-z^{-8}}{1-z^{-1}}}$$x(n) \zt \dfrac{1-z^{-8}}{1-z\inv}$.

   思路二：平移性质，$x(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z-z^{-7}}{z-1}}$$x(n) \zt \dfrac{z - z^{-7}}{z-1}$.

2. 思路一：利用定义——裂项相消，或者逐项求导

   思路二：微分性质，由 $u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }1+{\displaystyle \frac{1}{z-1}}$$u(n) \zt 1 + \dfrac{1}{z - 1}$，有

   $$\{\begin{array}{l}n\cdot u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z}{(z-1{)}^2}},\\ n^2\cdot u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z^2+z}{(z-1{)}^3}}.\end{array}\Rightarrow n(n-1)u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{2z}{(z-1{)}^3}}.$$

   思路三：由卷积定理可得. 实际上，由数学归纳法有

   $${\displaystyle \frac{z}{(z-a{)}^{k+1}}}\stackrel{\mathcal{Z}}{\leftrightarrow }{a}^{n-k}(\genfrac{}{}{0ex}{}{n}{k})u(n),$$

   取 $k=2$$k = 2$，即得 $n(n-1)u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{2z}{(z-1{)}^3}}$$n (n-1) u(n) \zt \dfrac{2z}{(z-1)^3}$.

3. 思路一：利用微分性质，$x(n)=(n+1)u(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z^2}{(z-1{)}^2}}$$x(n) = (n + 1) u(n) \zt \dfrac{z^2}{(z-1)^2}$.

   思路二：利用卷积定理，$x(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\left({\displaystyle \frac{z}{z-1}}\right)}^2$$x(n) \zt \pqty{\dfrac{z}{z-1}}^2$.

4. 见 6.10 备注，由幂级数展开得到 $x(n)\stackrel{\mathcal{Z}}{\leftrightarrow }\ln {\displaystyle \frac{z-b}{z-a}}$$x(n) \zt \ln\dfrac{z-b}{z-a}$.

5. 由 ${\displaystyle \frac{a^n}{n}}u(n-1)\stackrel{\mathcal{Z}}{\leftrightarrow }\ln {\displaystyle \frac{z}{z-a}}$$\dfrac{a^n}{n} u(n-1) \zt \ln\dfrac{z}{z-a}$ 与位移性质得 $x(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z}{a}}\ln {\displaystyle \frac{z}{z-a}}$$x(n) \zt \dfrac{z}{a} \ln\dfrac{z}{z-a}$.

6. 利用微分性质和尺度性质（见本题备注中的结论），有 $x(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z^2}{z^2+\frac{1}{4}}}$$x(n) \zt \dfrac{z^2}{z^2 + \cfrac{1}{4}}$.

7. 思路一：利用定义与差比数列求和公式.

   思路二：利用卷积定理（或直接由第二问中提到的结论）.

   思路三：利用微分性质，$x(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{-z}{(z+1{)}^2}}$$x(n) \zt \dfrac{-z}{(z+1)^2}$.

8. 思路一：利用定义，$x(n)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z^2+2z+3}{z^3}}$$x(n) \zt \dfrac{z^2 + 2z + 3}{z^3}$.

   思路二：利用时移性质.

备注

• 第二问标答分母次方应该为 3.

• 利用 6.1 备注中的结论和尺度性质（序列指数加权），$a^{n}f(n)\stackrel{\mathcal{Z}}{\leftrightarrow }F\left({\displaystyle \frac{z}{a}}\right)$$a^n f(n) \zt F\pqty{\dfrac{z}{a}}$，有

  $$\begin{array}{rl}{\beta }^n\cos \left(n{\omega }_0\right)u(n)& \stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{z(z-\beta \cos {\omega }_0)}{z^2-2\beta z\cos {\omega }_0+{\beta }^2}},\mathrm{ROC}:\left|z\right|>\left|\beta \right|,\\ {\beta }^n\sin \left(n{\omega }_0\right)u(n)& \stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{\beta z\sin {\omega }_0}{z^2-2\beta z\cos {\omega }_0+{\beta }^2}},\mathrm{ROC}:\left|z\right|>\left|\beta \right|.\end{array}$$

• 第八问标答有误.

6.12

6.13

1. $Y(z)={\displaystyle \frac{z}{z-a}}\cdot {\displaystyle \frac{b}{b-z}}={\displaystyle \frac{b}{b-a}}({\displaystyle \frac{z}{z-a}}-{\displaystyle \frac{z}{z-b}})$$Y(z) = \dfrac{z}{z-a} \cdot \dfrac{b}{b-z} = \dfrac{b}{b-a} \pqty{\dfrac{z}{z-a} - \dfrac{z}{z-b}}$.

   $y(n)={\displaystyle \frac{b}{b-a}}[a^{n}u(n)+b^{n}u(-n-1)]$$y(n) = \dfrac{b}{b-a} \bqty{a^n u(n) + b^n u(-n-1)}$.

2. $Y(z)={\displaystyle \frac{z^{-1}}{z-a}}={\displaystyle \frac{z}{a^2(z-a)}}-{\displaystyle \frac{1}{az}}-{\displaystyle \frac{1}{a^2}}$$Y(z) = \dfrac{z\inv}{z-a} = \dfrac{z}{a^2 (z - a)} - \dfrac{1}{az} - \dfrac{1}{a^2}$.

   $y(n)=a^{n-2}u(n)-a\delta (n-1)-a^{-2}\delta (n)=a^{n-2}u(n-2)$$y(n) = a^{n-2} u(n) - a \delta(n-1) - a^{-2} \delta(n) = a^{n-2} u(n-2)$.

3. $Y(z)={\displaystyle \frac{z}{z-a}}\cdot {\displaystyle \frac{z^2}{z-1}}=z+{\displaystyle \frac{a^2}{a-1}}{\displaystyle \frac{z}{z-a}}+{\displaystyle \frac{1}{1-a}}{\displaystyle \frac{z}{z-1}}$$Y(z) = \dfrac{z}{z-a} \cdot \dfrac{z^2}{z-1} = z + \dfrac{a^2}{a-1} \dfrac{z}{z-a} + \dfrac{1}{1-a} \dfrac{z}{z-1}$.

   $y(n)=\delta (n+1)+{\displaystyle \frac{a^{n+2}-1}{a-1}}u(n)={\displaystyle \frac{1-a^{n+2}}{1-a}}u(n+1)$$y(n) = \delta(n+1) + \dfrac{a^{n+2} - 1}{a-1} u(n) = \dfrac{1-a^{n+2}}{1-a} u(n+1)$.

备注 第一问应当注意收敛域；并不严谨.

6.14

3. ${\displaystyle \frac{i{e}^b(-1+e^{2i\text{w0}})z^2}{2(e^{b}z-1)(-z+e^{i\text{w0}})(-1+e^{i\text{w0}}z)}}$$\d\frac{i e^b \left(-1+e^{2 i \text{w0}}\right) z^2}{2 \left(e^b z-1\right) \left(-z+e^{i \text{w0}}\right) \left(-1+e^{i \text{w0}} z\right)}$.

备注 不想做这题.

6.15

1. $Y(z)-2.5(z^{-1}Y(z)-1)+z^{-2}Y(z)-z^{-1}+1=0$$Y(z) - 2.5 (z\inv Y(z) - 1) + z^{-2} Y(z) - z\inv + 1 = 0$,

   $Y(z)={\displaystyle \frac{z^{-1}-3.5}{1-2.5z^{-1}+z^{-2}}}={\displaystyle \frac{z(1-3.5z)}{(z-0.5)(z-2)}}={\displaystyle \frac{0.5z}{z-0.5}}-{\displaystyle \frac{4z}{z-2}}$$Y(z) = \dfrac{z\inv - 3.5}{1 - 2.5 z\inv + z^{-2}} = \dfrac{z (1 - 3.5 z)}{\pqty{z - 0.5} \pqty{z - 2}} = \dfrac{0.5z}{z-0.5} - \dfrac{4z}{z-2}$.

   $y(n)={0.5}^{n+1}-2^{n+2}$$y(n) = 0.5^{n+1} - 2^{n+2}$.

2. $Y(z)-z^{-1}Y(z)-2(z^{-2}Y(z)+3)=0$$Y(z) - z\inv Y(z) - 2 (z^{-2} Y(z) + 3) = 0$,

   $Y(z)={\displaystyle \frac{6}{1-z^{-1}-2z^{-2}}}={\displaystyle \frac{6z^2}{(z-2)(z+1)}}={\displaystyle \frac{4z}{z-2}}+{\displaystyle \frac{2z}{z+1}}$$Y(z) = \dfrac{6}{1 - z\inv - 2 z^{-2}} = \dfrac{6z ^2}{(z - 2) (z + 1)} = \dfrac{4z}{z - 2} + \dfrac{2z}{z + 1}$,

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

3. $Y(z)+0.1(z^{-1}Y(z)+4)-0.02(z^{-2}Y(z)+4z^{-1}+6)={\displaystyle \frac{10z}{z-1}}$$Y(z) + 0.1 (z\inv Y(z) + 4) -0.02 (z^{-2} Y(z) + 4 z\inv + 6) = \dfrac{10 z}{z - 1}$,

   ${\displaystyle Y(z)=z\frac{9.72z^2+0.36z-0.08}{(z-1)(z^2+0.1z+0.02)}}$$\d Y(z) = z \frac{9.72 z^2+0.36 z-0.08}{(z-1) \left(z^2+0.1 z+0.02\right)}$,

4. $Y(z)-0.9z^{-1}Y(z)={\displaystyle \frac{0.05z}{z-1}}$$Y(z) - 0.9 z\inv Y(z) = \dfrac{0.05 z}{z - 1}$,

   $Y(z)=z{\displaystyle \frac{0.05z}{(z-1)(z-0.9)}}={\displaystyle \frac{0.5z}{z-1}}-{\displaystyle \frac{0.45z}{z-0.9}}$$Y(z) = z \dfrac{0.05 z}{(z - 1) (z - 0.9)} = \dfrac{0.5z}{z - 1} - \dfrac{0.45 z}{z - 0.9}$,

   $y(n)=0.5-0.45\cdot {0.9}^n$$y(n) = 0.5 - 0.45 \cdot 0.9^n$.

5. $Y(z)+5z^{-1}Y(z)={\displaystyle \frac{z}{(z-1{)}^2}}$$Y(z) + 5 z\inv Y(z) = \dfrac{z}{(z - 1)^2}$,

   $Y(z)=z{\displaystyle \frac{z}{(z-1{)}^2(z+5)}}=-{\displaystyle \frac{5z}{36(z+5)}}+{\displaystyle \frac{z}{6(z-1{)}^2}}+{\displaystyle \frac{5z}{36(z-1)}}$$Y(z) = z \dfrac{z}{(z - 1)^2 (z + 5)} = -\dfrac{5z}{36 (z + 5)} + \dfrac{z}{6 (z - 1)^2} + \dfrac{5z}{36 (z - 1)}$,

   $y(n)=[{\displaystyle \frac{n}{6}}+{\displaystyle \frac{5}{36}}-{\displaystyle \frac{5}{36}}(-5{)}^n]u(n)$$y(n) = \bqty{\dfrac{n}{6} + \dfrac{5}{36} - \dfrac{5}{36} (-5)^n} u(n)$.

6. $(z^{2}Y(z)-z^2-z)-(zY(z)-z)-2Y(z)={\displaystyle \frac{z}{z-1}}$$(z^2 Y(z) - z^2 - z) - (z Y(z) - z) - 2 Y(z) = \dfrac{z}{z-1}$,

   $Y(z)={\displaystyle \frac{z(z^2-z+1)}{(z-1)(z-2)(z+1)}}={\displaystyle \frac{z}{z-2}}-{\displaystyle \frac{z}{2(z-1)}}+{\displaystyle \frac{z}{2(z+1)}}$$Y(z) = \dfrac{z (z^2 - z + 1)}{(z - 1) (z - 2) (z + 1)} = \dfrac{z}{z - 2} - \dfrac{z}{2 (z - 1)} + \dfrac{z}{2 (z + 1)}$,

   $y(n)=[2^n-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}(-1{)}^n]u(n)$$y(n) = \bqty{2^n - \dfrac{1}{2} + \dfrac{1}{2} (-1)^n} u(n)$.

备注

• 这里解出的 $y(n)$$y(n)$ 无需加上 $u(n)$$u(n)$，因为还需考虑负数的情况，并且上述答案对负数的情况也是成立的（非因果信号）.

• 第三问答案似乎有误，

  xxxxxxxxxx
  7
  1
  Apart[((10 z)/(z - 1) - 0.28 + 0.08/z)/((1 + 0.1 z^-1 + 0.02 z^-2) z), z]
  2
  
  3
  RSolveValue[{
  4
          y[n] + 0.1 y[n - 1] - 0.02 y[n - 2] == 10 UnitStep[n], 
  5
          y[-1] == 4, y[-2] == 6
  6
      }, y[n], n
  7
  ] // Simplify
  

6.16

1. $8z^2-2z-3=(2z+1)(4z-3)$$8z^2 - 2z - 3 = (2z + 1) (4z - 3)$，故为稳定.

2. $2z^2+5z+2=(2z+1)(z+2)$$2z^2 + 5z + 2 = (2z + 1) (z + 2)$，故为不稳定.

3. $2z^2+z-1=(2z-1)(z+1)$$2z^2 + z - 1 = (2z - 1) (z + 1)$，故为临界稳定.

4. $z^2-z+1=(z-{\displaystyle \frac{1-\mathrm{j}\sqrt{3}}{2}})(z-{\displaystyle \frac{1+\mathrm{j}\sqrt{3}}{2}})$$z^2 - z + 1 = \pqty{z - \dfrac{1 - \j \sqrt 3}{2}} \pqty{z - \dfrac{1 + \j \sqrt 3}{2}}$，故为临界稳定.

6.17

6.18

1. $Y(z)={\displaystyle \frac{z}{z+1}}$$Y(z) = \dfrac{z}{z + 1}$,

   $h(n)=(-1{)}^{n}u(n)$$h(n) = (-1)^n u(n)$.

   系统临界稳定.

2. 思路一（直接求卷积）：$y(n)=x(n)\ast h(n)=5[1+(-1{)}^n]u(n)$$y(n) = x(n) * h(n) = 5 \bqty{1 + (-1)^n} u(n)$.

   思路二（用卷积定理）

   $Y(n)={\displaystyle \frac{10z}{z-1}}{\displaystyle \frac{z}{z+1}}={\displaystyle \frac{5z}{z-1}}+{\displaystyle \frac{5z}{z+1}}$$Y(n) = \dfrac{10z}{z - 1} \dfrac{z}{z + 1} = \dfrac{5z}{z - 1} + \dfrac{5z}{z + 1}$,

   $y(n)=5[1+(-1{)}^n]u(n)$$y(n) = 5 \bqty{1 + (-1)^n} u(n)$.

6.19

1. $y(n)-y(n-1)+{\displaystyle \frac{1}{2}}y(n-2)=x(n-1)$$y(n) - y(n-1) + \dfrac{1}{2} y(n-2) = x(n-1)$.

2. $H(z)={\displaystyle \frac{z}{z^2-z+0.5}}$$H(z) = \dfrac{z}{z^2 - z + 0.5}$.

   

3. 由 ${\alpha }^n\sin \left(\omega n\right)\stackrel{\mathcal{Z}}{\leftrightarrow }{\displaystyle \frac{\alpha z\sin \omega }{z^2-2\alpha z\cos \omega +{\alpha }^2}}$$\alpha^n \sin(\omega n) \zt \dfrac{\alpha z \sin\omega}{z^2 - 2 \alpha z \cos\omega + \alpha^2}$ 得 $y(n)=2^{1-\frac{n}{2}}\sin {\displaystyle \frac{\pi }{4}}nu(n)$$y(n) = 2^{1 - \tfrac{n}{2}} \sin\dfrac{\pi}{4} n u(n)$.

6.20

1. $H(z)={\displaystyle \frac{1}{3-6z^{-1}}}={\displaystyle \frac{z}{3(z-2)}}$$H(z) = \dfrac{1}{3 - 6 z\inv} = \dfrac{z}{3 (z - 2)}$,

   $h(n)={\displaystyle \frac{2^n}{3}}u(n)$$h(n) = \dfrac{2^n}{3} u(n)$.

2. $H(z)=1-5z^{-1}+8z^{-3}$$H(z) = 1 - 5 z\inv + 8 z^{-3}$,

   $h(n)=\delta (n)-5\delta (n-1)+8\delta (n-3)$$h(n) = \delta(n) - 5 \delta(n - 1) + 8 \delta(n - 3)$.

3. $H(z)={\displaystyle \frac{1}{1-z^{-2}/4}}={\displaystyle \frac{z^2}{(z+0.5)(z-0.5)}}={\displaystyle \frac{0.5z}{z-0.5}}+{\displaystyle \frac{0.5z}{z+0.5}}$$H(z) = \dfrac{1}{1 - z^{-2}/4} = \dfrac{z^2}{(z + 0.5) (z - 0.5)} = \dfrac{0.5z}{z - 0.5} + \dfrac{0.5z}{z + 0.5}$,

   $h(n)=0.5[{0.5}^n+(-0.5{)}^n]u(n)$$h(n) = 0.5 \bqty{0.5^n + (-0.5)^n} u(n)$.

4. $H(z)={\displaystyle \frac{1}{1-3z^{-1}+3z^{-2}-z^{-3}}}={\displaystyle \frac{z^3}{(z-1{)}^3}}={\displaystyle \frac{z}{z-1}}+{\displaystyle \frac{2z}{(z-1{)}^2}}+{\displaystyle \frac{z}{(z-1{)}^3}}$$H(z) = \dfrac{1}{1 - 3 z^{-1} + 3 z^{-2} - z^{-3}} = \dfrac{z^3}{(z-1)^3} = \dfrac{z}{z-1} + \dfrac{2z}{(z-1)^2} + \dfrac{z}{(z-1)^3}$,

   $h(n)=[1+2n+{\displaystyle \frac{n(n-1)}{2}}]u(n)={\displaystyle \frac{(n+1)(n+2)}{2}}u(n)$$h(n) = \bqty{1 + 2n + \dfrac{n (n-1)}{2}} u(n) = \dfrac{(n + 1) (n + 2)}{2} u(n)$.

5. $H(z)={\displaystyle \frac{1-3z^{-2}}{1-5z^{-1}+6z^{-2}}}=1+{\displaystyle \frac{5z-9}{(z-2)(z-3)}}=1-{\displaystyle \frac{z}{z-2}}+{\displaystyle \frac{6z}{z-3}}$$H(z) = \dfrac{1 - 3 z^{-2}}{1 - 5 z^{-1} + 6 z^{-2}} = 1 + \dfrac{5z - 9}{(z - 2) (z - 3)} = 1 - \dfrac{z}{z - 2} + \dfrac{6z}{z - 3}$.

   $h(n)=\delta (n)-2^{n}u(n)+6\cdot 3^{n}u(n)$$h(n) = \delta(n) - 2^n u(n) + 6 \cdot 3^n u(n)$.

备注

• 考虑零状态响应；系统框图略.

• 第四问实际上有 ${\left({\displaystyle \frac{z}{z-a}}\right)}^k\stackrel{\mathcal{Z}}{\leftrightarrow }{a}^n{\displaystyle \frac{(n+k{)}_{(k)}}{n!}}u(n)$$\pqty{\dfrac{z}{z - a}}^k \zt a^n \dfrac{(n+k)_{(k)}}{n!} u(n)$.

  证明 已知 $k=1$$k = 1$ 时成立，由数学归纳法，若 $k$$k$ 时成立，则

  $$\begin{array}{rl}{\left({\displaystyle \frac{z}{z-a}}\right)}^{k+1}& \stackrel{\mathcal{Z}}{\leftrightarrow }\sum _{m=0}^n{a}^n{\displaystyle \frac{(m+k{)}_{(k)}}{m!}}u(n)\\ & =...\end{array}$$

  （等以后补充吧.）

6.21

0. $H(z)={\displaystyle \frac{z}{z-0.5}}-{\displaystyle \frac{z}{z-10}}$$H(z) = \dfrac{z}{z - 0.5} - \dfrac{z}{z - 10}$,

1. 当 $10<\left|z\right|\le \mathrm{\infty }$$10 < \vqty{z} \le \infty$ 时，

   $h(n)=({0.5}^n-{10}^n)u(n)$$h(n) = (0.5^n - 10^n) u(n)$.

   因果、不稳定.

2. 当 $0.5<\left|z\right|<10$$0.5 < \vqty{z} < 10$ 时，

   $h(n)={0.5}^{n}u(n)+{10}^{n}u(-n-1)$$h(n) = 0.5^n u(n) + 10^n u(-n-1)$.

   非因果、稳定.

6.22

0. $y(n)-ay(n-1)=x(n)$$y(n) - a y(n-1) = x(n)$,

   $H(z)={\displaystyle \frac{z}{z-a}}$$H(z) = \dfrac{z}{z - a}$,

1. $Y_1(z)={\displaystyle \frac{z^2}{(z-a)(z-1)}}={\displaystyle \frac{a}{a-1}}{\displaystyle \frac{z}{z-a}}-{\displaystyle \frac{1}{a-1}}{\displaystyle \frac{z}{z-1}}$$Y_1(z) = \dfrac{z^2}{(z - a) (z - 1)} = \dfrac{a}{a - 1} \dfrac{z}{z - a} - \dfrac{1}{a - 1} \dfrac{z}{z - 1}$,

   $y_1(n)=\underset{\text{瞬}\text{态}\text{响}\text{应}}{\underbrace{{\displaystyle \frac{a^{n+1}}{a-1}}u(n)}}+\underset{\text{稳}\text{态}\text{响}\text{应}}{\underbrace{{\displaystyle \frac{-1}{a-1}}u(n)}}$$y_1(n) = \underbrace{ \dfrac{a^{n+1}}{a-1} u(n) }_{瞬态响应} + \underbrace{ \dfrac{-1}{a-1} u(n) }_{稳态响应}$.

2. $Y_2(z)={\displaystyle \frac{z^2}{(z-a)(z-{\text{e}}^{\mathrm{j}\omega })}}={\displaystyle \frac{{\text{e}}^{\mathrm{j}\omega }}{{\text{e}}^{\mathrm{j}\omega }-a}}{\displaystyle \frac{z}{z-{\text{e}}^{\mathrm{j}\omega }}}+{\displaystyle \frac{a}{a-{\text{e}}^{\mathrm{j}\omega }}}{\displaystyle \frac{z}{z-a}}$$Y_2(z) = \dfrac{z^2}{(z-a) (z - \e^{\j\omega})} = \dfrac{\e^{\j\omega}}{\e^{\j\omega} - a} \dfrac{z}{z - \e^{\j\omega}} + \dfrac{a}{a - \e^{\j\omega}} \dfrac{z}{z - a}$,

   $y_2(n)=\underset{\text{稳}\text{态}\text{响}\text{应}}{\underbrace{{\displaystyle \frac{{\text{e}}^{\mathrm{j}\omega }}{a-{\text{e}}^{\mathrm{j}\omega }}}{\text{e}}^{\mathrm{j}n\omega }u(n)}}+\underset{\text{瞬}\text{态}\text{相}\text{应}}{\underbrace{{\displaystyle \frac{-a}{a-{\text{e}}^{\mathrm{j}\omega }}}{a}^{n}u(n)}}$$y_2(n) = \underbrace{ \dfrac{\e^{\j\omega}}{a - \e^{\j\omega}} \e^{\j n\omega} u(n) }_{稳态响应} + \underbrace{ \dfrac{-a}{a - \e^{\j\omega}} a^n u(n) }_{瞬态相应}$.

6.23

1. $H_1(z)={\displaystyle \frac{z}{z-1}}$$H_1(z) = \dfrac{z}{z - 1}$,

2. $H(z)=(H_1+H_2)H_3={\displaystyle \frac{2z}{(z+1)(z-1)}}$$H(z) = (H_1 + H_2) H_3 = \dfrac{2z}{(z + 1) (z - 1)}$,

3. $X(z)=1+{\displaystyle \frac{1}{z}}$$X(z) = 1 + \dfrac{1}{z}$,

4. $Y_{\text{zs}}(z)={\displaystyle \frac{2}{z-1}}$$Y_\text{zs}(z) = \dfrac{2}{z - 1}$,

5. $y_{\text{zs}}(n)=2u(n-1)$$y_\text{zs}(n) = 2 u(n-1)$.

6.24 ?

1. $X_1(z)={\displaystyle \frac{z}{z-0.5}}$$X_1(z) = \dfrac{z}{z - 0.5}$,

   $Y_1(z)=1+{\displaystyle \frac{az}{z-0.25}}$$Y_1(z) = 1 + \dfrac{az}{z - 0.25}$,

   $H(z)=1-{\displaystyle \frac{0.5}{z}}+a{\displaystyle \frac{z-0.5}{z-0.25}}$$H(z) = 1 - \dfrac{0.5}{z} + a \dfrac{z - 0.5}{z - 0.25}$,

   $X_2(z)={\displaystyle \frac{z}{z+2}}$$X_2(z) = \dfrac{z}{z + 2}$,

   $Y_2(z)={\displaystyle \frac{z-0.5}{z+2}}+a{\displaystyle \frac{z-0.5}{z-0.25}}{\displaystyle \frac{z}{z+2}}$$Y_2(z) = \dfrac{z - 0.5}{z + 2} + a \dfrac{z - 0.5}{z - 0.25} \dfrac{z}{z + 2}$,

6.25