第四章

4.1

1. $1-{\text{e}}^{-\alpha t}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{1}{s}}-{\displaystyle \frac{1}{s+\alpha }}={\displaystyle \frac{\alpha }{s(s+\alpha )}}$$1 - \e^{-\alpha t} \lt \dfrac{1}{s} - \dfrac{1}{s + \alpha} = \dfrac{\alpha}{s (s + \alpha)}$.

2. $t{\text{e}}^{-2t}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{1}{(s+2{)}^2}}$$t \e^{-2t} \lt \dfrac{1}{(s + 2)^2}$.

3. $[1-\cos \left(\alpha t\right)]{\text{e}}^{-\beta t}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{1}{s+\beta }}-{\displaystyle \frac{s+\beta }{(s+\beta {)}^2+{\alpha }^2}}$$\bqty{1 - \cos(\alpha t)} \e^{-\beta t} \lt \dfrac{1}{s + \beta} - \dfrac{s + \beta}{(s + \beta)^2 + \alpha^2}$.

4. $2\delta (t)-3{\text{e}}^{-7t}\stackrel{\mathcal{L}}{\leftrightarrow }2-{\displaystyle \frac{3}{s+7}}$$2 \delta(t) - 3\e^{-7t} \lt 2 - \dfrac{3}{s + 7}$.

5. ${\text{e}}^{-\alpha t}\sinh \left(\beta t\right)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{\beta }{(s+\alpha {)}^2-{\beta }^2}}$$\e^{-\alpha t} \sinh(\beta t) \lt \dfrac{\beta}{(s + \alpha)^2 - \beta^2}$.

6. ${\displaystyle \frac{{\text{e}}^{-\alpha t}-{\text{e}}^{-\beta t}}{\beta -\alpha }}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{1}{(s+\alpha )(s+\beta )}}$$\dfrac{\e^{-\alpha t} - \e^{-\beta t}}{\beta - \alpha} \lt \dfrac{1}{(s + \alpha) (s + \beta)}$.

7. ${\text{e}}^{-(t+a)}\cos \left(\omega t\right)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{(s+1){\text{e}}^{-a}}{(s+1{)}^2+{\omega }^2}}$$\e^{-(t + a)} \cos(\omega t) \lt \dfrac{(s + 1) \e^{-a}}{(s + 1)^2 + \omega^2}$.

8. 为了利用时移性质，先将函数拆开：

   $t{\text{e}}^{-(t-2)}u(t-1)=((t-1){\text{e}}^{-(t-1)}+{\text{e}}^{-(t-1)})\text{e}\cdot u(t-1)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{s+2}{(s+1{)}^2}}{\text{e}}^{-(s-1)}$$t \e^{-(t - 2)} u(t - 1) = \pqty{(t-1) \e^{-(t-1)} + \e^{-(t-1)}} \e \cdot u(t-1) \lt \dfrac{s + 2}{(s + 1)^2} \e^{-(s - 1)}$.

9. ${\text{e}}^{-at}f\left({\displaystyle \frac{t}{a}}\right)\stackrel{\mathcal{L}}{\leftrightarrow }aF(as+a^2)$$\e^{-a t} f\pqty{\dfrac{t}{a}} \lt a F(as + a^2)$.

10. 注：频域卷积难以求解，可利用频域微分性质.

    $t^2\cos \left(2t\right)\stackrel{\mathcal{L}}{\leftrightarrow }(-1{)}^2{\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{s}^2}{\displaystyle \frac{s}{s^2+4}}={\displaystyle \frac{2s^3-24s}{(s^2+4{)}^3}}}$$t^2 \cos(2t) \lt (-1)^2 \ddv[2]{s} \dfrac{s}{s^2 + 4} = \dfrac{2s^3 - 24 s}{(s^2 + 4)^3}$.

11. 法一：频域积分

    ${\displaystyle \frac{1-{\text{e}}^{-\alpha t}}{t}}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle {\int }_s^{+\mathrm{\infty }}({\displaystyle \frac{1}{s}}-{\displaystyle \frac{1}{s+\alpha }})\mathrm{d}s={\ln {\displaystyle \frac{s}{s+\alpha }}\phantom{\int }|}_s^{+\mathrm{\infty }}=\ln (1+{\displaystyle \frac{a}{s}})}$$\dfrac{1 - \e^{-\alpha t}}{t} \lt \dint_s^{+\infty} \pqty{\dfrac{1}{s} - \dfrac{1}{s + \alpha}} \ds = \eval{\ln\dfrac{s}{s + \alpha}}_s^{+\infty} = \ln \pqty{1 + \dfrac{a}{s}}$.

    法二：幂级数展开

    $$\begin{array}{rl}{\displaystyle \frac{1-{\text{e}}^{-\alpha t}}{t}}& =\alpha -{\displaystyle \frac{{\alpha }^{2}t}{2}}+\cdots +(-1{)}^{n+1}{\displaystyle \frac{{\alpha }^n{t}^{n-1}}{n!}}\\ & \stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{\alpha }{s}}-{\displaystyle \frac{{\alpha }^2}{2s^2}}+\cdots +(-1{)}^{n+1}{\displaystyle \frac{{\alpha }^n}{n{s}^n}}\\ & =\ln (1+{\displaystyle \frac{a}{s}}).\end{array}$$

    备注：使用幂级数展开法时，需要验证展开系数是否满足条件.

12. 与第 11 题相同，

    ${\displaystyle \frac{{\text{e}}^{-3t}-{\text{e}}^{-5t}}{t}}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle {\int }_s^{+\mathrm{\infty }}({\displaystyle \frac{1}{s+3}}-{\displaystyle \frac{1}{s+5}})\mathrm{d}s=\ln {\displaystyle \frac{s+5}{s+3}}}$$\dfrac{\e^{-3t} - \e^{-5t}}{t} \lt \dint_s^{+\infty} \pqty{\dfrac{1}{s + 3} - \dfrac{1}{s + 5}} \ds = \ln \dfrac{s + 5}{s + 3}$.

13. ${\displaystyle \frac{4}{2s+3}}\stackrel{\mathcal{L}}{\leftrightarrow }2{\text{e}}^{-\frac{3}{2}t}$$\dfrac{4}{2s + 3} \lt 2 \e^{-\tfrac{3}{2} t}$.

14. ${\displaystyle \frac{4}{s(2s+3)}}={\displaystyle \frac{4}{3s}}-{\displaystyle \frac{4}{3}}{\displaystyle \frac{1}{s+\frac{3}{2}}}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{4}{3}}(1-{\text{e}}^{-\frac{3}{2}t})$$\dfrac{4}{s (2s + 3)} = \dfrac{4}{3s} - \dfrac{4}{3} \dfrac{1}{s + \cfrac{3}{2}} \lt \dfrac{4}{3} \pqty{1 - \e^{-\tfrac{3}{2}t}}$.

15. ${\displaystyle \frac{1}{s(s^2+5)}}={\displaystyle \frac{1}{5s}}-{\displaystyle \frac{1}{5}}{\displaystyle \frac{s}{s^2+5}}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{1-\cos \left(\sqrt{5}t\right)}{5}}$$\dfrac{1}{s (s^2 + 5)} = \dfrac{1}{5s} - \dfrac{1}{5} \dfrac{s}{s^2 + 5} \lt \dfrac{1 - \cos(\sqrt 5 t)}{5}$.

16. ${\displaystyle \frac{3s}{(s+4)(s+2)}}={\displaystyle \frac{6}{s+4}}-{\displaystyle \frac{3}{s+2}}\stackrel{\mathcal{L}}{\leftrightarrow }6{\text{e}}^{-4t}-3{\text{e}}^{-2t}$$\dfrac{3s}{(s + 4) (s + 2)} = \dfrac{6}{s + 4} - \dfrac{3}{s + 2} \lt 6 \e^{-4t} - 3 \e^{-2t}$.

17. ${\displaystyle \frac{1}{s^2+1}}+1\stackrel{\mathcal{L}}{\leftrightarrow }\sin t+\delta (t)$$\dfrac{1}{s^2 + 1} + 1 \lt \sin t + \delta(t)$.

18. ${\displaystyle \frac{1-RCs}{s(1+RCs)}}={\displaystyle \frac{1}{s}}-{\displaystyle \frac{2}{s+\frac{1}{RC}}}\stackrel{\mathcal{L}}{\leftrightarrow }1-2{\text{e}}^{-\frac{t}{RC}}$$\dfrac{1 - RC s}{s (1 + RC s)} = \dfrac{1}{s} - \dfrac{2}{s + \cfrac{1}{RC}} \lt 1 - 2 \e^{-\tfrac{t}{RC}}$.

19. ${\displaystyle \frac{4s+5}{s^2+5s+6}}={\displaystyle \frac{7}{s+3}}-{\displaystyle \frac{3}{s+2}}\stackrel{\mathcal{L}}{\leftrightarrow }7{\text{e}}^{-3t}-3{\text{e}}^{-2t}$$\dfrac{4s + 5}{s^2 + 5s + 6} = \dfrac{7}{s + 3} - \dfrac{3}{s + 2} \lt 7 \e^{-3t} - 3 \e^{-2t}$.

20. 思路一：频域微分性质

    由 $\cos \left(\sqrt{3}t\right)u(t)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{s}{s^2+3}}$$\cos(\sqrt 3 t) u(t) \lt \dfrac{s}{s^2 + 3}$，有 $t\cos \left(\sqrt{3}t\right)u(t)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{s^2-3}{(s^2+3{)}^2}}$$t \cos(\sqrt 3 t) u(t) \lt \dfrac{s^2 - 3}{(s^2 + 3)^2}$，于是

    $$\begin{array}{rl}{\displaystyle \frac{1}{(s^2+3{)}^2}}& ={\displaystyle \frac{1}{6}}({\displaystyle \frac{1}{s^2+3}}-{\displaystyle \frac{s^2-3}{(s^2+3{)}^2}})\\ & \stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{\sin \left(\sqrt{3}t\right)}{6\sqrt{3}}}-{\displaystyle \frac{t\cos \left(\sqrt{3}t\right)}{6}}.\end{array}$$

    思路二：时域卷积性质

    $$\begin{array}{rl}{\displaystyle \frac{1}{(s^2+3{)}^2}}& \stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{1}{3}}\sin \left(\sqrt{3}t\right)u(t)\ast \sin \left(\sqrt{3}t\right)u(t)\\ & ={\displaystyle \frac{\sin \left(\sqrt{3}t\right)-\sqrt{3}t\cos \left(\sqrt{3}t\right)}{6\sqrt{3}}}.\end{array}$$

21. 利用极限法与特值代入法，计算量不大：

    $$\begin{array}{rl}& {\displaystyle \frac{s}{(s+a)[(s+\alpha {)}^2+{\beta }^2]}}\\ & ={\displaystyle \frac{-a}{(a-\alpha {)}^2+{\beta }^2}}({\displaystyle \frac{1}{s+a}}-{\displaystyle \frac{s+\alpha }{(s+\alpha {)}^2+{\beta }^2}}-{\displaystyle \frac{{\alpha }^2+{\beta }^2-a\alpha }{a\beta }}{\displaystyle \frac{\beta }{(s+\alpha {)}^2+{\beta }^2}})\\ & \stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{-a}{(a-\alpha {)}^2+{\beta }^2}}({\text{e}}^{-at}-(\cos \left(\beta t\right)+{\displaystyle \frac{{\alpha }^2+{\beta }^2-a\alpha }{a\beta }}\sin \left(\beta t\right)){\text{e}}^{-\alpha t}).\end{array}$$

22. $\ln {\displaystyle \frac{s}{s+9}}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{{\text{e}}^{-9t}-1}{t}}$$\ln \dfrac{s}{s + 9} \lt \dfrac{\e^{-9 t} - 1}{t}$.

备注

• 具体思路参考拉普拉斯变换的性质、拉普拉斯逆变换的求解的笔记.

• 更严谨而完整的，应该给出拉氏变换的收敛域. 注意有些结果看似不同，但在收敛域交定义域内确是相同的. 本题没有这样的例子.

• 第 9 题应说明 $a>0$$a > 0$.

• 第 11 题不能对两项分别求拉普拉斯变换，因为拉氏变换并不存在. 实际上，有

  $$\begin{array}{rl}{\displaystyle \frac{u(t)}{(t+a{)}^n}}& \stackrel{\mathcal{L}}{\leftrightarrow }\{\begin{array}{ll}{s}^{n-1}{\text{e}}^{as}{\displaystyle {\int }_{-\mathrm{\infty }}^{-as}{\displaystyle \frac{{\text{e}}^x\mathrm{d}x}{(-x{)}^n}},}& a>0,n\in \mathbb{R},\\ s^{n-1}{\text{e}}^{as}\mathrm{\Gamma }(1-n,as),& a\in \mathbb{R},n<1,\\ \text{不}\text{存}\text{在},& a\le 0,n\ge 1.\end{array}\end{array}$$

  其中上式不能用幂级数展开法求解，因为不满足条件.（11 题则满足条件）

  右式的积分为柯西主值积分. 由上，特殊的，有

  • 当 $n\in \mathbb{N}$$n \in \N$ 时，有 $t^{n}u(t)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{n!}{s^{n+1}}}$$t^n u(t) \lt \dfrac{n!}{s^{n+1}}$.

  • 当 $a>0$$a > 0$ 时，有 ${\displaystyle \frac{u(t)}{t+a}}\stackrel{\mathcal{L}}{\leftrightarrow }-{\text{e}}^{as}\mathrm{Ei}(-as)$$\dfrac{u(t)}{t+a} \lt -\e^{as} \operatorname{Ei}(-as)$.

• 诸如第 20 题的结论，可直接参考笔记. 了解思路后，使用时查表即可. 这里也直接列出吧：

  象函数 $F(s)$$F(s)$	原函数 $f(t)$$f(t)$	说明
  ${\displaystyle \frac{1}{(s^2+{\omega }^2{)}^2}}$$\dfrac{1}{(s^2 + \omega^2)^2}$	${\displaystyle \frac{\sin \left(\omega t\right)-\omega t\cos \left(\omega t\right)}{2{\omega }^3}}$$\dfrac{\sin(\omega t) - \omega t \cos(\omega t)}{2\omega^3}$	$\omega \in \mathbb{C}$$\omega \in \C$.
  ${\displaystyle \frac{s}{(s^2+{\omega }^2{)}^2}}$$\dfrac{s}{(s^2 + \omega^2)^2}$	${\displaystyle \frac{t\sin \left(\omega t\right)}{2\omega }}$$\dfrac{t\sin(\omega t)}{2\omega}$	$\omega \in \mathbb{C}$$\omega \in \C$.
  ${\displaystyle \frac{s^2}{(s^2+{\omega }^2{)}^2}}$$\dfrac{s^2}{(s^2 + \omega^2)^2}$	${\displaystyle \frac{\sin \left(\omega t\right)+\omega t\cos \left(\omega t\right)}{2\omega }}$$\dfrac{\sin(\omega t) + \omega t \cos(\omega t)}{2\omega}$	$\omega \in \mathbb{C}$$\omega \in \C$.
  ${\displaystyle \frac{s^3}{(s^2+{\omega }^2{)}^2}}$$\dfrac{s^3}{(s^2 + \omega^2)^2}$	$\cos \left(\omega t\right)-{\displaystyle \frac{\omega t}{2}}\sin \left(\omega t\right)$$\cos(\omega t) - \dfrac{\omega t}{2} \sin(\omega t)$	$\omega \in \mathbb{C}$$\omega \in \C$.
  ${\displaystyle \frac{1}{(s^2+{\omega }^2{)}^3}}$$\dfrac{1}{(s^2 + \omega^2)^3}$	${\displaystyle \frac{(3-t^2{\omega }^2)\sin \left(t\omega \right)-3t\omega \cos \left(t\omega \right)}{8{\omega }^5}}$$\d \frac{\left(3-t^2 \omega ^2\right) \sin (t \omega )-3 t \omega \cos (t \omega )}{8 \omega ^5}$	$\omega \in \mathbb{C}$$\omega \in \C$.
  ${\displaystyle \frac{s}{(s^2+{\omega }^2{)}^3}}$$\dfrac{s}{(s^2 + \omega^2)^3}$	${\displaystyle \frac{t(\sin \left(t\omega \right)-t\omega \cos \left(t\omega \right))}{8{\omega }^3}}$$\d\frac{t (\sin (t \omega )-t \omega \cos (t \omega ))}{8 \omega ^3}$	$\omega \in \mathbb{C}$$\omega \in \C$.

• 第 20 题中利用到的卷积的结论，也罗列如下：

  • 角频率不同

  $$\begin{array}{rl}\sin \left(at\right)u(t)\ast \sin \left(bt\right)u(t)& ={\displaystyle \frac{a\sin \left(bt\right)-b\sin \left(at\right)}{a^2-b^2}}u(t),\\ \sin \left(at\right)u(t)\ast \cos \left(bt\right)u(t)& ={\displaystyle \frac{a\cos \left(bt\right)-a\cos \left(at\right)}{a^2-b^2}}u(t),\\ \cos \left(at\right)u(t)\ast \cos \left(bt\right)u(t)& ={\displaystyle \frac{a\sin \left(at\right)-b\sin \left(bt\right)}{a^2-b^2}}u(t).\end{array}$$

  • 角频率相同

  $$\begin{array}{rl}\sin \left(\omega t\right)u(t)\ast \sin \left(\omega t\right)u(t)& ={\displaystyle \frac{\sin \left(\omega t\right)-\omega t\cos \left(\omega t\right)}{2\omega }}u(t),\\ \sin \left(\omega t\right)u(t)\ast \cos \left(\omega t\right)u(t)& ={\displaystyle \frac{t\sin \left(\omega t\right)}{2}}u(t),\\ \cos \left(\omega t\right)u(t)\ast \cos \left(\omega t\right)u(t)& ={\displaystyle \frac{\sin \left(\omega t\right)+\omega t\cos \left(\omega t\right)}{2\omega }}u(t).\end{array}$$

• 第 21 题的答案中，右中括号的位置错了.

• 第 22 题，一般的，有 ${\displaystyle \frac{{\text{e}}^{-\alpha t}-{\text{e}}^{-\beta t}}{t}}\stackrel{\mathcal{L}}{\leftrightarrow }\ln {\displaystyle \frac{s+\beta }{s+\alpha }}$$\dfrac{\e^{-\alpha t} - \e^{-\beta t}}{t} \lt \ln \dfrac{s + \beta}{s + \alpha}$，其中 $\alpha ,\beta \in \mathbb{C}$$\alpha, \beta \in \C$.

4.2

1. ${\text{e}}^{-t}u(t-2)={\text{e}}^{-2}\cdot {\text{e}}^{-(t-2)}u(t-2)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{{\text{e}}^{-2(s+1)}}{s+1}}$$\e^{-t} u(t - 2) = \e^{-2} \cdot \e^{-(t - 2)} u(t - 2) \lt \dfrac{\e^{-2 (s + 1)}}{s + 1}$.

2. ${\text{e}}^{-(t-2)}u(t-2)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{{\text{e}}^{-2s}}{s+1}}$$\e^{-(t - 2)} u(t - 2) \lt \dfrac{\e^{-2s}}{s + 1}$.

3. ${\text{e}}^{-(t-2)}u(t)={\text{e}}^2\cdot {\text{e}}^{-t}u(t)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{{\text{e}}^2}{s+1}}$$\e^{-(t - 2)} u(t) = \e^2 \cdot \e^{-t} u(t) \lt \dfrac{\e^2}{s + 1}$.

4. 先考虑

   $$\begin{array}{rl}\sin (2t+2)& =\sin 2t\cos 2+\cos 2t\sin 2\\ & \stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{2\cos 2+s\sin 2}{s^2+4}},\end{array}$$

   于是 $\sin \left(2t\right)u(t-1)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{2\cos 2+s\sin 2}{s^2+4}}{\text{e}}^{-s}$$\sin(2t) u(t - 1) \lt \dfrac{2 \cos 2 + s \sin 2}{s^2 + 4} \e^{-s}$.

5. 拆分后求解：

4.3

1. 拆分后求解：

2. $u({\displaystyle \frac{t}{2}}-1)=u(t-2)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{{\text{e}}^{-2s}}{s}}$$u\pqty{\dfrac{t}{2} - 1} = u(t - 2) \lt \dfrac{\e^{-2s}}{s}$.

3. 直接利用定义（也可以使用性质）

4. 拆开：

5. 法一：由 ${\text{e}}^{-t}\sin \left(t\right)u(t)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{1}{(s+1{)}^2+1}}$$\e^{-t} \sin(t) u(t) \lt \dfrac{1}{(s + 1)^2 + 1}$ 和时域微分性质，

   $${\displaystyle \frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}({\text{e}}^{-t}\sin \left(t\right)u(t))\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{s^2}{(s+1{)}^2+1}}.}$$

   法二：直接求导，没法一快，但计算量也不大.

6. 先求 $\mathcal{L}[\mathrm{Sa}(t)]$$\cal L\bqty{\Sa(t)}$.

   法一：频域积分

   $$\begin{array}{r}\mathrm{Sa}(t)\stackrel{\mathcal{L}}{\leftrightarrow }{\int }_s^{+\mathrm{\infty }}{\displaystyle \frac{\mathrm{d}s}{s^2+1}}={\displaystyle \frac{\pi }{2}}-\arctan \left(s\right).\end{array}$$

   法二：幂级数展开

   $$\begin{array}{rl}\mathrm{Sa}(t)& =1-{\displaystyle \frac{t^2}{3!}}+\cdots +(-1{)}^n{\displaystyle \frac{t^{2n}}{(2n+1)!}}\\ & \stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{1}{s}}-{\displaystyle \frac{1}{3s^3}}+\cdots +{\displaystyle \frac{(-1{)}^n}{(2n+1)s^{2n+1}}}\\ & =\arctan \left({\displaystyle \frac{1}{s}}\right)=\mathrm{arccot}\left(s\right)\end{array}$$

   法三：利用 $\mathrm{Sa}(t)u(t)\stackrel{\mathcal{F}}{\leftrightarrow }{\displaystyle \frac{\pi }{2}}-\mathrm{j}\mathrm{arth}(\omega )$$\Sa(t) u(t) \ft \dfrac{\pi}{2} - \j \arth(\omega)$，于是 $\mathrm{Sa}(t)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{\pi }{2}}-\arctan \left(s\right)$$\Sa(t) \lt \dfrac{\pi}{2} - \arctan(s)$.

   之后由尺度性质即有 ${\displaystyle \frac{\sin \left(at\right)}{t}}={\displaystyle \frac{\pi }{2}}-\arctan \left({\displaystyle \frac{s}{a}}\right)$$\dfrac{\sin(at)}{t} = \dfrac{\pi}{2} - \arctan\pqty{\dfrac{s}{a}}$.

7. 由 4.1 题第 22 问的备注中的结论，${\displaystyle \frac{{\text{e}}^{-3t}-{\text{e}}^{-5t}}{t}}\stackrel{\mathcal{L}}{\leftrightarrow }\ln {\displaystyle \frac{s+5}{s+3}}$$\dfrac{\e^{-3t} - \e^{-5t}}{t} \lt \ln\dfrac{s + 5}{s + 3}$.

备注 第六问中用三种方法得到的结果看似不同，但在收敛域与定义域的交集内是相同的.

4.4

1. $f(0_{+})=\underset{s\to \mathrm{\infty }}{lim}sF(s)=1,f(\mathrm{\infty })=\underset{s\to 0}{lim}sF(s)=0$$f(0_+) = \clim s s F(s) = 1, f(\infty) = \liml_{s\to0} s F(s) = 0$.

2. $f(0_{+})=1,f(\mathrm{\infty })=0$$f(0_+) = 1, f(\infty) = 0$.

3. $f(0_{+})=0,f(\mathrm{\infty })={\displaystyle \frac{1}{2}}$$f(0_+) = 0, f(\infty) = \dfrac{1}{2}$.

4. $f(0_{+})=0,f(\mathrm{\infty })=0$$f(0_+) = 0, f(\infty) = 0$.

5. $f(0_{+})=2,f(\mathrm{\infty })$$f(0_+) = 2, f(\infty)$ 不存在.

6. $f(0_{+})=0,f(\mathrm{\infty })$$f(0_+) = 0, f(\infty)$ 不存在.

备注

• 终值定理要求极点在左半平面，因此第五题不能使用终值定理.（或者是否可以用某种方式为所有发散的级数定义极限？似乎是一个在理论上有趣、在数学上有意义、在应用中有价值的问题）

• 对于初值定理，如果非真分式，则应化为真分式后再使用定理.

• 注意 $\delta (0_{-})=\delta (0_{+})=0$$\delta(0_-) = \delta(0_+) = 0$.

4.5

1. 法一（直接卷积）

   $$\begin{array}{rl}{y}_{\text{zs}}(t)& =tu(t)-(t-2)u(t-2)-{\displaystyle \frac{1-{\text{e}}^{-2t}}{2}}u(t)+{\displaystyle \frac{1-{\text{e}}^{-2(t-2)}}{2}}u(t-2)\\ & =(t+{\displaystyle \frac{{\text{e}}^{-2t}-1}{2}})u(t)+({\displaystyle \frac{5-{\text{e}}^{-2(t-2)}}{2}}-t)u(t-2)\end{array}$$

   法二（拉普拉斯变换）

   • $H(s)={\displaystyle \frac{1}{s}}-{\displaystyle \frac{1}{s+2}}$$H(s) = \dfrac{1}{s} - \dfrac{1}{s + 2}$.

   • $F(s)={\displaystyle \frac{1-{\text{e}}^{-2s}}{s}}$$F(s) = \dfrac{1 - \e^{-2s}}{s}$.

   • $Y_{\text{zs}}(s)={\displaystyle \frac{2(1-{\text{e}}^{-2s})}{s^2(s+2)}}$$Y_\text{zs}(s) = \dfrac{2 (1 - \e^{-2s})}{s^2 (s + 2)}$.

   从而部分分式展开求得逆变换.

2. $Y_{\text{zs}}(s)={\displaystyle \frac{2}{s^3}}$$Y_\text{zs}(s) = \dfrac{2}{s^3}$,

   $F(s)={\displaystyle \frac{s+2}{s^2}}$$F(s) = \dfrac{s+2}{s^2}$.

   $f(t)=(1+2t)u(t)$$f(t) = (1 + 2t) u(t)$.

备注

• 第一问答案有误，已由计算机验证：

1
Convolve[
2
    (1 - E^(-2 t)) UnitStep[t],
3
    UnitStep[t] - UnitStep[t - 2],
4
    t, x
5
] // TrigToExp

4.6

1. $F(s)={\displaystyle \frac{1}{s+1}}$$F(s) = \dfrac{1}{s + 1}$.

2. $Y_{\text{zs}}(s)={\displaystyle \frac{1}{s+1}}-{\displaystyle \frac{2}{s+2}}+{\displaystyle \frac{3}{s+3}}$$Y_\text{zs}(s) = \dfrac{1}{s + 1} - \dfrac{2}{s + 2} + \dfrac{3}{s + 3}$.

3. $H(s)=2+{\displaystyle \frac{2}{s+2}}-{\displaystyle \frac{6}{s+3}}$$H(s) = 2 + \dfrac{2}{s + 2} - \dfrac{6}{s + 3}$.

4. 后续思路一：

   1. $h(t)=2\delta (t)+(2{\text{e}}^{-2t}-6{\text{e}}^{-3t})u(t)$$h(t) = 2 \delta(t) + \pqty{2\e^{-2t} - 6\e^{-3t}} u(t)$.

   2. $g(t)=h^{(-1)}(t)=(1-{\text{e}}^{-2t}+2{\text{e}}^{-3t})u(t)$$g(t) = h^{(-1)}(t) = (1 - \e^{-2t} + 2\e^{-3t}) u(t)$.

5. 后续思路二：

   1. $U(s)={\displaystyle \frac{1}{s}}$$U(s) = \dfrac{1}{s}$.

   2. $G(s)={\displaystyle \frac{1}{s}}-{\displaystyle \frac{1}{s+2}}+{\displaystyle \frac{2}{s+3}}$$G(s)= \dfrac{1}{s} - \dfrac{1}{s + 2} + \dfrac{2}{s + 3}$.

   3. $g(t)=(1-{\text{e}}^{-2t}+2{\text{e}}^{-3t})u(t)$$g(t) = (1 - \e^{-2t} + 2\e^{-3t}) u(t)$.

4.7

1. $sY(s)-1+2Y(s)=F(s)={\displaystyle \frac{1}{s}}$$s Y(s) - 1 + 2 Y(s) = F(s) = \dfrac{1}{s}$.

2. $Y(s)={\displaystyle \frac{s+1}{s(s+2)}}={\displaystyle \frac{1}{2s}}+{\displaystyle \frac{1}{2(s+2)}}$$Y(s) = \dfrac{s + 1}{s (s + 2)} = \dfrac{1}{2s} + \dfrac{1}{2 (s + 2)}$.

3. $y(t)={\displaystyle \frac{1+{\text{e}}^{-2t}}{2}}u(t)$$y(t) = \dfrac{1 + \e^{-2t}}{2} u(t)$.

4.8

0. $H(s)={\displaystyle \frac{2s+2}{(s+2)(s+1)}}={\displaystyle \frac{2}{s+2}}$$H(s) = \dfrac{2s + 2}{(s + 2) (s + 1)} = \dfrac{2}{s + 2}$.

   $h(t)=2{\text{e}}^{-2t}u(t)$$h(t) = 2 \e^{-2t} u(t)$.

1. $Y_{\text{zs}}(s)=({\displaystyle \frac{1}{s}}-{\displaystyle \frac{1}{s+2}}){\text{e}}^{-2s}$$Y_\text{zs}(s) = \pqty{\dfrac{1}{s} - \dfrac{1}{s + 2}} \e^{-2s}$.

   $y_{\text{zs}}(t)=(1-{\text{e}}^{-2(t-2)})u(t-2)$$y_\text{zs}(t) = \pqty{1 - \e^{-2(t - 2)}} u(t - 2)$.

2. $Y_{\text{zs}}(s)={\displaystyle \frac{2}{s+1}}-{\displaystyle \frac{2}{s+2}}$$Y_\text{zs}(s) = \dfrac{2}{s + 1} - \dfrac{2}{s + 2}$.

   $y_{\text{zs}}(t)=2({\text{e}}^{-t}-{\text{e}}^{-2t})u(t)$$y_\text{zs}(t) = 2 (\e^{-t} - \e^{-2t}) u(t)$.

3. $Y_{\text{zs}}(s)={\displaystyle \frac{2}{s^2(s+2)}}={\displaystyle \frac{1}{2(s+2)}}+{\displaystyle \frac{1}{s^2}}-{\displaystyle \frac{1}{2s}}$$Y_\text{zs}(s) = \dfrac{2}{s^2 (s + 2)} = \dfrac{1}{2 (s + 2)} + \dfrac{1}{s^2} - \dfrac{1}{2s}$.

   $y_{\text{zs}}(t)=[t+{\displaystyle \frac{{\text{e}}^{-2t}-1}{2}}]u(t)$$y_\text{zs}(t) = \bqty{t + \dfrac{\e^{-2t} - 1}{2}} u(t)$.

备注 以上三题直接卷积其实来的更快. 因为这里每题单独算的话，二者计算量差不多. 但是单位冲激响应只需要计算一次，而用拉氏变换的话需要为每个信号计算拉氏变换. 对于 $f(t)=t^{n}u(t)$$f(t) = t^n u(t)$，使用卷积的计算量也更小，因为可以直接使用不完全伽马函数的结论.（需要注意大部分情况下是恰好相反的）

4.9

1. $h(t)=2{\text{e}}^{-2t}$$h(t) = 2 \e^{-2t}$.

2. $H(s)={\displaystyle \frac{2}{s+2}}$$H(s) = \dfrac{2}{s + 2}$.

3. $R(s)={\displaystyle \frac{1}{s}}-{\displaystyle \frac{1}{s+2}}-{\displaystyle \frac{1}{(s+2{)}^2}}$$R(s) = \dfrac{1}{s} - \dfrac{1}{s + 2} - \dfrac{1}{(s + 2)^2}$.

4. $E(s)={\displaystyle \frac{1}{s}}-{\displaystyle \frac{1}{2(s+2)}}$$E(s) = \dfrac{1}{s} - \dfrac{1}{2 (s + 2)}$.

5. $e(t)=(1-{\displaystyle \frac{1}{2}}{\text{e}}^{-2t})u(t)$$e(t) = \pqty{1 - \dfrac{1}{2} \e^{-2t}} u(t)$.

4.10

1. $I(s)={\displaystyle \frac{E}{s}}{(R+sL//{\displaystyle \frac{1}{sC}})}^{-1}={\displaystyle \frac{E}{s}}{\displaystyle \frac{s^{2}LC+1}{RLC{s}^2+Ls+R}}$$I(s) = \dfrac{E}{s} \pqty{R + sL \parallel \dfrac{1}{sC}}\inv = \dfrac{E}{s} \dfrac{s^2 LC + 1}{RLC s^2 + L s + R}$.

2. ${\displaystyle i(t)=E(\frac{L(\exp \left(t(-\frac{\sqrt{-L(4C{R}^2-L)}}{2CLR}-\frac{1}{2CR})\right)-\exp \left(t(\frac{\sqrt{-L(4C{R}^2-L)}}{2CLR}-\frac{1}{2CR})\right))}{R\sqrt{-L(4C{R}^2-L)}}+\frac{1}{R})}$$\d i(t) = E \left(\frac{L \left(\exp \left(t \left(-\frac{\sqrt{-L \left(4 C R^2-L\right)}}{2 C L R}-\frac{1}{2 C R}\right)\right)-\exp \left(t \left(\frac{\sqrt{-L \left(4 C R^2-L\right)}}{2 C L R}-\frac{1}{2 C R}\right)\right)\right)}{R \sqrt{-L \left(4 C R^2-L\right)}}+\frac{1}{R}\right)$.

备注 不给出数值写着怪麻烦的，因此本题由 mathematica 求解.

4.11

1. 如下：

   $$\begin{array}{rl}H(s)& ={\displaystyle \frac{{(\frac{1}{R_0}+sC+\frac{1}{sL})}^{-1}}{R+{(\frac{1}{R_0}+sC+\frac{1}{sL})}^{-1}}}={\displaystyle \frac{1}{RC}}{\displaystyle \frac{s}{s^2+\frac{R+R_0}{R{R}_{0}C}s+\frac{1}{LC}}},\\ h(t)& =\frac{(\cosh \left(\frac{t(R+R_0)}{2CR{R}_0}\right)-\sinh \left(\frac{t(R+R_0)}{2CR{R}_0}\right))(\sqrt{L(L(R+R_0{)}^2-4C{R}^2{R_0}^2)}\cosh \left(\frac{t\sqrt{L(L(R+R_0{)}^2-4C{R}^2{R_0}^2)}}{2CLR{R}_0}\right)-L(R+R_0)\sinh \left(\frac{t\sqrt{L(L(R+R_0{)}^2-4C{R}^2{R_0}^2)}}{2CLR{R}_0}\right))}{CR\sqrt{L(L(R+R_0{)}^2-4C{R}^2{R_0}^2)}}\end{array}$$

2. 如下：

​x
1
H = Apart[((
2
    Subscript[R, 1] + Subscript[R, 2] + 1/(s Subscript[C, 2]))/(
3
    Subscript[R, 1] Subscript[R, 2] s Subscript[C, 1]) + 1/(
4
    Subscript[R, 2] Subscript[C, 2] s) + 1)^-1, s
5
]
6
H // TeXForm
7

8
h = InverseLaplaceTransform[H, s, t]
9
h // TeXForm

4.12

1. $u_{\text{C}}(0_{-})={\displaystyle \frac{E}{2}}$$u_\text{C}(0_-) = \dfrac{E}{2}$.

2. $I(s)={\displaystyle \frac{E}{2L}}{\displaystyle \frac{1}{s^2+\frac{1}{LC}}}$$I(s) = \dfrac{E}{2L} \dfrac{1}{s^2 + \cfrac{1}{LC}}$.

3. $i(t)={\displaystyle \frac{E}{2}}\sqrt{{\displaystyle \frac{C}{L}}}\sin \left({\displaystyle \frac{t}{\sqrt{LC}}}\right)u(t)$$i(t) = \dfrac{E}{2} \sqrt{\dfrac{C}{L}} \sin\pqty{\dfrac{t}{\sqrt{LC}}} u(t)$.

4.13

1. $e(t)=E(1-{\displaystyle \frac{t}{T}})u(t)+{\displaystyle \frac{Et}{T}}u(t-T)$$e(t) = E (1 - \dfrac{t}{T}) u(t) + \dfrac{Et}{T} u(t - T)$.

2. $E(s)={\displaystyle \frac{E}{s}}-{\displaystyle \frac{E}{T{s}^2}}+{\displaystyle \frac{s+1}{s^2}}E{\text{e}}^{-sT}$$E(s) = \dfrac{E}{s} - \dfrac{E}{T s^2} + \dfrac{s+1}{s^2} E \e^{-sT}$.

3. $H(s)={\displaystyle \frac{s}{2(s+20)}}$$H(s) = \dfrac{s}{2(s + 20)}$.

4. ${\displaystyle v_2(t)=\frac{E}{2}(e^{-20(t-T)}u(t-T)+(\frac{1}{20}-\frac{1}{20}{e}^{-20(t-T)})u(t-T)-\frac{\frac{1}{20}-\frac{e^{-20t}}{20}}{T}+e^{-20t})}$$\d v_2(t) = \frac{E}{2} \left(e^{-20 (t-T)} u (t-T)+\left(\frac{1}{20}-\frac{1}{20} e^{-20 (t-T)}\right) u (t-T)-\frac{\frac{1}{20}-\frac{e^{-20 t}}{20}}{T}+e^{-20 t}\right)$.

4.14

1. 由 $(V_1-(s+1)V_2-s{V}_2){\displaystyle \frac{1}{s}}=(s+1)V_2$$\pqty{ V_1 - (s + 1) V_2 - s V_2 } \dfrac{1}{s} = (s + 1) V_2$，得 $H(s)={\displaystyle \frac{k}{s^2+(3-k)s+1}}$$H(s) = \dfrac{k}{s^2 + (3 - k) s + 1}$.

2. $h(t)={\displaystyle \frac{4}{\sqrt{3}}}{\text{e}}^{-\frac{t}{2}}\sin {\displaystyle \frac{\sqrt{3}}{2}}tu(t)$$h(t) = \dfrac{4}{\sqrt 3} \e^{-\tfrac{t}{2}} \sin\dfrac{\sqrt3}{2} t u(t)$.

备注 相比于前几题，这题是真的善良（各种意义上）.

4.15

1. $F_{\text{a}}(s)={\displaystyle \frac{1-{\text{e}}^{-\frac{T}{2}s}}{s(1-{\text{e}}^{-sT})}}={\displaystyle \frac{1}{s(1+{\text{e}}^{-\frac{sT}{2}})}}$$F_\text a(s) = \dfrac{1 - \e^{-\tfrac{T}{2}s}}{s\pqty{1 - \e^{-sT}}} = \dfrac{1}{s \pqty{1 + \e^{-\tfrac{sT}{2}}}}$.

2. $F_{\text{b}}(s)={\displaystyle \frac{\omega }{s^2+{\omega }^2}}{\displaystyle \frac{1+{\text{e}}^{-\frac{sT}{2}}}{1-{\text{e}}^{-\frac{sT}{2}}}},\omega ={\displaystyle \frac{2\pi }{T}}$$F_\text b(s) = \dfrac{\omega}{s^2 + \omega^2} \dfrac{1 + \e^{-\tfrac{sT}{2}}}{1 - \e^{-\tfrac{sT}{2}}}, \omega = \dfrac{2\pi}{T}$.

备注 周期函数可直接使用结论；周期函数与其他函数相乘，可以从定义与性质出发.

4.16

1. 思路一

   $$f(t){\delta }_{T_{\text{s}}}(t)=f(t){\displaystyle \sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}\delta (t-n{T}_{\text{s}})\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{F(s)}{2\pi \mathrm{j}}}\ast {\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\text{e}}^{-s\cdot n{T}_{\text{s}}}.}}$$

   思路二

   $$\begin{array}{rl}f(t){\delta }_{T_{\text{s}}}(t)& ={\displaystyle \sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}f(n{T}_{\text{s}})\delta (t-n{T}_{\text{s}})\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \sum _{n=0}^{\mathrm{\infty }}f(n{T}_{\text{s}}){\text{e}}^{-s\cdot n{T}_{\text{s}}}.}}\end{array}$$

   思路三

   $$f(t){\delta }_{T_{\text{s}}}(t)=f(t){\displaystyle \sum _{n=-\mathrm{\infty }}^{+\mathrm{\infty }}{\displaystyle \frac{{\text{e}}^{-\mathrm{j}n{\omega }_{\text{s}}t}}{T_{\text{s}}}}\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{F(s)}{2\pi \mathrm{j}\cdot T_{\text{s}}}}\ast {\displaystyle \sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{1}{s+\mathrm{j}n{\omega }_{\text{s}}}}.}}$$

   思路四

   $$f(t){\delta }_{T_{\text{s}}}(t)\stackrel{\mathcal{L}}{\leftrightarrow }{\displaystyle \frac{F(s)}{2\pi \mathrm{j}}}\ast \mathcal{L}[{\delta }_{T_{\text{s}}}(t)]={\displaystyle \frac{F(s)}{2\pi \mathrm{j}}}\ast {\displaystyle \frac{1}{1-{\text{e}}^{-st}}}.$$

   以上四种形式是等价的.

2. $F_{\text{s}}(s)={\displaystyle \frac{1}{1-{\text{e}}^{-(s+a)T}}}$$F_\text s(s) = \dfrac{1}{1 - \e^{-(s + a) T}}$.

4.17

1. $H_3(s)={\displaystyle \frac{1}{s}}$$H_3(s) = \dfrac{1}{s}$.

   $H(s)={\displaystyle \frac{1}{s}}+{\displaystyle \frac{1}{s+1}}-{\displaystyle \frac{1}{s+2}}$$H(s) = \dfrac{1}{s} + \dfrac{1}{s + 1} - \dfrac{1}{s + 2}$.

   $h(t)=(1+{\text{e}}^{-t}-{\text{e}}^{-2t})u(t)$$h(t) = \pqty{1 + \e^{-t} - \e^{-2t}} u(t)$.

2. 法一（直接卷积）：$y_{\text{zs}}(t)=(t+{\displaystyle \frac{1}{2}}-{\text{e}}^{-t}+{\displaystyle \frac{{\text{e}}^{-2t}}{2}})u(t)$$y_\text{zs}(t) = \pqty{t + \dfrac{1}{2} - \e^{-t} + \dfrac{\e^{-2t}}{2}} u(t)$.

   法二（拉氏变换）：

   $Y_{\text{zs}}(s)={\displaystyle \frac{1}{s^2}}+{\displaystyle \frac{1}{2s}}-{\displaystyle \frac{1}{s+1}}+{\displaystyle \frac{1}{2(s+2)}}$$Y_\text{zs}(s) = \dfrac{1}{s^2} + \dfrac{1}{2s} - \dfrac{1}{s + 1} + \dfrac{1}{2 (s + 2)}$.

   $y_{\text{zs}}(t)=(t+{\displaystyle \frac{1}{2}}-{\text{e}}^{-t}+{\displaystyle \frac{{\text{e}}^{-2t}}{t}})u(t)$$y_\text{zs}(t) = \pqty{t + \dfrac{1}{2} - \e^{-t} + \dfrac{\e^{-2t}}{t}} u(t)$.

4.18

1. $H(s)=(1+H_1(s))H_2(s)={\displaystyle \frac{1-{\text{e}}^{-2s}}{s}}$$H(s) = (1 + H_1(s)) H_2(s) = \dfrac{1 - \e^{-2s}}{s}$。

   $h(t)=u(t)-u(t-2)$$h(t) = u(t) - u(t - 2)$.

2. 法一（直接卷积）：$y_{\text{zs}}(t)={\displaystyle \frac{t^2}{2}}u(t)-{\displaystyle \frac{(t-2{)}^2}{2}}u(t-2)$$y_\text{zs}(t) = \dfrac{t^2}{2} u(t) - \dfrac{(t - 2)^2}{2} u(t - 2)$.

   法二（拉氏变换）：

   $Y_{\text{zs}}(t)={\displaystyle \frac{1-{\text{e}}^{-2s}}{s^3}}$$Y_\text{zs}(t) = \dfrac{1 - \e^{-2s}}{s^3}$.

   $y_{\text{zs}}(t)={\displaystyle \frac{t^2}{2}}u(t)-{\displaystyle \frac{(t-2{)}^2}{2}}u(t-2)$$y_\text{zs}(t) = \dfrac{t^2}{2} u(t) - \dfrac{(t - 2)^2}{2} u(t - 2)$.

4.19

备注 第四问很重要 ⭐️，应当认为右侧为开路电压，这是合理且默认的条件，否则缺少条件而无法计算. 具体有如下思路：

1. 直接从耦合电感的公式出发.

2. 采用并联电感的互感消去法.

3. 使用空心变压器的等效公式.（T 形电路）.

附 MATLAB 绘制图像的函数编写如下，上述问题给出特值后调用即可.

xxxxxxxxxx
35
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

4.20

1. $H(s)={\displaystyle \frac{5}{s^2+s+5}}$$H(s) = \dfrac{5}{s^2 + s + 5}$.

2.