第二章作业

2.1 $r_0(t)=e_0(t)\ast h_0(t)$$r_0(t) = e_0(t) * h_0(t)$. 以下图略.

1. $r_1(t)=2r_0(t)$$r_1(t) = 2r_0(t)$.

2. $r_2(t)=r_0(t)-r_0(t-2)$$r_2(t) = r_0(t) - r_0(t-2)$.

3. $r_3(t)=r_0(t-1)$$r_3(t) = r_0(t-1)$.

4. $r_4(t)=e_0(-t)\ast h_0(t)$$r_4(t) = e_0(-t) * h_0(t)$ 与信号和冲激响应有关, 故不能确定.

5. $r_5(t)=r_0(-t)$$r_5(t) = r_0(-t)$.

6. $r_6(t)=r_0^{″}(t)$$r_6(t) = r_0''(t)$.

2.2 $f(t)=(t+1)u(t+1)-tu(t)-u(t-1)$$f(t) = (t+1) u(t+1) - t u(t) - u(t-1)$.

1. $f^{\prime }(t)=u(t+1)-u(t)-\delta (t-1)$$f'(t) = u(t+1) - u(t) - \delta(t-1)$.

2. $f^{″}(t)=\delta (t+1)-\delta (t)-{\delta }^{\prime }(t-1)$$f''(t) = \delta(t+1) - \delta(t) - \delta'(t-1)$.

注 标答非本题.

2.3

1. $f^{″}(t)=\delta (t)-\delta (t-1)-\delta (t-2)$$f''(t) = \delta(t) - \delta(t-1) - \delta(t-2)$.

2. $f(t)=tu(t)-(t-1)u(t-1)-(t-2)u(t-2)$$f(t) = t u(t) - (t-1)u(t-1) - (t-2)u(t-2)$.

注

• ⭐️ 所有的积分默认为定积分, 从 $-\mathrm{\infty }$$-\infty$ 积到 $t$$t$.

• 标答有误.

• 图中冲激函数的值标为 (-1) 或 (1) 均可.

2.4

1. $y_{\text{zs}}(t)={\displaystyle \frac{t}{4}}[u(t)-u(t-4)]$$y_\text{zs}(t) = \dfrac{t}{4} [u(t) - u(t-4)]$.

2. 思路一: 傅里叶变换

   1. $y_{\text{zs}}(t)=f(t)\ast h(t)$$y_\text{zs}(t) = f(t) * h(t)$, 于是 $\mathcal{F}{y}_{\text{zs}}(t)=\mathcal{F}f(t)\mathcal{F}h(t)$$\cal F y_\text{zs}(t) = \cal F f(t) \cal F h(t)$.

   2. $$\begin{array}{rl}\mathcal{F}f(t)& ={\displaystyle \frac{1}{2\pi }}\mathrm{j}\pi [\delta (\omega +1)-\delta (\omega -1)]\ast [\pi \delta (\omega )+\frac{1}{\mathrm{j}\omega }]\\ & ={\displaystyle \frac{\mathrm{j}\pi }{2}}[\delta (\omega +1)-\delta (\omega -1)]+{\displaystyle \frac{1}{2}}[{\displaystyle \frac{1}{\omega +1}}-{\displaystyle \frac{1}{\omega -1}}],\end{array}$$

3. 思路二: 拉普拉斯变换

   $$\begin{array}{rl}\mathcal{L}f(t)& ={\displaystyle {\int }_0^{+\mathrm{\infty }}\sin t\cdot u(t){\text{e}}^{-st}\mathrm{d}t={\displaystyle \frac{1}{1+s^2}}}\\ \mathcal{L}{y}_{\text{zs}}(t)& ={\displaystyle \frac{1}{4s^2}}-({\displaystyle \frac{1}{4s^2}}+{\displaystyle \frac{1}{s}}){\text{e}}^{-4s}\\ \mathcal{L}h(t)& ={\displaystyle \frac{\mathcal{L}{y}_{\text{zs}}(t)}{\mathcal{L}f(t)}}={\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4s^2}}-({\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{4s^2}}+{\displaystyle \frac{1}{s}}+s){\text{e}}^{-4s}\\ h(t)& ={\displaystyle \frac{tu(t)+\delta (t)}{4}}-{\displaystyle \frac{tu(t-4)+\delta (t-4)+4{\delta }^{\prime }(t-4)}{4}}\\ & ={\displaystyle \frac{\delta (t)-\delta (t-4)}{4}}-{\delta }^{\prime }(t-4)+{\displaystyle \frac{t}{4}}[u(t)-u(t-4)].\end{array}$$

注

1. 这样简写一般不会引起歧义.

2. 用傅里叶变换的话, 计算非常复杂.

3. 拉普拉斯变换我直接用计算机算的.

4. 标答最后两项丢了一个 t.

2.5

1. 无冲激信号及其高阶导数, 故无跳变. $r(0_{+})=r(0_{-})=0$$r(0_+) = r(0_-) = 0$.

   另外，$r^{\prime }(t)$$r'(t)$ 包含的最高阶冲激信号为 $u(t)$$u(t)$，从而 $r(t)$$r(t)$ 中包含的最高阶冲激信号为 $tu(t)$$t u(t)$.

2. 有冲激信号, 有跳变.

   1. 思路一: $r^{\prime }(t)$$r'(t)$ 包含的最高阶冲激信号为 $\delta (t)$$\delta(t)$, 于是对微分方程两端积分, 得 $r(0_{+})=r(0_{-})+3=3$$r(0_+) = r(0_-) + 3 = 3$.

   2. 思路二: $r(t)$$r(t)$ 包含的最高阶冲激信号为 $u(t)$$u(t)$ (即 ${\delta }^{(-1)}(t)$$\delta^{(-1)}(t)$), 设系数为 $a$$a$, 代入方程得 $a=3$$a = 3$, 从而 $r(0_{+})=r(0_{-})+3=3$$r(0_+) = r(0_-) + 3 = 3$.

   3. 思路三: $r(t)={\displaystyle \frac{3}{p+2}}\delta (t)=3{\text{e}}^{-2t}u(t)$$r(t) = \dfrac{3}{p+2} \delta(t) = 3 \e^{-2 t} u(t)$, 于是 $r(0_{+})=3$$r(0_+) = 3$.

3. 有冲激信号, 有跳变.

   1. 思路一: $r^{″}(t)$$r''(t)$ 包含的最高阶冲激信号为 $\delta (t)$$\delta(t)$, 系数为 1,

      1. 对两端积分, 得 $r^{\prime }(0_{+})=r^{\prime }(0_{-})+1$$r'(0_+) = r'(0_-) + 1$.

      2. 再次积分, 有 $r(0_{+})=r(0_{-})=0$$r(0_+) = r(0_-) = 0$.

注 第三问忘了给出 $r^{\prime }(0_{-})$$r'(0_-)$.

2.6

1. 特征方程为 $p^3+7p^2+15p+9=(p+1)(p+3{)}^2$$p^3 + 7p^2 + 15p + 9 = (p+1) (p+3)^2$.

2. 于是齐次通解为 $y_{\text{zi}}(t)=C_1{\text{e}}^{-t}+(C_2+C_{3}t){\text{e}}^{-3t}$$y_\text{zi}(t) = C_1 \e^{-t} + (C_2 + C_3 t) \e^{-3 t}$.

3. 代入初值解得 $y_{\text{zi}}(t)=6{\text{e}}^{-t}-4{\text{e}}^{-3t}-5t{\text{e}}^{-3t}$$y_\text{zi}(t) = 6\e^{-t} - 4\e^{-3t} - 5t \e^{-3t}$.

注 若要计算零状态响应, 可用传输算子 $H(p)=-{\displaystyle \frac{3}{4}}{\displaystyle \frac{1}{p+1}}+{\displaystyle \frac{11}{4}}{\displaystyle \frac{1}{p+3}}+{\displaystyle \frac{3}{2}}{\displaystyle \frac{1}{(p+3{)}^2}}$$H(p) = -\dfrac{3}{4} \dfrac{1}{p + 1} + \dfrac{11}{4} \dfrac{1}{p + 3} + \dfrac{3}{2} \dfrac{1}{(p+3)^2}$.

2.7

注

• 标答有误 (已用计算机验证).

• 计算诸如 ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}({\text{e}}^{-t}-{\text{e}}^{-2t})u(t)}$$\d\dv{t} (\e^{-t} - \e^{-2t}) u(t)$ 的式子时, 完全不需要全部展开. 这是连续的, 忽略 $u(t)$$u(t)$ 直接导就行了. 之前一个个展开真的是憨憨做法.

2.8 首先解得零状态响应与零输入响应:

1. $y_1(t)=(5.5e^{-3t}+0.5\sin 2t)u(t)$$y_1(t) = (5.5 e^{-3t} + 0.5 \sin 2t) u(t)$.

2. $y_2(t)=3{\text{e}}^{-3t}u(t)+(\sin 2(t-t_0)-{\text{e}}^{-3(t-t_0)})u(t-t_0)$$y_2(t) = 3\e^{-3t} u(t) + (\sin2(t-t_0) - \e^{-3(t-t_0)}) u(t-t_0)$.

注 标答漏写了 $u(t-t_0)$$u(t-t_0)$.

2.9

注

• 下面这个积分很容易展开, 但是挺麻烦的. 可以用下不完全伽马函数表示, 但是没必要.

• 我没有继续计算了, 但我不相信可以化成标答的形式.

2.10

1. $e(t)=tu(t)-(t-1)u(t-1)-(t-3)u(t-3)+(t-4)u(t-4)$$e(t) = tu(t) - (t-1) u(t-1) - (t-3) u(t-3) + (t-4) u(t-4)$.

2. 暂记 $r_1(t)={\text{e}}^{-t}u(t)\ast tu(t)=({\text{e}}^{-t}+t-1)u(t)$$r_1(t) = \e^{-t} u(t) * tu(t) = (\e^{-t} + t - 1) u(t)$.

3. 于是 $r(t)=r_1(t)-r_1(t-1)-r_1(t-3)+r_1(t-4)$$r(t) = r_1(t) - r_1(t-1) - r_1(t-3) + r_1(t-4)$.

2.11

1. 思路一: 传输算子法.

   1. $iR+L{\displaystyle \frac{\mathrm{d}i}{\mathrm{d}t}={\displaystyle \frac{1}{C}}{\int }_{-\mathrm{\infty }}^{t}i\mathrm{d}t=f(t)}$$iR + L \ddv{i}{t} = \dfrac{1}{C} \int_{-\infty}^t i \dt = f(t)$.

   2. $(p^2+5p+6)i(t)=\delta (t)$$(p^2 + 5p + 6) i(t) = \delta(t)$.

   3. $i_{\text{zs}}(t)={\displaystyle \frac{\delta (t)}{p+2}}-{\displaystyle \frac{\delta (t)}{p+3}}=({\text{e}}^{-2t}-{\text{e}}^{-3t})u(t)$$i_\text{zs}(t) = \dfrac{\delta(t)}{p+2} - \dfrac{\delta(t)}{p+3} = (\e^{-2t} - \e^{-3t}) u(t)$.

2. 思路二: 拉普拉斯变换法.

   1. $i(s)={\displaystyle \frac{\frac{1}{s}}{5+s+\frac{6}{s}}}={\displaystyle \frac{1}{s^2+5s+6}}={\displaystyle \frac{1}{s+2}}-{\displaystyle \frac{1}{s+3}}$$i(s) = \dfrac{\cfrac{1}{s}}{5 + s + \cfrac{6}{s}} = \dfrac{1}{s^2 + 5s +6} = \dfrac{1}{s+2} - \dfrac{1}{s+3}$.

   2. $i_{\text{zs}}(t)=({\text{e}}^{-2t}-{\text{e}}^{-3t})u(t)$$i_\text{zs}(t) = (\e^{-2t} - \e^{-3t}) u(t)$.

注 标答有误 (甚至方向都错了).

2.12

1. $r(t)={\displaystyle \frac{2p}{p+3}}\delta (t)=2\delta (t)-{\displaystyle \frac{6}{p+3}}\delta (t)=2\delta (t)--6{\text{e}}^{-3t}u(t)$$r(t) = \dfrac{2p}{p+3} \delta(t) = 2 \delta(t) - \dfrac{6}{p+3} \delta(t) = 2\delta(t) -- 6\e^{-3t} u(t)$.

2. $r(t)={\displaystyle \frac{p^2+3p+3}{p+2}}=(p+1+{\displaystyle \frac{1}{p+2}})\delta (t)={\delta }^{\prime }(t)+\delta (t)+{\text{e}}^{-2t}u(t)$$r(t) = \dfrac{p^2+3p+3}{p+2} = \pqty{p+1 + \dfrac{1}{p+2}} \delta(t) = \delta'(t) + \delta(t) + \e^{-2t} u(t)$.

2.13

注

• 我不理解为什么这么喜欢出不完全伽马函数的题, 表示起来枯燥无趣.

• 关系式: $\gamma (s,x)+\mathrm{\Gamma }(s,x)=\mathrm{\Gamma }(s)$$\gamma(s, x) + \Gamma(s, x) = \Gamma(s)$.

2.14

代入 $a=2,b=3$$a = 2, b = 3$ 即得原式 $={\displaystyle \frac{\mathrm{\Gamma }(3)\mathrm{\Gamma }(4)}{\mathrm{\Gamma }(7)}}{t}^{6}u(t)={\displaystyle \frac{t^{6}u(t)}{60}}$$= \dfrac{\Gamma(3) \Gamma(4)}{\Gamma(7)} t^6 u(t) = \dfrac{t^6 u(t)}{60}$.

注 好, 这下欧拉积分都齐了.

2.15

具体就不展开了, 参考第二次翻转课堂练习题第五题.

2.16

注 我是显微镜, 标答把 $t$$t$ 写成 $\tau$$\tau$ 了.

2.17

1. 当 $t=0_{-}$$t = 0_-$ 时, 电流源被电感短路.

2. 当 $t>0$$t > 0$ 时,

   1. 求 $u_{\text{C}}$$u_\text C$.

      1. 思路一: 解 $2={\displaystyle \frac{u_{\text{C}}}{3}}+{\displaystyle \frac{\mathrm{d}{u}_{\text{C}}}{\mathrm{d}t}}$$2 = \dfrac{u_\text C}{3} + \ddv{u_\text C}{t}$.

      2. 思路二: 由三要素法, $u_{\text{C}}=6(1-{\text{e}}^{-\frac{t}{3}})u(t)$$u_\text C = 6 \pqty{1 - \e^{-\tfrac{t}{3}}} u(t)$.

   2. 求 $u_{\text{L}}$$u_\text L$.

      1. 思路一: 解微分方程.

      2. 思路二: 由三要素法, $u_{\text{L}}=-{\text{e}}^{-\frac{t}{4}}u(t)$$u_\text L = -\e^{-\tfrac{t}{4}} u(t)$.

   3. 于是 $u_{\text{ac}}(t)=6(1-{\text{e}}^{-\frac{t}{3}})u(t)+{\text{e}}^{-\frac{t}{4}}u(t)$$u_\text{ac}(t) = 6\pqty{1 - \e^{-\tfrac{t}{3}}} u(t) + \e^{-\tfrac{t}{4}} u(t)$.

2.18

以下只考虑 $t>0$$t > 0$ 的情况.

1. 齐次通解为 $y(t)=C{\text{e}}^{-3t}$$y(t) = C \e^{-3t}$, 代入初值 $y(0_{-})=1.5$$y(0_-) = 1.5$, 得 $y_{\text{zi}}(t)=1.5{\text{e}}^{-3t}$$y_\text{zi}(t) = 1.5 \e^{-3t}$.

2. 全解为 $y(t)=C{\text{e}}^{-3t}+1$$y(t) = C \e^{-3t} + 1$, 代入初值 $y(0_{-})=0$$y(0_-) = 0$, 得 $y_{\text{zs}}(t)=1-{\text{e}}^{-3t}$$y_\text{zs}(t) = 1 - \e^{-3t}$.

3. 全响应为 $y(t)=1+0.5{\text{e}}^{-3t}$$y(t) = 1 + 0.5 \e^{-3t}$.

2.19

2.20

2.21

1. $0<t<3$$0 < t < 3$,

   1. $u_{\text{C}}(t)=10(1-{\text{e}}^{-\frac{t}{5}})u(t)\mathrm{V}$$u_\text C(t) = 10 \pqty{1 - \e^{-\tfrac{t}{5}}} u(t) \ \mathrm{V}$.

   2. $i_{\text{C}}(t)={\text{e}}^{-\frac{t}{5}}u(t)\mathrm{A}$$i_\text C(t) = \e^{-\tfrac{t}{5}} u(t) \ \mathrm{A}$.

2. $t=3$$t = 3$,

   1. $u_{\text{C}}(t)\approx 4.51188\mathrm{V}$$u_\text C(t) \approx 4.51188\ \mathrm{V}$.

   2. $i_{\text{C}}(t)\approx 0.548812\mathrm{A}$$i_\text C(t) \approx 0.548812 \mathrm{A}$.

3. $t\ge 3$$t \ge 3$,

   1. $u_{\text{C}}(t)\approx ({\displaystyle \frac{10}{3}}+1.17855{\text{e}}^{-\frac{3(t-3)}{5}})u(t-3)\mathrm{V}$$u_\text C(t) \approx \pqty{\dfrac{10}{3} + 1.17855 \e^{-\tfrac{3(t-3)}{5}}} u(t-3)\ \mathrm{V}$.

   2. $i_{\text{C}}(t)\approx -0.353565{\text{e}}^{-\frac{3(t-3)}{5}}u(t-3)$$i_\text C(t) \approx -0.353565 \e^{-\tfrac{3(t-3)}{5}} u(t-3)$.

注

• 标答只给了 $t\ge 3$$t \ge 3$ 时的结果.

• $t\ge 3$$t \ge 3$ 时不是直接对 $u_{\text{C}}(t)$$u_\text C(t)$ 的表达式求导, 那样会多出一项 $4.51188\delta (t-3)$$4.51188 \delta(t-3)$.

2.22

$\sin \left(t\right)u(t)\ast h(t)=2tu(t)-4(t-1)u(t-1)$$\sin(t) u(t) * h(t) = 2t u(t) - 4(t-1) u(t-1)$.

1. 思路一: 傅里叶变换

2. 思路二: 拉普拉斯变换

   1. $\mathcal{L}[\sin \left(t\right)](s)={\displaystyle {\int }_0^{+\mathrm{\infty }}\sin \left(t\right){\text{e}}^{-st}\mathrm{d}t={\displaystyle \frac{1}{s^2+1}}}$$\cal L\bqty{\sin(t)}(s) = \intoi \sin(t) \e^{-st} \dt = \dfrac{1}{s^2 + 1}$.

   2. $\mathcal{L}[y_{\text{zs}}(t)](s)={\displaystyle \frac{2-4{\text{e}}^{-s}}{s^2}}$$\cal L\bqty{y_\text{zs}(t)}(s) = \dfrac{2-4\e^{-s}}{s^2}$.

   3. $\mathcal{L}[h(t)](s)=2-4{\text{e}}^{-s}+{\displaystyle \frac{2-4{\text{e}}^{-s}}{s^2}}$$\cal L\bqty{h(t)}(s) = 2 - 4\e^{-s} + \dfrac{2-4\e^{-s}}{s^2}$.

   4. $h(t)=2\delta (t)-4\delta (t-1)+2tu(t)-4(t-1)u(t-1)$$h(t) = 2\delta(t) - 4\delta(t-1) + 2t u(t) - 4(t-1) u(t-1)$.

提问 标答的思路是什么? ⭐️

注 恼恼, 怎么又要求卷积的逆. 好, 我推一下.

1. 三角函数的拉普拉斯变换

于是有

2. 幂函数的拉普拉斯变换

3. 时移性质

若 $F(s)=\mathcal{L}[f(t)u(t)]$$F(s) = \cal L[f(t)u(t)]$, 则

4. 频移性质

2.23

1. $t=0_{-}$$t = 0_-$,

   1. $i_{\text{L}}(0_{-})=1\mathrm{A}$$i_\text L(0_-) = 1\ \mathrm{A}$.

   2. $u_{\text{C}}(0_{-})=6\mathrm{V}$$u_\text C(0_-) = 6\ \mathrm{V}$.

2. $t>0$$t > 0$,

   1. $\begin{array}{rl}10& =u_{\text{C}}+4({\displaystyle \frac{1}{5}}{u}_{\text{C}}^{\prime }+i_{\text{L}})=i_{\text{L}}^{\prime }+6i_{\text{L}}+{\displaystyle \frac{4}{5}}(i_{\text{L}}^{″}+6i_{\text{L}}^{\prime }+5i_{\text{L}})\end{array}$$\begin{align} 10 &= u_\text C + 4 \pqty{ \dfrac{1}{5} u'_\text C + i_\text L } = i'_\text L + 6 i_\text L + \dfrac{4}{5} \pqty{i''_\text L + 6 i'_\text L + 5 i_\text L} \end{align}$.

      $4i^{″}+29i^{\prime }+50i=50$$4 i'' + 29 i' + 50 i = 50$.

   2.  

注 这里算的有问题了. ⭐️

2.24

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

   $y_{\text{zs}}(t)=x_1(t)\ast h(t)=2({\text{e}}^{-t}-{\text{e}}^{-2t})u(t)$$y_\text{zs}(t) = x_1(t) * h(t) = 2\pqty{\e^{-t} - \e^{-2t}} u(t)$.

   $y_{\text{zi}}(t)=y_1(t)-y_{\text{zs}}(t)=2{\text{e}}^{-2t}u(t)$$y_\text{zi}(t) = y_1(t) - y_\text{zs}(t) = 2\e^{-2t} u(t)$.

2. $y_{\text{zi}}(t)=4{\text{e}}^{-2t}u(t)$$y_\text{zi}(t) = 4\e^{-2t} u(t)$.

   $y_{\text{zs}}(t)=2\delta (t)-4{\text{e}}^{-2t}u(t)$$y_\text{zs}(t) = 2 \delta(t) - 4 \e^{-2t} u(t)$.

   $y_2(t)=2\delta (t)$$y_2(t) = 2\delta(t)$.

2.25

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

2. $r(t)=(4\cos 2t-{\text{e}}^{-t})u(t)$$r(t) = \pqty{4 \cos 2t - \e^{-t}} u(t)$.

2.26

1. $g^{\prime }(t)+r_{\text{zi}}(t)=\delta (t)+{\text{e}}^{-t}u(t)$$g'(t) + r_\text{zi}(t) = \delta(t) + \e^{-t} u(t)$.

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

3. $g^{\prime }(t)-g(t)=\delta (t)-2{\text{e}}^{-t}u(t)$$g'(t) - g(t) = \delta(t) - 2\e^{-t} u(t)$.

4. $g(t)={\text{e}}^{-t}u(t)$$g(t) = \e^{-t} u(t)$. (一般方法?)

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

6. $r_{\text{zi}}(t)=2{\text{e}}^{-t}u(t)$$r_\text{zi}(t) = 2\e^{-t} u(t)$.

7. $x_3(t)=t(u(t)-u(t-1))$$x_3(t) = t \pqty{u(t) - u(t-1)}$.

8. $y_3(t)=x_3(t)\ast h(t)+r_{\text{zi}}(t)$$y_3(t) = x_3(t) * h(t) + r_\text{zi}(t)$.

9. $y_3(t)=(1+{\text{e}}^{-t})u(t)-u(t-1)$$y_3(t) = (1 + \e^{-t}) u(t) - u(t-1)$.