第三章练习解答

一、填空题

1. $f_1(t)\ast f_2(t)={\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}2{\text{e}}^{-2x}{I}_{\{0<x<t-2\}}\mathrm{d}x=(1-{\text{e}}^{-2(t-2)})u(t-2)}$$f_1(t) * f_2(t) = \inti 2\e^{-2x} I_\Bqty{0 < x < t-2} \dx = (1 - \e^{-2(t-2)}) u(t -2)$.

   备注: 设 $\alpha ,\beta >0$$\alpha, \beta > 0$, 当 $\alpha \ne \beta$$\alpha \ne \beta$ 时,

   $$\begin{array}{rl}{\text{e}}^{-\alpha t}u(t)\ast {\text{e}}^{-\beta t}u(t)& ={\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\text{e}}^{-\alpha x-\beta (t-x)}{I}_{\{0<x<t\}}\mathrm{d}x={\displaystyle \frac{{\text{e}}^{-\alpha t}-{\text{e}}^{-\beta t}}{\beta -\alpha }}u(t).}\end{array}$$

   当 $\alpha =\beta$$\alpha = \beta$ 时, ${\text{e}}^{-\alpha t}u(t)\ast {\text{e}}^{-\alpha t}u(t)={\displaystyle {\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\text{e}}^{-\alpha t}{I}_{\{0<x<t\}}\mathrm{d}x=t{\text{e}}^{-\alpha t}}$$\e^{-\alpha t} u(t) * \e^{-\alpha t} u(t) = \inti \e^{-\alpha t} I_\Bqty{0 < x < t} \dx = t \e^{-\alpha t}$.

2. $f(2t-7)\leftrightarrow {\displaystyle \frac{1}{2}}F\left({\displaystyle \frac{\omega }{2}}\right){\text{e}}^{-\mathrm{j}\omega \frac{7}{2}}$$f(2t-7) \rla \dfrac{1}{2} F\pqty{\dfrac{\omega}{2}} \e^{-\j\omega \tfrac{7}{2}}$.

3. $H(\omega )={\text{e}}^{-\mathrm{j}\omega t_0}{I}_{\{\left|\omega \right|<{\omega }_{\text{c}}\}},t_0\ge 0$$H(\omega) = \e^{-\j\omega t_0} I_\Bqty{\vqty{\omega} < \omega_\text c}, t_0 \ge 0$.

4. $T[\{k_1{f}_1(t)+k_2{f}_2(t)\},\left\{0\right\}]=k_{1}T[\{f_1(t)\},\left\{0\right\}]+k_{2}T[\{f_2(t)\},\left\{0\right\}]$$T\bqty{ \Bqty{k_1 f_1(t) + k_2 f_2(t)}, \Bqty{0} } = k_1 T\bqty{ \Bqty{f_1(t)}, \Bqty{0} } + k_2 T\bqty{ \Bqty{f_2(t)}, \Bqty{0} }$.

5. 原式 $=-9{\displaystyle {\int }_{0_{-}}^t\delta (t)\mathrm{d}t=-9u(t)}$$= -9 \dint_{0_-}^t \delta(t) \dt = -9 u(t)$.

二、选择题

1. D.

   $ke(t)$$k e(t)$ 的响应为 $k^{2}r(t)$$k^2 r(t)$, 故为非线性.

2. A,

   $f(t)={\displaystyle {\int }_{-1}^1{\displaystyle \frac{\delta (t-2)+\delta (t+2)}{4}}\mathrm{d}t=0}$$f(t) = \dint_{-1}^1 \dfrac{\delta(t - 2) + \delta(t+2)}{4} \dt = 0$.

3. C,

   $\mathrm{Sa}(100t)+\mathrm{Sa}(50t)\leftrightarrow \pi I_{\{-100<\omega <100\}}+\pi I_{\{-50<\omega <50\}}$$\Sa(100t) + \Sa(50t) \rla \pi I_\Bqty{-100 < \omega < 100} + \pi I_\Bqty{-50 < \omega < 50}$,

   故 $T_{\text{s}}={\displaystyle \frac{1}{2f_{\text{m}}}}={\displaystyle \frac{\pi }{{\omega }_{\text{m}}}}={\displaystyle \frac{\pi }{100}}$$T_\text s = \dfrac{1}{2 f_\text m} = \dfrac{\pi}{\omega_\text m} = \dfrac{\pi}{100}$.

4. C, 由卷积定理与卷积的性质即得.

5. B

   若..., 则 $k_1{f}_1(t)+k_2{f}_2(t)$$k_1 f_1(t) + k_2 f_2(t)$ 的响应为 $k_1{y}_1(t)+k_2{y}_2(t)$$k_1 y_1(t) + k_2 y_2(t)$, 故为线性.

6. A, 由卷积定理与频移性质即得.

7. D, 由性质即得.

8. B,

   由微分性质与 $\mathrm{sgn}(t)\leftrightarrow {\displaystyle \frac{2}{\mathrm{j}\omega }}$$\sgn(t) \rla \dfrac{2}{\j\omega}$ 得 $t\mathrm{sgn}(t)\leftrightarrow -{\displaystyle \frac{2}{{\omega }^2}}$$t \sgn(t) \rla -\dfrac{2}{\omega^2}$, 从而得到 $f(t)=-{\displaystyle \frac{t}{2}}\mathrm{sgn}(t)$$f(t) = -\dfrac{t}{2} \sgn(t)$.

三、计算题

1. 最低抽样率 $f_{\text{s}}=2f_{\text{m}}$$f_\text s = 2 f_\text m$.

   1. $f(t):f_{\text{s}}=2a$$f(t) \colon f_\text s = 2a$.

   2. $f(2t-1):f_{\text{s}}=4a$$f(2t-1) \colon f_\text s = 4a$.

   3. $3f^2(t):f_{\text{s}}=4a$$3f^2(t) \colon f_\text s = 4 a$.

   4. $f(t)\cos \left(2a\pi t\right):f_{\text{s}}=4a$$f(t) \cos(2a\pi {\color{red} t}) \colon f_\text s = 4a$.

2. 求冲激响应、系统函数及其响应.

   1. 思路一: 先求 $h(t)$$h(t)$, 首先有 $\{\begin{array}{l}{h}_1(t)={\displaystyle \frac{{\omega }_0}{2\pi }}{\mathrm{Sa}}^{\prime }({\omega }_{0}t),\\ h_2(t)=\delta (t-{\displaystyle \frac{2\pi }{{\omega }_0}}),\\ h_3(t)=u(t).\end{array}$$\begin{cases} h_1(t) = \dfrac{\omega_0}{2\pi} \Sa'(\omega_0 t), \\ h_2(t) = \delta\pqty{t - \dfrac{2\pi}{\omega_0}}, \\ h_3(t) = u(t). \end{cases}$

      于是

      $$\begin{array}{rl}h(t)& =[h_1(t)-h_1(t)\ast h_2(t)]\ast h_3(t)\\ & ={\displaystyle \frac{{\omega }_0}{2\pi }}[\mathrm{Sa}({\omega }_{0}t)-\mathrm{Sa}({\omega }_{0}t-2\pi )],\\ H(\omega )& =I_{\{-{\omega }_0<\omega <{\omega }_0\}}{\displaystyle \frac{1-{\text{e}}^{-\mathrm{j}\omega \frac{2\pi }{{\omega }_0}}}{2}}.\end{array}$$

      ---

      思路二, 先求 $H(t)$$H(t)$, 首先有 $\{\begin{array}{l}{H}_1(\omega )={\displaystyle \frac{\mathrm{j}\omega }{2}}{I}_{\{-{\omega }_0<\omega <{\omega }_0\}},\\ H_2(\omega )={\text{e}}^{-\mathrm{j}\omega \frac{2\pi }{{\omega }_0}},\\ H_3(\omega )=\pi \delta (\omega )+{\displaystyle \frac{1}{\mathrm{j}\omega }}.\end{array}$$\begin{cases} H_1(\omega) = \dfrac{\j\omega}{2} I_\Bqty{-\omega_0 < \omega < \omega_0}, \\ H_2(\omega) = \e^{-\j\omega \tfrac{2\pi}{\omega_0}}, \\ H_3(\omega) = \pi\delta(\omega) + \dfrac{1}{\j\omega}. \end{cases}$

      于是

      $$\begin{array}{rl}H(\omega )& =[H_1(\omega )-H_1(\omega )H_2(\omega )]H_3(\omega )\\ & =I_{\{-{\omega }_0<\omega <{\omega }_0\}}{\displaystyle \frac{1-{\text{e}}^{-\mathrm{j}\omega \frac{2\pi }{{\omega }_0}}}{2}},\\ h(t)& ={\displaystyle \frac{{\omega }_0}{2\pi }}[\mathrm{Sa}({\omega }_{0}t)-\mathrm{Sa}({\omega }_{0}t-2\pi )].\end{array}$$

   2. 利用系统函数, 本质上是用了一次卷积定理,

      即 $y(t)={\mathcal{F}}^{-1}Y(\omega )={\mathcal{F}}^{-1}\mathcal{F}[x(t)\ast h(t)]={\mathcal{F}}^{-1}[X(\omega )H(\omega )]$$y(t) = \cal F\inv Y(\omega) = \cal F\inv \cal F[x(t) * h(t)] = \cal F\inv[X(\omega) H(\omega)]$.

      $$\begin{array}{rl}X(\omega )& =\mathrm{j}\pi \delta (\omega +2{\omega }_0)-\mathrm{j}\pi \delta (\omega -2{\omega }_0)+\\ & \pi \delta (\omega +{\displaystyle \frac{{\omega }_0}{2}})+\pi \delta (\omega -{\displaystyle \frac{{\omega }_0}{2}}),\\ Y(\omega )& =X(\omega )H(\omega )={\displaystyle \frac{\pi }{2}}[\delta (\omega +{\displaystyle \frac{{\omega }_0}{2}})+\delta (\omega -{\displaystyle \frac{{\omega }_0}{2}})],\\ y(t)& =\cos \left({\displaystyle \frac{{\omega }_{0}t}{2}}\right).\end{array}$$