第五章

5 - 1 $ r(t) = (\mathrm{e}^{-2t} - \mathrm{e}^{-3t})u(t) $

$$ \begin{array}{r l}{5\sim2}&{{}r\left(t\right)=\frac{1}{\sqrt{2}}\mathrm{s i n}\left(t-45^{\circ}\right)+\frac{1}{\sqrt{10}}\mathrm{s i n}\left(3t-72^{\circ}\right)}\end{array} $$

$$ \begin{array}{r l}{\mathbf{5}-\mathbf{3}}&{{}H(\mathrm{j}\omega)=\frac{\mathrm{j}\omega}{C(\omega_{0}^{2}-\omega^{2})}+\frac{\pi}{2C}\big[\delta\big(\omega+\omega_{0}\big)+\delta\big(\omega-\omega_{0}\big)\big]}\end{array} $$

$$ h\left(t\right)=\frac{1}{C}\cos\left(\omega_{0}t\right)u\left(t\right) $$

$$ \begin{aligned}&5-4\quad H(s)=\frac{C_{1}}{C_{1}+C_{2}}\frac{s+\frac{1}{R_{1}C_{1}}}{s+\frac{R_{1}+R_{2}}{R_{1}R_{2}(C_{1}+C_{2})}}\end{aligned} $$

无失真条件 $ R_{1}C_{1}=R_{2}C_{2} $

$$5-5\quad H(s)=\frac{R_{2}s^{2}+(1+R_{1}R_{2})s+R_{1}}{s^{2}+(R_{1}+R_{2})s+1}$$

无失真条件 $ R_{1}=R_{2}=1\Omega $，无延迟

5-6 对两种信号的响应均为 $ \mathrm{Sa}[\omega_{c}(t-t_{0})] $

5 - 7 $ r(t) = \mathrm{Sa}[\omega_0(t - t_0)] $

$$ \begin{array}{r l}{\textbf{5-9}}&{{}r\left(t\right)=\frac{1}{\pi}\Bigg\{\mathrm{S i}\bigg[\frac{2\pi}{\tau}\bigg(t+\frac{\tau}{2}\bigg)\bigg]-\mathrm{S i}\bigg[\frac{2\pi}{\tau}\bigg(t-\frac{\tau}{2}\bigg)\bigg]\Bigg\}}\end{array} $$

5-10

$$ h\left(t\right)=\frac{2\omega_{\mathrm{c}}}{\pi}\mathrm{S a}\left[\omega_{\mathrm{c}}\left(t-t_{0}\right)\right]\mathrm{c o s}\left(\omega_{0}t\right) $$

非因果，不能实现

5-11 (1)

$$ \upsilon_{2}(t)=\frac{1}{\pi}\big[\mathrm{S i}(t-t_{0}-T)-\mathrm{S i}(t-t_{0})\big] $$

(2)

$$ v_{2}(t)=\mathrm{S a}\bigg[\frac{1}{2}(t-t_{0}-T)\bigg]-\mathrm{S a}\bigg[\frac{1}{2}(t-t_{0})\bigg] $$

$$ \begin{array}{r l}{\textbf{5-12}}&{{}y(t)=\frac{1}{T}[\ensuremath{t u}(t)-\left(t-T\right)u(t-T)-\left(t-\tau\right)u(t-\tau)+\left(t-T-\tau\right)u(t-T-\tau)]}\end{array} $$

$$ \begin{array}{r l}{\textbf{5-13}}&{h\left(t\right)=\frac{\omega_{\mathrm{c}}}{2\pi}\Bigg\{\mathrm{S a}\left[\omega_{\mathrm{c}}\left(t-t_{0}\right)\right]+\frac{1}{2}\mathrm{S a}\Bigg[\omega_{\mathrm{c}}\left(t-t_{0}+\frac{\pi}{\omega_{\mathrm{c}}}\right)\Bigg]+\frac{1}{2}\mathrm{S a}\Bigg[\omega_{\mathrm{c}}\left(t-t_{0}-\frac{\pi}{\omega_{\mathrm{c}}}\right)\Bigg]}\end{array}\Bigg\} $$

$$ \begin{array}{r l}{\textbf{5-14}}&{h\left(t\right)=h_{i}\left(t\right)+\displaystyle\sum_{k=1}^{m}\frac{a_{k}}{2}\Bigg[h_{i}\left(t-\frac{k}{\omega_{1}}\right)-h_{i}\left(t+\frac{k}{\omega_{1}}\right)\Bigg]}\end{array} $$

其中 $ h_{i}(t)=\frac{\omega_{c}}{\pi}\mathrm{Sa}\left[\omega_{c}(t-t_{0})\right] $

5-17 将 $ F_{1}(\omega) $ 与本地载波信号之频谱（冲激函数）进行卷积（频域），即可恢复含有 $ G(\omega) $ 之频谱。再经低通滤波取出 $ G(\omega) $

$$ \textcircled{5}-18\quad\textcircled{V}(\omega)=G\left(\omega+\omega_{0}\right)u\left(-\omega-\omega_{0}\right)+G\left(\omega-\omega_{0}\right)u\left(\omega-\omega_{0}\right) $$

5 - 19 $ \mathrm{Sa}[\omega_{c}(t-t_{0})]\cos(\omega_{0}t) $

5-20 (1) $ h(t)=\frac{\sin 2\Omega(t-t_{0})}{\pi(t-t_{0})} $

(2)

$$ r\left(t\right)=\frac{1}{2}\left[\frac{\sin\Omega\left(t-t_{0}\right)}{\Omega\left(t-t_{0}\right)}\right]^{2} $$

(3) $ r(t) = 0 $

（4）是线性时变系统

$$ \begin{aligned}5-25\quad&\triangle<\frac{T}{4\pi},a=\frac{\Delta}{T+\Delta},k=\frac{1}{T+\Delta}\end{aligned} $$