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多径信道冲激响应

假设发送信号为

\[x\left( t \right) = {\mathop{\rm Re}\nolimits} \left\{ {s\left( t \right){e^{j2\pi {f_c}t}}} \right\} \]

其中 ${s\left( t \right)} $ 是 \({x\left( t \right)}\) 的等效基带信号,\(f_{c}\) 是载波频率。忽略噪声,则接收信号是直射信号和所有可分辨多径分量之和

\[y\left( t \right) = {\mathop{\rm Re}\nolimits} \left\{ {\sum\limits_{i = 0}^{N\left( t \right)} {{\beta_i}\left( t \right)s\left( {t - {\tau _i}\left( t \right)} \right){e^{j2\pi {f_c }\left( {t - {\tau _i}\left( t \right)} \right) }}} } \right\} \]

其中可分辨多径数目为 \({N\left( t \right)}\)\(i=0\) 时对应着直射信号。各径的长度为 \({{d_i}\left( t \right)}\),则对应的时延为 \({\tau _i}\left( t \right) = \frac{{{d_i}\left( t \right)}}{c}\). \({{\beta_i}\left( t \right)}\) 表示大尺度衰落系数。

\({\phi _i}\left( t \right) = 2\pi {f_c}{\tau _i}\left( t \right)\),则可以把接收信号简化为

\[\begin{aligned}y(t) &=\operatorname{Re}\left\{\left[\sum_{i=0}^{N(t)} \beta_{i}(t) e^{-j \phi_{i}(t)} s\left(t-\tau_{i}(t)\right)\right] e^{j 2 \pi f_{c} t}\right\} \\&=\operatorname{Re}\left\{r(t) e^{j 2 \pi f_{c} t}\right\}\end{aligned} \]

其中

\[r(t)={\sum\limits_{i = 0}^{N\left( t \right)} {{\beta_i}\left( t \right){e^{ -j{\phi _i}\left( t \right)}}s\left( {t - {\tau _i}\left( t \right)} \right){}} } \]

这部分可以看作是接收信号 \(y\left( t \right)\) 的等效基带信号.由于 \({{\beta_i}\left( t \right)}\) 取决于路径损耗和阴影衰落,而 \({{\phi _i}\left( t \right)}\) 取决于时延,则一般可假设这两个随机过程是相互独立的。

等效基带输入信号 \(s\left( t \right)\) 与时变信道的等效基带冲激响应 \(h\left( {t,\tau } \right)\) 卷积再上变频到载波频率 \(f_{c}\) 即可得到接收信号 \(y\left( t \right)\). 这里直接写出冲激响应函数的表达式「\(r(t)=s(t)*h(t,\tau)\)

\[\begin{aligned} h\left( {t,\tau } \right) &= \sum\limits_{i = 0}^{N\left( t \right)} {{\beta_i}\left( t \right){e^{- j{\phi _i}\left( t \right)}}\delta \left( {t - {\tau _i}\left( t \right)} \right)} \\&= \sum\limits_{i = 0}^{N\left( t \right)} \alpha_i(t) \delta \left( {t - {\tau _i}\left( t \right)} \right)\end{aligned} \]

通常把\(\alpha_i(t)={\beta_i}\left( t \right){e^{ j{\phi _i}\left( t \right)}}\)当成信道衰落系数,注意这是基带的信道。

当接收端移动的时候,令\({{\theta _i}\left( t \right)}\) 表示第 \(i\) 径信道信号的到达方向与接收机运动方向的夹角,此时接收信号可以写为

\[\begin{aligned} y\left( t \right) &= {\mathop{\rm Re}\nolimits} \left\{ {\sum\limits_{i = 0}^{N\left( t \right)} {{\beta_i}\left( t \right)s\left( {t - {\tau _i} \left( t \right)-\frac{v\mathrm{cos}\theta_i(t)\cdot t}{C}} \right){e^{j2\pi {f_c }\left( {t - {\tau _i}\left( t \right)}-\frac{v\mathrm{cos}\theta_i(t)\cdot t}{C} \right) }}} } \right\} \\ &=\operatorname{Re}\left\{\left[\sum_{i=0}^{N(t)} \beta_{i}(t) e^{-j \phi_{i}(t)} e^{-j2\pi f_m \mathrm{cos} \theta_i(t)\cdot t} s\left(t-\tau_{i}(t)-\frac{v\mathrm{cos}\theta_i(t)\cdot t}{C}\right)\right] e^{j 2 \pi f_{c} t}\right\} \\ &\approx \operatorname{Re}\left\{\left[\sum_{i=0}^{N(t)} \alpha_{i}(t) e^{-j2\pi f_m \mathrm{cos} \theta_i(t)\cdot t} s\left(t-\tau_{i}(t)\right)\right] e^{j 2 \pi f_{c} t}\right\} \\\end{aligned} \]

其中\(f_m=\frac{vf_c}{C}=\frac{v}{\lambda}\)称为最大多普勒频移。

此时等效基带信道冲激响应可以写为

\[\begin{aligned} h\left( {t,\tau } \right) &= \sum\limits_{i = 0}^{N\left( t \right)} \alpha_i(t)e^{-j2\pi f_m \mathrm{cos} \theta_i(t)\cdot t} \delta \left( {t - {\tau _i}\left( t \right)} \right)\end{aligned} \]

posted @ 2022-04-08 16:20  MayeZhang  阅读(3443)  评论(0编辑  收藏  举报