Jakes 模型
前面我們介紹了多徑信道合成信號可表示為：
r(t)=Re{∑i=0N(t)?1ai(t)u(t?τi(t))ej2πfc(t?τi(t))+?Di(t)}
r(t)=Re \left\{\sum_{i=0}^{N(t)-1}a_{i}(t)u(t-\tau_{i}(t))e^{j2\pi f_{c}(t-\tau_{i}(t))+\phi_{D_{i}}(t)} \right\}
r(t)=Re????i=0∑N(t)?1?ai?(t)u(t?τi?(t))ej2πfc?(t?τi?(t))+?Di??(t)????
假設窄帶模型這個在OFDM系統中是可以滿足的即時延擴展遠小于符號長度，這樣符號時間內u(t?τi(t))=u(t)u(t-\tau_{i}(t))=u(t)u(t?τi?(t))=u(t),為了更好的描述基帶信號我們假設發射信號只是相位為?0\phi_{0}?0?未調制信號。
r(t)=Re{∑i=0N(t)?1ai(t)u(t)e?j2πfcτi(t)+?Di(t)}=∑i=0N(t)?1ai(t)e?j2πfcτi(t)+?Di(t)+?0
\begin{align*}r(t)&=Re \left\{\sum_{i=0}^{N(t)-1}a_{i}(t)u(t)e^{-j2\pi f_{c}\tau_{i}(t)+\phi_{D_{i}}(t)}\right\}\\&=\sum_{i=0}^{N(t)-1}a_{i}(t)e^{-j2\pi f_{c}\tau_{i}(t)+\phi_{D_{i}}(t)+\phi_{0}}
\end{align*}
r(t)?=Re????i=0∑N(t)?1?ai?(t)u(t)e?j2πfc?τi?(t)+?Di??(t)????=i=0∑N(t)?1?ai?(t)e?j2πfc?τi?(t)+?Di??(t)+?0??
這里我們假設觀察時間內變化足夠緩慢：ai(t)=ai,τi(t)=τi,fDi(t)=fDi，?Di(t)=2πfDita_{i}(t)=a_{i},\tau_{i}(t)=\tau_{i},f_{D_{i}}(t)=f_{D_{i}}，\phi_{D_{i}}(t)=2\pi f_{D_{i}}tai?(t)=ai?,τi?(t)=τi?,fDi??(t)=fDi??，?Di??(t)=2πfDi??t，則可使用信道沖擊響應替換如下：
h(t)=∑i=0N(t)?1aie?j2πfcτi+2πfDit+?0=∑i=0N(t)?1aie?j?i(t)
\begin{align*}h(t) &=\sum_{i=0}^{N(t)-1}a_{i}e^{-j2\pi f_{c}\tau_{i}+2\pi f_{D_{i}}t+\phi_{0}}\\&=\sum_{i=0}^{N(t)-1}a_{i}e^{-j\phi_{i}(t)}\\\end{align*}
h(t)?=i=0∑N(t)?1?ai?e?j2πfc?τi?+2πfDi??t+?0?=i=0∑N(t)?1?ai?e?j?i?(t)?
其中?i(t)=2πfcτi?2πfDit??0\phi_{i}(t)=2\pi f_{c}\tau_{i}-2\pi f_{D_{i}}t-\phi_{0}?i?(t)=2πfc?τi??2πfDi??t??0?
Rh(Δt)=E[h(t)h?(t+Δt)]=E[∑iN(t)?1∑jN(t)?1aie?j?i(t)aj?ej?j(t+Δt)]=E[∑iN(t)?1∑jN(t)?1aiaj?e?j?i(t)ej?j(t+Δt)]=E[∑iN(t)?1∑jN(t)?1aiaj?e?j(2πfcτi?2πfDit??0)+j(2πfcτj?2πfDj(t+Δt)??0)]=E[∑i=0N(t)∣ai∣2e?j2πfDiΔt]=E[∑i=0N(t)∣ai∣2e?j2πfmcos?θiΔt]
\begin{align*}
R_{h}(\Delta t)&=E\left[ h(t)h^{*}(t+\Delta t)\right]\\
&=E\left[\sum_{i}^{N(t)-1}\sum_{j}^{N(t)-1}a_{i}e^{-j\phi_{i}(t)}a_{j}^{*}e^{j\phi_{j}(t+\Delta t)} \right]\\
&=E\left[\sum_{i}^{N(t)-1}\sum_{j}^{N(t)-1}a_{i}a_{j}^{*}e^{-j\phi_{i}(t)}e^{j\phi_{j}(t+\Delta t)} \right]\\
&=E\left[\sum_{i}^{N(t)-1}\sum_{j}^{N(t)-1}a_{i}a_{j}^{*}e^{-j(2\pi f_{c}\tau_{i}-2\pi f_{D_{i}}t-\phi_{0})+j(2\pi f_{c}\tau_{j}-2\pi f_{D_{j}}(t+\Delta t)-\phi_{0})} \right]\\
&=E\left[ \sum_{i=0}^{N(t)}|a_{i}|^2e^{-j2\pi f_{D_{i}}\Delta t} \right]\\
&=E\left[ \sum_{i=0}^{N(t)}|a_{i}|^2e^{-j2\pi f_{m}\cos \theta_{i}\Delta t} \right]
\end{align*}
Rh?(Δt)?=E[h(t)h?(t+Δt)]=E?i∑N(t)?1?j∑N(t)?1?ai?e?j?i?(t)aj??ej?j?(t+Δt)?=E?i∑N(t)?1?j∑N(t)?1?ai?aj??e?j?i?(t)ej?j?(t+Δt)?=E?i∑N(t)?1?j∑N(t)?1?ai?aj??e?j(2πfc?τi??2πfDi??t??0?)+j(2πfc?τj??2πfDj??(t+Δt)??0?)?=E?i=0∑N(t)?∣ai?∣2e?j2πfDi??Δt?=E?i=0∑N(t)?∣ai?∣2e?j2πfm?cosθi?Δt??
當i≠ji\neq ji=j時ai,aj,?i,?ja_{i},a_{j},\phi_{i},\phi_{j}ai?,aj?,?i?,?j?相互獨立,fDi=fmcos?θif_{D_{i}}=f_{m}\cos\theta_{i}fDi??=fm?cosθi? ，功率歸一化E{∣ai∣2}=1E\{|a_{i}|^2\}=1E{∣ai?∣2}=1,當存在N條路徑N→∞，θ∽(0,2π)N \to \infty，\theta \backsim(0,2\pi)N→∞，θ∽(0,2π)。看文獻好像大于8條多徑就可以了。
Rh(Δt)=12π∫?ππe?j2πfmcos?θΔtdθ=1π∫0πe?j2πfmcos?θΔtdθ=J0(2πfmΔt)
\begin{align*}R_{h}(\Delta t)&=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-j2\pi f_{m}\cos\theta\Delta t} d\theta\\&=\frac{1}{\pi}\int_{0}^{\pi}e^{-j2\pi f_{m}\cos\theta\Delta t} d\theta\\&=J_{0}(2\pi f_{m}\Delta t)
\end{align*}
Rh?(Δt)?=2π1?∫?ππ?e?j2πfm?cosθΔtdθ=π1?∫0π?e?j2πfm?cosθΔtdθ=J0?(2πfm?Δt)?
0 階貝塞爾函數性質
J0(x)=1π∫0πe?jxcos?θdθJn(?x)=(?1)nJn(x)J0(x)=J0(?x)J_{0}(x)=\frac{1}{\pi}\int_{0}^{\pi}e^{-jx\cos\theta}d\theta\\J_{n}(-x)=(-1)^{n}J_{n}(x)\\J_{0}(x)=J_{0}(-x)
J0?(x)=π1?∫0π?e?jxcosθdθJn?(?x)=(?1)nJn?(x)J0?(x)=J0?(?x)
上面我的得到了多徑多普勒信道下的時域相關性描述。
