带通和低通随机过程

带通随机过程定义
带通随机过程\(X(t)\)为实随机过程过并且其功率谱分布在中心频率\(f_{0}\)附近，也就是说：

\[S(f) = 0, \quad |f\pm f_{0}|>W \]

基本假设
带通随机过程\(X(t)\)是广义平稳的。并且，其复等效低通随机过程\(X_{l}(t)\)也是广义平稳的。换句话说：

• \(X_{i}(t),X_{q}(t)\)是广义平稳的；
• \(X_{i}(t),X_{q}(t)\)联合广义平稳；

基于上述基本假设，可以得到如下性质：

P1)
如果带通随机过程\(X(t)\)的均值为零，那么其复等效低通随机过程的实部(\(X_{i}(t)\))和虚部(\(X_{q}(t)\))的均值也为零。
证明：
根据等效低通与带通信号的关系：

\[X(t)=\text{Re}[X_{l}(t)e^{j2\pi f_{c}t}]=\text{Re}\big[\big(X_{i}(t)+jX_{q}(t)\big)\big(\cos(2\pi f_{c}t)+j\sin(2\pi f_{c}t)\big)\big] \]

可知

\[X(t) = X_{i}(t)\cos(2\pi f_{c}t)-X_{q}(t)\sin(2\pi f_{c}t) \]

如果\(E[X(t)]=0\)，那么

\[\begin{aligned} &E[X_{i}(t)\cos(2\pi f_{c}t)-X_{q}(t)\sin(2\pi f_{c}t)]\\ &=E[X_{i}(t)]\cos(2\pi f_{c}t)-E[X_{q}(t)]\sin(2\pi f_{c}t)=0 \end{aligned}\]

上式中由于\(\cos(2\pi f_{c}t),\sin(2\pi f_{c}t)\)不恒定为零，而整体恒等于零。可以得出：

\[E[X_{i}(t)]=E[X_{q}(t)]=0 \]

P2) \(\left \{ \begin{aligned}&R_{X_{i}}(\tau)=R_{X_{q}}(\tau)\\&R_{X_{i},X_{q}}(\tau) =-R_{X_{q},X_{i}}(\tau)\end{aligned}\right.\)
证明：

\[\begin{aligned} R_{X}(\tau) &= E[X(t+\tau)X(t)]\\ &=E\left[\text{Re}[X_{l}(t+\tau)e^{j2\pi f_{c}(t+\tau)}]\text{Re}[X_{l}(t)e^{j2\pi f_{c}t}]\right]\\ &=E\big[[X_{i}(t+\tau)\cos(2\pi f_{c}(t+\tau))-X_{q}(t+\tau)\sin(2\pi f_{c}(t+\tau))][X_{i}(t)\cos(2\pi f_{c}t)-X_{q}(t)\sin(2\pi f_{c}t)]\big]\\ &=E[X_{i}(t+\tau)X_{i}(t)\cos(2\pi f_{c}(t+\tau))\cos(2\pi f_{c}t)]-E[X_{i}(t+\tau)\cos(2\pi f_{c}(t+\tau))X_{q}(t)\sin(2\pi f_{c}t)]\\ &\quad -E[X_{q}(t+\tau)\sin(2\pi f_{c}(t+\tau))X_{i}(t)\cos(2\pi f_{c}t)]+E[X_{q}(t+\tau)\sin(2\pi f_{c}(t+\tau))X_{q}(t)\sin(2\pi f_{c}t)]\\ &= R_{X_{i}}(\tau)\left[\frac{1}{2}\big(\cos(2\pi f_{c}\tau)+\cos(2\pi f_{c}(2t+\tau))\big)\right]-\\ &\quad R_{X_{i},X_{q}}(\tau)\left[\frac{1}{2}\big(\sin(2\pi f_{c}(2t+\tau))-\sin(2\pi f_{c}\tau)\big)\right]-\\ &\quad R_{X_{q},X_{i}}(\tau)\left[\frac{1}{2}\big(\sin(2\pi f_{c}(2t+\tau))+\sin(2\pi f_{c}\tau)\big)\right]+\\ &\quad R_{X_{q}}(\tau)\left[\frac{1}{2}\big(\cos(2\pi f_{c}\tau)-\cos(2\pi f_{c}(2t+\tau))\big)\right]\\ &=\frac{1}{2}\cos(2\pi f_{c}\tau)[R_{X_{i}}(\tau)+R_{X_{q}}(\tau)]+\color{blue}{\frac{1}{2}\cos(2\pi f_{c}(2t+\tau))[R_{X_{i}}(\tau)-R_{X_{q}}(\tau)]}+\\ &\quad \frac{1}{2}\sin(2\pi f_{c}\tau)[R_{X_{i},X_{q}}(\tau)-R_{X_{q},X_{i}}(\tau)]-\color{blue}{\frac{1}{2}\sin(2\pi f_{c}(2t+\tau))[R_{X_{i},X_{q}}(\tau)+R_{X_{q},X_{i}}(\tau)]} \end{aligned} \]

上等式的左边是关于\(\tau\)的函数，那么右边也应该只是\(\tau\)的函数，也就是说含有变量\(t\)的项应当为0。那么可以得出

\[\begin{aligned} \frac{1}{2}\cos(2\pi f_{c}(2t+\tau))[R_{X_{i}}(\tau)-R_{X_{q}}(\tau)] &= 0\\ \frac{1}{2}\sin(2\pi f_{c}(2t+\tau))[R_{X_{i},X_{q}}(\tau)+R_{X_{q},X_{i}}(\tau)] & =0 \end{aligned} \]

从而有

\[\left \{ \begin{aligned}&R_{X_{i}}(\tau)=R_{X_{q}}(\tau)\\&R_{X_{i},X_{q}}(\tau) =-R_{X_{q},X_{i}}(\tau)\end{aligned}\right. \]

并且有：

\[R_{X}(\tau) = R_{X_{i}}(\tau)\cos(2\pi f_{c}\tau)+R_{X_{i},X_{q}}(\tau)\sin(2\pi f_{c}\tau) \]

\[R_{X}(\tau) = R_{X_{q}}(\tau)\cos(2\pi f_{c}\tau)-R_{X_{q},X_{i}}(\tau)\sin(2\pi f_{c}\tau) \]

P3) \(R_{X}(\tau)=\frac{1}{2}\text{Re}[R_{X_{l}}(\tau)e^{j2\pi f_{c}t}]\)
证明：
根据带通信号与等效低通信号之间的关系，我们直接猜想

\[R_{X}(\tau)=\text{Re}[R_{X_{l}}(\tau)e^{j2\pi f_{c}\tau}] \]

下面进行验证。根据自相关函数的定义式有

\[\begin{aligned} R_{X_{l}}(\tau)&=E[X_{l}(t+\tau)X_{l}^{*}(t)]\\ &=E\big[\big(X_{i}(t+\tau)+jX_{q}(t+\tau)\big)\big(X_{i}(t)-jX_{q}(t)\big)\big]\\ &=R_{X_{i}}(\tau)+R_{X_{q}}(\tau)-jR_{X_{i},X_{q}}(\tau)+jR_{X_{q},X_{i}}(\tau)\\ &\overset{P2}{=}2R_{X_{i}}(\tau)-j2R_{X_{i},X_{q}}(\tau) \end{aligned} \]

那么

\[\begin{aligned} \text{Re}[R_{X_{l}}(\tau)e^{j2\pi f_{c}\tau}]&=\text{Re}\Big[\Big(2R_{X_{i}}(\tau)-j2R_{X_{i},X_{q}}(\tau)\Big)\Big(\cos(2\pi f_{c}\tau)+j\sin(2\pi f_{c}\tau)\Big)\Big]\\ &=2R_{X_{i}}(\tau)\cos(2\pi f_{c}\tau)+2R_{X_{i},X_{q}}(\tau)\sin(2\pi f_{c}\tau) \end{aligned}\]

结合P2证明中的

\[R_{X}(\tau) = R_{X_{i}}(\tau)\cos(2\pi f_{c}\tau)+R_{X_{i},X_{q}}(\tau)\sin(2\pi f_{c}\tau) \]

可以发现，我们的猜想是错误的，猜想的结果比正确的结果大了一倍。因此需要乘以因子\(1/2\)。从而可以发现

\[R_{X}(\tau)=\frac{1}{2}\text{Re}[R_{X_{l}}(\tau)e^{j2\pi f_{c}\tau}] \]

P4) \(S_{X}(f)=\frac{1}{4}[S_{X_{l}}(f-f_{c})+S_{X_{l}}^{*}(-f-f_{c})]\)
证明：
由P3可知

\[R_{X}(\tau) = \frac{1}{2}\text{Re}[R_{X_{l}}(\tau)e^{j2\pi f_{c}\tau}] \]

进而可得：

\[R_{X}(\tau) = \frac{1}{4}[R_{X_{l}}(\tau)e^{j2\pi f_{c}\tau}+R_{X_{l}}^{*}(\tau)e^{-j2\pi f_{c}\tau}] \]

根据傅里叶变换的性质：

\[x^{*}(t) \leftrightarrow X^{*}(-f)\quad \text{共轭性} \]

\[x(t)e^{j2\pi f_{0}t} \leftrightarrow X(f-f_{0})\quad \text{频移性(调制性)} \]

频移性以及共轭性，可以得到：

\[\begin{aligned} S_{X}(f)&=\frac{1}{4}[S_{X_{l}}(f-f_{c})+S_{X_{l}}(-(f+f_{0}))]\\ &=\frac{1}{4}[S_{X_{l}}(f-f_{c})+S_{X_{l}}(-f-f_{0})] \end{aligned}\]

P5) 若\(X_{i}(t)\)或\(X_{q}(t)\)的均值为零，那么\(X_{i}(t)\)和\(X_{q}(t)\)不相关。
证明：
根据互相关的定义，有

\[R_{X_{i},X_{q}}(\tau)=E[X_{i}(t+\tau)X_{q}(t)]=E[X_{q}(t)X_{i}(t+\tau)]=R_{X_{q},X_{i}}(-\tau) \]

根据P2)有

\[R_{X_{i},X_{q}}(\tau)=-R_{X_{q},X_{i}}(\tau)=-R_{X_{i},X_{q}}(-\tau) \]

因此，

\[R_{X_{i},X_{q}}(0) = 0 \]

从而

\[E[X_{i}(t)X_{q}(t)] = 0 = E[X_{i}(t)]E[X_{q}(t)] \]

P6) 若\(S_{X_{l}}(-f)=S_{X_{l}}^{*}(f)\)，并且\(X_{i}(t+\tau)\)或\(X_{q}(t)\)的均值为零，其中\(\tau\)为任意值，那么\(X_{i}(t+\tau)\)不相关\(X_{q}(t)\)。
证明：
由P3可知

\[R_{X_{l}}(\tau) = 2R_{X_{i}}(\tau)-j2R_{X_{i},X_{q}}(\tau) \]

同时\(S_{X_{l}}(-f)=S_{X_{l}}^{*}(f)\)意味着\(S_{X_{l}}(f)\)具有共轭对称性，那么\(R_{X_{l}}(\tau)\)为实数，从而得出

\[R_{X_{i},X_{q}}(\tau)=0=E[X_{i}(t+\tau)X_{q}(\tau)]=E[X_{i}(t+\tau)]E[X_{q}(t)] \]