CPM信号的功率谱密度
1. CPM的功率谱密度
恒定幅度CPM信号可以表示为：
\[s(t;\bold{I})=\sqrt{\frac{2\mathcal{E}}{T}}\text{Re}[e^{j\phi(t;\bold{I})}e^{j2\pi f_{c}t}]\tag{1.1}
\]
其中
\[\phi(t;\bold{I})=2\pi h \sum_{k=-\infty}^{\infty}I_{k}q(t-kT)\tag{1.2}
\]
序列\(\{I_{n}\}\)中每一个符号取M个电平值\(\{\pm1,\pm3,\cdots,\pm(M-1)\}\)之一。这些符号统计独立并具有相同的分布，且具有先验概率
\[P_{n} = P(I_{k}=n),\quad n=\pm1,\pm3,\cdots,\pm(M-1)\tag{1.3}
\]
CPM带通信号的等效低通为：
\[v_{l}(t) = e^{j\phi(t;\bold{I})}\tag{1.4}
\]
等效低通信号的自相关函数为：
\[\begin{aligned}
R_{v_{l}}(t+\tau,t)&=E\left[e^{j\phi(t+\tau;\bold{I})}e^{-j\phi(t;\bold{I})}\right]\\
&=E\left[\exp\left(j2\pi h\sum_{k=-\infty}^{\infty}I_{k}[q(t+\tau-kT)-q(t-kT)]\right)\right]\\
&=E\left[\prod_{k=-\infty}^{\infty}\exp\big(j2\pi h\color{blue}{I_{k}}[q(t+\tau-kT)-q(t-kT)]\big)\right]\\
&=\prod_{k=-\infty}^{\infty}\left[\sum_{\color{blue}{n}=-(M-1)\\ \quad\text{n is odd}}^{M-1}\color{blue}{P_{n}}\exp\Big(j2\pi h\color{blue}{n}[q(t+\tau-kT)-q(t-kT)]\Big)\right]
\end{aligned}\tag{1.5}\]
那么平均自相关函数为：
\[\begin{aligned}
\overline{R}_{v_{l}}(\tau) &= \frac{1}{T}\int_{0}^{T}R_{v_{l}}(t+\tau,t)dt\\
&=\frac{1}{T}\int_{0}^{T} \prod_{k=-\infty}^{\infty}\left[\sum_{n=-(M-1)\\ \quad\text{n is odd}}^{M-1}P_{n}e^{j2\pi hn[\color{red}{q(t+\tau-kT)-q(t-kT)}]}\right] dt
\end{aligned}\tag{1.6}\]
式(5),(6)中看似有无穷多项相乘，实则不然，因为\(q(t)\)具有如下性质，
\[q(t) = \left \{ \begin{aligned} &0,\quad &t\le0\\ &\text{Value that increase with t}, \quad &0<t<LT \\ &\frac{1}{2}, \quad &t\ge LT\end{aligned} \right.\tag{1.7}
\]
取不同\(k\)值，观察(1.6)式中红色部分的值：
-
\(q(t-kT)\)
- 观察(1.6)式的积分范围，可知\(t\)的取值范围为\([0,T)\)；
- \(k\ge 1\)，则\(t-kT<0\)，那么\(q(t-kT)=0\);
- \(k=0\)， \(t-kT=t\in[0,T)<LT\)，那么\(0\le q(t-kT)<\frac{1}{2}\)；
- \(k\le-L\)，\(t-kT=t+LT\ge LT\)，那么\(q(t-kT)=\frac{1}{2}\);
- \(k=-L+1\)，\(t-kT=t+(L-1)T< LT\)，那么\(0<q(t-kT)<\frac{1}{2}\);
-
\(q(t+\tau-kT)\)
- 假设\(\color{red}{\tau=\xi+mT\ge0,\xi\in[0,T),m=0,1,\cdots}\)(后面解释为什么假设\(\tau>0\)是合理的)，即，将\(\tau\)分解为\(T\)的整数倍部分以及小数倍部分；
- \(t+\tau-kT=t+\xi+mT-kT=t+\xi-(k-m)T\)，\(t+\xi\in[0,2T)\)
- \(k-m\ge 2\rightarrow k\ge m+2\rightarrow t+\xi-(k-m)T<0\)，那么\(q(t+\tau-kT)=0\)
- \(k-m = 1\rightarrow k=m+1\rightarrow t+\xi-(k-m)T=t+\xi-T\in[-T,T]\le LT\)，那么\(q(t+\tau-kT)=0或者0<q(t+\tau-kT)<\frac{1}{2}\)
- \(k=m-L\rightarrow t+\xi-(k-m)T=t+\xi+LT\ge LT\)，那么\(q(t+\tau-kT)=\frac{1}{2}\)
- \(k=m-L+1\rightarrow t+\xi-(k-m)T=t+\xi+(L-1)T\in[(L-1)T,(L+1)T)\)，那么\(q(t+\tau-kT)=\frac{1}{2}或者0<q(t+\tau-kT)<\frac{1}{2}\)
将上述讨论结果总结如下图：
可以发现，当\(k\ge m+1\)时，\(q(t+\tau-kT)-q(t-kT)=0-0=0\)，此时
\[\sum_{n=-(M-1)\\ \quad\text{n is odd}}^{M-1}P_{n}e^{j2\pi hn[\color{red}{q(t+\tau-kT)-q(t-kT)}]}=\sum_{n=-(M-1)\\ \quad\text{n is odd}}^{M-1}P_{n}=1
\]
同理，当\(k<1-L\)时，\(q(t+\tau-kT)-q(t-kT)=\frac{1}{2}-\frac{1}{2}=0\)，此时
\[\sum_{n=-(M-1)\\ \quad\text{n is odd}}^{M-1}P_{n}e^{j2\pi hn[\color{red}{q(t+\tau-kT)-q(t-kT)}]}=\sum_{n=-(M-1)\\ \quad\text{n is odd}}^{M-1}P_{n}=1
\]
因此，式(1.6)简化为
\[\begin{aligned}
\overline{R}_{v_{l}}(\tau) &= \frac{1}{T}\int_{0}^{T}R_{v_{l}}(t+\tau,t)dt\\
&=\frac{1}{T}\int_{0}^{T} \prod_{k=1-L}^{m+1}\left[\sum_{n=-(M-1)\\ \quad\text{n is odd}}^{M-1}P_{n}e^{j2\pi hn[\color{red}{q(t+\tau-kT)-q(t-kT)}]}\right] dt
\end{aligned}\tag{1.8}\]
上面假设了\(\tau\ge 0\)，这样做是合理的，因为\(\overline{R}_{v_{l}}(t)\)具有共轭对称性。那么就可以根据\(\tau\ge0\)直接得到\(\tau<0\)的部分。\(\overline{R}_{v_{l}}(t)\)的共轭对称性证明如下：
\[\begin{aligned}
\overline{R}^{*}_{v_{l}}(\tau)&=\frac{1}{T}\int_{0}^{T}E\left[e^{-j2\pi h\sum_{k=-\infty}^{\infty}I_{k}[q(t+\tau-kT)-q(t-kT)]}\right]dt\\
&=\frac{1}{T}\int_{0}^{T}E\left[e^{j2\pi h\sum_{k=-\infty}^{\infty}I_{k}[q(t-kT)-q(t+\tau-kT)]}\right]dt\\
&\overset{v=t+\tau}{=}\frac{1}{T}\int_{\tau}^{T+\tau}E\left[e^{j2\pi h\sum_{k=-\infty}^{\infty}I_{k}[q(v-\tau-kT)-q(v-kT)]}\right]dv\\
&=\frac{1}{T}\int_{0}^{T}E\left[e^{j2\pi h\sum_{k=-\infty}^{\infty}I_{k}[q(v-\tau-kT)-q(v-kT)]}\right]dv\\
&=\overline{R}_{v_{l}}(-\tau)
\end{aligned}
\]
上面得到CPM信号的平均自相关函数如(1.8)式所示，下面对其进行傅里叶变换得到平均功率谱密度：
\[\begin{aligned}
\overline{S}_{v_{l}}(f) &=\int_{-\infty}^{\infty}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau\\
&=\int_{-\infty}^{0}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau+\int_{0}^{\infty}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau\\
&=\int_{0}^{\infty}\overline{R}_{v_{l}}(-\tau)e^{j2\pi f\tau}d\tau+\int_{0}^{\infty}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau\\
&=\int_{0}^{\infty}[\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}]^{*}d\tau+\int_{0}^{\infty}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau\\
&=2\text{Re}\left[\color{red}{\int_{0}^{\infty}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau}\right]
\end{aligned}
\tag{1.9}
\]
下面对(1.9)式中的红色部分进行求解：
\[\begin{aligned}
&\int_{0}^{\infty}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau\\
&=\int_{0}^{LT}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau+\int_{LT}^{\infty}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau\\
&=\color{blue}{\int_{0}^{LT}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau}+\color{red}{\sum_{m=L}^{\infty}\int_{mT}^{(m+1)T}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau}
\end{aligned}
\tag{1.10}
\]
上式对积分区间的划分方式，是因为当\(\tau>LT\)时，\(\overline{R}_{v_{l}}(\tau)\)中的\(q(t+\tau-kT)\)和\(q(t-kT)\)交叠的区间易于计算，如下图所示：
式(1.10)中的蓝色部分为有限积分，能够轻易获得。下面重点考虑红色部分，此时的\(\tau>LT\)，平均自相关函数可以进一步简化：
\[\begin{aligned}
\overline{R}_{v_{l}}(\tau)&=\frac{1}{T}\int_{0}^{T} \prod_{k=1-L}^{m+1}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[q(t+\tau-kT)-q(t-kT)]}\right]dt\\
&=\frac{1}{T}\int_{0}^{T}\left(\prod_{k=1-L}^{0}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[q(t+\tau-kT)-q(t-kT)]}\right]\times \\
\prod_{k=1}^{m-L}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[q(t+\tau-kT)-q(t-kT)]}\right]\times\\
\prod_{k=m+1-L}^{m+1}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[q(t+\tau-kT)-q(t-kT)]}\right]\right)dt\\
&=\frac{1}{T}\int_{0}^{T}\left(\prod_{k=1-L}^{0}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[1/2-q(t-kT)]}\right]\times \\
\prod_{k=1}^{m-L}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[1/2-0]}\right]\times\\
\prod_{k=m+1-L}^{m+1}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[q(t+\tau-kT)-0]}\right]\right)dt\\
&=\frac{1}{T}\int_{0}^{T}\left(\prod_{k=1-L}^{0}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[1/2-q(t-kT)]}\right]\times\\
\prod_{k'=1-L}^{1}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[q(t+\xi-k'T)]}\right]\quad (k'=k-m)\\
\prod_{k=1}^{m-L}\left[\sum_{n\in S}P_{n}e^{j\pi hn}\right]\right)dt\\
&=[\Phi_{I}(h)]^{m-L}\lambda(\xi)
\end{aligned}
\]
其中：
\[\Phi_{I}(h) = \sum_{n\in S} P_{n}e^{j\pi hn}
\]
\[\lambda(\xi)=\frac{1}{T}\int_{0}^{T}\left(\prod_{k=1-L}^{0}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[1/2-q(t-kT)]}\right]\prod_{k'=1-L}^{1}\left[\sum_{n\in S}P_{n}e^{j2\pi hn[q(t+\xi-k'T)]}\right]\right)dt
\]
从而可以计算(1.10)式的红色部分：
\[\begin{aligned}
&\sum_{m=L}^{\infty}\int_{mT}^{(m+1)T}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau\\
&=\sum_{m=L}^{\infty}\int_{mT}^{(m+1)T}[\Phi_{I}(h)]^{m-L}\lambda(\xi)e^{-j2\pi f\tau}d\tau\\
&=\sum_{m=L}^{\infty}\int_{0}^{T}[\Phi_{I}(h)]^{m-L}\lambda(\xi)e^{-j2\pi f(\xi+mT)}d\xi\quad (\xi = \tau -mT)\\
&=\color{blue}{\left(\sum_{m=L}^{\infty}[\Phi_{I}(h)]^{m-L}e^{-j2\pi fmT}\right)}\left(\int_{0}^{T}\lambda(\xi)e^{-j2\pi f\xi}d\xi\right)\\
\end{aligned}
\tag{1.11}
\]
式(1.11)中的蓝色部分其实是一个等比级数的和式，当\(|\Phi_{I}(h)|<1\)时，其结果为：
\[\begin{aligned}
\sum_{m=L}^{\infty}[\Phi_{I}(h)]^{m-L}e^{-j2\pi fmT}&=\sum_{m=L}^{\infty}\left([\Phi_{I}(h)]^{m-L}e^{-j2\pi f(m-L)T}\right)e^{-j2\pi fLT}\\
&=e^{-j2\pi fLT}\sum_{n=0}^{\infty}[\Phi_{I}(h)]^{n}e^{-j2\pi fnT}\\
&=\frac{e^{-j2\pi fLT}}{1-\Phi_{I}(h)e^{-j2\pi fT}}
\end{aligned}
\tag{1.12}\]
当\(|\Phi_{I}(h)|=1\)时，令
\[\Phi_{I}(h) = e^{j2\pi \nu}, \quad 0\le \nu <1
\]
此时
\[\begin{aligned}
\sum_{m=L}^{\infty}[\Phi_{I}(h)]^{m-L}e^{-j2\pi fmT}&=e^{-j2\pi fLT}\sum_{n=0}^{\infty}[\Phi_{I}(h)]^{n}e^{-j2\pi fnT}\\
&=e^{-j2\pi fLT}\sum_{n=0}^{\infty}[e^{j2\pi \nu}]^{n}e^{-j2\pi fnT}\\
&=e^{-j2\pi fLT}\sum_{n=0}^{\infty}e^{-j2\pi T(f-\nu/T)n}\\
&=e^{-j2\pi fLT}\left(\frac{1}{2}+\frac{1}{2T}\sum_{n=-\infty}^{\infty}\left[\delta(f-\frac{\nu+n}{T})-j\frac{1}{\pi(f-(\nu+n)/T)}\right]\right)
\end{aligned}
\tag{1.13}\]
式(1.13)的最后一行利用了模为1的等比复级数的和。
将式(1.12)和(1.13)带入式(1.11)可以得到：
\[\begin{aligned}
&\sum_{m=L}^{\infty}\int_{mT}^{(m+1)T}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}d\tau\\
&=\left\{\begin{aligned}
&\left(\frac{e^{-j2\pi fLT}}{1-\phi_{I}(h)e^{-j2\pi fT}}\right)\left(\int_{0}^{T}\lambda(\xi)e^{-j2\pi f\xi}d\xi\right),\quad &|\Phi_{I}(h)|<1\\
&e^{-j2\pi fLT}\left(\frac{1}{2}+\frac{1}{2T}\sum_{n=-\infty}^{\infty}\left[\delta(f-\frac{\nu+n}{T})-j\frac{1}{\pi(f-(\nu+n)/T)}\right]\right)\left(\int_{0}^{T}\lambda(\xi)e^{-j2\pi f\xi}d\xi\right),\quad &|\Phi_{I}(h)|=1
\end{aligned}\right.
\end{aligned}
\tag{1.14}
\]
将式(1.14)带入式(1.10)，然后再将式(1.10)带入式(1.9)即可得出CPM信号的平均功率谱密度。例如当\(\Phi_{I}(h)|<1\)时，平均功率谱密度为：
\[\overline{S}_{v_{l}}(f)=2\text{Re}\left[\int_{0}^{LT}\overline{R}_{v_{l}}(\tau)e^{-j2\pi f\tau}+\left(\frac{e^{-j2\pi fLT}}{1-\phi_{I}(h)e^{-j2\pi fT}}\right)\left(\int_{0}^{T}\lambda(\xi)e^{-j2\pi f\xi}d\xi\right)\right]
\]
其中\(\overline{R}_{v_{l}}(\tau),\tau\in[0,LT)\)表达式为：
\[\overline{R}_{v_{l}}(\tau)=\frac{1}{T}\int_{0}^{T} \prod_{k=1-L}^{m+1}\left[\sum_{n=-(M-1)\\ \quad\text{n is odd}}^{M-1}P_{n}e^{j2\pi hn[\color{red}{q(t+\tau-kT)-q(t-kT)}]}\right] dt
\]
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