相位调制(PM)

在数字相位调制中，M个信号波形可表示为

\[\begin{aligned} s_{m}(t) &= \text{Re}\left[g(t)e^{j\frac{2\pi(m-1)}{M}}e^{j2\pi f_{c}t}\right],\quad m= 1,2,\cdots,M\\ &= g(t)\cos\left[2\pi f_{c}t + \frac{2\pi}{M}(m-1)\right]\\ &= g(t)\cos\left[\frac{2\pi}{M}(m-1)\right]\cos(2\pi f_{c}t)-g(t)\sin\left[\frac{2\pi}{M}(m-1)\right]\sin(2\pi f_{c}t)\\ &=\cos\left[\frac{2\pi}{M}(m-1)\right]\color{red}{g(t)\cos(2\pi f_{c}t)}-\sin\left[\frac{2\pi}{M}(m-1)\right]\color{red}{g(t)\sin(2\pi f_{c}t)}\\ &= \cos(\theta_{m})\color{red}{y_{1}(t)}-\sin(\theta_{m})\color{red}{y_{2}(t)} \end{aligned}\tag{1}\]

• \(g(t)\)为实信号脉冲
• \(\theta_{m}=2\pi\frac{(m-1)}{M}, m=1,2,\cdots,M\)是载波的M个可能的相位，用于传输要发送的信息

1. 正交展开

由(1)式的最后两行可以看出，\(s_{m}(t)\)被表示成了两个基函数\(\color{blue}{y_{1}(t)=g(t)\cos(2\pi f_{c}t),y_{2}(t)= -g(t)\sin(2\pi f_{c}t)}\)的线性组合，其中的两个系数分别为\(\color{blue}{\cos(\theta_{m}),\sin(\theta_{m})}\)。

又由于

\[\begin{aligned} &\quad\int_{-\infty}^{\infty}y_{1}(t)y_{2}^{*}(t)dt\\ &=-\int_{-\infty}^{\infty}g(t)\cos(2\pi f_{c}t)g(t)\sin(2\pi f_{c}t)\\ &= -\int_{-\infty}^{\infty}g^{2}(t)\cos(2\pi f_{c}t)\sin(2\pi f_{c}t)dt\\ &=-\int_{-\infty}^{\infty}g^{2}(t)\frac{1}{2}\sin(4\pi f_{c}t)\\ &\approx 0 \end{aligned} \]

可见\(y_{1}(t),y_{2}(t)\)是相互正交的。对他们分别进行能量归一化，可得到对\(s_{m}(t)\)进行正交展开的标准正交基。

\[\begin{aligned} \mathcal{E}_{y_{1}(t)}&=\int_{-\infty}^{\infty}y_{1}^{2}(t)dt\\ &=\int_{-\infty}^{\infty}g^{2}(t)\cos^{2}(2\pi f_{c}t)dt\\ &=\int_{-\infty}^{\infty}g^{2}(t)\frac{1}{2}[1+\cos(4\pi f_{c}t)]dt\\ &=\frac{1}{2}\mathcal{E}_{g}+\frac{1}{2}\int_{-\infty}^{\infty}g^{2}(t)\cos(4\pi f_{c}t)dt\\ &\approx\frac{1}{2}\mathcal{E}_{g} \end{aligned}\]

\[\begin{aligned} \mathcal{E}_{y_{2}(t)}&=\int_{-\infty}^{\infty}y_{2}^{2}(t)dt\\ &=\int_{-\infty}^{\infty}g^{2}(t)\sin^{2}(2\pi f_{c}t)dt\\ &=\int_{-\infty}^{\infty}g^{2}(t)\frac{1}{2}[1-\cos(4\pi f_{c}t)]dt\\ &=\frac{1}{2}\mathcal{E}_{g}-\frac{1}{2}\int_{-\infty}^{\infty}g^{2}(t)\cos(4\pi f_{c}t)dt\\ &\approx\frac{1}{2}\mathcal{E}_{g} \end{aligned}\]

可以得到标准正交基为：

\[\begin{aligned} \phi_{1}(t)&=\frac{y_{1}(t)}{\sqrt{\mathcal{E}_{y_{1}(t)}}}=\sqrt{\frac{2}{\mathcal{E}_{g}}}g(t)\cos(2\pi f_{c}t)\\ \phi_{2}(t)&=\frac{y_{2}(t)}{\sqrt{\mathcal{E}_{y_{2}(t)}}}=-\sqrt{\frac{2}{\mathcal{E}_{g}}}g(t)\sin(2\pi f_{c}t) \end{aligned} \tag{2} \]

将(1)式进行等价改写，并将(2)式带入可得：

\[\begin{aligned} s_{m}(t) &=\cos\theta_{m}\sqrt{\frac{\mathcal{E}_{g}}{2}}\color{red}{\sqrt{\frac{2}{\mathcal{E}_{g}}}g(t)\cos(2\pi f_{c}t)}+\sin\theta_{m}\sqrt{\frac{\mathcal{E}_{g}}{2}}\color{red}{\left(-\sqrt{\frac{2}{\mathcal{E}_{g}}}g(t)\sin(2\pi f_{c}t)\right)}\\ &=\cos\theta_{m}\sqrt{\frac{\mathcal{E}_{g}}{2}}\color{red}{\phi_{1}(t)}+\sin\theta_{m}\sqrt{\frac{\mathcal{E}_{g}}{2}}\color{red}{\phi_{2}(t)}\\ \end{aligned} \tag{3} \]

(3)式即为相位调制信号的正交展开式，相应的矢量表达式为：

\[\begin{aligned} \bold{s_{m}} &= \left[\cos\theta_{m}\sqrt{\frac{\mathcal{E}_{g}}{2}}, \sin\theta_{m}\sqrt{\frac{\mathcal{E}_{g}}{2}}\right]\\ &=\left[\sqrt{\frac{\mathcal{E}_{g}}{2}}\cos\left(\frac{2\pi}{M}(m-1)\right),\sqrt{\frac{\mathcal{E}_{g}}{2}}\sin\left(\frac{2\pi}{M}(m-1)\right)\right],\quad m=1,\cdots,M\\ \end{aligned} \tag{4} \]

2. 能量计算

信号\(s_{m}(t)\)的能量为：

\[\begin{aligned} \mathcal{E}_{m}&=\int_{-\infty}^{\infty} g^{2}(t)\cos^{2}\left[2\pi f_{c}t + \frac{2\pi}{M}(m-1)\right]dt\\ &\approx\frac{1}{2}\mathcal{E}_{g} \end{aligned} \tag{5} \]

也可以使用矢量表达式计算获得：

\[\begin{aligned} \mathcal{E}_{m} &=\left(\cos(\theta_{m})\sqrt{\frac{\mathcal{E}_{g}}{2}}\right)^{2}+\left(\sin(\theta_{m})\sqrt{\frac{\mathcal{E}_{g}}{2}}\right)^{2}\\ &=\cos^{2}(\theta_{m})\frac{\mathcal{E}_{g}}{2}+ \sin^{2}(\theta_{m})\frac{\mathcal{E}_{g}}{2}\\ &\approx\frac{1}{2}\mathcal{E}_{g} \end{aligned} \tag{6} \]

(5)式和(6)式的结果均与m无关，可见相位调制信号具有等能量的特性。

平均信号能量(Average Signal Energy)为：

\[\begin{aligned} \mathcal{E}_{avg}=\mathcal{E}_{m}=\frac{1}{2}\mathcal{E}_{g} \end{aligned} \]

平均比特信号能量为：

\[\mathcal{E}_{bavg}=\frac{\mathcal{E}_{avg}}{k}=\frac{\mathcal{E}_{g}}{2\log_{2}{M}} \]

3. 信号点距离

任意一对信号点之间的欧氏距离为：

\[\begin{aligned} d_{mn}&=\sqrt{\Vert \bold{s_{m}-s_{n}} \Vert}\\ &=\sqrt{\left(\sqrt{\frac{\mathcal{E}_{g}}{2}}(\cos\theta_{m}-\cos\theta_{n})\right)^{2}+\left(\sqrt{\frac{\mathcal{E}_{g}}{2}}(\sin\theta_{m}-\sin\theta_{n})\right)^{2}}\\ &=\sqrt{\frac{\mathcal{E}_{g}}{2}(\cos^{2}\theta_{m}+\cos^{2}\theta_{n}-2\cos\theta_{m}\cos\theta_{n}+\sin^{2}\theta_{m}+\sin^{2}\theta_{n}-2\sin\theta_{m}\sin\theta_{n})}\\ &=\sqrt{\frac{\mathcal{E}_{g}}{2}[2-2(cos\theta_{m}\cos\theta_{n}+\sin\theta_{m}\sin\theta_{n})]}\\ &=\sqrt{\mathcal{E}_{g}[1-\cos(\theta_{m}-\theta_{n})]}\\ &=\sqrt{\mathcal{E}_{g}\left[1-\cos\left(\frac{2\pi}{M}(m-n)\right)\right]} \end{aligned} \tag{7} \]

最小距离为\(|m-n|=1\)，由此可以得到：

\[d_{\min}=\sqrt{\mathcal{E}_{g}\left[1-\cos\left(\frac{2\pi}{M}\right)\right]}=\sqrt{2\mathcal{E}_{g}\sin^{2}\left(\frac{\pi}{M}\right)}\tag{8} \]

由平均比特信号能量与脉冲信号能量的关系可得：

\[\mathcal{E}_{g}=2\log_{2}{M}\mathcal{E}_{bavg} \]

带入(8)式可得：

\[d_{\min}=\sqrt{4\log_{2}{M}\times\sin^{2}(\frac{\pi}{M})\times\mathcal{E}_{bavg}}\approx2\sqrt{\frac{\pi^{2}\log_{2}M}{M^{2}}\mathcal{E}_{bavg}} \]