调频：双频调制的推导

设双频信号为
$$f(t)=A_{m1}\cos {\omega }_{m1}t+A_{m2}\cos {\omega }_{m2}t$$
由调频信号的一般表达式可得：
$$\begin{array}{rl}{S}_{FM}(t)& =A\cos \left[{\omega }_{c}t+\frac{K_{FM}{A}_{m1}}{{\omega }_{m1}}\sin {\omega }_{m1}t+\frac{K_{FM}{A}_{m2}}{{\omega }_{m2}}\sin {\omega }_{m2}t\right]\\ & =A\cos \left[{\omega }_{c}t+{\beta }_{FM1}\sin {\omega }_{m1}t+{\beta }_{FM2}\sin {\omega }_{m2}t\right]\end{array}$$
其中调频指数
$$\begin{gathered} {\beta }_{FM1}=\frac{K_{FM}{A}_{m1}}{{\omega }_{m1}} \\ {\beta }_{FM2}=\frac{K_{FM}{A}_{m2}}{{\omega }_{m2}} \end{gathered}$$
引入相量：
$$\begin{array}{rl}{\dot{S}}_{FM}(t)& =A\exp j\left[{\omega }_{c}t+{\beta }_1\sin {\omega }_{1}t+{\beta }_2\sin {\omega }_{2}t\right]\\ & =A\exp (j{\omega }_{c}t)\cdot \exp ({\beta }_1\sin {\omega }_{1}t)\cdot \exp ({\beta }_2\sin {\omega }_{2}t)\end{array}$$
对第二项展开，可由第一类Bessel函数表示
$$\begin{array}{rl}\exp ({\beta }_1\sin {\omega }_{1}t)& =\cos ({\beta }_1\sin {\omega }_{1}t)+j\sin ({\beta }_1\sin {\omega }_{1}t)\\ & =\sum _{n=-\infty }^{\infty }{J}_{2n}({\beta }_1)\cos 2n{\omega }_{1}t+j\cdot \sum _{n=-\infty }^{\infty }{J}_{2n-1}({\beta }_1)\sin (2n-1){\omega }_{1}t\end{array}$$

结合Bessel函数性质：
$$J_{-n}(x)=(-1{)}^n{J}_n(x)$$
和三角函数奇偶性，得：
$$\begin{gathered} J_{-(2n-1)}({\beta }_1)\cos (2n-1){\omega }_{1}t=-J_{2n-1}({\beta }_1)\cos (2n-1){\omega }_{1}t \\ {J}_{-2n}({\beta }_1)\sin (-2n{\omega }_{1}t)=-J_{2n}({\beta }_1)\sin (2n{\omega }_{1}t) \end{gathered}$$
在原式基础上添加一些和为0的项，化为：
$$\begin{array}{rl}\exp ({\beta }_1\sin {\omega }_{1}t)=& \sum _{n=-\infty }^{\infty }{J}_{2n}({\beta }_1)\cos 2n{\omega }_{1}t+\stackrel{这一项正负相消为0}{\overbrace{\sum _{n=-\infty }^{\infty }{J}_{2n-1}({\beta }_1)\cos (2n-1){\omega }_{1}t}}+\\ & \underset{这一项正负抵消为0}{\underbrace{j\cdot \sum _{n=-\infty }^{\infty }{J}_{2n}({\beta }_1)\sin 2n{\omega }_{1}t}}+j\cdot \sum _{n=-\infty }^{\infty }{J}_{2n-1}({\beta }_1)\sin (2n-1){\omega }_{1}t\\ =& \sum _{n=-\infty }^{\infty }{J}_n({\beta }_1)\cos n{\omega }_{1}t+j\cdot \sum _{n=-\infty }^{\infty }{J}_n({\beta }_1)\sin {\omega }_{1}t\\ =& \sum _{n=-\infty }^{\infty }{J}_n({\beta }_1)\exp {\omega }_{1}t\end{array}$$

$$\begin{array}{rl}{\dot{S}}_{FM}(t)=& A{e}^{j{\omega }_{c}t}\left[\sum _{n=-\infty }^{\infty }{J}_n({\beta }_1)e^{jn{\omega }_{1}t}\right]\left[\sum _{k=-\infty }^{\infty }{J}_k({\beta }_2)e^{jk{\omega }_{2}t}\right]\\ & =A\sum _{n=-\infty }^{\infty }\sum _{k=-\infty }^{\infty }{J}_n({\beta }_1)J_k({\beta }_2)\exp j({\omega }_c+n{\omega }_{m1}+k{\omega }_{m2})t\end{array}$$

相量还原得到：
$$S_{FM}(t)=A\sum _{n=-\infty }^{\infty }\sum _{k=-\infty }^{\infty }{J}_n({\beta }_{FM1})J_k({\beta }_{FM2})\cos ({\omega }_c+n{\omega }_{m1}+k{\omega }_{m2})t$$