多普勒频率的推导(纯公式版)

多普勒频率的推导(纯公式版)

前奏

$$d=v\Delta t$$
$$c\tau =c\Delta t+v\Delta t$$
$$c{\tau }^{\prime }=c\Delta t-v\Delta t$$
$$\frac{c{\tau }^{\prime }}{c\tau }=\frac{c\Delta t-v\Delta t}{c\Delta t+v\Delta t}$$
$${\tau }^{\prime }=\frac{c-v}{c+v}\tau$$

方法一

$$d=v\Delta t$$
$$\frac{c}{f_r}-d=c\Delta t$$
$$\Delta t=\frac{1}{c+v}\frac{c}{f_r}$$
$$d=\frac{1}{f_r}\frac{cv}{c+v}$$
$$\frac{c}{f_r^{\prime }}=c\Delta t-v\Delta t$$
$$f_r^{\prime }=\frac{c+v}{c-v}{f}_r$$
$$f_0^{\prime }=\frac{c+v}{c-v}{f}_0$$
$$f_d=f_0^{\prime }-f_0=\frac{c+v}{c-v}{f}_0-f_0$$
$$v<<c$$
$$c=\lambda f_0$$
$$f_d=\frac{2v}{c}{f}_0=\frac{2v}{\lambda }$$

方法二

$$R(t)=R_0-v\ast (t-t_0)$$
$$x_r(t)=x(t-\varphi (t))$$
$$\varphi (t)=\frac{2}{c}R(t)=\frac{2}{c}(R_0-vt+v{t}_0)$$
$$x_r(t)=x((1+\frac{2v}{c})t-{\varphi }_0)$$
$${\varphi }_0=\frac{2R_0}{c}+\frac{2v}{c}{t}_0$$
$$\gamma =1+\frac{2v}{c}$$
$$x_r(t)=x(\gamma t-{\varphi }_0)$$
$$x(t)=y(t)cos{\omega }_{0}t$$
$$x_r(t)=y(\gamma t-{\varphi }_0)cos(\gamma {\omega }_{0}t-{\omega }_0{\varphi }_0)$$
$$X_r(\omega )=\frac{1}{2\gamma }(Y(\frac{\omega }{\gamma }-{\omega }_0)+Y(\frac{\omega }{\gamma }+{\omega }_0))$$
$${\omega }_d={\omega }_0-\gamma {\omega }_0$$
$$f_d=f_0-\gamma f_0$$
$$f_d=\frac{2v}{c}{f}_0=\frac{2v}{\lambda }$$