LFM信号模型-01

LFM信号由于其具有较大的带宽时宽积，广泛应用于现代雷达系统中。频率线性地向上或向下扫过脉冲宽带，以向上线性调频为例，LFM 信号可以表示为

\begin{matrix} (1) & x (t) = e x p (j 2 π (f_{0} t + \frac{μ}{2} t^{2})) Rect (\frac{t}{τ_{0}}) \end{matrix}

$x\left(t\right)= exp(j2\pi(f_0t+\frac{\mu}{2}t^2))\text{Rect}\left(\frac{t}{\tau_0}\right) \tag{1}$

其中， $\text{Rect}\left(\frac{t}{\tau_0}\right)$ 表示宽带为 $\tau_0$ 的矩形脉冲，该式为线性调频信号的解析信号。其频谱主要由 $exp(j\pi\mu t^2)\text{Rect}\left(\frac{t}{\tau_0}\right)$ 决定,而 $exp(j2\pi f_0t)$ 相当于对频谱引入了关于中心频率为 $f_0$ 的频移。

考虑基带LFM信号，调频信号的频率范围从 $f_0\to f_0+\mu\tau_0$ ,故 $f_0$ 可设置为 $-\frac{\mu\tau_0}{2}$ .设LFM信号的带宽为 $B$ ,求得下列信号的傅里叶变换，即

\begin{matrix} (2) & \begin{aligned} \hat{X} (f) & = \int_{- \infty}^{+ \infty} e x p (j π μ t^{2}) Rect (\frac{t}{τ_{0}}) e^{- j 2 π f t} d t \\ = \int_{- \frac{τ_{0}}{2}}^{\frac{τ_{0}}{2}} e^{j π μ t^{2}} e^{- j 2 π f t} d f \end{aligned} \end{matrix}

$\begin{aligned} \hat{X}(f)&=\int_{-\infty}^{+\infty}exp(j\pi\mu t^2)\text{Rect}\left(\frac{t}{\tau_0}\right)e^{-j2\pi ft}dt\\ &=\int_{-\frac{\tau_0}{2}}^{\frac{\tau_0}{2}}e^{j\pi\mu t^2}e^{-j2\pi ft}df \end{aligned} \tag{2}$

考虑菲涅耳积分的形式，进行变量代换，设 $\mu'=\pi\mu$ ,令 $z=\sqrt{\frac{2}{\pi}}\left(\sqrt{\mu'}t-\frac{\pi f}{\sqrt{\mu'}}\right)$ ,并且有 $dt=\sqrt{\frac{\pi}{2\mu'}}dz$ ,式(2)可以改写为

\begin{matrix} (3) & \begin{aligned} \hat{X} (f) & = \sqrt{\frac{π}{2 μ^{'}}} e^{- j (π f)^{2} / μ^{'}} \int_{- z_{1}}^{z_{2}} e^{j π z^{2} / 2} d z \\ = \sqrt{\frac{π}{2 μ^{'}}} e^{- j (π f)^{2} / μ^{'}} {\int_{0}^{z_{1}} e^{j π z^{2} / 2} + \int_{0}^{z_{2}} e^{j π z^{2} / 2}} \\ z_{1} & = - \sqrt{\frac{2 μ^{'}}{π}} (\frac{τ_{0}}{2} + \frac{π f}{μ^{'}}) = \sqrt{\frac{B τ_{0}}{2}} (1 + \frac{f}{B / 2}) \\ z_{2} & = \sqrt{\frac{2 μ^{'}}{π}} (\frac{τ_{0}}{2} - \frac{f}{μ}) = \sqrt{\frac{B τ_{0}}{2}} (1 - \frac{f}{B / 2}) \end{aligned} \end{matrix}

$\begin{aligned} \hat{X}(f)&=\sqrt{\frac{\pi}{2\mu'}}e^{-j(\pi f)^2/\mu'}\int_{-z_1}^{z_2}e^{j\pi z^2/2}dz\\ &=\sqrt{\frac{\pi}{2\mu'}}e^{-j(\pi f)^2/\mu'}\left\{\int_{0}^{z_1}e^{j\pi z^2/2}+\int_0^{z_2}e^{j\pi z^2/2}\right\}\\ z_1 &=-\sqrt{\frac{2\mu'}{\pi}}\left(\frac{\tau_0}{2}+\frac{\pi f}{\mu'}\right)=\sqrt{\frac{B\tau_0}{2}}\left(1+\frac{f}{B/2}\right)\\ z_2 &=\sqrt{\frac{2\mu'}{\pi}}\left(\frac{\tau_0}{2}-\frac{f}{\mu}\right)=\sqrt{\frac{B\tau_0}{2}}\left(1-\frac{f}{B/2}\right)\\ \end{aligned} \tag{3}$

利用 $C(z),S(z)$ 表示菲涅尔(Fresnel)积分，定义为

\begin{matrix} (4) & C (z) = \int_{0}^{z} c o s (\frac{π v^{2}}{2}) d v S (z) = \int_{0}^{z} s i n (\frac{π v^{2}}{2}) d v \end{matrix}

$C(z)=\int_0^zcos(\frac{\pi v^2}{2})dv\\ S(z)=\int_0^zsin(\frac{\pi v^2}{2})dv \tag{4}$

有 $C(-z)=-C(z),S(-z)=-S(z)$ ,其中，若 $z >> 1$ ,菲涅尔积分可近似表示为

\begin{matrix} (5) & C (z) \approx \frac{1}{2} + \frac{1}{π z} s i n (\frac{π}{2} z^{2}) S (z) \approx \frac{1}{2} - \frac{1}{π z} c o s (\frac{π}{2} z^{2}) \end{matrix}

$C(z)\approx\frac{1}{2}+\frac{1}{\pi z}sin(\frac{\pi}{2}z^2)\\ S(z)\approx\frac{1}{2}-\frac{1}{\pi z}cos(\frac{\pi}{2}z^2) \tag{5}$

进而可得到 $\hat{X}(f)$ 的频谱为

\begin{matrix} (6) & \hat{X} (f) = \sqrt{\frac{π}{2 μ^{'}}} e^{- j (π f)^{2} / (μ^{'})} {[C (z_{1}) + C (z_{2})] + j [S (z_{1}) + S (z_{2})]} = \sqrt{\frac{1}{2 μ}} e^{- j (π f)^{2} / (μ^{'})} {[C (z_{1}) + C (z_{2})] + j [S (z_{1}) + S (z_{2})]} \end{matrix}

$\hat{X}(f)=\sqrt{\frac{\pi}{2\mu'}}e^{-j(\pi f)^2/(\mu')}\left\{[C(z_1)+C(z_2)]+j[S(z_1)+S(z_2)]\right\}\\ =\sqrt{\frac{1}{2\mu}}e^{-j(\pi f)^2/(\mu')}\left\{[C(z_1)+C(z_2)]+j[S(z_1)+S(z_2)]\right\} \tag{6}$

令时宽带宽积为 $Q=B\tau_0$ ,若该值较大，从式(6)中可以推导，当频段处于调频信号的频段，即 $-B/2\sim B/2$ ,此时 $z_1$ 和 $z_2$ 近似相等，且此时 $C(z_1),C(z_2),S(z_1),S(z_2)$ 均近似等于1/2, LFM信号的频谱幅值近似等于 $\sqrt{\frac{1}{\mu}}$ ,当频段远大于B/2或远小于-B/2时， $z_1，z_2$ 可近似认为互为相反数，于是其频谱幅值近似为0.后续仿真可验证此结论。