quartz/content/synthetic_aperture_radar_imaging/Chirp.md
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---
title: Chirp - 啁啾
tags:
- basic
- signal
date: 2023-06-30
---
啁啾Chirp是指频率随时间而改变增加或减少的信号。其名称来源于这种信号听起来类似鸟鸣的啾声。
![](synthetic_aperture_radar_imaging/attachments/Linear-chirp.svg)
Chirp常常被用在sonar, radar, laser systems里。其中为了能够测量长距离又保留时间的分辨率雷达需要短时间的派冲波但是又要持续的发射信号啁啾信号可以同时保留连续信号和脉冲的特性因此被应用在雷达和声纳探测上。
# Definition
## 瞬时频率 (instantaneous angular frequency)
有一信号,$x(t)=A\sin{(\phi(t))}$,其瞬时角频率为
$$
\omega(t)=\frac{d\phi(t)}{dt}
$$
经适当归一化后得到瞬时频率
$$
f(t)=\frac{1}{2\pi}\frac{d\phi(t)}{dt}
$$
## 啁啾度
对前两式再求导,得到瞬时角频率的变化速率为**瞬时角啁啾度**(instantaneous angular chirpyness)
$$
\gamma(t)=\frac{d^2\phi(t)}{dt^2}
$$
类似有**瞬时(普通)啁啾度**(instantaneous ordinary chirpyness)
$$
c(t)=\frac{1}{2\pi}\gamma(t)=\frac{1}{2\pi}\frac{d^2\phi(t)}{dt^2}
$$
# Types
## Linear
![](synthetic_aperture_radar_imaging/attachments/Pasted%20image%2020230418110700.png)
啁啾的瞬时频率$f(t)$呈线性变化
$$f(t)=f_0 + ct$$
$$
c = \frac{f_1-f_0}{T}
$$
c是一个常值
Also
$$
\phi(t)=\phi_0 + 2\pi \int_{0}^t f(\tau)d\tau =\phi_0 = 2\pi(\frac{c}{2}t^2 + f_0 t)
$$
相位为t的二次函数从而可以继续推导出信号在time domain
$$
x(t)=A \cos{(\phi_0 + 2\pi (\frac{c}{2}t^2 + f_0 t))}
$$
这种Linear Chirp信号也被称为二次相位讯号(**quadratic-phase signal**)
## Exponential
![](synthetic_aperture_radar_imaging/attachments/Pasted%20image%2020230418111708.png)
Exponential chirp也叫geometric chirp瞬时频率以指数变化即$f(t_2)/f(t_1)$会是常数
signal frequency:
$$
f(t)=f_0 k^t
$$
$$
k = (\frac{f(T)}{f_0})^{\frac{1}{T}} = \text{constant}
$$
相位:
$$
\phi(t)=\phi_0 + 2\pi\int_0^t f(\tau)d\tau = \phi_0 + 2\pi f_0 (\frac{k^t - 1}{\ln(k)})
$$
time-domain:
$$
x(t) = \sin{[\phi_0 + 2\pi f_0(\frac{k^t - 1}{\ln(k)})]}
$$
## Hyperbolic
双曲线线性调频用于雷达应用,因为它们在被多普勒效应([Doppler Effect](physics/wave/doppler_effect.md))扭曲后显示出最大的匹配滤波器([Matched filter](https://en.wikipedia.org/wiki/Matched_filter))响应。
signal frequency:
$$
f(t) = \frac{f_0 f_1 T}{(f_0 - f_1)t + f_1T}
$$
Phase:
$$
\phi(t) = \phi_0 + 2\pi \int_0^t f(\tau)d\tau = \phi_0 + 2\pi \frac{-f_0f_1 T}{f_1 - f_0}\ln(1 - \frac{f_1-f_0}{f_1 T}t)
$$
time-domain:
$$
x(t) = \sin{[\phi_0 + 2\pi \frac{-f_0f_1 T}{f_1 - f_0}\ln(1 - \frac{f_1-f_0}{f_1 T}t)]}
$$