TI mmWave radar sensors Tutorial 笔记 | Module 3: Velocity Estimation

本系列为TI(Texas Instruments) mmWave radar sensors 系列视频公开课 的学习笔记。

• 视频网址: https://training.ti.com/intro-mmwave-sensing-fmcw-radars-module-1-range-estimation?context=1128486-1139153-1128542

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FMCW Radars – Module 1 : Range Estimation
FMCW Radars – Module 2 : The Phase of the IF Signal
FMCW Radars – Module 3 : Velocity Estimation
FMCW Radars – Module 4 : Some System Design Topics
FMCW Radars – Module 5 : Angle Estimation

Module 3 Velocity Estimation

• Content
  • Qucik review of background material on FFT’s
  • Measuring Velociy
  • Maximum measurable velocity
  • Velocity Resolution and concept of frame

This module : to answer the following questions

• How does the radar estimate velocity(v) of an object?

• What if there are multiple objects at the same range? (with different relative volocities)

• How close? (velocity resolution )
• Is there a limit on the maximum velocity that a radar can measure?
  • maximum velocity

FFT’s on a complex sequence

• Consider a discrete signal corresponding to a phasor rotating a constant rate of $\omega$ radians per sample
  • An FFT on these series of samples produces a peak with the location of the peak $\omega$

Note: 下图中的${\omega }_1=\omega$

• If the signal consists of the sum of two phasors:
  • $\Rightarrow$ The FFT has two peaks
  • each phasor rotating at the rate of ${\omega }_1$ and ${\omega }_2$ radians per sample, respectively

Question: How far apart the two frequencies ${\omega }_1$ and ${\omega }_2$ have to be for them to show up as separate peaks in the FT?

• Case 1: ${\omega }_1=0$, ${\omega }_2=\pi /N$

  • Over N samples , the 2nd phasor has traversed 0.5 cycle ($\pi$ rads) more than the 1st phasor

    ❌ not suficient to resolve the two objects in the frequency domain

  • But over 2N smaples , the 2nd phasor has traversed 1 cycle ($\pi$ rads) more than the 1st phasor

    ✅ ❌ able to resolve the two objects in the frequency domain

结论: Longer the sequence length $\Rightarrow$ better resolution

• 当 序列长度为N 时，样本间的$\omega$差异至少(分辨率 )为 $2\pi /N$ ($rad/sample$)

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总结: 离散信号 VS 连续信号 频率分辨率

• Continuous signals:
  • $\Delta f = \frac{1}{T} $ cycles/sec
• Discrete signals:
  • $\Delta \omega =\frac{2\pi }{N}$ radians/sample =$ \frac{1}{N} $cycles/sample
  • (1 cycle = 2 $\pi$ radians )

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How to measure the velocity (v) of an object using 2 chirps

• Process:
  • 1 Transmit two chirps separated by $T_c$
  • 2 The range-FFTs corresponding to each chirp will have peaks in the same location but with differing phase
  • 3 The measured phase difference ($\omega$) corresponds to a motion in the object of $v{T}_c$
• 速度计算：
  • $\omega =\frac{4\pi v{T}_c}{\lambda }$
  • $\Rightarrow$ $v=\frac{\lambda \omega }{4\pi T_c}$

结论: The phase difference measured across two consecutive chirps can be used to estimate the velocity of the object

Maximum measurable velocity

• Unambiguous :
  • Movement away from radar $\Rightarrow$ $\omega >0$
  • Movement towards the radar $\Rightarrow$ $\omega <0$
  • The measurement is unambiguous only if $|\omega| < \pi $(180 ^\circ)

• 因此，最大measurable速度：
  • $\frac{4\pi v{T}_c}{\lambda }<\pi$ $\Rightarrow$ $v<\frac{\lambda }{4T_c}$

结论 : The maximum relative speed ($v_{max}$) that can be measured by 2 chirps spaced $T_c$ apart is:

• $v_{max}=\frac{\lambda }{4T_c}$

• 因此，Higher $v_{max}$ requires closely spaced chirps

Measuring velocity with multiple objects at the same range

目前，已经知道了如何测量 the velocity of a single object

• 该方法同样适用于multiple objects at different ranges

• What if there are multiple objects at the same range?

• consider two objects equidistant from the radar approaching the radar at speeds $v_1$ and $v_2$
• Solution: Transmit N equi-spaced chirps
  • (a frame )

• 即：在 Range-FFT peaks对应的sequence of phasors 再做FFT ，就可以 resolves the two objects
  • 该FFT称为: Doppler-FFT

结论 : ${\omega }_1$ and ${\omega }_2$ correspond to the phase difference between consecutive chirps for the respectively objects

• $v_1 = \frac{\lambda \omega_1}{4 \pi T_c} $
• $v_2 = \frac{\lambda \omega_2}{4 \pi T_c} $

Velocity resolution

• What is the velocity resolution ($v_{res}$) capability of the “doppler-FFT”?
  • 即doppler-FFT结果中的两个peak能够被区分
• 推导：
  • 假设两个物体的速度差为$\Delta v$, they will have their respective angular frequencies separated by $\Delta \omega =\frac{4\pi \Delta v{T}_c}{\lambda }$
  • 若观察sequence的长度为N, 则要求$\Delta \omega =2\pi /N$
  • $\Rightarrow$ $\Delta v>\frac{\lambda }{2N{T}_c}$

注意: 这里的 N: number of chirps in one frame !

• 因此，也可按照下图表示为$v_{res}=\frac{\lambda }{2T_f}$
• $T_f$: frame time!

Question: 比较下述两个雷达的maximum measurable velocity ($v_{max}$) 和 velocity resolution ($v_{res}$)

• $v_{max}=\frac{\lambda }{4T_c}$ $\Rightarrow$ Radar A更好
• $v_{res}$二者相同

Next module:

• 使用目前关于range 和 velocity 的知识
• design a transmitted which meets certain specified requirements