• TI mmWave radar sensors Tutorial 笔记 | Module 2: The phase of the IF signal


    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 2: The Phase of the IF Signal

    Last Module: the frequency of IF signal
    • f I F = S 2 d / c f_{IF} = S2d/c fIF=S2d/c

    This Module:

    • look into the phase of the IF signal
    • The Phase is very important if we wish to understand the capability of FMCW radar to:
    1. respond to very small displacements displacements 位移 in objects

    2. measure the velocity very quickly and accurately

    • 是heartbeat monitoring and vibration detection的基础

    Fourier Transforms: A quick review

    • Fourier Transform converts a time domain signal into the frequency domain
    • A sinusoid in the time domain produces a peak in the frequency domain
    • The signal in the Frequency domian is complex
      • each value is a phasor with a amplitude and a phase: A e j θ Ae^{j\theta} Aejθ
      • 可用图形表示(如下图)

    picture 20

    • Phase of the peak is equal to the initial phase of the sinusoid

    picture 21

    Note: The above is strictly true only for a complex input tone (in the form of e j ω t e^{j\omega t} et)

    • 但对于real input概念上也equally applicable with a few mathematical modifications
    • 此处为了从概念上进行理解,忽略这些修正

    Frequency of the IF signal: Recap from module 1

    • 两种观察方式:
      • A-t plot
      • f-t plot
    • In module 1, focus on f-t plot to understand the IF signal
      • A single object in front of the radar produces an IF signal with a constant frequency of S 2 d / c S2d/c S2d/c

    picture 22

    • This module:
      • use the A-t plot to analyze the relationship between the Phase and the Distance

    Phase of the IF signal

    • Let’s look at the ‘A-t’ plot

      • To get more intuition into the nature of the IF signal
    • IF signal:

      • For an object at a distance d from the radar, the IF signal will be a sinusoid:

      • A s i n ( 2 π f t + ϕ 0 ) Asin(2\pi f t + \phi_0) Asin(2πft+ϕ0)

      • 其中 f = S 2 d / c f = S2d/c f=S2d/c

      • What is the ϕ 0 \phi_0 ϕ0 :

        🚩 The phase of point C in the following image C点相位

        🚩 Recall: mixer输出信号的initial phase就是 the difference of the initial phase of the two inputs ⇒ \Rightarrow (A点相位 - B点相位)

        🚩 也是 the peak point in FFT(IF signal) 的相位

    • 相移 VS 位移 :what happens to the phase fo the IF signal if the object moves by a small ditance Δ τ \Delta \tau Δτ?

      • 灰色曲线: before the movement

      • 蓝色曲线: after the movement

      • The phase of the TX: delay

        🚩 Phase diference between A and D: Δ Φ = 2 π f c Δ τ = 4 π Δ d λ \Delta \Phi = 2 \pi f_c \Delta \tau = \frac{4\pi \Delta d}{\lambda} ΔΦ=2πfcΔτ=λ4πΔd

      • The phase of the RX: 不变 (注: 因为根据电磁场与波,垂直入射相位改变 π \pi π, 所以不变)

      • Therefore, Δ ϕ \Delta \phi Δϕ of IF signal = Δ ϕ \Delta \phi Δϕ of TX signal = Δ Φ = 2 π f c Δ τ = 4 π Δ d λ =\Delta \Phi = 2 \pi f_c \Delta \tau = \frac{4\pi \Delta d}{\lambda} =ΔΦ=2πfcΔτ=λ4πΔd

      • 最终结论: 4 π Δ d λ \frac{4\pi \Delta d}{\lambda} λ4πΔd

    picture 23

    Sensitivity of the IF signal for small displacements in the object

    • Recall: the IF signal is A s i n ( 2 π f t + ϕ 0 ) Asin(2\pi ft + \phi_0) Asin(2πft+ϕ0)

      • f = S 2 d / c f = S2d/c f=S2d/c
      • Δ ϕ = 4 π Δ d λ \Delta \phi = \frac{4\pi \Delta d}{\lambda} Δϕ=λ4πΔd
    • Now: 如果目标物体位移了一小段,IF signal的频率和相位将会如何变化?

      • 一小段(small): compared to the range resolution of the radar

    • An example:
      • S = 50 M H z / u s S = 50MHz/us S=50MHz/us, T c = 40 u s T_c = 40us Tc=40us, 77 G H z 77GHz 77GHz, 1 m m = λ / 4 1mm = \lambda/4 1mm=λ/4

      • What happens if an object in front of the randar changes its position by 1 m m 1mm 1mm

      • Phase: Δ ϕ = 4 π Δ d λ = π = 18 0 ∘ \Delta \phi=\frac{4 \pi \Delta d}{\lambda}=\pi=180^{\circ} Δϕ=λ4πΔd=π=180

      • Frequency: by Δ f = S 2 Δ d c = 50 × 1 0 12 × 2 × 1 × 1 0 − 3 3 × 1 0 8 = 333   H z \Delta \mathrm{f}=\frac{\mathrm{S} 2 \Delta d}{c}=\frac{50 \times 10^{12} \times 2 \times 1 \times 10^{-3}}{3 \times 10^{8}}=333 \mathrm{~Hz} Δf=cSd=3×10850×1012×2×1×103=333 Hz

        🚩 但 尽管 333 Hz looks like a big number ⇒ \Rightarrow But in the observation window , this corresponds to only additional Δ f T c = 333 × 40 × 1 0 − 6 = 0.013 \Delta f T_c = 333 \times 40 \times 10^{-6} = 0.013 ΔfTc=333×40×106=0.013 cycles

        ⇒ \Rightarrow 在频谱图中不会被体现


    结论 : The phase of the IF signal is very sensitive to small changes in object range.

    • 下图所示
      • An object at certain distance produces an IF signal with a certain frequency and phase (上图)
      • small motion in the object changes the phase of the IF signal but not the frequency

    picture 24

    picture 25

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

    • Transmit two chirps separated by T c T_c Tc
    • The range-FFTs corresponding to each chirp will have peaks in the same location but with different phase
    • The measured phase difference ω \omega ω corresponds to a motion in the object of v T c vT_c vTc
      • ω = 4 π v T c λ \omega = \frac{4\pi v T_c}{\lambda} ω=λ4πvTc
      • ⇒ \Rightarrow v = λ ω 4 π T c v=\frac{\lambda \omega}{4 \pi T_c} v=4πTcλω

    picture 26

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

    Measurements on a Vibrating Object

    • blue block:
      • Small (amplitude ~1mm) vibrations over time
      • Δ d \Delta d Δd is a fraction of a wavelength

    picture 28

    • measure the time evolution of phase
      • obtained by FFT
      • can be used to estimate both the amplitude and periodicity of the vibration

    picture 29

    Epilogue

    • What we learned in this module:

      • The phase of IF signal is very sensitve to small changes in the range of the object
      • we can exploit it to meaure the velocity
      • Δ ϕ = 4 π Δ d λ \Delta \phi = \frac{4\pi \Delta d}{\lambda} Δϕ=λ4πΔd
      • v = λ Δ ϕ 4 π T c v=\frac{\lambda \Delta \phi}{4 \pi T_c} v=4πTcλΔϕ
    • 下一个目标:

      • How to separate Multiple objects equidistant from the radar, but with differing velocities relative to the radar

        🚩 Equi-range objects which have differing velocities relative to the radar can be separated out using a “Doppler-FFT”

        ✅ see in the next module

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  • 原文地址:https://blog.csdn.net/qazwsxrx/article/details/126056814