本系列为TI(Texas Instruments) mmWave radar sensors 系列视频公开课 的学习笔记。
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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
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:
respond to very small displacements displacements 位移 in objects
measure the velocity very quickly and accurately
- 是heartbeat monitoring and vibration detection的基础
Note: The above is strictly true only for a complex input tone (in the form of e j ω t e^{j\omega t} ejωt)
- 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
Let’s look at the ‘A-t’ plot
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 ⇒ 即 (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
Recall: the IF signal is A s i n ( 2 π f t + ϕ 0 ) Asin(2\pi ft + \phi_0) Asin(2πft+ϕ0)
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?
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=cS2Δd=3×10850×1012×2×1×10−3=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=333×40×10−6=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.
结论 : The phase difference measured across two consecutive chirps can be used to estimate the velocity of the object
What we learned in this module:
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