Let’s start from the frame at the end of the last lecture on distance measurement . We know that a chirp corresponds to a sampled IF signal, and we arrange these sampled IF signals into a frame (frame) in the order of chirps, which obtains the FMCW signal that we actually receive and process.
Since the emission and return time of chirp is very short, we call its time dimension fast time (fast time) , and there is a chirp repetition time (CRT) between adjacent chirps that is relatively slow, so we call its time dimension is the slow time dimension . Borrowing a picture in the article "Soli: Ubiquitous Gesture Sensing with Millimeter", you can have an intuitive understanding of the raw signal.
Periodicity of phase difference
We first analyze the frequency spectrum obtained by FFT, which is divided into two parts, the amplitude part and the phase part. The amplitude part can represent the strength of the frequency here, and the phase part represents the phase corresponding to this frequency . Then, for the frame matrix after the range FFT, the fast time dimension is converted to the range dimension.
For an object on a certain range bin, we already know that its distance is expressed as
dtarget = c 2 K fpeak d_{target} = \frac{c}{2K}f_{peak}dtarget=2K _cfpeak
The solution of this distance we know is through the frequency part of the IF signal 2 π K τ 2 \pi K \tau2 π K τ obtained, and we now focus on its phase part2 π f 0 τ 2 \pi f_0 \tau2πf0t .
x I F ( t ) = A cos ( 2 π K τ t + 2 π f o τ ) x_{\tiny{IF}}(t) = A \cos(2\pi K\tau t+2\pi f_o \tau ) xIF(t)=Acos(2πKτt+2πfot )
Since
τ = 2 dc \tau = \frac{2d}{c}t=c2 d
So the phase ϕ \phiϕ
ϕ = 2 π fo 2 dc = 4 π focd \phi = 2\pi f_o \frac{2d}{c}=\frac{4\pi f_o}{c}dϕ=2πfoc2 d=c4 p fod
If the object in this range bin is moving, then every other chirp cycle CRT CRTCRT , the object will have a micro-displacement, and this micro-displacement will cause a relatively drastic change in phase, that is,
Δ ϕ = 4 π foc Δ d = 4 π f 0 cv ⋅ CRT \Delta \phi = \frac{4 \pi f_o}{c} \Delta d =\frac{4\pi f_0}{c}v \cdot CRTD ϕ=c4 p foΔd=c4 p f0v⋅CRT
If we put this CRT CRTCRT is regarded as a kind of sampling, then, forϕ \phiϕ changes will be able to extract the effective velocityvvThe information of v , which is why we use frame transmission - to obtain speed information. This point of view is not shown first, and will be described at the end.
We can also regard this process as a periodic motion of the phase difference, then we will get the periodic phase difference information by performing FFT analysis on it .
Further converted to the velocity dimension, we have
v = c 4 π fo ⋅ CRT Δ ϕ = λ 4 π ⋅ CRT Δ ϕ v = \frac{c}{4\pi f_o \cdot CRT}\Delta \phi =\frac{\lambda}{4 \pi \ cdot CRT}\Delta \phiv=4 p fo⋅CRTcD ϕ=4 p.m⋅CRTlD ϕ
So the Doppler FFT or Velocity FFT we want to do is to take out a column of slow time data corresponding to a range bin of Range FFT and perform FFT.
Doppler effect
So the question is, why is it called Doppler FFT? In basic physics, we have studied the basic Doppler effect. To give an example in life, you hear a police car roaring towards you on the street, and you hear the sound of the siren getting more urgent (this corresponds to the higher and higher frequency of the sound wave), and When the police car gets farther and farther away, the siren you hear is getting thinner (this corresponds to the lower and lower frequency of the sound wave).
Here, we use a mobile communication to describe the Doppler frequency offset formula caused by the mobile station (see page 123 in the Rappaport book), that is,
fd = v λ cos θ f_d = \frac{v}{\lambda}\ cos \thetafd=lvcosi
In the scenario considered by FMCW radar, the radial velocity is taken, that is, cos θ = 1 \cos \theta = 1cosi=1 , at the same time due to the sending and receiving of electric waves, the resultingfd f_dfdfor
f d = 2 v λ f_d = 2\frac{v}{\lambda} fd=2lv
Further substituting v into the formula converts to
fd = Δ ϕ 2 π ⋅ CRT f_d = \frac{\Delta \phi}{2 \pi \cdot CRT}fd=2 p.m⋅CRTD ϕ
It is worth pointing out that the main frequency part will also produce frequency deviation due to the movement of the object. But when the distance d of the object changes slightly, the phase change of the IF signal signal is very obvious, but the frequency change is not significant, which is far from reaching the frequency of the signal in the time of the CRT. That is, the phase change is sensitive to the fretting displacement .
We might as well use the example in the TI tutorial to understand perceptually: take λ = 4 mm \lambda = 4mml=4mm， C R T = 40 μ s CRT = 40 \mu s CRT=40μs， K = 50 M H z / μ s K = 50MHz/\mu s K=50 M Hz / μ s , when the object has a 1mm micro-motion displacement, there are:
相位性性 Δ ϕ = 4 π Δ d λ = π = 18 0 ∘ 相位性性 \ \Delta \phi = \frac{4 \pi \Delta d}{\lambda} =\pi =180^{\circ}Phase change Δ ϕ =l4 p D d=Pi=180∘
Frequency change Δ f = 2 K c Δ d = 333 H z Frequency change\ \Delta f = \frac{2K}{c} \Delta d=333HzFrequency change Δ f =c2 KΔd=333 Hz
And the change caused by this frequency offset on the frequency axis of slow time is actually not large, that is,
Δ f ⋅ CRT = 333 × 40 × 1 0 − 6 = 0.013 cycles \Delta f \cdot CRT=333\times 40 \times 10 ^{ -6} = 0.013 \cyclesf _⋅CRT=333×40×10−6=0.013 cycles
Maximum speed and speed resolution
Maximum speed
Since Δ ϕ \Delta \phiThe limit of Δ ϕ , gives the limit of the maximum speed, that is
− π < Δ ϕ < π -\pi < \Delta \phi < \pi− p<D ϕ<λ 4 ⋅ CRT < v < λ 4 ⋅ CRT -\frac {
\lambda}{4\ cdot
CRT} < v <\frac{\lambda}{4\cdot CRT}−4⋅CRTl<v<4⋅CRTl
Perceptual understanding, for example, use 5mm 5mm5 mm millimeter wave radar, then use100 μ s 100 \mu sFor a 100 μ s CRT, the maximum speed that can be achieved at this time is
vmax = λ 4 ⋅ CRT = 12.5 m / s v_{max} = \frac{\lambda}{4 \cdot CRT} =12.5m/svmax=4⋅CRTl=12.5m/s
speed resolution
Continue to borrow a picture from the TI tutorial (here define ω = Δ ϕ \omega = \Delta \phioh=Δ ϕ ), it is easy to find that the velocity resolution is related to our angular velocity resolution in the digital domain, because
Δ ω = 2 π N radians / sample = 1 N cycles / sample \Delta \omega = \frac{2\pi }{N} \ radians/sample=\frac{1}{N} \ cycles/sampleD o=N2 p.m radians/sample=N1 cycles/sample
Default
Δ v = λ 4 π ⋅ CRT Δ ω = λ 2 N ⋅ CRT \Delta v = \frac{\lambda}{4 \pi\cdot CRT} \Delta \omega = \frac{\lambda}{2N \cdot CRT}v _=4 p.m⋅CRTlD o=2N _⋅CRTl
Still using the measured data in the maximum speed, and taking N = 512, we perceptually realize that the speed resolution at this time is:
vres = λ 2 N ⋅ CRT = 0.0488 m / s v_{res} = \frac{\lambda} {2N \cdot CRT}=0.0488m/svres=2N _⋅CRTl=0.0488m/s
CRT-based sampling perspective
If we understand this phase change based on the sampling perspective of CRT, then for the formula
Δ ϕ = 4 π foc Δ d = 4 π f 0 cv ⋅ CRT \Delta \phi = \frac{4\pi f_o}{c} \ Delta d =\frac{4\pi f_0}{c}v \cdot CRTD ϕ=c4 p foΔd=c4 p f0v⋅CRT
We are the same on both sides exceptCRT CRTCRT , then:
Δ ϕ CRT = 4 π f 0 cv \frac{\Delta \phi}{CRT}=\frac{4\pi f_0}{c}vCRTD ϕ=c4 p f0vAccording
to the knowledge of differential calculus, we know that the left side can be understood as the pairϕ \phiϕ的微分，即
w = d ϕ dt = 2 π fpeakw = \frac{d\phi}{dt} = 2\pi f_{peak}w=dtdϕ=2πfpeak
Then there is:
fpeak = 2 v λ f_{peak} = 2\frac{v}{\lambda}fpeak=2lv
This formula shows that from the perspective of the frequency axis, what is directly measured at this time is the Doppler frequency offset. Further, there is:
v = λ 2 fpeakv =\frac{\lambda}{2 } f_{peak}v=2lfpeak
Since at this time CRT CRTThe reciprocal of the CRT is our equivalent sampling rate. Therefore, the range of frequency resolution is
− 1 2 ⋅ CRT < fpeak < 1 2 ⋅ CRT -\frac{1}{2 \cdot CRT} <f_{peak}<\frac{1}{2\cdot CRT}−2⋅CRT1<fpeak<2⋅CRT1
Then, the measurement range of available velocity is
− λ 4 ⋅ CRT < v < λ 4 ⋅ CRT -\frac{\lambda}{4 \cdot CRT} < v <\frac{\lambda}{4 \cdot CRT}−4⋅CRTl<v<4⋅CRTl
and velocity resolution
vres = λ 2 fres = λ 2 N ⋅ CRT v_{res} =\frac{\lambda}{2 } f_{res} = \frac{\lambda}{2N \cdot CRT}vres=2lfres=2N _⋅CRTl
This kind of perspective is of personal interest, so as to increase reference. Finally, a picture is also used to end the content of this section.
