Taking over the intermediate frequency IF signal from the previous blog post , we already know that the obtained intermediate frequency IF signal is in the form of:
x IF ( t ) = A ′ ′ cos ( 2 π K τ t + 2 π fo τ ) x_{\tiny{IF}}(t) = A^{\prime \prime} \cos(2\pi K \tau t+2\pi f_o \tau )xIF(t)=A′′cos(2πKτt+2πfot )
where time delay τ = 2 dc \tau = \frac{2d}{c}t=c2 d， d d d is the distance of the detection target.
In practice, we perform ADC sampling on the obtained IF signal, that is, convert the analog signal (analog signal) into a digital signal (digital signal), and then perform digital signal processing on the sampled signal .
Maximum detection distance
Suppose the sampling rate of the ADC is fs f_sfs, the number of sampling points is M, and the number of FFT points is N (N≥M). According to the knowledge of FFT spectrum analysis, we know that the maximum resolvable frequency component at this time is fs 2 \frac{f_s}{2}2fs。
Perform FFT on the sampled IF signal to complete its spectrum analysis, assuming we measure at frequency ffThere is a peak at f , which corresponds to the distance ddThere is a target at d . Right now:
f p e a k = K τ = K 2 d t a r g e t c f_{peak}=K\tau=K\frac{2d_{target}}{c} fpeak=K sq=Kc2 dtarget
To convert the distance, we have:
d t a r g e t = c 2 K f p e a k = c ⋅ C R T 2 B W f p e a k d_{target} = \frac{c}{2K}f_{peak}=\frac{c \cdot CRT }{2BW}f_{peak} dtarget=2K _cfpeak=2 B Wc⋅CRTfpeak
Among them, CRT refers to the chirp repetition period, that is, chirp repetition time, and BW refers to the sweep bandwidth of chirp, that is, band width.
Now we look at fpeak f_{peak}fpeakrange of values. Since the sampling rate of the ADC is fs f_sfs, the number of FFT points is N, that is, on the frequency axis ( fs 2 fs 2 \frac{f_s}{2}~ \frac{f_s}{2}2fs 2fs) is divided into N frequency bins, and the fpeak f_{peak} corresponding to the targetfpeakIt falls in one of these bins. Considering that the real signal has an image frequency, if we only take the positive frequency part
f p e a k = f s N k ， 0 ≤ k < N 2 f_{peak} = \frac{f_s}{N}k， \ 0≤k<\frac{N}{2} fpeak=Nfsk， 0≤k<2N
It is easy to see from the previous derivation that dtarget d_{target}dtarget 与 f p e a k f_{peak} fpeakinto a linear relationship, that is:
dmax = c 2 K fmax d_{max} = \frac{c}{2K} f_{max}dmax=2K _cfmax
Due to the positive frequency part, the maximum fpeak f_{peak}fpeakThat is fs 2 \frac{f_s}{2}2fs, so
dmax = c 4 K fs d_{max} = \frac{c}{4K}f_sdmax=4K _cfs
distance resolution
Two distinguishable range bins correspond to two distinguishable frequency bins, that is,
Δ d = d 1 − d 2 = c 2 K ( f 1 − f 2 ) = c 2 K Δ f = c 2 K fs N = c ⋅ CRT 2 BW fs N \Delta d = d_1 -d _2 = \frac{c}{2K}(f_1-f_2)=\frac{c}{2K}\Delta f=\frac{c}{2K} \frac{f_s}{N}=\frac{c \cdot CRT }{2BW}\frac{f_s}{N}Δd=d1−d2=2K _c(f1−f2)=2K _cf _=2K _cNfs=2 B Wc⋅CRTNfs
In the context of continuous sampling (where N denotes the number of sampling points in a CRT), we have:
N fs = CRT \frac{N}{f_s} = CRTfsN=CRT
thenΔ d \Delta dΔ d can be further simplified as:
Δ d = c 2 BW \Delta d = \frac{c}{2 BW}Δd=2 B Wc
An intuitive conclusion is that the distance resolution that can be achieved is only as good as BW BWB W , which is related to the sweep bandwidth.
Unable to sample the full cycle
But in fact, due to the settings of some devices, we often do not complete the sampling in the entire CRT cycle (here N represents the number of sampling points in a CRT), that is,
N fs = α CPT ≤ CRT , where α ≤ 1 \ frac{N}{f_s}=\alpha CPT \le CRT , where \alpha \le 1fsN=αCPT≤CRT , where α≤1
So in this case, the actual distance resolution is not as fine as expected, but slightly coarser.
Δ d = c 2 BW 1 α ≥ c 2 BW \Delta d = \frac{c}{2BW} \frac{1}{\alpha} \ge \frac{c}{2BW}Δd=2 B Wca1≥2 B Wc
Although there are smart kids here who said, I can fill in the zeros in the time domain, so it’s better to look at it more carefully. But it is a pity that " zero padding in the time domain is equivalent to interpolation in the frequency domain ", which only improves the visual resolution of FFT, but the actual wavelength resolution is only determined by the number of samples (that is, the number of sampling points M). If two objects fall into a certain range bin converted from the same original sampling points, then these two objects cannot be distinguished in the FFT spectrum map we finally obtained.
Let us have a perceptual understanding of this resolution, assuming that the frequency sweep bandwidth is 3GHz, then:
Δ d = c 2 BW = 3 × 1 0 8 2 × 3 × 1 0 9 m = 5 mm \Delta d= \frac {c}{2 BW}=\frac{3 \times 10^8}{2 \times 3 \times 10^9}m=5mmΔd=2 B Wc=2×3×1093×108m=5mm
That is, a millimeter-level distance resolution.
Impact of CRTs
In TI's tutorial video on FMCW radar, there was an interesting question about chirp 1 and chirp 2, they have the same sweep bandwidth, but chirp 1 has twice the period of chirp 2.
According to the previous derivation, we know that the distance resolutions corresponding to these two chirps should be the same .
But if we use the same ADC for sampling now: when the sampling frequency is the same, chirp 1 can obtain more sampling points, and its corresponding wavelength resolution is also higher. In the case of the same number of sampling points, chirp 1 will require a lower sampling rate than chirp 2, so the tolerance for ADC is improved; but at the same time, chirp 1 also takes longer to complete sampling. So, what is presented here will be a trade-off for ADC .
Finally, use a picture to summarize the content of range FFT:
