2022 China Postgraduate Mathematical Contest in Modeling Question A - Mobile scene super-resolution positioning problem

1. Original title

1. Background

In daily family life, people may need to spend a lot of time looking for small items randomly placed in some corners of the home. However, if some important items are affixed with circuit labels, and then use the full house coverage ability of a sweeping robot, these objects can be accurately located, which will greatly improve the convenience of people's lives. In the field of intelligent assisted driving or automatic driving, it is even more necessary to accurately detect the position and speed of adjacent vehicles and pedestrians to control vehicle speed, steering and braking to avoid accidents. These are all mobile scene positioning problems. Obviously, the higher the positioning accuracy, the greater the application value, especially the super-resolution positioning, which has broad application prospects.

Mobile scene super-resolution positioning refers to : in the above-mentioned mobile scene, design a robust low-complexity online algorithm to locate objects in real-time super-resolution. (Let's assume the antenna radius is much smaller than the object's distance).

2. Research Status

(1) The baseline algorithm in existing products is to measure distance and angle by adding Hamming window and then doing FFT. The advantage is low complexity, and the disadvantage is low resolution.
(2) Traditional algorithms, such as the MUSIC algorithm, separate the signal space and the noise space through spatial smoothing filtering and the decomposition of the characteristic subspace, but it will also cause a decrease in resolution and greater interference by noise.
(3) Existing compressive sensing algorithms take advantage of the sparsity of spatial object distribution, which can effectively improve the resolution, but it is a huge challenge to deal with this continuous Fourier dictionary scene and design a low-complexity algorithm.

3. Problems to be solved

(1) Based on the provided noise-free simulation data, establish a positioning model, calculate the relative position of the object, and display it in a two-dimensional polar coordinate diagram (the abscissa indicates the distance, and the ordinate indicates the angle).
(2) Based on the provided Gaussian noise simulation data, use the IF signal within a chirp period to design a super-resolution algorithm to accurately locate multiple objects.
(3) Design an online low-complexity algorithm, use a frame of intermediate frequency signals for super-resolution positioning, and verify the performance of the algorithm through numerical experiments. Based on a frame of data provided, the relative trajectory of the object is calculated and displayed in a two-dimensional graph (the abscissa represents the distance, and the ordinate represents the angle).
(4) Considering that due to aging and other reasons in the actual scene, the antenna array will also have errors in its own positioning. Based on the provided simulation data, an improved algorithm is designed to improve the robustness of the positioning algorithm.

2. Basic knowledge of radar

3. Solutions

According to the data given in the first question, data_q1 is a 86×256 matrix, in which 86 rows represent the number of antennas, and 256 columns represent the sampling points of a single chirp.
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insert image description hereThe code for the first question is as follows. If you have any questions, you can wait to refer to the second part of the blog post for basic knowledge of radar.

clc;
clear;
close all;

load data_q1;
% -----—— Copyright By ZhiZhao,20230510 ---------- %
%% 雷达参数设置
c = physconst('LightSpeed');    % 光速
f0 = 78.8e9;                    % 载频
Lambda = c/f0;                  % 波长

T = 32e-6;                  % 单个chirp的周期
Ts = 0.125e-6;              % 采样间隔
Fs  = 1/Ts;                 % 单个chirp的采样频率
Nf = 32;                    % 一帧的脉冲数
Na = 86;                    % 天线数量
L = 0.0815;                 % 天线孔径
u = 78.986e12;              % 调频斜率
NumADC = fix(Fs*T);         % 单个chirp的采样点
Nfft1 = NumADC;             % 距离维做FFT的点数
Fs_Ant = 1/T;               % 空间FFT采样频率
Nfft2 = 128;                % 角度做FFT的点数
d_lambda = 1/2;             % 阵元间距与波长之比

%% 计算过程
dataIn = Z;
% 目标回波信号时域
t = (0:1/Fs:T-1/Fs);
figure;
plot(t*1e6,real(dataIn(1,:)));
xlabel('时间/us');ylabel('幅值');title('目标回波信号时域');

% 求目标回波信号距离
[m,n] = size(dataIn);
fft1_Data = zeros(m,n);
for ii = 1:m
    fft1_Data(ii,:) = fft(dataIn(ii,:),Nfft1);
end

f1 = (0:Nfft1-1)*Fs/Nfft1;     % 真实频率刻度
figure;
plot(f1*c/(2*u),abs(fft1_Data(1,:)));
xlabel('距离/m');ylabel('幅值');title('目标回波信号距离维');

fft2_Data = zeros(Nfft2,Nfft1);
win2 = hamming(Na);
for jj = 1:n
    fft2_Data(:,jj) = fftshift(fft(fft1_Data(:,jj).*win2,Nfft2));
end

f2 = (-Nfft2/2:Nfft2/2-1)*180/pi;
figure;
mesh(f1*c/(2*u),f2*2/Nfft2,abs(fft2_Data));
xlabel('距离/m');ylabel('角度/°');title('目标距离和角度');

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Origin blog.csdn.net/weixin_45317919/article/details/127193369