使用时差(TDOA)对信号进行三边测量 [英] Trilateration of a signal using Time Difference(TDOA)

问题描述

我在查找或实施算法来查找信号源时遇到了一些麻烦.我的工作目的是找到发声器的位置.

I am having some trouble to find or implement an algorithm to find a signal source. The objective of my work is to find the sound emitter position.

为此，我使用了三个振动传感器.我使用的技术是基于到达时间差的多边测量.

To accomplish this I am using three vibration sensors. The technique that I am using is multilateration that is based on the time difference of arrival.

每个传感器之间的到达时间差通过接收信号的互相关来找到.

The time difference of arrival between each sensor are found using Cross Correlation of the received signals.

我已经实现了该算法来找到到达时间差，但是我的问题更多是在多纬度的工作原理上，根据我的参考资料我还不清楚，因此我找不到其他免费的良好参考资料/打开.

I already implemented the algorithm to find the time difference of arrival, but my problem is more on how multilateration works, it's unclear for me based on my reference, and I couldn't find any other good reference for this that are free/open.

我看到了这篇帖子使用TDOA进行三边测量 但由于我只有三个TDOA，我无法弄清楚如何解出多边维基百科页面的等式(7).

I saw this post Trilateration using TDOA But I can't figure out how to solve the set of equations(7) of the wikipedia page of multilateration as i have only the three TDOA.

在这方面的任何帮助将不胜感激

Any help on this would be much appreciated

推荐答案

您有三个传感器坐标A,B,C，信号源P的未知坐标，信号开始时间t0的未知时间和信号注册的三倍ta, tb, tc.

You have three sensor coordinates A,B,C, unknown coordinate of signal source P, unknown time of signal start t0, and three times of signal registration ta, tb, tc.

示例:让我们传感器A在12:00:05捕获信号，传感器B-在12:00:00，传感器C-12:00:07捕获信号.因此，请分配时间差:ta=5, tb=0, tc=7

Example: Let's sensor A caught a signal in 12:00:05, sensor B - in 12:00:00, sensor C - 12:00:07. So assign time differences: ta=5, tb=0, tc=7

从传感器到源的平方距离对应于速度为v的信号传播时间(声速在空中或其他环境中

Squared distances from sensors to source correspond to times of signal walk with velocity v (speed of sound in air or another environment)

(Ax-Px)^2 + (Ay-Py)^2 = (v*(ta-t0))^2   {1}
(Bx-Px)^2 + (By-Py)^2 = (v*(tb-t0))^2   {2}
(Cx-Px)^2 + (Cy-Py)^2 = (v*(tc-t0))^2   {3}

打开方括号，然后减去方程式({2}-{1}，{3}-{2}，{1}-{3})以丢弃未知项的平方.

Open the brackets, then subtract equations ({2}-{1}, {3}-{2},{1}-{3}) to discard squares of unknown terms.

 Ax^2-2*Ax*Px + Px^2 + Ay^2-2*Ay*Py + Py^2 = v^2*(ta^2 - 2*ta*t0 + t0^2)
 Bx^2-2*Bx*Px + Px^2 + By^2-2*By*Py + Py^2 = v^2*(tb^2 - 2*tb*t0 + t0^2)
 Cx^2-2*Cx*Px + Px^2 + Cy^2-2*Cy*Py + Py^2 = v^2*(tc^2 - 2*tc*t0 + t0^2)

 Bx^2-Ax^2 -2*(Bx-Ax)*Px + By^2-Ay^2 -2*(By-Ay)*Py = v^2*(tb^2-ta^2 -2*(tb-ta)*t0)    {1'}
 Cx^2-Bx^2 -2*(Cx-Bx)*Px + Cy^2-By^2 -2*(Cy-By)*Py = v^2*(tc^2-tb^2 -2*(tc-tb)*t0)    {2'}
 Ax^2-Cx^2 -2*(Ax-Cx)*Px + Ay^2-Cy^2 -2*(Ay-Cy)*Py = v^2*(ta^2-tc^2 -2*(ta-tc)*t0)    {3'}

现在您有了三个线性方程组的系统，其中三个未知数.可以使用一些广泛的算法来解决问题-高斯消除，

Now you have system of three linear equations with three unknowns. It might be solved with some widespread algorithms - Gauss elimination, LU decomposition etc.

请注意，解决方案的精度很大程度上取决于坐标和时间测量中的小误差(此方法不是很可靠).

Note that solution precision strongly depends on small errors in coordinates and time measurements (this method is not very robust).

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