给定三个接收器的位置以及接收器接收信号的时间(到达时间延迟)，如何定位信号? [英] How to localize a signal given the location of three receivers and the times at which when they receive the signal (Time Delay of Arrival)?
问题描述
我有3个接收器(A，B和C)，还有一些信号产生源(比如声音或光)，位置未知.给定A，B和C的位置，以及每个接收器听到"的时间.信号，我想确定信号源的方向.
我知道可以使用TDoA多边测量/三边测量来进行此操作，但是我在实现计算时遇到了麻烦.对于那些刚接触该主题的人，这里没有很多清晰，详细的信息.对于我来说，那里的内容比较模糊，理论性更高或有些深奥.
关于SO的一些类似帖子(但不完全是我所追求的):
蓝点是接收器，红点是对象的真实位置，黄叉是其估计位置.该区域为 512x512 m ，我将其与初始步骤 32 m 和 6 递归拟合，导致错误〜0.005 m
我对结果感到满意...您可以更改接收器的数量 n ，而无需对源或算法进行任何实际更改.我开始将接收器位置均匀分布在圆上，但这些位置可能是其他任何位置(并非全部都在粗略的单行上)
I have 3 receivers (A, B and C), and some signal producing source (let's say sound or light) with an unknown location. Given the locations of A,B and C, and the time at which each receiver "heard" the signal, I'd like to determine the direction of the source.
I understand there are ways to do so with TDoA multilateration/trilateration, however I'm having trouble implementing the calculation. There isn't a lot of clear, detailed information on this out there for those entirely new to the subject. What is out there is vague, more theoretical, or a bit too esoteric for me.
Some similar posts on SO (but not quite what I'm after): TDOA multilateration to locate a sound source Trilateration of a signal using Time Difference(TDOA)
This is also interesting, but assumes we have some boundaries: Multiliteration implementation with inaccurate distance data
@Dave also commented an excellent and fairly accessible resource https://sites.tufts.edu/eeseniordesignhandbook/files/2017/05/FireBrick_OKeefe_F1.pdf, but it falls short of going into enough depth that one might be able to actually implement this in code (at least, for someone without deep knowledge of regression, finding the intersection of the resulting hyperbolas, etc).
[EDIT]: I should add that I can assume the 3 sensors and the source are on the surface of the Earth, and the effects of the curvature of the Earth are negligible (i.e. we can work in 2-dimensions).
Interesting problem. I am too lazy to derive equations for algebraic solution. Instead why not fit the result?
So simply fit the 2D (or higher) position using any fitting method capable of finding local solution (using optimization/minimization of some error value). When I use mine simple approximation search to fit the position the results looks pretty good.
The algorithm is:
iterate through "all" positions on your range
of coarse not all the heuristics of the fitting will reduce the problem a lot.
on each tested position compute the delta times that would be measured
simple time of travel from tested position into your receiver stations.
normalize all delta times so the start from zero
so substract the smallest time of arrival from all the recivers times. The same do for the real measured times. This assures the times does not involve relative offset.
compute the difference between real measured times and computed ones
simple abs difference is enough. Use this value as fitting parameter (optimization).
Here small C++ example doing this using my approx class from link above:
//---------------------------------------------------------------------------
// TDoA Time Difference of Arrival
//---------------------------------------------------------------------------
const int n=3;
double recv[n][3]; // (x,y) [m] receiver position,[s] time of arrival of signal
double pos0[2]; // (x,y) [m] object's real position
double pos [2]; // (x,y) [m] object's estimated position
double v=340.0; // [m/s] speed of signal
double err=0.0; // [m] error between real and estimated position
//---------------------------------------------------------------------------
void compute()
{
int i;
double x,y,a,da,t0;
//---------------------------------------------------------
// init positions
da=2.0*M_PI/(n);
for (a=0.0,i=0;i<n;i++,a+=da)
{
recv[i][0]=256.0+(220.0*cos(a));
recv[i][1]=256.0+(220.0*sin(a));
}
pos0[0]=300.0;
pos0[1]=220.0;
// simulate measurement
t0=123.5; // some start time
for (i=0;i<n;i++)
{
x=recv[i][0]-pos0[0];
y=recv[i][1]-pos0[1];
a=sqrt((x*x)+(y*y)); // distance to receiver
recv[i][2]=t0+(a/v); // start time + time of travel
}
//---------------------------------------------------------
// normalize times into deltas from zero
a=recv[0][2]; for (i=1;i<n;i++) if (a>recv[i][2]) a=recv[i][2];
for (i=0;i<n;i++) recv[i][2]-=a;
// fit position
int N=6;
approx ax,ay;
double e,dt[n];
// min, max,step,recursions,&error
for (ax.init( 0.0,512.0, 32.0 ,N, &e);!ax.done;ax.step())
for (ay.init( 0.0,512.0, 32.0 ,N, &e);!ay.done;ay.step())
{
// simulate measurement -> dt[]
for (i=0;i<n;i++)
{
x=recv[i][0]-ax.a;
y=recv[i][1]-ay.a;
a=sqrt((x*x)+(y*y)); // distance to receiver
dt[i]=a/v; // time of travel
}
// normalize times dt[] into deltas from zero
a=dt[0]; for (i=1;i<n;i++) if (a>dt[i]) a=dt[i];
for (i=0;i<n;i++) dt[i]-=a;
// error
e=0.0; for (i=0;i<n;i++) e+=fabs(recv[i][2]-dt[i]);
}
pos[0]=ax.aa;
pos[1]=ay.aa;
//---------------------------------------------------------
// compute error
x=pos[0]-pos0[0];
y=pos[1]-pos0[1];
err=sqrt((x*x)+(y*y)); // [m]
}
//---------------------------------------------------------------------------
Here preview:
Blue dots are the receivers, Red dot is real position of object and yellow cross is its estimated position. The area is 512x512 m and I fit it with initial step 32 m and 6 recursions leading to error ~0.005 m
I am more than happy with the result... You can change the number of receivers n without any real change to source or algorithm. I initiated the receiver positions as uniformly distributed on circle but the positions might be any other (not all on single line of coarse)
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