Solving a System of Quadratic Equations for Sound TriangulationSound TriangulationTriangulation mathSolve $b_1 e^-a_1x^2-b_2 e^-a_2x^2-b_3 e^-a_3x^2=0, forall x$System of linear equations with repeated equationsTrying to show $|overrightarrowatimesoverrightarrowb|^2=|overrightarrowa|^2|overrightarrowb|^2-(overrightarrowa⋅overrightarrowb)^2$Closed-form formula for system of two bivariate quadratic polynomialsShow that three circles are coaxalSystem of differential equations with three unknown functionsProve that the given determinant is not zeroSystem of linear equations with positive solutions / intersection of quadrics in $mathbbR^6$System of coupled differential equations of complex functions.Quadratic simultaneous equations with three variables
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Solving a System of Quadratic Equations for Sound Triangulation
Sound TriangulationTriangulation mathSolve $b_1 e^-a_1x^2-b_2 e^-a_2x^2-b_3 e^-a_3x^2=0, forall x$System of linear equations with repeated equationsTrying to show $|overrightarrowatimesoverrightarrowb|^2=|overrightarrowa|^2|overrightarrowb|^2-(overrightarrowa⋅overrightarrowb)^2$Closed-form formula for system of two bivariate quadratic polynomialsShow that three circles are coaxalSystem of differential equations with three unknown functionsProve that the given determinant is not zeroSystem of linear equations with positive solutions / intersection of quadrics in $mathbbR^6$System of coupled differential equations of complex functions.Quadratic simultaneous equations with three variables
$begingroup$
I am currently attempting to solve a system of quadratic (and linear) systems that I have run into while trying to triangulate sound.
My hypothetical setup includes 3 sensors on a perfectly equilateral triangle, with one sensor located at $(0,0)$ and the other two located below it. (The specifics don't matter, as I am simply referring to the sensor locations using $a_1,a_2,a_3$ for the x-coordinates of the sensors, and $b_1,b_2,b_3$ for the y-coordinates of the sensors, with $r_1,r_2,r_3$ being the radii of the circles from each respective sensor to the sound point)
I am trying to specify equations for the x position of the sound, the y position of the sound, and finally the radius of the incident sensor to the sound (the sensor that picks up the sound wave first).
My equations are as follows:
$$(x - a_1)^2 + (y - b_1)^2 = r_1^2$$
$$(x - a_2)^2 + (y - b_2)^2 = r_2^2$$
$$(x - a_3)^2 + (y - b_3)^2 = r_3^2$$
$$r_3 = r_1 + (t_3 * textspeed of sound)$$
$$r_2 = r_1 + (t_2 *textspeed of sound)$$
In this example, I am assuming that the sound reaches sensor 1 first. I understand that a true solution requires 3 discrete solutions, one for each sensor being the "incident sensor". (assuming that there cannot be a scenario where sound perfectly reaches multiple sensors at the same time)
My known variables: $a_1,a_2,a_3,b_1,b_2,b_3,textspeed of sound, t_1,t_2,t_3$
My Unknown variables: $x,y,r_1,r_2,r_3$.
Now I understand that I can just substitute in the three linear equations, but that leaves me with three quadratic equations that I am unsure of how to solve and obtain a meaningful answer from.
I tried searching for revelant topics, and the closest I could come was this:
https://math.stackexchange.com/a/187858/656339
Which has the same setup as I, but doesn't detail how to solve it.
Any help would be appreciated.
systems-of-equations circles quadratics nonlinear-system
edited Mar 21 at 5:50
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 5:01
ReThinkReThink
61
$endgroup$
|
show 4 more comments
$begingroup$
I am currently attempting to solve a system of quadratic (and linear) systems that I have run into while trying to triangulate sound.
My hypothetical setup includes 3 sensors on a perfectly equilateral triangle, with one sensor located at $(0,0)$ and the other two located below it. (The specifics don't matter, as I am simply referring to the sensor locations using $a_1,a_2,a_3$ for the x-coordinates of the sensors, and $b_1,b_2,b_3$ for the y-coordinates of the sensors, with $r_1,r_2,r_3$ being the radii of the circles from each respective sensor to the sound point)
I am trying to specify equations for the x position of the sound, the y position of the sound, and finally the radius of the incident sensor to the sound (the sensor that picks up the sound wave first).
My equations are as follows:
$$(x - a_1)^2 + (y - b_1)^2 = r_1^2$$
$$(x - a_2)^2 + (y - b_2)^2 = r_2^2$$
$$(x - a_3)^2 + (y - b_3)^2 = r_3^2$$
$$r_3 = r_1 + (t_3 * textspeed of sound)$$
$$r_2 = r_1 + (t_2 *textspeed of sound)$$
In this example, I am assuming that the sound reaches sensor 1 first. I understand that a true solution requires 3 discrete solutions, one for each sensor being the "incident sensor". (assuming that there cannot be a scenario where sound perfectly reaches multiple sensors at the same time)
My known variables: $a_1,a_2,a_3,b_1,b_2,b_3,textspeed of sound, t_1,t_2,t_3$
My Unknown variables: $x,y,r_1,r_2,r_3$.
Now I understand that I can just substitute in the three linear equations, but that leaves me with three quadratic equations that I am unsure of how to solve and obtain a meaningful answer from.
I tried searching for revelant topics, and the closest I could come was this:
https://math.stackexchange.com/a/187858/656339
Which has the same setup as I, but doesn't detail how to solve it.
Any help would be appreciated.
systems-of-equations circles quadratics nonlinear-system
edited Mar 21 at 5:50
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 5:01
ReThinkReThink
61
$endgroup$
$begingroup$
What you’re trying to do is called “TDOA multilateration.” Technically speaking, triangulation involves angles, not arrival time differences.
$endgroup$
– amd
Mar 21 at 7:53
$begingroup$
You can convert the system of circle equations into a system of linear equations: the intersections of a pair of circles lie on their radical axis, an equation for which you can obtain by subtracting one circle equation from another. However, since you don’t know $r_1$, what you’re really going to end up having to do is compute intersections of hyperbolas.
$endgroup$
– amd
Mar 21 at 7:55
$begingroup$
The problem would be easier with a fourth sensor.
$endgroup$
– Claude Leibovici
Mar 21 at 9:16
$begingroup$
@amd could you expound a little on how to go about doing that? Converting to a system of linear equations would be preferred
$endgroup$
– ReThink
Mar 21 at 13:16
$begingroup$
Have a look at math.stackexchange.com/questions/2462039/triangulation-math/…
$endgroup$
– Claude Leibovici
Mar 21 at 14:46
|
show 4 more comments
$begingroup$
I am currently attempting to solve a system of quadratic (and linear) systems that I have run into while trying to triangulate sound.
My hypothetical setup includes 3 sensors on a perfectly equilateral triangle, with one sensor located at $(0,0)$ and the other two located below it. (The specifics don't matter, as I am simply referring to the sensor locations using $a_1,a_2,a_3$ for the x-coordinates of the sensors, and $b_1,b_2,b_3$ for the y-coordinates of the sensors, with $r_1,r_2,r_3$ being the radii of the circles from each respective sensor to the sound point)
I am trying to specify equations for the x position of the sound, the y position of the sound, and finally the radius of the incident sensor to the sound (the sensor that picks up the sound wave first).
My equations are as follows:
$$(x - a_1)^2 + (y - b_1)^2 = r_1^2$$
$$(x - a_2)^2 + (y - b_2)^2 = r_2^2$$
$$(x - a_3)^2 + (y - b_3)^2 = r_3^2$$
$$r_3 = r_1 + (t_3 * textspeed of sound)$$
$$r_2 = r_1 + (t_2 *textspeed of sound)$$
In this example, I am assuming that the sound reaches sensor 1 first. I understand that a true solution requires 3 discrete solutions, one for each sensor being the "incident sensor". (assuming that there cannot be a scenario where sound perfectly reaches multiple sensors at the same time)
My known variables: $a_1,a_2,a_3,b_1,b_2,b_3,textspeed of sound, t_1,t_2,t_3$
My Unknown variables: $x,y,r_1,r_2,r_3$.
Now I understand that I can just substitute in the three linear equations, but that leaves me with three quadratic equations that I am unsure of how to solve and obtain a meaningful answer from.
I tried searching for revelant topics, and the closest I could come was this:
https://math.stackexchange.com/a/187858/656339
Which has the same setup as I, but doesn't detail how to solve it.
Any help would be appreciated.
systems-of-equations circles quadratics nonlinear-system
edited Mar 21 at 5:50
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 5:01
ReThinkReThink
61
$endgroup$
I am currently attempting to solve a system of quadratic (and linear) systems that I have run into while trying to triangulate sound.
My hypothetical setup includes 3 sensors on a perfectly equilateral triangle, with one sensor located at $(0,0)$ and the other two located below it. (The specifics don't matter, as I am simply referring to the sensor locations using $a_1,a_2,a_3$ for the x-coordinates of the sensors, and $b_1,b_2,b_3$ for the y-coordinates of the sensors, with $r_1,r_2,r_3$ being the radii of the circles from each respective sensor to the sound point)
I am trying to specify equations for the x position of the sound, the y position of the sound, and finally the radius of the incident sensor to the sound (the sensor that picks up the sound wave first).
My equations are as follows:
$$(x - a_1)^2 + (y - b_1)^2 = r_1^2$$
$$(x - a_2)^2 + (y - b_2)^2 = r_2^2$$
$$(x - a_3)^2 + (y - b_3)^2 = r_3^2$$
$$r_3 = r_1 + (t_3 * textspeed of sound)$$
$$r_2 = r_1 + (t_2 *textspeed of sound)$$
In this example, I am assuming that the sound reaches sensor 1 first. I understand that a true solution requires 3 discrete solutions, one for each sensor being the "incident sensor". (assuming that there cannot be a scenario where sound perfectly reaches multiple sensors at the same time)
My known variables: $a_1,a_2,a_3,b_1,b_2,b_3,textspeed of sound, t_1,t_2,t_3$
My Unknown variables: $x,y,r_1,r_2,r_3$.
Now I understand that I can just substitute in the three linear equations, but that leaves me with three quadratic equations that I am unsure of how to solve and obtain a meaningful answer from.
I tried searching for revelant topics, and the closest I could come was this:
https://math.stackexchange.com/a/187858/656339
Which has the same setup as I, but doesn't detail how to solve it.
Any help would be appreciated.
systems-of-equations circles quadratics nonlinear-system
systems-of-equations circles quadratics nonlinear-system
edited Mar 21 at 5:50
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 5:01
ReThinkReThink
61
edited Mar 21 at 5:50
Rodrigo de Azevedo
13.1k41960
asked Mar 21 at 5:01
ReThinkReThink
61
edited Mar 21 at 5:50
Rodrigo de Azevedo
13.1k41960
edited Mar 21 at 5:50
Rodrigo de Azevedo
13.1k41960
edited Mar 21 at 5:50
Rodrigo de Azevedo
13.1k41960
13.1k41960
asked Mar 21 at 5:01
ReThinkReThink
61
asked Mar 21 at 5:01
ReThinkReThink
61
asked Mar 21 at 5:01
ReThinkReThink
61
61
$begingroup$
What you’re trying to do is called “TDOA multilateration.” Technically speaking, triangulation involves angles, not arrival time differences.
$endgroup$
– amd
Mar 21 at 7:53
$begingroup$
You can convert the system of circle equations into a system of linear equations: the intersections of a pair of circles lie on their radical axis, an equation for which you can obtain by subtracting one circle equation from another. However, since you don’t know $r_1$, what you’re really going to end up having to do is compute intersections of hyperbolas.
$endgroup$
– amd
Mar 21 at 7:55
$begingroup$
The problem would be easier with a fourth sensor.
$endgroup$
– Claude Leibovici
Mar 21 at 9:16
$begingroup$
@amd could you expound a little on how to go about doing that? Converting to a system of linear equations would be preferred
$endgroup$
– ReThink
Mar 21 at 13:16
$begingroup$
Have a look at math.stackexchange.com/questions/2462039/triangulation-math/…
$endgroup$
– Claude Leibovici
Mar 21 at 14:46
|
show 4 more comments
$begingroup$
What you’re trying to do is called “TDOA multilateration.” Technically speaking, triangulation involves angles, not arrival time differences.
$endgroup$
– amd
Mar 21 at 7:53
$begingroup$
You can convert the system of circle equations into a system of linear equations: the intersections of a pair of circles lie on their radical axis, an equation for which you can obtain by subtracting one circle equation from another. However, since you don’t know $r_1$, what you’re really going to end up having to do is compute intersections of hyperbolas.
$endgroup$
– amd
Mar 21 at 7:55
$begingroup$
The problem would be easier with a fourth sensor.
$endgroup$
– Claude Leibovici
Mar 21 at 9:16
$begingroup$
@amd could you expound a little on how to go about doing that? Converting to a system of linear equations would be preferred
$endgroup$
– ReThink
Mar 21 at 13:16
$begingroup$
Have a look at math.stackexchange.com/questions/2462039/triangulation-math/…
$endgroup$
– Claude Leibovici
Mar 21 at 14:46
$begingroup$
What you’re trying to do is called “TDOA multilateration.” Technically speaking, triangulation involves angles, not arrival time differences.
$endgroup$
– amd
Mar 21 at 7:53
$begingroup$
What you’re trying to do is called “TDOA multilateration.” Technically speaking, triangulation involves angles, not arrival time differences.
$endgroup$
– amd
Mar 21 at 7:53
$begingroup$
You can convert the system of circle equations into a system of linear equations: the intersections of a pair of circles lie on their radical axis, an equation for which you can obtain by subtracting one circle equation from another. However, since you don’t know $r_1$, what you’re really going to end up having to do is compute intersections of hyperbolas.
$endgroup$
– amd
Mar 21 at 7:55
$begingroup$
You can convert the system of circle equations into a system of linear equations: the intersections of a pair of circles lie on their radical axis, an equation for which you can obtain by subtracting one circle equation from another. However, since you don’t know $r_1$, what you’re really going to end up having to do is compute intersections of hyperbolas.
$endgroup$
– amd
Mar 21 at 7:55
$begingroup$
The problem would be easier with a fourth sensor.
$endgroup$
– Claude Leibovici
Mar 21 at 9:16
$begingroup$
The problem would be easier with a fourth sensor.
$endgroup$
– Claude Leibovici
Mar 21 at 9:16
$begingroup$
@amd could you expound a little on how to go about doing that? Converting to a system of linear equations would be preferred
$endgroup$
– ReThink
Mar 21 at 13:16
$begingroup$
@amd could you expound a little on how to go about doing that? Converting to a system of linear equations would be preferred
$endgroup$
– ReThink
Mar 21 at 13:16
$begingroup$
Have a look at math.stackexchange.com/questions/2462039/triangulation-math/…
$endgroup$
– Claude Leibovici
Mar 21 at 14:46
$begingroup$
Have a look at math.stackexchange.com/questions/2462039/triangulation-math/…
$endgroup$
– Claude Leibovici
Mar 21 at 14:46
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Solving system of quadratic and indeed, general polynomial equations is possible with techniques like Buchberger's algorithm. See the first two chapters of the book by Cox et.al. There are also many numerical tools to assist you with this. Sympy in python is one alternative (the one I'd recommend) and there is also one I wrote in C# following the textbook cited.
answered Mar 21 at 5:48
Rohit PandeyRohit Pandey
1,6581024
$endgroup$
add a comment |
$begingroup$
After comments, let us work with $4$ sonsors. So we have
$$(x - a_1)^2 + (y - b_1)^2 = c^2(t_1-tau)^2tag 1$$
$$(x - a_2)^2 + (y - b_2)^2 = c^2(t_2-tau)^2tag 2$$
$$(x - a_3)^2 + (y - b_3)^2 = c^2(t_3-tau)^2tag 3$$
$$(x - a_4)^2 + (y - b_4)^2 = c^2(t_4-tau)^2tag 4$$ where $c$ is the speed of sound and $tau$ the time at which was produced the sound.
Now, subtract $(1)$ from $(2)$, $(3)$ and $(4)$ to get
$$2(a_1-a_2)x+2(b_1-b_2)y+2c^2(t_2-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_2^2+b_2^2-c^2t_2^2)tag 5$$
$$2(a_1-a_3)x+2(b_1-b_3)y+2c^2(t_3-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_3^2+b_3^2-c^2t_3^2)tag 6$$
$$2(a_1-a_4)x+2(b_1-b_4)y+2c^2(t_4-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_4^2+b_4^2-c^2t_4^2)tag 7$$
Define, for more simplicity,
$$alpha_i=2(a_1-a_i)qquad beta_i=2(b_1-b_i)qquad gamma_i=2c^2(t_i-t_1)$$ $$k_i=(a_1^2+b_1^2-c^2t_1^2)-(a_i^2+b_i^2-c^2t_i^2)$$ $(i=2,3,4)$ to make
$$alpha_2x+beta_2y+gamma_2 tau=k_2tag 8$$
$$alpha_3x+beta_3y+gamma_3 tau=k_3tag 9$$
$$alpha_4x+beta_4y+gamma_4 tau=k_4tag 10$$
So, three linear equations in $(x,y,tau)$ (easy to solve - use matrix calculations or successive elimination as I did below) and the explicit solutions are
$$tau=frack_4 (alpha_3 beta_2-alpha_2 beta_3)+k_3 (alpha_2 beta_4-alpha_4 beta_2)+k_2 (alpha_4 beta_3-alpha_3 beta_4) alpha_4 (beta_3 gamma_2-beta_2 gamma_3)+alpha_3 (beta_2 gamma_
4-beta_4 gamma_2)+alpha_2 (beta_4 gamma_3-beta_3 gamma_4) tag 11$$
$$y=fracalpha_3 (gamma_2 colorredtau -k_2)+alpha_2 (k_3-gamma_3 colorredtau )alpha_
2 beta_3-alpha_3 beta_2tag 12$$
$$x=frack_2-beta_2 colorredy-gamma_2 colorredtau alpha_2tag 13$$
answered Mar 25 at 5:10
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
$begingroup$
To clarify, is tau the time elapsed from the sound occurring relative to the first sensor that detects the sound wave?
$endgroup$
– ReThink
Mar 25 at 13:29
$begingroup$
@ReThink. Suppose that the first sensor receives the information at 1:34.05, the second at 1:34.06, the third at 1:34.08 and the fourth at 1:34.08 and you find $tau=1:31.01$; this is the time at which the sound was produced.
$endgroup$
– Claude Leibovici
Mar 25 at 17:16
$begingroup$
Thanks a ton! One more question: When you define alpha, beta, gamma, and k, you use subscript '1', does this mean that these only hold if sensor #1 is the first to receive the sound? i.e., would I have to rearrange and replace with a subscript of 2,3,4 if the respective sensor is the first to receive the sound wave? ex: αi=2(a2−ai) if sensor #2 is the incident sensor? Or does your solution hold regardless of which sensor is the incident sensor?
$endgroup$
– ReThink
Mar 25 at 18:12
$begingroup$
@ReThink. There is no incident sensor. You have sensors located at $x_i,y_i$ which at time $t_i$ receive an information. If you want to use this concept (which is dangerous at least to me), start defining $t_min$ and redefine the $t_i$ as $t_i-t_min$.
$endgroup$
– Claude Leibovici
Mar 26 at 5:36
$begingroup$
I ran everything in Matlab just to visualize what's going on, and I understand now. Tau can be either an absolute or a relative time (it does not matter to the sensors) and t1 - t4 are offsets from this reference point (tau). The incident sensor does not matter, as that information is a part of the arrival times of the sound waves to the respective sensors, which is a fundamental part of t1-t4. Thank you very much for taking the time to explain this concept! It's fascinating that the problem can be simplified by just adding one more sensor.
$endgroup$
– ReThink
Mar 27 at 2:12
|
show 1 more comment
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
Solving system of quadratic and indeed, general polynomial equations is possible with techniques like Buchberger's algorithm. See the first two chapters of the book by Cox et.al. There are also many numerical tools to assist you with this. Sympy in python is one alternative (the one I'd recommend) and there is also one I wrote in C# following the textbook cited.
answered Mar 21 at 5:48
Rohit PandeyRohit Pandey
1,6581024
$endgroup$
add a comment |
$begingroup$
Solving system of quadratic and indeed, general polynomial equations is possible with techniques like Buchberger's algorithm. See the first two chapters of the book by Cox et.al. There are also many numerical tools to assist you with this. Sympy in python is one alternative (the one I'd recommend) and there is also one I wrote in C# following the textbook cited.
answered Mar 21 at 5:48
Rohit PandeyRohit Pandey
1,6581024
$endgroup$
add a comment |
$begingroup$
Solving system of quadratic and indeed, general polynomial equations is possible with techniques like Buchberger's algorithm. See the first two chapters of the book by Cox et.al. There are also many numerical tools to assist you with this. Sympy in python is one alternative (the one I'd recommend) and there is also one I wrote in C# following the textbook cited.
answered Mar 21 at 5:48
Rohit PandeyRohit Pandey
1,6581024
$endgroup$
Solving system of quadratic and indeed, general polynomial equations is possible with techniques like Buchberger's algorithm. See the first two chapters of the book by Cox et.al. There are also many numerical tools to assist you with this. Sympy in python is one alternative (the one I'd recommend) and there is also one I wrote in C# following the textbook cited.
answered Mar 21 at 5:48
Rohit PandeyRohit Pandey
1,6581024
edited Mar 21 at 5:55
answered Mar 21 at 5:48
Rohit PandeyRohit Pandey
1,6581024
answered Mar 21 at 5:48
Rohit PandeyRohit Pandey
1,6581024
answered Mar 21 at 5:48
Rohit PandeyRohit Pandey
1,6581024
1,6581024
add a comment |
add a comment |
$begingroup$
After comments, let us work with $4$ sonsors. So we have
$$(x - a_1)^2 + (y - b_1)^2 = c^2(t_1-tau)^2tag 1$$
$$(x - a_2)^2 + (y - b_2)^2 = c^2(t_2-tau)^2tag 2$$
$$(x - a_3)^2 + (y - b_3)^2 = c^2(t_3-tau)^2tag 3$$
$$(x - a_4)^2 + (y - b_4)^2 = c^2(t_4-tau)^2tag 4$$ where $c$ is the speed of sound and $tau$ the time at which was produced the sound.
Now, subtract $(1)$ from $(2)$, $(3)$ and $(4)$ to get
$$2(a_1-a_2)x+2(b_1-b_2)y+2c^2(t_2-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_2^2+b_2^2-c^2t_2^2)tag 5$$
$$2(a_1-a_3)x+2(b_1-b_3)y+2c^2(t_3-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_3^2+b_3^2-c^2t_3^2)tag 6$$
$$2(a_1-a_4)x+2(b_1-b_4)y+2c^2(t_4-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_4^2+b_4^2-c^2t_4^2)tag 7$$
Define, for more simplicity,
$$alpha_i=2(a_1-a_i)qquad beta_i=2(b_1-b_i)qquad gamma_i=2c^2(t_i-t_1)$$ $$k_i=(a_1^2+b_1^2-c^2t_1^2)-(a_i^2+b_i^2-c^2t_i^2)$$ $(i=2,3,4)$ to make
$$alpha_2x+beta_2y+gamma_2 tau=k_2tag 8$$
$$alpha_3x+beta_3y+gamma_3 tau=k_3tag 9$$
$$alpha_4x+beta_4y+gamma_4 tau=k_4tag 10$$
So, three linear equations in $(x,y,tau)$ (easy to solve - use matrix calculations or successive elimination as I did below) and the explicit solutions are
$$tau=frack_4 (alpha_3 beta_2-alpha_2 beta_3)+k_3 (alpha_2 beta_4-alpha_4 beta_2)+k_2 (alpha_4 beta_3-alpha_3 beta_4) alpha_4 (beta_3 gamma_2-beta_2 gamma_3)+alpha_3 (beta_2 gamma_
4-beta_4 gamma_2)+alpha_2 (beta_4 gamma_3-beta_3 gamma_4) tag 11$$
$$y=fracalpha_3 (gamma_2 colorredtau -k_2)+alpha_2 (k_3-gamma_3 colorredtau )alpha_
2 beta_3-alpha_3 beta_2tag 12$$
$$x=frack_2-beta_2 colorredy-gamma_2 colorredtau alpha_2tag 13$$
answered Mar 25 at 5:10
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
$begingroup$
To clarify, is tau the time elapsed from the sound occurring relative to the first sensor that detects the sound wave?
$endgroup$
– ReThink
Mar 25 at 13:29
$begingroup$
@ReThink. Suppose that the first sensor receives the information at 1:34.05, the second at 1:34.06, the third at 1:34.08 and the fourth at 1:34.08 and you find $tau=1:31.01$; this is the time at which the sound was produced.
$endgroup$
– Claude Leibovici
Mar 25 at 17:16
$begingroup$
Thanks a ton! One more question: When you define alpha, beta, gamma, and k, you use subscript '1', does this mean that these only hold if sensor #1 is the first to receive the sound? i.e., would I have to rearrange and replace with a subscript of 2,3,4 if the respective sensor is the first to receive the sound wave? ex: αi=2(a2−ai) if sensor #2 is the incident sensor? Or does your solution hold regardless of which sensor is the incident sensor?
$endgroup$
– ReThink
Mar 25 at 18:12
$begingroup$
@ReThink. There is no incident sensor. You have sensors located at $x_i,y_i$ which at time $t_i$ receive an information. If you want to use this concept (which is dangerous at least to me), start defining $t_min$ and redefine the $t_i$ as $t_i-t_min$.
$endgroup$
– Claude Leibovici
Mar 26 at 5:36
$begingroup$
I ran everything in Matlab just to visualize what's going on, and I understand now. Tau can be either an absolute or a relative time (it does not matter to the sensors) and t1 - t4 are offsets from this reference point (tau). The incident sensor does not matter, as that information is a part of the arrival times of the sound waves to the respective sensors, which is a fundamental part of t1-t4. Thank you very much for taking the time to explain this concept! It's fascinating that the problem can be simplified by just adding one more sensor.
$endgroup$
– ReThink
Mar 27 at 2:12
|
show 1 more comment
$begingroup$
After comments, let us work with $4$ sonsors. So we have
$$(x - a_1)^2 + (y - b_1)^2 = c^2(t_1-tau)^2tag 1$$
$$(x - a_2)^2 + (y - b_2)^2 = c^2(t_2-tau)^2tag 2$$
$$(x - a_3)^2 + (y - b_3)^2 = c^2(t_3-tau)^2tag 3$$
$$(x - a_4)^2 + (y - b_4)^2 = c^2(t_4-tau)^2tag 4$$ where $c$ is the speed of sound and $tau$ the time at which was produced the sound.
Now, subtract $(1)$ from $(2)$, $(3)$ and $(4)$ to get
$$2(a_1-a_2)x+2(b_1-b_2)y+2c^2(t_2-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_2^2+b_2^2-c^2t_2^2)tag 5$$
$$2(a_1-a_3)x+2(b_1-b_3)y+2c^2(t_3-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_3^2+b_3^2-c^2t_3^2)tag 6$$
$$2(a_1-a_4)x+2(b_1-b_4)y+2c^2(t_4-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_4^2+b_4^2-c^2t_4^2)tag 7$$
Define, for more simplicity,
$$alpha_i=2(a_1-a_i)qquad beta_i=2(b_1-b_i)qquad gamma_i=2c^2(t_i-t_1)$$ $$k_i=(a_1^2+b_1^2-c^2t_1^2)-(a_i^2+b_i^2-c^2t_i^2)$$ $(i=2,3,4)$ to make
$$alpha_2x+beta_2y+gamma_2 tau=k_2tag 8$$
$$alpha_3x+beta_3y+gamma_3 tau=k_3tag 9$$
$$alpha_4x+beta_4y+gamma_4 tau=k_4tag 10$$
So, three linear equations in $(x,y,tau)$ (easy to solve - use matrix calculations or successive elimination as I did below) and the explicit solutions are
$$tau=frack_4 (alpha_3 beta_2-alpha_2 beta_3)+k_3 (alpha_2 beta_4-alpha_4 beta_2)+k_2 (alpha_4 beta_3-alpha_3 beta_4) alpha_4 (beta_3 gamma_2-beta_2 gamma_3)+alpha_3 (beta_2 gamma_
4-beta_4 gamma_2)+alpha_2 (beta_4 gamma_3-beta_3 gamma_4) tag 11$$
$$y=fracalpha_3 (gamma_2 colorredtau -k_2)+alpha_2 (k_3-gamma_3 colorredtau )alpha_
2 beta_3-alpha_3 beta_2tag 12$$
$$x=frack_2-beta_2 colorredy-gamma_2 colorredtau alpha_2tag 13$$
answered Mar 25 at 5:10
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
$begingroup$
To clarify, is tau the time elapsed from the sound occurring relative to the first sensor that detects the sound wave?
$endgroup$
– ReThink
Mar 25 at 13:29
$begingroup$
@ReThink. Suppose that the first sensor receives the information at 1:34.05, the second at 1:34.06, the third at 1:34.08 and the fourth at 1:34.08 and you find $tau=1:31.01$; this is the time at which the sound was produced.
$endgroup$
– Claude Leibovici
Mar 25 at 17:16
$begingroup$
Thanks a ton! One more question: When you define alpha, beta, gamma, and k, you use subscript '1', does this mean that these only hold if sensor #1 is the first to receive the sound? i.e., would I have to rearrange and replace with a subscript of 2,3,4 if the respective sensor is the first to receive the sound wave? ex: αi=2(a2−ai) if sensor #2 is the incident sensor? Or does your solution hold regardless of which sensor is the incident sensor?
$endgroup$
– ReThink
Mar 25 at 18:12
$begingroup$
@ReThink. There is no incident sensor. You have sensors located at $x_i,y_i$ which at time $t_i$ receive an information. If you want to use this concept (which is dangerous at least to me), start defining $t_min$ and redefine the $t_i$ as $t_i-t_min$.
$endgroup$
– Claude Leibovici
Mar 26 at 5:36
$begingroup$
I ran everything in Matlab just to visualize what's going on, and I understand now. Tau can be either an absolute or a relative time (it does not matter to the sensors) and t1 - t4 are offsets from this reference point (tau). The incident sensor does not matter, as that information is a part of the arrival times of the sound waves to the respective sensors, which is a fundamental part of t1-t4. Thank you very much for taking the time to explain this concept! It's fascinating that the problem can be simplified by just adding one more sensor.
$endgroup$
– ReThink
Mar 27 at 2:12
|
show 1 more comment
$begingroup$
After comments, let us work with $4$ sonsors. So we have
$$(x - a_1)^2 + (y - b_1)^2 = c^2(t_1-tau)^2tag 1$$
$$(x - a_2)^2 + (y - b_2)^2 = c^2(t_2-tau)^2tag 2$$
$$(x - a_3)^2 + (y - b_3)^2 = c^2(t_3-tau)^2tag 3$$
$$(x - a_4)^2 + (y - b_4)^2 = c^2(t_4-tau)^2tag 4$$ where $c$ is the speed of sound and $tau$ the time at which was produced the sound.
Now, subtract $(1)$ from $(2)$, $(3)$ and $(4)$ to get
$$2(a_1-a_2)x+2(b_1-b_2)y+2c^2(t_2-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_2^2+b_2^2-c^2t_2^2)tag 5$$
$$2(a_1-a_3)x+2(b_1-b_3)y+2c^2(t_3-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_3^2+b_3^2-c^2t_3^2)tag 6$$
$$2(a_1-a_4)x+2(b_1-b_4)y+2c^2(t_4-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_4^2+b_4^2-c^2t_4^2)tag 7$$
Define, for more simplicity,
$$alpha_i=2(a_1-a_i)qquad beta_i=2(b_1-b_i)qquad gamma_i=2c^2(t_i-t_1)$$ $$k_i=(a_1^2+b_1^2-c^2t_1^2)-(a_i^2+b_i^2-c^2t_i^2)$$ $(i=2,3,4)$ to make
$$alpha_2x+beta_2y+gamma_2 tau=k_2tag 8$$
$$alpha_3x+beta_3y+gamma_3 tau=k_3tag 9$$
$$alpha_4x+beta_4y+gamma_4 tau=k_4tag 10$$
So, three linear equations in $(x,y,tau)$ (easy to solve - use matrix calculations or successive elimination as I did below) and the explicit solutions are
$$tau=frack_4 (alpha_3 beta_2-alpha_2 beta_3)+k_3 (alpha_2 beta_4-alpha_4 beta_2)+k_2 (alpha_4 beta_3-alpha_3 beta_4) alpha_4 (beta_3 gamma_2-beta_2 gamma_3)+alpha_3 (beta_2 gamma_
4-beta_4 gamma_2)+alpha_2 (beta_4 gamma_3-beta_3 gamma_4) tag 11$$
$$y=fracalpha_3 (gamma_2 colorredtau -k_2)+alpha_2 (k_3-gamma_3 colorredtau )alpha_
2 beta_3-alpha_3 beta_2tag 12$$
$$x=frack_2-beta_2 colorredy-gamma_2 colorredtau alpha_2tag 13$$
answered Mar 25 at 5:10
Claude LeiboviciClaude Leibovici
125k1158136
$endgroup$
After comments, let us work with $4$ sonsors. So we have
$$(x - a_1)^2 + (y - b_1)^2 = c^2(t_1-tau)^2tag 1$$
$$(x - a_2)^2 + (y - b_2)^2 = c^2(t_2-tau)^2tag 2$$
$$(x - a_3)^2 + (y - b_3)^2 = c^2(t_3-tau)^2tag 3$$
$$(x - a_4)^2 + (y - b_4)^2 = c^2(t_4-tau)^2tag 4$$ where $c$ is the speed of sound and $tau$ the time at which was produced the sound.
Now, subtract $(1)$ from $(2)$, $(3)$ and $(4)$ to get
$$2(a_1-a_2)x+2(b_1-b_2)y+2c^2(t_2-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_2^2+b_2^2-c^2t_2^2)tag 5$$
$$2(a_1-a_3)x+2(b_1-b_3)y+2c^2(t_3-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_3^2+b_3^2-c^2t_3^2)tag 6$$
$$2(a_1-a_4)x+2(b_1-b_4)y+2c^2(t_4-t_1)tau=(a_1^2+b_1^2-c^2t_1^2)-(a_4^2+b_4^2-c^2t_4^2)tag 7$$
Define, for more simplicity,
$$alpha_i=2(a_1-a_i)qquad beta_i=2(b_1-b_i)qquad gamma_i=2c^2(t_i-t_1)$$ $$k_i=(a_1^2+b_1^2-c^2t_1^2)-(a_i^2+b_i^2-c^2t_i^2)$$ $(i=2,3,4)$ to make
$$alpha_2x+beta_2y+gamma_2 tau=k_2tag 8$$
$$alpha_3x+beta_3y+gamma_3 tau=k_3tag 9$$
$$alpha_4x+beta_4y+gamma_4 tau=k_4tag 10$$
So, three linear equations in $(x,y,tau)$ (easy to solve - use matrix calculations or successive elimination as I did below) and the explicit solutions are
$$tau=frack_4 (alpha_3 beta_2-alpha_2 beta_3)+k_3 (alpha_2 beta_4-alpha_4 beta_2)+k_2 (alpha_4 beta_3-alpha_3 beta_4) alpha_4 (beta_3 gamma_2-beta_2 gamma_3)+alpha_3 (beta_2 gamma_
4-beta_4 gamma_2)+alpha_2 (beta_4 gamma_3-beta_3 gamma_4) tag 11$$
$$y=fracalpha_3 (gamma_2 colorredtau -k_2)+alpha_2 (k_3-gamma_3 colorredtau )alpha_
2 beta_3-alpha_3 beta_2tag 12$$
$$x=frack_2-beta_2 colorredy-gamma_2 colorredtau alpha_2tag 13$$
answered Mar 25 at 5:10
Claude LeiboviciClaude Leibovici
125k1158136
answered Mar 25 at 5:10
Claude LeiboviciClaude Leibovici
125k1158136
answered Mar 25 at 5:10
Claude LeiboviciClaude Leibovici
125k1158136
answered Mar 25 at 5:10
Claude LeiboviciClaude Leibovici
125k1158136
125k1158136
$begingroup$
To clarify, is tau the time elapsed from the sound occurring relative to the first sensor that detects the sound wave?
$endgroup$
– ReThink
Mar 25 at 13:29
$begingroup$
@ReThink. Suppose that the first sensor receives the information at 1:34.05, the second at 1:34.06, the third at 1:34.08 and the fourth at 1:34.08 and you find $tau=1:31.01$; this is the time at which the sound was produced.
$endgroup$
– Claude Leibovici
Mar 25 at 17:16
$begingroup$
Thanks a ton! One more question: When you define alpha, beta, gamma, and k, you use subscript '1', does this mean that these only hold if sensor #1 is the first to receive the sound? i.e., would I have to rearrange and replace with a subscript of 2,3,4 if the respective sensor is the first to receive the sound wave? ex: αi=2(a2−ai) if sensor #2 is the incident sensor? Or does your solution hold regardless of which sensor is the incident sensor?
$endgroup$
– ReThink
Mar 25 at 18:12
$begingroup$
@ReThink. There is no incident sensor. You have sensors located at $x_i,y_i$ which at time $t_i$ receive an information. If you want to use this concept (which is dangerous at least to me), start defining $t_min$ and redefine the $t_i$ as $t_i-t_min$.
$endgroup$
– Claude Leibovici
Mar 26 at 5:36
$begingroup$
I ran everything in Matlab just to visualize what's going on, and I understand now. Tau can be either an absolute or a relative time (it does not matter to the sensors) and t1 - t4 are offsets from this reference point (tau). The incident sensor does not matter, as that information is a part of the arrival times of the sound waves to the respective sensors, which is a fundamental part of t1-t4. Thank you very much for taking the time to explain this concept! It's fascinating that the problem can be simplified by just adding one more sensor.
$endgroup$
– ReThink
Mar 27 at 2:12
|
show 1 more comment
$begingroup$
To clarify, is tau the time elapsed from the sound occurring relative to the first sensor that detects the sound wave?
$endgroup$
– ReThink
Mar 25 at 13:29
$begingroup$
@ReThink. Suppose that the first sensor receives the information at 1:34.05, the second at 1:34.06, the third at 1:34.08 and the fourth at 1:34.08 and you find $tau=1:31.01$; this is the time at which the sound was produced.
$endgroup$
– Claude Leibovici
Mar 25 at 17:16
$begingroup$
Thanks a ton! One more question: When you define alpha, beta, gamma, and k, you use subscript '1', does this mean that these only hold if sensor #1 is the first to receive the sound? i.e., would I have to rearrange and replace with a subscript of 2,3,4 if the respective sensor is the first to receive the sound wave? ex: αi=2(a2−ai) if sensor #2 is the incident sensor? Or does your solution hold regardless of which sensor is the incident sensor?
$endgroup$
– ReThink
Mar 25 at 18:12
$begingroup$
@ReThink. There is no incident sensor. You have sensors located at $x_i,y_i$ which at time $t_i$ receive an information. If you want to use this concept (which is dangerous at least to me), start defining $t_min$ and redefine the $t_i$ as $t_i-t_min$.
$endgroup$
– Claude Leibovici
Mar 26 at 5:36
$begingroup$
I ran everything in Matlab just to visualize what's going on, and I understand now. Tau can be either an absolute or a relative time (it does not matter to the sensors) and t1 - t4 are offsets from this reference point (tau). The incident sensor does not matter, as that information is a part of the arrival times of the sound waves to the respective sensors, which is a fundamental part of t1-t4. Thank you very much for taking the time to explain this concept! It's fascinating that the problem can be simplified by just adding one more sensor.
$endgroup$
– ReThink
Mar 27 at 2:12
$begingroup$
To clarify, is tau the time elapsed from the sound occurring relative to the first sensor that detects the sound wave?
$endgroup$
– ReThink
Mar 25 at 13:29
$begingroup$
To clarify, is tau the time elapsed from the sound occurring relative to the first sensor that detects the sound wave?
$endgroup$
– ReThink
Mar 25 at 13:29
$begingroup$
@ReThink. Suppose that the first sensor receives the information at 1:34.05, the second at 1:34.06, the third at 1:34.08 and the fourth at 1:34.08 and you find $tau=1:31.01$; this is the time at which the sound was produced.
$endgroup$
– Claude Leibovici
Mar 25 at 17:16
$begingroup$
@ReThink. Suppose that the first sensor receives the information at 1:34.05, the second at 1:34.06, the third at 1:34.08 and the fourth at 1:34.08 and you find $tau=1:31.01$; this is the time at which the sound was produced.
$endgroup$
– Claude Leibovici
Mar 25 at 17:16
$begingroup$
Thanks a ton! One more question: When you define alpha, beta, gamma, and k, you use subscript '1', does this mean that these only hold if sensor #1 is the first to receive the sound? i.e., would I have to rearrange and replace with a subscript of 2,3,4 if the respective sensor is the first to receive the sound wave? ex: αi=2(a2−ai) if sensor #2 is the incident sensor? Or does your solution hold regardless of which sensor is the incident sensor?
$endgroup$
– ReThink
Mar 25 at 18:12
$begingroup$
Thanks a ton! One more question: When you define alpha, beta, gamma, and k, you use subscript '1', does this mean that these only hold if sensor #1 is the first to receive the sound? i.e., would I have to rearrange and replace with a subscript of 2,3,4 if the respective sensor is the first to receive the sound wave? ex: αi=2(a2−ai) if sensor #2 is the incident sensor? Or does your solution hold regardless of which sensor is the incident sensor?
$endgroup$
– ReThink
Mar 25 at 18:12
$begingroup$
@ReThink. There is no incident sensor. You have sensors located at $x_i,y_i$ which at time $t_i$ receive an information. If you want to use this concept (which is dangerous at least to me), start defining $t_min$ and redefine the $t_i$ as $t_i-t_min$.
$endgroup$
– Claude Leibovici
Mar 26 at 5:36
$begingroup$
@ReThink. There is no incident sensor. You have sensors located at $x_i,y_i$ which at time $t_i$ receive an information. If you want to use this concept (which is dangerous at least to me), start defining $t_min$ and redefine the $t_i$ as $t_i-t_min$.
$endgroup$
– Claude Leibovici
Mar 26 at 5:36
$begingroup$
I ran everything in Matlab just to visualize what's going on, and I understand now. Tau can be either an absolute or a relative time (it does not matter to the sensors) and t1 - t4 are offsets from this reference point (tau). The incident sensor does not matter, as that information is a part of the arrival times of the sound waves to the respective sensors, which is a fundamental part of t1-t4. Thank you very much for taking the time to explain this concept! It's fascinating that the problem can be simplified by just adding one more sensor.
$endgroup$
– ReThink
Mar 27 at 2:12
$begingroup$
I ran everything in Matlab just to visualize what's going on, and I understand now. Tau can be either an absolute or a relative time (it does not matter to the sensors) and t1 - t4 are offsets from this reference point (tau). The incident sensor does not matter, as that information is a part of the arrival times of the sound waves to the respective sensors, which is a fundamental part of t1-t4. Thank you very much for taking the time to explain this concept! It's fascinating that the problem can be simplified by just adding one more sensor.
$endgroup$
– ReThink
Mar 27 at 2:12
|
show 1 more comment
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$begingroup$
What you’re trying to do is called “TDOA multilateration.” Technically speaking, triangulation involves angles, not arrival time differences.
$endgroup$
– amd
Mar 21 at 7:53
$begingroup$
You can convert the system of circle equations into a system of linear equations: the intersections of a pair of circles lie on their radical axis, an equation for which you can obtain by subtracting one circle equation from another. However, since you don’t know $r_1$, what you’re really going to end up having to do is compute intersections of hyperbolas.
$endgroup$
– amd
Mar 21 at 7:55
$begingroup$
The problem would be easier with a fourth sensor.
$endgroup$
– Claude Leibovici
Mar 21 at 9:16
$begingroup$
@amd could you expound a little on how to go about doing that? Converting to a system of linear equations would be preferred
$endgroup$
– ReThink
Mar 21 at 13:16
$begingroup$
Have a look at math.stackexchange.com/questions/2462039/triangulation-math/…
$endgroup$
– Claude Leibovici
Mar 21 at 14:46