Re: Localization - System of two unknowns.

On Aug 28, 7:39 pm, "maz_p5" <maz...@xxxxxxxxxxx> wrote:
On Aug 27, 3:42=A0am, "maz_p5" <maz...@xxxxxxxxxxx> wrote:
Hi,

I have to derive a location using the below formulas:

d12=3Dsqrt((x1-xs)^2+(y1-ys)^2)-sqrt((x2-xs)^2+(y2-ys)^2)
d13=3Dsqrt((x1-xs)^2+(y1-ys)^2)-sqrt((x3-xs)^2+(y3-ys)^2)

d12=3D0.085, d13=3D0.56. (x1,y1)=3D(0,0), (x2,y2)=3D(0,6),
(x3,y3)=3D(4,6=
)

I have to find the point (xs,ys). How can I do this in MATLAB. What
technique can be applied. PLEASE HELP.
If possible, Please explain with the help of a code. Its urgent.

Thank you.

Hello Maz,

There are several ways to attack this kind of problem. I typically use
a differential correction method. The following link is to a MathCad
paper of mine showing how to do this. Even though it is not MatLab, I
think you will see how to do this from my paper.

http://www.claysturner.com/dsp/gps%20eqn%20solve.pdf

IHTH,
Clay

Hello Clay,

Thanks for your paper. I will try out the solution and get back to you.

But, the point at which i am stuck is the square root. There are various
systems done by least square method but these equations have square roots
which is causing me a problem and it is worrying me. There is nothing of
that sort in your method.

Also, my d12 is not the difference between measured and predicted value as
discussed in your paper. This is a sound localization system and d12 is the
difference between the distance form microphone 1 to the source(sound
source) and from microphone 2 to the source.

Do you think your method would work on this system?

Thanks,
-Maz- Hide quoted text -

- Show quoted text -

Hello Maz,

The implicit differentiation reduces the exponents by one, so we end
up with linear equations - no squares and square roots. But I do have
square roots in other parts of the problem, but they shouldn't be an
issue. Numerically finding a square root is of the same order of
difficulty as division.

Just rewrite the basic equations using the distances as you have them
defined. You should be able to apply the method I outlined to your
equations. The method of differential correction is quite generic and
may be applied to all sorts of problems.

IHTH,
Clay


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