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jtiernan2008

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Can someone please help answer this hyperbola question

pdf attached...  page 4, third paragraph on the left

ok I have drawen the triangle. I just do not know what to do next... I do not need the matlab code. I just need to understand this so I want this is fine to do by hand I just dont understand what I have to do next.

Can someone please take a look.

thanks in advance
screener.png
toa.pdf
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aburr
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"Finally, plot the hyperbolae
corresponding to each of the three baselines to
see the solution."
----
The three base lines are the three sides of your triangle.
If I read the tutorial correctly, you have a program which will plot a hyperbola based on each pair of points. The one which is based on the bottom of your triangel will have to be rotates 90 degrees.
The ones bases on the sides of your triangle will have to be rotated 45 degrees one clockwise, the other counterclockwise. Plot these three graphs on the same screen and note that the three curves intersect at one (almost) point. That is the desired location. (see figure 6)
The vertices of the triangle tell you where your RX's are located.

But there is nothing to plot until you have DTOA data.

Do you understand how the DTOA data affects the shape of the hyperbola?


===================================================
Make sure you really understand the two RX, 2D case.

With RX1 at (0,0) and RX2 at (0,10):

    What does it mean if  T1-T2 =  0.00 ns?   What is the shape of the TX locus?

    What does it mean if  T1-T2 = 33.33 ns?  What is the shape of the TX locus?

    What does it mean if  T1-T2 = 40.00 ns?   Is this even possible.

    What does the family of TX locus curves look like for the following cases:

                                   T1-T2 =    0.00 ns
                                   T1-T2 =    5.00 ns
                                   T1-T2 =  10.00 ns
                                   T1-T2 =  15.00 ns
                                   T1-T2 =  20.00 ns
                                               :
                                               :

Avatar of jtiernan2008
jtiernan2008

ASKER

"Do you understand how the DTOA data affects the shape of the hyperbola?" - this is based on the definition of the hyperbola that the locus of the points of the hyperbola correspond to a constant distance difference between the distances d1 and d2 which also corresponds to a constant or given time difference of arrival at the two locations.

"What does it mean if  T1-T2 =  0.00 ns?   What is the shape of the TX locus?"
This means the point is exactly in the middle. as DTOA = 0 There is no shape on the locus as you only have one point.

"What does it mean if  T1-T2 = 33.33 ns?  What is the shape of the TX locus?"
c(ti-tj)=sqrt((xi-x)^2+(yi-y)^2)-sqrt(xj-x)^2+(yj-y)^2)

but it is not possible to solve this equation unless a third RX is introduced?

Is this correct?

>>This means the point is exactly in the middle. as DTOA = 0 There is no shape on the locus as you only have one point.

That is not correct.    

If T1-T2 = 0.00 ns, then the transmitter is equidistant from both RX's.
In 2D, the transmitter can be anywhere on the LINE y=5.
This line is a degenerate hyperbola.

If  T1-T2 = 33.33 ns, it means that the TX is d=c*33.33ns =10 meters further from
RX1 than RX2.  This is another very different degenerate case of a hyperbola.

>> but it is not possible to solve this equation unless a third RX is introduced?

You can't solve for the TX position with two RX's,  but you CAN solve for the locus
of possible positions.  An once you found those loci, you can find their intersection.  That's what this exercise/paper is about.
>>If  T1-T2 = 33.33 ns, it means that the TX is 0 meters further from RX1 than RX2.  
 
Do you see there is  only one way for this to be possible.

What does it mean if  T1-T2 = 40.00 ns?   Is this even possible.
>>If  T1-T2 = 33.33 ns, it means that the TX is 10 meters further from RX1 than RX2.  
 
Do you see there is  only one way for this to be possible.
But one way does not mean one point.
There is still a very well defined locus.

What does it mean if  T1-T2 = 40.00 ns?   Is this even possible.
ok but how do you solve for the locus of possible positions if you are unable to find the points?


"What does it mean if  T1-T2 = 40.00 ns?   Is this even possible"
This brings the distance to 12 metres.. why would it not be possible?
>>If  T1-T2 = 33.33 ns, it means that the TX is 10 meters further from RX1 than RX2.  

Draw a 1 m circle around RX1.  Draw an 11 m circle around RX2.  Find the intersection.
Draw a 5 m circle around RX1.  Draw a   15 m circle around RX2.  Find the intersection.
Do as many cases as necessary to understand what's going on.
That's one way to find the locus.
Do you see how this is a degenerate case?

>>What does it mean if  T1-T2 = 40.00 ns?   Is this even possible?
    This brings the distance to 12 metres.. why would it not be possible?

Draw a 1 m circle around RX1.  Draw a 13 m circle around RX2.  Find the intersection.
Draw a 5 m circle around RX1.  Draw a 15 m circle around RX2.  Find the intersection.
Do you see what's happening?

You should definitely draw 3 to 5 circles for several intermediate cases as well:

                                   T1-T2 =    0.00 ns  ==>  0 m difference
                                   T1-T2 =    6.66 ns  ==>  2 m difference
                                   T1-T2 =  20.00 ns  ==>  6 m difference
                                   T1-T2 =  30.00 ns  ==>  9 m difference

I have drawn the first part of it and placed the RX1 at 0,0 as requested.. the circles intersect and draw down the way
screener2.jpg
The examples and number I have been giving are for the two RX, 2D case with time
in ns and distances in meters.

The Matlab example has time in us and distances in kilometers.  

And you need to understand the basics of the hyperbolae before you worry about
rotating them.

>>===================================================
>>Make sure you really understand the two RX, 2D case.
>>With RX1 at (0,0) and RX2 at (0,10):

You should definitely draw 3 to 5 pairs of circles for several cases:

                                   T1-T2 =    0.00 ns  ==>   0 m difference
                                   T1-T2 =    6.66 ns  ==>   2 m difference
                                   T1-T2 =  20.00 ns  ==>   6 m difference
                                   T1-T2 =  30.00 ns  ==>   9 m difference
                                   T1-T2 =  33.33 ns  ==> 10 m difference
                                  T1-T2 =   40.00 ns  ==> 12 m difference

To find point on a particular locus, you have to draw one circle around RX1 and
and one around RX2.  A pair of circle can intersect in 0, 1, or 2 points.

Your goal is come up with the family of hyperbolae that are characteristic of the
DTOA method.
Your drawing is sort of correct, except I think the larger circle should be around (0,0).

Where do the circle intersect, and what is the locus for  

                     T1-T2 =  33.33 ns  ==> 10 m difference
>> Draw a 1 m circle around RX1.  Draw an 11 m circle around RX2.  Find the intersection.
Draw a 5 m circle around RX1.  Draw a   15 m circle around RX2.  Find the intersection.

ok I have redrawn as follows;
screener.jpg
ASKER CERTIFIED SOLUTION
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d-glitch
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I better get drawing..
D-Glitch as always thanks a million for your help
you have been an enormous help with my project
thanks a million