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Posted on 2009-04-19
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The pictures below are the result of the matlab code uploaded which draws and rotates the matlab code to find the intersections of the hyperbolas and therefore TX.
Everything is fone as long as none of the RX have negative values.
This can be seen from the screenshots.
Why is this? Why do the hyperbolas not intersect when negative values are introduced?
mscreener1.jpg
mscreener2.jpg
matlab3.jpg
matlab4.jpg
toa.txt
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Question by:jtiernan2008
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Expert Comment

ID: 24179639
jtiernan2008,

"Why do the hyperbolas not intersect when negative values are introduced?"

I don't know that that is a general rule for hyperbolas, or if it just happens that way for your specific examples.

Let's see what others say.
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Author Comment

ID: 24179711
tutorial attached here which explains the concept
toa.pdf
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Expert Comment

ID: 24180459
"Everything is fone as long as none of the RX have negative values"
It seemd to me that in allyour screen shots at least some of the RX have negative values. In fact the RX are the same in all the shots.
What IS the difference between screener 2 and 3?
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Author Comment

ID: 24180484
I meant TX can have negative values. Sorry about the typo. RX can have negative without any problems.
The TX coordinates are different in each of the screenshots
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Accepted Solution

aburr earned 1600 total points
ID: 24180498
There is quite a bit of difference between RX and RX.
Note that each hyperpola consists of two parts.
Are you sure that in the difficult cases you are using the correct part of each of the two part hyperbolas?
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Expert Comment

ID: 24180500
between RX and TX (typos are easy)
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Author Comment

ID: 24180507
There is nothing wrong with the code. It has been uploaded there so you can check it if you have Matlab.
I can guarantee that no matter what RX or TX values are used whenever eighter or both TX x and y values is a negative number the hyperbolas do not intersect however if TX are all positive then the hyperbolas will intersect. The above screenshots are just examples but this happens everytime. I can guarantee you I have tried a lot of values at this point.
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Expert Comment

ID: 24180982
"I can guarantee you I have tried a lot of values at this point."
I am sure, but values of what? The RX have not been changed. The TX are output values.
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Author Comment

ID: 24180992
I tried a large number of different TX values. There is no reason to change the RX values.
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Expert Comment

ID: 24181577
"I tried a large number of different TX values."
Does not the program output the TX values. The TX values are what you are looking for. They are not something you make up.
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Author Comment

ID: 24181584
The program is not looking for TX values nor RX values. It is a proof of concept that the hyperbolas intersect at TX. Tx is provided at the start of the Matlab code and can therefore be changed as needed.
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Assisted Solution

yuk99 earned 400 total points
ID: 24181626
As aburr noted, each hyperbola has actually 2 parts. Remember, when you played with hyperbolas, you draw 2 parts - above and under x axes. You can draw another part, if you change y12, y23, y 13 in lines 59, 61, 63 to negatives (-y12 etc). Which part of hyperbola to use does not depend on negative TX coordinates, but on which RX is closer to TX. You can probably find the correct part doing y12sign=sign(df1-df2);
[xout,yout] = xfm1(x,y12*y12sign,theta12,xoff12,yoff12);
This is of course for hyperbola betwween f1 and f2. Make appropriate changes for other pair of stations as well.
Try and see if it helps.
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Expert Comment

ID: 24181635
jtiernan2008,
BTW congratulations to become one of the best CLOSER! :)
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Author Comment

ID: 24181640
one of the best CLOSER? what do you mean?
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Author Closing Comment

ID: 31571987
Thanks a million
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Expert Comment

ID: 24181902
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Expert Comment

ID: 24212428
Yes, congratulations,  jtiernan2008

Look under "The Closer" at that link.  You are #3.
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