Silicon Investor (SI) -- The First Internet Community

STOCKTALK

We've detected that you're using an ad content blocking browser plug-in or feature. Ads provide a critical source of revenue to the continued operation of Silicon Investor. We ask that you disable ad blocking while on Silicon Investor in the best interests of our community. If you are not using an ad blocker but are still receiving this message, make sure your browser's tracking protection is set to the 'standard' level.

Technology Stocks : Qualcomm Moderated Thread - please read rules before posting -- Ignore unavailable to you. Want to Upgrade?

To: pheilman_ who wrote (23200)	6/1/2002 1:12:26 PM
From: rkral	Read Replies (2) \| Respond to of 196649

The mobile transmits, both base stations receive and measure time difference of arrival (t1). Since the exact time of transmission from the mobile is not known, all that can be concluded is the relative distances(?) from the two base stations. d1 = d2 + ct1. So for your case of two BTS and the mobile in between, the estimated location would be a line perpendicular to the highway ... Given the scenario you presented, I agree with your equation. d1 - d2 = ct1. (I repositioned one term.) However, this is the eq'n of a hyperbola, not a straight line. To quote my geometry book: "a hyperbola is defined as the locus of a point which moves in a plane in such a way that the difference of its distance from two fixed points is a constant." I'm sure you recognize the BTSs as the two fixed points .. (ct1) as the constant .. and (d1 - d2) as the difference. The hyperbola does become a "line perpendicular to the highway" in the special case where t1 = 0. That is, with zero time difference, the mobile is equidistant between the BTSs. But the scenario you presented is not the way EOTD works. (E-OTD really, but I dislike typing hyphens when they don't seem necessary.) An EOTD measurement for one BTS yields an absolute distance of the mobile from the BTS, not a relative distance. Hence the circles in the Cambridge site graphic.* Stated another way, the use of circles in the EOTD method is proof that the calculated distance is absolute. So let's look at the "co-linear case" again. As I originally presented, "Visualize two stations as 10 km apart. The E-OTD method shows the mobile to be 6 km from one BTS, and 4 km from the other." Graphical solution: On a horizontal line, mark two points 10 centimeters (or any unit you like) apart. Draw a 6 cm circle centered on the left point. Draw a 4 cm circle centered on the right point. If you were careful with your construction, the circles will intersect. Algebraic solution: For simplicity, let's assume the co-linearity is along an east-west line. The west BTS is at (x,y) of (-5,0) .. the east at (+5,0) .. that is, 10 km apart. As before, EOTD tells us the mobile is 6 km from the west BTS and 4 km from the east BTS. (West BTS circle): x^2 + y^2 + 10x = 11 (plot it if you wish) (East BTS circle): x^2 + y^2 - 10x = (-9) (West minus East): 20x = 20, or x = 1 (Subst. x=1 in West): 1 + y^2 + 10 = 11, or y^2=0, or y=0 This (1,0) algebraic solution also locates the mobile co-linearly, specifically along the east-west line between the stations. I concede, as I did in my original post, that there can be a large error here. This is because the two circles create a tangential intersection, not a "perpendicular" one. I additionally concede that, with the mobile exactly 6 km east of the west BTS and exactly 4 km west of the east BTS, the EOTD measurements might come up as 5.9 km and 3.9 km, in which case the circles don't intersect. But it doesn't take much of a software algorithm (a little AI maybe) to determine the most probable position of the mobile. Similar if the measurements come up 6.1 km and 4.1 km. The mobile could be either to the north or the south of the highway .. but it would be a reasonably small search area. If that's your wife out there on that east-west highway, calling 911, not knowing where she is, bleeding to death, and the EOTD only gets data from two BTSs, per the examples above ... do you really want the GSM guys to say "the colinear case does not work in general", and report only the BTS locations to emergency services? I hope this puts the "co-linear case" to bed. [edit: the Cambridge] Web site is either intentionally or unintentionally wrong.* What specifically do you think is technically incorrect on the site? Ron P.S. I favor CDMA and A-GPS. I am also long QCOM, although a very small position at this point in the market cycle.