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Interference : GPS Transmission

  1. Mar 20, 2009 #1
    The GPS (Global Positioning System) satellites are approximately 5.18 m across and transmit two low-power signals, one of which is at 1575.42 MHz (in the UHF band). In a series of laboratory tests on the satellite, you put two 1575.42 MHz UHF transmitters at opposite ends of the satellite. These broadcast in phase uniformly in all directions. You measure the intensity at points on a circle that is several hundred meters in radius and centered on the satellite. You measure angles on this circle relative to a point that lies along the centerline of the satellite (that is, the perpendicular bisector of a line which extends from one transmitter to the other). At this point on the circle, the measured intensity is 2.00 W/m^2

    a)At how many other angles in the range 0 < theta < 90 is the intensity also 2.00 W/m^2?

    b) Find the four smallest( positive) angles in the range 0 < theta < 90 for which the intensity is 2.00 W/m^2.

    c) What is the intensity at a point on the circle at an angle of 4.65 degree from the centerline?

    Θ = arcsin(m*lambda/(d))
    lambda = c/f = 0.190294 m.

    I = 2 * cos^s (( pi*5.18 / 0.190) sin 4.65
    = 1.97

    but,my calculation is wrong :cry:

    anyone knows why my calculation is wrong??
     
    Last edited: Mar 20, 2009
  2. jcsd
  3. Mar 20, 2009 #2
    First, since they don't give you the exact radius of the circle I'm guessing that they're only interested in power based on the phase relationship of the two signals, not their relative distance.

    What is the phase relationship of the two sources at the reference point? At what other points around the quarter circle will the two signals have the same phase relationship?
     
  4. Mar 20, 2009 #3
    the phase difference between two sources is alpha=((2*pi*d)/lambda) * sin theta

    then???
     
  5. Mar 20, 2009 #4
    how about this one
    what i did is
    deg = d/lambda = 27.22,
    then i take the integer value, so the answer is 27..
    but again, it turned out to be wrong..
     
  6. Mar 21, 2009 #5
    please advise me how to do part A??
     
  7. Mar 21, 2009 #6
  8. Mar 21, 2009 #7
    "What is the phase relationship of the two sources at the reference point? At what other points around the quarter circle will the two signals have the same phase relationship?"

    "the phase difference between two sources is alpha=((2*pi*d)/lambda) * sin theta
    then??? "

    Wrong answer. I'm looking for a number, the difference in phase of the two signals when they reach a point on the circle that is perpendicular to the line connecting the two sources and halfway between them. Don't forget that one wavelength is equivalent to 2*pi radians so finding the phase difference between the two sources is the same as finding the distance difference between the point and each source and dividing by lambda.
     
  9. Mar 21, 2009 #8
    How did you determine your answer in #4, 27, was wrong?
     
  10. Mar 21, 2009 #9
    yup, that's what i did rite??
    divide the distance by lambda..

    i am doing this assignment on mastering physic..
    when i entered my answer equal to 27, it turned to be wrong.. :cry:
     
  11. Mar 21, 2009 #10
    A) may be a little deceptive. Note the word "other".

    a)At how many OTHER angles in the range 0 < theta < 90 is the intensity also 2.00 W/m^2?
     
  12. Mar 21, 2009 #11
    so???
    i don't understand.. :shy:

    what do they mean by "other" here??
     
  13. Mar 21, 2009 #12
    What is theta for the point in the example? It says, "You measure angles on this circle relative to a POINT that lies along the centerline of the satellite (that is, the perpendicular bisector of a line which extends from one transmitter to the other). At this point on the circle, the measured intensity is 2.00 W/m^2" Does your answer, 27, include this point or not?

    Your formula "the phase difference between two sources is alpha=((2*pi*d)/lambda) * sin theta" gives the phase, not the phase difference.
     
  14. Mar 21, 2009 #13
    so, the phase difference will be equal to (2*pi*d)/lambda
    rite?
     
  15. Mar 21, 2009 #14
    You have two paths, one from one source to the point and the other from the other source to the point. To find phase difference you must consider the difference in distance of both paths but you have only one "d" in your formula. What is the difference in distance from each source to the point on the circle given in the problem?
     
  16. Mar 21, 2009 #15
    yup,, i know that the formula should be (2*pi*/lambda) * (r2-r1)
    why don't we just assume that (r2-r1) is equal to d ?
    can i?
     
  17. Mar 21, 2009 #16
    Okay so what is d at the point given in the problem?
     
  18. Mar 21, 2009 #17
    is it 5.18 ???
     
  19. Mar 21, 2009 #18
    The diagram I'm picturing is a line 5.18 meters long from one source to the other. At the midpoint of that line, a perpendicular is drawn several hundred meters to a point on the circle. Now if you draw a line from each source to the point on the circle, what is the difference in length between both lines?
     
  20. Mar 21, 2009 #19
    then it would be equilateral triangle??
    then..the distance will be (5.18/2) / sin 30 ?
     
  21. Mar 21, 2009 #20
    Let's call it an isosceles triangle. What is the difference in length between the two equal sides of an isosceles triangle?
     
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