Interference : GPS Transmission

In summary: At this point on the circle, the measured intensity is 2.00 W/m^2" Does your answer, 27, include this point or not? No, it does not include the point. What you're looking for is the phase difference between the two sources when they reach a point on the circle that is perpendicular to the line connecting the two sources and halfway between them.
  • #1
cupcake
76
0
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??
 
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  • #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?
 
  • #3
skeptic2 said:
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?

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

then?
 
  • #4
cupcake said:
a)At how many other angles in the range 0 < theta < 90 is the intensity also 2.00 W/m^2?

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..
 
  • #5
please advise me how to do part A??
 
  • #6
:cry:
 
  • #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.
 
  • #8
How did you determine your answer in #4, 27, was wrong?
 
  • #9
skeptic2 said:
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.

yup, that's what i did rite??
divide the distance by lambda..

skeptic2 said:
How did you determine your answer in #4, 27, was wrong?

i am doing this assignment on mastering physic..
when i entered my answer equal to 27, it turned to be wrong.. :cry:
 
  • #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?
 
  • #11
skeptic2 said:
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?

so?
i don't understand.. :shy:

what do they mean by "other" here??
 
  • #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.
 
  • #13
skeptic2 said:
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.

so, the phase difference will be equal to (2*pi*d)/lambda
rite?
 
  • #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?
 
  • #15
skeptic2 said:
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?

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?
 
  • #16
Okay so what is d at the point given in the problem?
 
  • #17
is it 5.18 ?
 
  • #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?
 
  • #19
skeptic2 said:
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?

then it would be equilateral triangle??
then..the distance will be (5.18/2) / sin 30 ?
 
  • #20
Let's call it an isosceles triangle. What is the difference in length between the two equal sides of an isosceles triangle?
 
  • #21
skeptic2 said:
Let's call it an isosceles triangle. What is the difference in length between the two equal sides of an isosceles triangle?

i guess the r1=r2, so the difference will be zero?
 
  • #22
Good, and this point is on the circle at a theta of zero degrees. But remember part a) says it wants the number of points with the same power level from 0 < theta < 90. How many would that be?
 
  • #23
skeptic2 said:
Good, and this point is on the circle at a theta of zero degrees. But remember part a) says it wants the number of points with the same power level from 0 < theta < 90. How many would that be?

hmm...in order to have the same Intensity, does it mean i have to find the point when cos theta is max or equal to 1? but between 0 and 90, there's only one point when cos theta is equal to 1..
 
  • #24
No, It means that because the distances back to the two sources are equal and because the two sources are in phase, the two signals must be in phase at that point. At how many OTHER points along the circle from 0 < theta < 90 will the two signals be exactly in phase?
 
  • #25
skeptic2 said:
No, It means that because the distances back to the two sources are equal and because the two sources are in phase, the two signals must be in phase at that point. At how many OTHER points along the circle from 0 < theta < 90 will the two signals be exactly in phase?

i think if they are in phase, they will be form a maximum fringe..
and to determine a maximum fringe we use..
deg = d/lambda = 27.2

omg..! i don't understand... :cry:
 
  • #26
Good. Remember that the reference point was at 0 degrees and they wanted the points at thetas from 0 < theta < 90 so how many points remain?
 
  • #27
skeptic2 said:
good. Remember that the reference point was at 0 degrees and they wanted the points at thetas from 0 < theta < 90 so how many points remain?

26 ?
 
  • #28
That's my guess.
 
  • #29
omg... noo...
it's still wrong... :cry: :cry:
 
  • #30
Somebody else will have to find a better solution then.
 
  • #31
i hope so..

please somebody enlighten me..

though, i guessed nothing wrong about the answer 27 :cry: :cry:
 
  • #32
somebody..
please advise me how to part A...
i don't know why my answer is wrong.. :( :(
 
  • #33
In my opinion 27 was the correct answer. I only started looking at other possibilities after you said it was wrong. I wish you would post the correct answer when you find out.
 

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