Interference : GPS Transmission

Click For Summary

Homework Help Overview

The discussion revolves around the interference patterns created by GPS satellite signals, specifically focusing on the intensity measurements at various angles from the satellite. The problem involves understanding the phase relationship of two transmitters broadcasting at 1575.42 MHz and how this affects the intensity measured at points on a circle around the satellite.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the phase relationship of the signals and its impact on intensity measurements. Questions arise regarding the calculation of angles where intensity equals a specific value and the interpretation of the term "other" in the context of the problem. There is also discussion on the correct application of formulas for phase difference and intensity.

Discussion Status

The discussion is ongoing, with participants offering various interpretations and calculations. Some have provided insights into the phase relationship and its implications, while others seek clarification on specific points and calculations. There is no explicit consensus yet, as participants continue to explore different angles and approaches.

Contextual Notes

Participants note the lack of specific information regarding the radius of the circle and the implications of this on the calculations. There is also a focus on the need to consider the geometry of the situation, particularly in relation to the distances from the transmitters to the measurement points.

cupcake
Messages
73
Reaction score
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??
 
Last edited:
Physics news on Phys.org
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?
 
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?
 
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..
 
please advise me how to do part A??
 
:cry:
 
"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.
 
How did you determine your answer in #4, 27, was wrong?
 
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.
 

Similar threads

How Many Angles Maintain 2.00 W/m² Intensity in GPS Satellite Tests?
  • · Replies 2 ·
Replies
2
Views
4K
Problem involving interference and medium with index n
  • · Replies 1 ·
Replies
1
Views
1K
Distance needed to walk in order to hear local maximum
  • · Replies 27 ·
Replies
27
Views
5K
Solve Refractive Index Homework: Order of Interference & n
  • · Replies 5 ·
Replies
5
Views
3K
Interference Problem: Finding Maximum Radiation Intensity
  • · Replies 10 ·
Replies
10
Views
2K
Graduate Interference Intensity: Questions & Answers
  • · Replies 7 ·
Replies
7
Views
2K
Interference of light, emission spectroscopy prac. report
  • · Replies 1 ·
Replies
1
Views
2K
Interference of waves - 2 point source Question
  • · Replies 2 ·
Replies
2
Views
7K
A 3rd method of finding wavelength in a double slit
  • · Replies 6 ·
Replies
6
Views
3K
What Is the Optimal Angle Between Polarizers to Maximize Light Transmission?
  • · Replies 6 ·
Replies
6
Views
3K