MATH
CHALLENGING ASSIGNMENT
BY :- AMIT GARG
USE OF LINEAR ALGEBRA IN GPS SYSTEM
GPS (GLOBAL POSTIONING SYSTEM……..)
The Global Positioning system is a space-based satellite navigation system
that provides location and time information in all weather conditions,
WORKING
• A GPS system consist of transmitter (GPS satellite ) and receiver .
• A GPS system uses 4 satellites at a time to find the location of any given
point .
• A GPS system uses a concept of TRILATERATION .
• 4th satellite is only used to increase the accuracy of the position (so
minimum 3 satellites are required to find the location of any given point )
ReceiverTransmitter
TRILATERATION
Intersection
of all three
circles .
Satellite A
Satellite B
Satellite C
Required
Position of
the receiver .
Distance
between satellite
A and receiver
A Satellite measures two things at a time…..
Time….
Location….
• Every Satellite measures the time taken to reach the electromagnetic signal from it
to the receiver .
• By multiplying it with the speed of the electromagnetic waves we can calculate the
distance between the satellite and the receiver (Radius of the circle ) .
• Satellite also sends its exact location to the receiver so that it can be taken as the
centre of the circle .
Solutions using linear algebra……
• Let us assume that the co-ordinates of the receiver location is (x,y,z) .
Now , creating the equations…..
• Time will be measured in milli-seconds ……
Any point above sea level satisfies the equation :-
x2+y2+z2 = 1
• All four satellites will sent the time(when the data was sent to the
receiver) and location .
• Let the data from the four satelltite is given as :-
SATELLITE X,Y,Z- POSITION TIME
1 . (1,2,0) 19.9
2. (2,0,2) 2.4
3 . (1,1,1) 32.6
4 . (2,1,0) 19.9
• Now , the distance between receiver and satellite is measured in terms of earth
radii per second (c = 0.047 earth radii per millisecond.)
• D=c(t-tsent) ….where t is the time when the signal is received by the receiver .
• Let (x; y; z) be the position of your GPS receiver on Earth and t be the time the
signals arrive.
• For satellite 1….
d=0.047*(t-19.9)…………(1)
By using pythagorus theorem in three dimensions…
d = 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧 − 0 2 ………(2)
• Simillarly , equations can be made for all four satellites …..
now ,
equation (1) and (2) are equal ,therefore ………..
0.047*(t-19.9)= 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧 − 0 2
squaring both sides…..we get…
(0.047*(t-19.9))2 =(x-1)2 +(y-2)2 +(z-0)2
=>2x + 4y + 0z − 2(.0472)(19.9)t = 12 + 22 + 02 − .0472(19.9)2+ x2 + y2 + z2 − .0472t2
………….(3)
• Simillarly , for all four satellites equations can be made but all contains same type of
squared terms of x,y,z,t.
• Therefore , these variables can be easily removed by subtracting equation 1 from
equations 2,3,4 ……
• So we get four linear equations…..which can be solved easily with the help of
matrices….
2x − 4y + 4z + 2(.0472)(17.5)t = 8 − 5 + .0472(19.92 − 2.42)
0x − 2y + 2z − 2(.0472)(12.7)t = 3 − 5 + .0472(19.92 − 32.62)
2x − 2y + 0z + 2(.0472)(0)t = 5 − 5 + .0472(19.92 − 19.92)
Other three equations are……
CONVERSION OF LINEAR EQUATIONS IN MATRIX FORM……..
2 -4 4 .077 3.86
0 -2 2 -.056 -3.47
2 -2 0 0 0
Augmented
form
Continuing with the solution of the system, the reduced row echelon form for
the augmented matrix is……
1 0 0 .095 5.41
0 1 0 -095 5.41
0 0 1 0.67 3.67
Reduced ech
form .
From above method we can easily calculate the position of the
receiver……
This is how a GPS network work……
References :- 1). https://www.youtube.com/watch?v=4O3ZVHVFhes&spfreload=1
2) Research paper of sir kalmen .

How GPS works

  • 1.
    MATH CHALLENGING ASSIGNMENT BY :-AMIT GARG USE OF LINEAR ALGEBRA IN GPS SYSTEM
  • 2.
    GPS (GLOBAL POSTIONINGSYSTEM……..) The Global Positioning system is a space-based satellite navigation system that provides location and time information in all weather conditions, WORKING • A GPS system consist of transmitter (GPS satellite ) and receiver . • A GPS system uses 4 satellites at a time to find the location of any given point . • A GPS system uses a concept of TRILATERATION . • 4th satellite is only used to increase the accuracy of the position (so minimum 3 satellites are required to find the location of any given point ) ReceiverTransmitter
  • 3.
    TRILATERATION Intersection of all three circles. Satellite A Satellite B Satellite C Required Position of the receiver . Distance between satellite A and receiver
  • 4.
    A Satellite measurestwo things at a time….. Time…. Location…. • Every Satellite measures the time taken to reach the electromagnetic signal from it to the receiver . • By multiplying it with the speed of the electromagnetic waves we can calculate the distance between the satellite and the receiver (Radius of the circle ) . • Satellite also sends its exact location to the receiver so that it can be taken as the centre of the circle .
  • 5.
    Solutions using linearalgebra…… • Let us assume that the co-ordinates of the receiver location is (x,y,z) . Now , creating the equations….. • Time will be measured in milli-seconds …… Any point above sea level satisfies the equation :- x2+y2+z2 = 1 • All four satellites will sent the time(when the data was sent to the receiver) and location . • Let the data from the four satelltite is given as :- SATELLITE X,Y,Z- POSITION TIME 1 . (1,2,0) 19.9 2. (2,0,2) 2.4 3 . (1,1,1) 32.6 4 . (2,1,0) 19.9
  • 6.
    • Now ,the distance between receiver and satellite is measured in terms of earth radii per second (c = 0.047 earth radii per millisecond.) • D=c(t-tsent) ….where t is the time when the signal is received by the receiver . • Let (x; y; z) be the position of your GPS receiver on Earth and t be the time the signals arrive. • For satellite 1…. d=0.047*(t-19.9)…………(1) By using pythagorus theorem in three dimensions… d = 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧 − 0 2 ………(2) • Simillarly , equations can be made for all four satellites ….. now , equation (1) and (2) are equal ,therefore ……….. 0.047*(t-19.9)= 𝑥 − 1 2 + 𝑦 − 2 2 + 𝑧 − 0 2 squaring both sides…..we get… (0.047*(t-19.9))2 =(x-1)2 +(y-2)2 +(z-0)2 =>2x + 4y + 0z − 2(.0472)(19.9)t = 12 + 22 + 02 − .0472(19.9)2+ x2 + y2 + z2 − .0472t2 ………….(3)
  • 7.
    • Simillarly ,for all four satellites equations can be made but all contains same type of squared terms of x,y,z,t. • Therefore , these variables can be easily removed by subtracting equation 1 from equations 2,3,4 …… • So we get four linear equations…..which can be solved easily with the help of matrices…. 2x − 4y + 4z + 2(.0472)(17.5)t = 8 − 5 + .0472(19.92 − 2.42) 0x − 2y + 2z − 2(.0472)(12.7)t = 3 − 5 + .0472(19.92 − 32.62) 2x − 2y + 0z + 2(.0472)(0)t = 5 − 5 + .0472(19.92 − 19.92) Other three equations are…… CONVERSION OF LINEAR EQUATIONS IN MATRIX FORM…….. 2 -4 4 .077 3.86 0 -2 2 -.056 -3.47 2 -2 0 0 0 Augmented form
  • 8.
    Continuing with thesolution of the system, the reduced row echelon form for the augmented matrix is…… 1 0 0 .095 5.41 0 1 0 -095 5.41 0 0 1 0.67 3.67 Reduced ech form . From above method we can easily calculate the position of the receiver…… This is how a GPS network work……
  • 9.
    References :- 1).https://www.youtube.com/watch?v=4O3ZVHVFhes&spfreload=1 2) Research paper of sir kalmen .