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PROJECT

GPS POSITIONING

EE4853-5853 ENAV
Module 13 – GNSS – Part 2: User Equipment

In this project, you will be determining the receiver position based on GPS pseudorange
measurements. You are provided with the satellite positions (corrected for earth rotation,
in the ECEF coordinate frame at the time of reception), dual-frequency GPS code-phase
measurements, and the position truth of the user.

First, you are going to calculate the position without taking ionosphere and troposphere
into consideration. Next, you determine the ionosphere delay using the dual frequency
measurements, and apply the appropriate ionosphere correction to get a better positioning
accuracy. Then you will calculate and apply the approximate troposphere delay based on
the satellite elevation and user height. Finally, you will be further enhancing positioning
performance by applying precise clock and orbit corrections.

Provide write-up and Matlab code

Task 1: GPS error sources (10%)
List all relevant GPS error sources, and their mitigation strategies (maximum 2 lines per
error source)

For the remaining questions, use Matlab.

Load the attached EE4853-5853_GPS_PR.mat into Matlab. You should now have the
following variables:

PRNs List of the PRNs in the datasets
L1PSR GPS L1 pseudorange measurements [m]
L2PSR GPS L2 pseudorange measurements [m]
SatBC Satellite clock and orbit – from broadcast ephemeris.
SatPPP Satellite clock and orbit – from precise ephemeris
TruthLLH Truth position in [Lat;Lon;Height]. Lat and Lon are in radians,

Height in meters

The following constants may be useful:
Speed of light: 299792458.0 m/s
GPS L1 frequency: 1575.42e6 Hz
GPS L2 frequency: 1227.60e6 Hz

See also the following m-files for coordinate transformations:
LLH2ECEF
ECEF2LLH

ECEF2ENU
ENU2ECEF

Task 2: Position calculation using ILS (20%)
Calculate the GPS position using the measured pseudoranges (L1PSR), and the broadcast
ephemeris satellite clock and orbit (SatBC.ECEF and SatBC.Clock)

– Add the satellite clock correction SatBC.Clock to the measured pseudorange.
Note: this correction is already in meters.

– Do not apply any iono and tropo corrections to the pseudoranges yet.
– Use the Iterative Least Squares (ILS) method as explained in Module 13 to solve

for user position and receiver clock error. Implement the following function for
this:

[EstPos,Residuals,H]=CalcGPSPos(PR,SatECEF)
with

PR = [Nx1] pseudoranges, for example from L1PSR
SatECEF = [3xN] sat positions in ECEF, for example from SatBC.ECEF

Provide the following:
a) User position in Lat (degrees), Lon (degrees), Height (meters, ellipsoidal)
b) Error in user position with respect to truth in meters
c) The norm of the range residuals
d) The HDOP, PDOP, and VDOP of the solution. Note: The DOP calculations need

to be with respect to the locally-level frame. For this, you need to calculate the H-
matrix for the satellites in an ENU frame, with the user at (0,0,0).

Hint: in your newton iteration, your pseudorange residuals are calculated by subtracting
the estimated pseudorange from the measured pseudorange. The estimated pseudorange
is the estimated range + the estimated receiver clock offset.

Next, we are going to mitigate some of the GPS error sources. Fill in the following table
using your calculations in the next tasks:

PRNs 26 2 5 4 25 29 10 13 12
Elevation
Iono delay
Tropo delay
Satellite Clock error
Satellite Orbit error

Task 3: Iono correction (20%)
Using the L1 and L2 pseudorange measurements, calculate the iono delay (expressed in
meters) for all satellites.

a) Calculate the Iono delays and fill them into the table

b) Correct the L1 pseudorange measurements with the iono delays, and recalculate
the user position. Provide

i. The improved user position in Lat, Lon, Height
ii. Error in user position with respect to truth in meters

iii. The norm of the range residuals

Task 4: Tropo correction (20%)
To calculate the tropo corrections, you first need the satellite elevation angles. For this,
convert the satellite ECEF coordinates into an ENU coordinate frame, with the position
found at 3 as the origin. Using the satellite coordinates in the local-level ENU coordinate
frame you now can calculate the satellite elevation angles.

a) Calculate the satellite elevation angles and fill them into the table
b) Calculate the satellite tropospheric delays, and put those in the table as well
c) Correct the L1 pseudorange measurements with the iono and tropo delays, and

recalculate the user position. Provide
i. The improved user position in Lat, Lon, Height

ii. Error in user position with respect to truth in meters
iii. The norm of the range residuals

Task 5: Precise clock and orbit (20%)
Using an extensive network of ground-based reference stations it is possible to calculate
the satellite clock and orbit parameters much more precisely than those provided in the
GPS broadcast message. See for example
http://igscb.jpl.nasa.gov/components/usage.html. The struct SatPPP contains those
precise clock and orbit parameters for this time epoch.

a) Calculate the error of the broadcast ephemeris clock, compared to the precise
ephemeris. Put your result in the table.

b) Calculate the error of the broadcast ephemeris orbit, compared to the precise
ephemeris, projected on the Line-Of-Sight vector. Put your result in the table.

c) Calculate the improved position using iono and tropo corrections, and the precise
clock and orbit (SatPPP). Provide:

i. The improved user position in Lat, Lon, Height
ii. Error in user position with respect to truth in meters

iii. The norm of the range residuals

Task 6: Remaining errors & Conclusions (10%)
The position solution is still not perfect. What remaining errors do you expect there to be
in the measurements, and how do you suggest they can be mitigated?

Draw your conclusions about the magnitude of the various errors, and how easy or hard it
is to mitigate them. Take into consideration the potential receiver complexity (for
example dual frequency vs single frequency), required ground infrastructure,
communication, etc.

GPS Project – Data and Functions/ECEF2ENU.m
function enu = ECEF2ENU(ece,orgece,orgllh)
%ENU = ECEF2ENU(ECEF,orgECEF,orgLLH)
%
% EC2ENU: Convert ECEF coordinates to East-North-Up with
% respect to orgECEF and orgLLH (orgECEF is the same
% location as orgLLH, orgLLH in radians).
%
% VARIABLES:
%
% OU-ECE-AEC Oct. 1988 FvG
% WJP2011: Vectorized. Now accepts ece as a 3×1, a 3xM, 3xMxN, etc matrix

% Rotate the difference vector into ENU coordinates
sla = sin(orgllh(1,1)); cla = cos(orgllh(1,1));
slo = sin(orgllh(2,1)); clo = cos(orgllh(2,1));
C = [ -slo clo 0 ; …
-sla*clo -sla*slo cla; …
cla*clo cla*slo sla];

if size(ece,2)==1 && ndims(ece)==2; %3×1 vector
% enu = [ -slo clo 0 ; …
% -sla*clo -sla*slo cla; …
% cla*clo cla*slo sla] * difece;
difece = ece – orgece; % difference between coordinate origins
enu= C * difece;
else %3xM or 3xMxN etc
%Vectorized ECEF2ENU function:
enu=NaN(size(ece));
%reshape ece to 3xN matrix
%ece=reshape(ece,3,size(ece,2)*size(ece,3)*size(ece,4));
ece=reshape(ece,3,[]);
difece=zeros(size(ece));
difece(1,:)=ece(1,:)-orgece(1);
difece(2,:)=ece(2,:)-orgece(2);
difece(3,:)=ece(3,:)-orgece(3);
for i=1:3
enu(i,:)=C(i,1)*difece(1,:)+C(i,2)*difece(2,:)+C(i,3)*difece(3,:);
end
end

__MACOSX/GPS Project – Data and Functions/._ECEF2ENU.m

GPS Project – Data and Functions/ECEF2LLH.m
function wgs = ECEF2LLH(xyz)
%wgs = ECEF2LLH(xyz)
%
% This function converts cartesian (x,y,z) coordinates of a reference
% point in ECEF to lat, lon, height coordinates in the WGS 84 system
%
% VARIABLES:
% wgs – 3 by 1 array containing position lat, lon, height
% (rad, rad, m)
% xyz – 3 by 1 array containing position as x,y,z
% (m, m, m)
%
% a – semi-major axis
% f – spheroidal flattening
% esq – eccentricity squared
% sp – sine of lattitude
% cp – cosine of latitude
% sl – sine of longitude
% cl – cosine of longitude
% gsq – intermediate temp variable
% x – cartesian coordinte in feet
% r – intermediate temp variable
%
% This function is based on one developed by the
% National Geodetic Survey, Rockville, Maryland.
%
% WJP2010: Vectorized. Now accepts xyz as a 3×1, a 3xM, 3xMxN, or a 3xMxNxO matrix

wgs=zeros(size(xyz));
f=1/298.257223563;
esq=f*(2-f);
a=6378137;
x = xyz(1,:,:,:);
y = xyz(2,:,:,:);
z = xyz(3,:,:,:);
rsq = (x.*x) + (y.*y);
h = esq.*z;
for i = 1:6;
zp = z + h;
r = sqrt(rsq + (zp.*zp));
sp = zp./r;
gsq = 1.0 – (esq.*sp.*sp);
en = a./sqrt(gsq);
p = en.*esq.*sp;
%if abs(h-p) < 0.0005,break,end h = p; end; p = atan2(zp,sqrt(rsq)); h = r - en; wgs(2,:,:,:) = atan2(y, x);%*180/pi; wgs(1,:,:,:) = p;%*180/pi; wgs(3,:,:,:) = h; % wgs=zeros(3,1); % f=1/298.257223563; % esq=f*(2-f); % a=6378137; % x = xyz(1); % y = xyz(2); % z = xyz(3); % rsq = (x*x) + (y*y); % h = esq*z; % for i = 1:6; % zp = z + h; % r = sqrt(rsq + (zp*zp)); % sp = zp/r; % gsq = 1.0 - (esq*sp*sp); % en = a/sqrt(gsq); % p = en*esq*sp; % if abs(h-p) < 0.0005,break,end % h = p; % end; % p = atan2(zp,sqrt(rsq)); % h = r - en; % wgs(2) = atan2(y, x);%*180/pi; % wgs(1) = p;%*180/pi; % wgs(3) = h; GPS Project - Data and Functions/EE4853-5853_GPS_PR.mat L1PSR:[9x1 double array] L2PSR:[9x1 double array] PRNs:[1x9 double array] SatPPP:[1x1 struct array] [3x9 double array] [9x1 double array] SatBC:[1x1 struct array] [3x9 double array] [9x1 double array] TruthLLH:[3x1 double array] GPS Project - Data and Functions/ENU2ECEF.m function ecef = ENU2ECEF(enu,orgece,orgllh) % ecef = ENU2ECEF(enu,orgece,orgllh) % % ENU2EC: Convert ENU coordinates to ECEF with respect to % the origin (orgece is the same location as orgllh). % % OU-ECE-AEC Oct. 1988 FvG sla = sin(orgllh(1)); cla = cos(orgllh(1)); slo = sin(orgllh(2)); clo = cos(orgllh(2)); ROT = [-slo clo 0; -sla*clo -sla*slo cla; cla*clo cla*slo sla]; difece = inv(ROT) * enu; % add the difference between the enu and ecef origins ecef = orgece + difece; GPS Project - Data and Functions/LLH2ECEF.m function ecef = LLH2ECEF(llh) % % LLH2EC: Convert lat/lon/height coordinates to Earth-Centered % Earth-Fixed (ECEF) coordinates (WGS72). % % Input: llh = lat,long,height location (rad,rad,user units) % Output: ecef = x,y,z (user units) % % OU-ECE-AEC Oct. 1988 FvG % WJP2010: Vectorized. Now accepts llh as a 3x1, a 3xM, 3xMxN, or a 3xMxNxO matrix %A = 6378135.0; % Earth's radius (m), old number from 1997 A = 6378137.0; % Earth's radius (m) per WGS84, CB %E = 8.181881066e-02; % Eccentricity, old number from 1997 E = 8.18191908e-2; % Eccentricity per WGS84, CB ESQ = E * E; % ecef=zeros(3,1); % % SP = sin(llh(1)); % CP = cos(llh(1)); % SL = sin(llh(2)); % CL = cos(llh(2)); % GSQ = 1.0 - (ESQ*SP*SP); % EN = A / sqrt(GSQ); % Z = (EN + llh(3)) * CP; % ecef(1) = Z * CL; % ecef(2) = Z * SL; % EN = EN - (ESQ * EN); % ecef(3) = (EN + llh(3)) * SP; ecef=zeros(size(llh)); SP = sin(llh(1,:,:,:)); CP = cos(llh(1,:,:,:)); SL = sin(llh(2,:,:,:)); CL = cos(llh(2,:,:,:)); GSQ = 1.0 - (ESQ.*SP.*SP); EN = A ./ sqrt(GSQ); Z = (EN + llh(3,:,:,:)) .* CP; ecef(1,:,:,:) = Z .* CL; ecef(2,:,:,:) = Z .* SL; EN = EN - (ESQ .* EN); ecef(3,:,:,:) = (EN + llh(3,:,:,:)) .* SP;

Michael S. Braasch, Ph.D., P.E., Fellow Ion
Ohio University Avionics Engineering Center, Director

GPS: Theory & Practice

Part

: Ordinary Least Squares Solution

by Michael S. BraaschGPS: Theory & Practice: Part 9 – OLS Solution

STAND-ALONE GPS POSITIONING

ERROR CONSIDERATIONS
The pseudorange observation equation shows us
there are two primary sources for error:

We will get into the details of the errors in the
remainder of the course

PRi = xu − xi( )

2
+ yu − yi( )

2
+ zu − zi( )

2
+ bu

Receiver measurement
Satellite position

calculation

STAND-ALONG POSITIONING (CONT’D)

In the absence of errors

PRi = pseudorange to the i-th satellite
bu = receiver (user) clock offset or bias (converted to units of
distance through multiplication by the speed of light)
Four unknowns ( xu, yu, zu, bu), thus we need at least four
independent measurements. However…
Highly non-linear equation!
Traditionally solved via linearization of a Taylor series
expansion

PRi = xu − xi( )
2
+ yu − yi( )
2
+ zu − zi( )
2
+ bu

LINEARIZATION

Recall the Taylor series for a scalar function, f(x)

The series is valid ‘in the neighborhood’ of x = c
We can linearize the above equation by only retaining the
first two terms in the series
For the pseudorange equation, however, we must deal
with the fact that it is a function of four variables, not one

f x( ) = f c( )+ ′f c( )⋅ x − c( )+ ′′f c( )

2!
⋅ x − c( )2 +

PSEUDORANGE EQN LINEARIZATION

PRi = PRio

xo − xi

PRio − bo
xu − xo( )+

yo − yi
PRio − bo

yu − yo( )+
zo − zi

PRio − bo
zu − zo( )+1⋅ bu − bo( )

where :
xo , yo , zo , bo = initial guess of the user state

PRio = pseudorange from the i
th satellite to

the initial guess state

Let : δ x = xu − xo ; δ y = yu − yo ;
δ z = zu − zo ; δb = bu − bo ; δ PRi = PRi − PRio

Then : δ PRi

xo − xi

PRio − bo
δ x +

yo − yi
PRio − bo

δ y +
zo − zi

PRio − bo
δ z +1⋅δb

ORDINARY LEAST SQUARES (OLS) SOLUTION

The last equation on the previous slide is linear in the four
unknowns
Standard linear algebra can then be applied to the solution

where

δ PR1
δ PR2
δ PR3


δ PRN

⎡

⎣

⎢
⎢
⎢

⎢

⎢
⎢
⎢

⎤

⎦

⎥
⎥
⎥

⎥

⎥
⎥
⎥

H21 H22 H23 H24
H31 H32 H33 H34
   

H N 1 H N 2 H N 3 H N 4

⎡
⎣

⎢
⎢
⎢
⎢
⎢
⎢

⎢
⎤
⎦

⎥
⎥
⎥
⎥
⎥
⎥

⎥

δ x
δ y
δ z
δb

⎡
⎣
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥

⇒ y = Hβ

H k1 =

xo − xk
PRio − bo

; H k 2 =
yo − yk

PRio − bo
; H k 3 =

zo − zk
PRio − bo

; H k 4 = 1

OLS SOLUTION (CONT’D)

In reality, the measurements contain errors and
thus a more complete representation would be

where epsilon is simply the vector of
measurement errors
The OLS solution is then given by:

y = Hβ + ε

βOLS = H

T H( )−1 H T y
8

OLS SOLUTION (CONT’D)
At this point we must remember that we have only
solved for the deltas
Our total solution, then is given by:

Note the hat or carrot symbol simply means ‘estimate’

x̂u
ŷu
ẑu
b̂u

⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
=

xo
yo
zo
bo

⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
+

δ xOLS
δ yOLS
δ zOLS
δbOLS

⎡
⎣
⎢
⎢
⎢
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎥
⎥
⎥
9

Two questions that need to be answered:
#1) Where did the initial guess (xo, yo, zo, bo) come from in
the first place?
#2) What happens if the initial guess is not close to the truth
and hence the linearized pseudorange equation is not a good
approximation?

For near earth users, we can simply set all four initial
guess components equal to zero (i.e., center of the
earth and zero clock bias)

When initializing everything to zero, the first total
solution will NOT give the correct values
However, the first solution WILL be closer to the truth
than the initial guess. This new solution can then serve as
the initial guess and the whole process can be iterated
This process can simply be iterated until the total solution
converges (usually five to seven times)
Once the solution has converged, then the solution at time
k can serve as the initial guess for the solution at time k+1
and only a single solution step is required

EFFECT OF SV CLOCK OFFSE

The SV clock offset must be computed and the corresponding
pseudorange must be compensated accordingly
Position solution will be off, typically, by tens of kilometers if
this is not done
The SV clock offset is multiplied by the WGS-84 speed-of-
light and then added to the pseudorange measurement
Note the subtraction indicated in IS-GPS-200 (see the last
page of Part 3 ‘Time References’) is used in computing the
time-of-transmission. An earlier TOT yields a larger PR
This correction to the pseudorange needs to be performed,
obviously, before the user position solution is computed

EFFECT OF EARTH ROTATION

The aforementioned position solution would be sufficient
were it not for the fact that the earth rotates
The equations require knowledge of the satellite position at
the time-of-transmission (TOT). However, we want to solve
for the user position at the time-of-reception (TOR).
The problem is that both user and satellite positions are
described in earth-centered earth-fixed (ECEF) coordinates
and the ECEF coordinate frame is not in the same orientation
at TOR that it was at TOT
If this phenomenon is not taken into account, the horizontal
position can be off by as much as 30 meters

EARTH ROTATION CORRECTION

Estimate signal-transit-time by dividing the pseudorange
by the WGS-84 speed-of-light (the pseudorange needs to
be corrected for SV clock offset and, optionally, for
receiver clock offset)
Compute the amount of earth rotation by multiplying the
signal-transit-time by the WGS-84 earth angular rate
Compute a rotation matrix (around the z-axis) for the
earth rotation angle
Multiply the rotation matrix times the SV position vector
to get the corrected SV position vector

EARTH ROTATION CORRECTION (CONTINUED)

ttransit
i =

PRi − b̂u
c

+ Δtsv

θearth = ω ie ⋅ttransit
i

Ri =
cθ sθ 0
−sθ cθ 0
0 0 1

⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥

corrected _ sv i _ pos = Ri * x
svi

y
svi

z
svi

⎡
⎣⎢

⎤
⎦⎥

OLS VELOCITY SOLUTION
• The ordinary least squares velocity solution is

similar to the position solution. It’s actually
easier since the line-of-sight velocity equation is
linear from the start

Where:
= measured delta-range to satellite i
= velocity vector of satellite i
= velocity vector of the receiver
= unit vector from the receiver to satellite i
= receiver clock offset rate (i.e., frequency bias)

DRi = vi ⋅ûi − v ⋅ûi +
b

vi
v
ûi
b

The first term on the right-hand side is known and can be
incorporated into the ‘measurement’:

Now define:

Thus:

Or:
17

! DRi = DRi ” vi #ûi = “v #ûi +
!b

i1 = “uxi ; !i 2 = “uyi ; !i3 = “uzi ; !i 4 = 1

! DR1
! DR2
! DR3
!

! DR
N

”

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(11 (12 (13 (14
(21 (22 (23 (24
(31 (32 (33 (34
! ! ! !

(
N 1 ( N 2 ( N 3 ( N 4

”
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v
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v
z

“b

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‘

y = H! Thus : !OLS = H

T H( )”1 H T y

OLS VELOCITY SOLUTION (CONT’D)

TIME AND FREQUENCY ESTIMATION

So far we have focused on determination of user
position and velocity
However, in the process we have also solved for
the user’s clock offset (bias) and clock drift (or
equivalently, the clock’s frequency offset)
The receiver’s estimate of time can then be
corrected to an accuracy on the order of tens of
nanoseconds

tcorrected = traw ! b̂u

EE5853M13 Errata and Additional Information

% % %

Some typos, etc have been pointed out to me in M13:

Slide 10: In the last equation, every “1” should be replaced with “i”

Slide 25: At the very end, it should say “… and rho_L2 the L2 pseudorange”

Slide 11: The hat symbols seem to come out of nowhere. In every case, it is referring to the
initial guess. What, then, does R_hat mean? It is simply the distance computed from the satellite
to the initial guess user position (NOTE: distance! not pseudorange; do not include the guess
user clock offset)

GPS_09_Slides_OLS_Solution: I am providing some slides I have from a different course that I
have taught on GPS. You only need slides 1 – 12 (ignore the later stuff since, for example, the
effects of Earth rotation have already been accounted for in the satellite position data that you
have been given). This material is obviously similar to what has been presented in the screencast
but seeing it from a different perspective may be helpful.

Typo in supplemental GPS_09_Slides_OLS_Solution: Slide 7: every “k” should be an “i”

HINT: If your first solution (i.e., before any atmospheric corrections) yields horizontal errors on
the order of tenths of meters and vertical error on the order of tens of meters … you’re spot
on! As you implement corrections, you will see the horizontal errors increase some but the
vertical error decreasing significantly.

% % %

I’ve received several questions related to Task 2b. I have realized that I was not completely
clear in describing that subtask. As currently written, it simply says, “Error in user position with
respect to truth in meters”. However, it does not specify what coordinate frame should be
used. Obviously LLH is in appropriate since the subtask specifies to provide the results in
meters. Thus you could either use ECEF or ENU. The most insightful frame is the ENU frame
since it gives you horizontal and vertical components directly (all SatNav systems are weakest in
the vertical). The easiest way to compute position error in the ENU frame is to use the function
ECEF2ENU but use the truth position as the origin of the ENU frame. For example:

pos_ecef = % your computed position solution in its original ECEF form

pos_enu_err = ECEF2ENU(pos_ecef, TruthECEF, TruthLLH)

% % %
Let me clarify what is meant by ‘range residuals’. The residuals are just the delta_PR vector after
the solution has converged (i.e., you iterate the solution until the magnitude of the delta_x vector

is less than 1e-3). If there were no measurement errors, the delta_PR vector would converge to
zeros as the solution iterates. However, due to the presence of measurement errors, the elements
of the delta_PR vector will not converge to zero even though the delta_x vector does. The
residuals are thus a measure of the inconsistency of the measurements which is an indication of
their errors. As you progress through the various tasks of the project, the magnitude of the range
residuals should decrease.

% % %

It was pointed out to me that the link given in the Task 5 portion of the project description is
broken. Here’s an update:

Products

The page has a number of items on it. The part that is relevant to the project is the “GPS
Satellite Ephemerides / Satellite & Station Clocks” and specifically the last entry in the table
which is labeled “Final”. You see from the table that a number of products are provided and
generally the ‘later’ they are, the more accurate they are. E.g., with more post-processing
data/time it is possible to get a better result. The ‘Final’ satellite clock and orbit products take a
couple of weeks to generate but they are the most accurate.

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