# Understanding Doppler Shifts

page in work

Doppler shifting is caused by compression of the waves emitted from a source. For audio waves it is the physical objecting 'catching up' with waves emitted at a constant velocity (speed of sound) in front and 'leaving behind' the waves emitted backwards. Wikipedia has several articles to make this more understandable. For electromagnetic waves, it is actually time compression of the source, but it can be modeled as the same effect with the same equations.

We need to model the Doppler shifts of a radio signal in order to match real world data to an actual orbit (or position along a ground track of a known orbit).

## Linear Case

Let's start simple. We have a road and there is an observer (A) standing at some distance d from this road. A car (B) is traveling at a constant velocity v. Let us assume that it is emitting engine noise at frequency ftx. The road is straight, and so we will use Cartesian coordinates for this example.

$B_{x}(t)=v*t+B_{x0}&$
$B_y(t)=B_{y0}$
$A_{x,y}=(A_{x0},A_{y0})$

The distance between them is just the Pythagorean Theorem, which is the distance formula.
$D_{AB}(t)=\sqrt{(A_x-B_x(t)))^2+(A_y-B_y))^2}$
Or in the full expanded form
$D_{AB}(t)=\sqrt{(A_x0-(v*t+B_x0))^2+(A_y0-B_y0)^2}$

Now what is the relative velocity between these two points? It is simply the rate of change of the distance. If you don't know calculus, this is a GREAT example of just how powerful it can be. Luckily, we can take what is called a derivative -- which is exactly the rate of change of a formula and we can use Wolfram Alpha to do the hard work for us. Seriously. Click the link or type "derivative of sqrt{(A1-(v*t+B1))^2+(A2-B2)^2} with respect to t" into the box on the main page. It will spit the following back at you.

${dD_{AB} \over dt} = \sqrt{(A_x0-(v*t+B_x0))^2+(A_y0-B_y0)^2} {d \over dt}$
${dD_{AB} \over dt} = D'(t)=V_{AB}(t) = {{v*(A_{x0}-B_{x0}-t v)} \over \sqrt((A_{x0}-B_{x0}-t v)^2+(A_{y0}-B_{y0})^2)}$

Now that we have the formula for relative velocity of an object moving at a constant speed along the x axis, let's plot an example with the following parameters (you are standing 5 meters from the road and the car moving from -10 to +10 meters along the road at 1 meter per second).

$A_{x0,y0} = (0,0)$ and $B_{x0,y0} = (5,-10)$
$v=1$
Relative Distance Between the car and the observer as a function of time:
http://www.wolframalpha.com/input/?i=plot+sqrt%7B%280-%281*t%2B-10%29%29%5E2%2B%280-5%29%5E2%7D+from+0+to+20

Relative Velocity Between the car and the observer as a function of time:
http://www.wolframalpha.com/input/?i=plot+%28%280-%28-10%29-x%29%29%2Fsqrt%28%280-%28-10%29-x%29%5E2%2B%280-5%29%5E2%29+x+from+0+to+20

We're almost there. The formula for the doppler shift of an emitted frequency is on the wikipedia article where ft is the transmitted frequency and c is the speed of waves in the medium (in this case, the speed of sound in air)

So this makes our final (huff puff!) equation!

$f_{rx}(t) = {{2*f_{tx}*v (A_{x0}-B_{x0}-t v)} \over {c * \sqrt((A_{x0}-B_{x0}-t v)^2+(A_{y0}-B_{y0})^2)}}$

Assume engine noise Ftx = 3khz and c = 343 m/sec (at standard conditions)
Plotting this (Wolfram Alpha "plot (2*3000*((0-(-10)-x))/sqrt((0-(-10)-x)^2+(0-5)^2))/343 x from 0 to 20") shows the final received frequency over time! YAY! NEEEEEEEEOoowwwww!!! Except our car is traveling a whopping 1 meter per second, so the doppler is only 15Hz. Feel free to imagine that this is 150Hz and the car is moving at 10m/sec.

http://www.wolframalpha.com/input/?i=plot+%282*3000*%28%280-%28-10%29-x%29%29%2Fsqrt%28%280-%28-10%29-x%29%5E2%2B%280-5%29%5E2%29%29%2F343+x+from+0+to+20

This relationship is linear as long as the car stays below some fraction of the speed of sound (Hopefully!). If we wish to assume that the car can go REALLY fast, we need to use the expanded form of the doppler equation which is left up to the reader and Wolfram Alpha:

## Circular Case

It is clear that this picture does not work for a satellite. A satellite is not traveling along a linear path, instead, a circle. Let's follow the same idea but close the track from a country road into a racetrack where the cars are traveling in a circle.

We can use the same variables, but now we can reference everyone against the center of the track instead of using the observer as the origin. The same car B is traveling at the same constant velocity v on a track with a radius Rb while our observer A is at distance Ra from the center of the track, and an angle Atheta. We are doing away with the above notation because it takes forever and Matlab is so nice! We can still use Wolfram Alpha for calculus and plotting. Note that from here on out we will switch from audio to radio waves (from speed of sound in air to speed of light) and we will be using the expanded doppler formula because satellites move REALLY fast.

```%% 2D circular Relative Velocity
%using kilometers and seconds
clear all;

ObsAlta = 6378;       %sea level
SatAlt = 6378+850;   %850km orbit
OrbitalPeriod = 6120;       %102 minutes
SatAngularVel = 2*pi/OrbitalPeriod;

t=0:OrbitalPeriod/100:2*OrbitalPeriod; %1000 seconds
InitialAngle = 0;

Vt = -(ObsAlta*SatAlt*SatAngularVel.*sin(InitialAngle+SatAngularVel.*t))./(sqrt(ObsAlta^2+SatAlt^2-2*ObsAlta*SatAlt.*cos(InitialAngle+SatAngularVel.*t)));
subplot(3,1,1);
plot(t,Vt_den);
title(['Distance to Observer: ' num2str(min(Vt_den)) 'km at ' num2str(TcrossingInterp2) 's']);
subplot(3,1,2);
plot(t,Vt);
title('Velocity to Observer');
```

## Polar Spherical Case

The circular case models what happens when a satellite passes DIRECTLY over you, but cannot account for what happens when there is an offset to one side of the pass. We can only model this case by moving forward to spherical coordinates. Things get a little messy here, but we will use the exact same approaches we used before -- the equations are just a little longer because we have to account for the new unit vector Phi. We will use the standard where phi represents rotation along the equator and theta represents the angle between the north pole and the vector while r is again the radius.

This case works quite well, and in fact is almost enough to solve our problem. We can model a 90 degree inclination orbit which is representative of any other circular orbit. We can probably even use this model to solve for geo-locations within about 250km which isn't bad for a 'rough guess'.

What is that 250km all about? Well unfortunately this is the distance that a point will move along the equator during a 15 minute satellite pass (assuming a sat altitude of 850km). The effect will get better at the pole and worse at the equator, but we really can't ignore this.

```%% 3D Relative Velocity
%using kilometers and seconds
%ObsPhi = 20*pi/180;
%ObsTheta = pi/2;

ObsTheta =     1.74;
ObsPhi =     0.135;

SatPhi = 0; %inclination of 90
SatTheta = pi/2;; %Initial position of 0 (equator)
OrbitalPeriod = 6127.2;       %102 minutes
SatAngularVel = 2*pi/OrbitalPeriod;

Fscale = 4.9;
Foffset = 2048;

t=-.07*OrbitalPeriod:OrbitalPeriod/300:.07*OrbitalPeriod; %1000 seconds

Vt = Vt_num ./ Vt_den;

subplot(3,1,1);
plot(t,Vt_den);
title(['Distance to Observer: ' num2str(min(Vt_den)) 'km at ' num2str(TcrossingInterp2) 's']);

subplot(3,1,2);
plot(t,Vt);
title('Velocity to Observer');

%%Doppler
c = 299792; %in km per second
Ft = 401.65e6; %401.65Mhz uplink freq
F =(2*Vt*Ft)./(c-Vt);
subplot(3,1,3);
plot(t,F,xdata,Fscale*ydata+Foffset,'-o');
title('Doppler Shift of Observer (at 401Mhz)');
max(F)
```

## Polar Spherical Case with Rotating Earth

This is a simple modification. We can just make phi of the observer also time dependent then re-calculate the derivative. The curve changes shape noticeably, which is expected because 250 miles is no small distance.

```%% 3D Relative Velocity with Earth's Rotation
%derivative of sqrt(A1^2+B1^2-2*A1*B1*(sin (v*t+A2)*sin (B2)*cos (A3-(v2*t+B3))+cos (v*t+A2)*cos (B2))) with respect to t
%(d)/(dt)(sqrt(A1^2+B1^2-2 A1 B1 (sin(v t+A2) sin(B2) cos(A3-(v2 t+B3))+cos(v t+A2) cos(B2)))) = -(A1 B1 (v2 sin(B2) sin(A2+t v) sin(A3-B3-t v2)+v sin(B2) cos(A2+t v) cos(A3-B3-t v2)-v cos(B2) sin(A2+t v)))/sqrt(A1^2-2 A1 B1 (sin(B2) sin(A2+t v) cos(A3-B3-t v2)+cos(B2) cos(A2+t v))+B1^2)
%v is sat angular velocity = SatAngularVel
%A2 is sat theta start = SatTheta
%A3 is sat phi => SatPhi

%B2 is obs theta => ObsTheta
%B3 is obs phi start => ObsPhi
%v2 is obs angular velocity => ObsAngularVel

ydata = [1506, 1022, 493, -61, -618, -1154, -1648, -2089, -2472, -2798, -3070, -3296, -3482];
xdata = [42.311,72.211,102.4552,132.2255,162.215,192.2178,222.314,252.2178,282.2255,312.2255,342.2255,372.2255,402.2255];

ObsRad = 6368; %center of European pass (50 degrees)
ObsTheta = 1.74;
ObsPhi = 0.135;
SidrealPeriod = 86164; %23 hours 56 minutes and 4 seconds => seconds
ObsAngularVel = 2*pi/SidrealPeriod;

SatPhi = 0; %inclination of 90 (really its 98, but we don't realy have a term for this
SatTheta = pi/2; %Initial position of 0 (equator)
OrbitalPeriod = 6127.2; %102.12 minutes => seconds
SatAngularVel = 2*pi/OrbitalPeriod;

Fscale = 4.9;
Foffset = 2048;

t = -0.07*OrbitalPeriod : OrbitalPeriod/1000 : 0.07*OrbitalPeriod; %1000 seconds
Vt = Vt_num ./ Vt_den;

subplot(3,1,1);
plot(t,Vt_den);

TcrossingInterp2= interp1(Vt,t,0);

title(['Distance to Observer: ' num2str(min(Vt_den)) 'km at ' num2str(TcrossingInterp2) 's']);

subplot(3,1,2);
plot(t,Vt);
title('Velocity to Observer');

%%Doppler
c = 299792; %in km per second
Ft = 401.65e6; %401.65Mhz uplink freq
F =(2*Ft*Vt)./(c-Vt);
subplot(3,1,3);
plot(t,F,xdata,Fscale*ydata+Foffset,'-o');
title('Doppler Shift of Observer (at 401Mhz)');
max(F);
```

## Variable Inclination Spherical Case with Rotating Earth

Whittling away errors, we have now exposed the next problem. Orbits have inclinations, which means that the earth's angular velocity vector and the satellite's motion vector are no longer orthagonal. During ascending nodes and descending nodes the velocity of the earth slightly adds and subtracts from the relative velocity, and thus each node will have a slightly different shape based on this larger or smaller velocity.

Luckily this just means adding a time dependence to the satellite's phi variable, while also remembering that theta is 90 degrees at the equator but inclination is 0 degrees at the equator. We can model this by adding i as inclination and setting phi(t) = cos(i)(omega*t+phi0) and theta(t) = sin(i)(omega*t+theta0) + pi/2. Notice that theta0 and phi0 are now scaled by the inclination. This means that they are now relative to the position along the orbit and NOT absolute in terms of earth coordinates (but could be transformed to absolute by the inclination factor, limiting them to valid possible values)

I believe this THIS model finally has enough fidelity to use for geolocation based on orbital data and doppler shift!