# Sol3_2012b.docx

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Homework #4
Due Date: March 9, 2011

Problem 1 (20 points)

You have been retained by a mapping company to do a quick assessment of the suitability of radar
interferometry for floodplain mapping. The required height accuracy is 30 cm at a horizontal spacing of
1 meter. They own their own corporate jet, which has a wingspan of 20 m that could be used for the
baseline of the interferometer. They also believe that they can buy an X-Band radar (3.2 cm
wavelength) at a reasonable price, with the option of buying a Ka-Band radar (0.8 cm wavelength) at a
considerably higher price. Finally, they plan to set the data recording window such that they will acquire
data between 25 and 60 degrees angle of incidence.

(a) They would like to put GPS antennas at each wingtip to determine the position of
each of the two antennas. According to their GPS expert, the position of each GPS
antenna can be reconstructed to an accuracy of 1 cm. What would be the
contribution to the height error at either frequency if they fly the jet at 5 km and 10
km altitudes above the terrain?
(b) What should the signal-to-noise ratio be for each radar if the phase noise
contribution to the height error should be less than 10 cm?
(c) Would you recommend radar interferometry to them as a viable solution to their
problem? If so, which radar should they buy?

Solution

In reality the GPS error is more than likely distributed uniformly in three dimensions around a point
centered on the GPS antenna phase center. To be able to calculate the errors in the height, we need to
know the standard deviation of the baseline length and tilt angle errors. We can model this as follows.

We shall assume that the baseline can be described by a coordinate system in which the baseline is
along the x-axis with end points at  10,0,0 and 10,0,0  . To this baseline we shall add two
independent identically distributed random variables centered at each end point. These random
variables themselves are the errors in the position of the GPS antenna phase centers, and are made up
of errors in the three dimensions. We shall assume that the errors in each dimension are identically
distributed and independent Gaussian random variables with zero mean and standard deviation
  1 cm .
Baseline length and tilt error.

We calculate the baseline length error by assuming the baseline ends are
 10  x1, y1, z1  , 10  x2 , y2 , z2  .   The six random variables are all identically distributed Gaussian
variables with zero mean and 1 cm standard deviations. The baseline length is then

B       20  x2  x1    y2  y1    z2  z1 
2                2             2

To find the standard deviation of the baseline length, we performed a Monte Carlo simulation using
100,000 realizations and found the standard deviation to be 1.56 cm.

Now, for the baseline tilt angle, we need to find the angle between the baseline and the x-y plane. This
angle is

                                                      
  tan  1                        z2  z1                  ,
                                                    2 
 20  x2  x1    y2  y1    z2  z1  
2             2


A Monte Carlo simulation gives the standard deviation of the baseline tilt angle to be .7 milliradians, or
.04 degrees.

To find the height error contribution of the GPS error, we need to combine these errors. From equation
6-138, the elevation of the point is (see the geometry in the figure)

z  y   h   cos

The error in the height estimate, assuming independent error sources for the baseline length and tilt
angle, is

z 2  z  2
2                2

z             B
         B

From equations 6-137 and 6-138, we find that

z
   sin   h tan 

and
z                        h
  tan     sin   tan 2 
B    B                    B

Therefore

2
h         2
 z   h tan       tan 2    B
2   2

B        

Here, H is the flight altitude, B is the nominal baseline length of 20 meters, and we assume the
baseline to be horizontal. Note that both errors are independent of the radar frequency.

A2
z
B
A1






h

Z(y)

y

The figure on the next page shows the errors as a function of the incidence angle for the two operating
altitudes. Note that the errors are worst at the far end of the swath in both cases, and worse at the
higher altitude. At this point, we can already stop and recommend the company finds a better way to
monitor the baseline length and tilt angle.
30

25
5 km
Height Error in meters

10 km
20

15

10

5

0
25   30      35            40           45     50   55   60
Incidence Angle

Signal-to-Noise Ratio.

We want to keep the height error contribution from the phase noise to less than 10 cm. From equations
6-137 and 6-138

h tan       1
 z 
2 B cos     SNR

where we have used the fact that

1
 
SNR

Therefore, we should have
2
 h tan          
SNR                    
 2 B cos       
             h   

Using the four cases, (two altitudes and two wavelengths), we find the results below.

40

5 km, X-Band
10 km, X-Band
5 km, Ka-Band
Signal-to-Noise Ratio in dB

30        10 km, Ka-Band

20

10

0
25    30          35          40         45             50   55   60
Incidence Angle

Note that everything else being equal, the Ka-band radar would have a slight advantage. Not only is the
required SNR lower, but in general the radar backscatter would be larger.

Based on these results, however, we cannot recommend radar interferometry unless they find a way to
measure the baseline length and angle more accurately. From these results, they probably need to do
better by a factor of 100 or so to make the required accuracy at the far end of the swath.
Problem 2. (10 points)

A C-band radar is flying at an altitude of 600 km and image a surface at a look angle of 35 degrees.
Assume the radius of the earth to be 6380 km. We plan to use this system to study surface subsidence
resulting from the extraction of groundwater from wells. Calculate:

1. The range to the earth surface assuming the earth to be spherical.
2. The expected phase shift for a vertical subsidence of 10 cm.
3. What is the index of refraction change due to atmospheric moisture that would give an
equivalent phase shift?
4. Assume the subsidence as a function of cross-track distance y is given by

Here z is the subsidence measured in the vertical direction, and is measured in centimeters. Plot
the expected differential phase as a function of y. Note that y is measured along the surface of
the sphere. Take to be that cross-track distance at which the look angle is 35 degrees.

Solution

Consider the geometry of the problem shown on the next page. The range is measured from the radar
platform to the point we are studying. If we apply the law of cosines, we can write

We can re-arrange this into a quadratic in R as follows:

The solutions are

Using the values for our problem, we find that
            R


G

Re+h

Re

g

To find the phase shift, we first have to find the component of the subsidence that is in the range
direction. If the subsidence is s, then the component in the direction of the radar range is

To find we use the law of cosines again to write

Therefore, we can write

Using our values, we find that

The phase shift due to the subsidence is then
The absolute phase can be written as

Here n is the index of refraction of the medium that the wave is propagating through, and is the
wavenumber in the absence of atmospheric moisture. The change in phase can then be written as

In the case of varying moisture but no subsidence, the second term on the right is zero. Therefore, the
change in index of refraction for a given phase change is

For our radar system, the change in index of refraction is

The cross-track distance at which the look angle is 35 degrees is given by

We can find    from the law of cosines:

From which we find

We can now vary the value of y along the earth surface according to the function given. For each value
of y, we can calculate a corresponding value of as follows

Once g is know, we can calculate the range to the point from

To find the component of the subsidence in the direction of the line of sight, we use

The differential phase due to the subsidence is then
The resulting phase is shown below. The phase is displayed in radians.

20

18

16

14

12

10

8

6

4

2

0
-10        -8        -6       -4        -2          0        2          4   6   8   10
(y-y0) in km

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