Brian Howard (#0240783) EE 501, Homework #5 - …dairyroadsolutions.com/signalprocessing/Homework5(Project).pdf · FM transmitter signal; the ﬁle sca.dat was collected at the Manastash

Brian Howard (#0240783) EE 501, Homework #5 - Manastash Ridge Radar Data Processing

Page 1

To do this project, you will need to download two data files:

http://rrsl.ee.washington.edu/ee501/ref.dat

http://rrsl.ee.washington.edu/ee501/sca.dat

Each of these files represents 10 seconds of IQ data collected on 29 October 2003, UT 10:23:00 with the

radar receivers set to 96.5 MHz. The file ref.dat was collected at the UW, and is an accurate copy of the

FM transmitter signal; the file sca.dat was collected at the Manastash Ridge Radar (near Ellensburg,

WA), and contains the scattered signals.

Problem 1. Data integrity. Load the file ref.dat into your computer analysis tool.

a) Verify that the first four samples are correctly loaded.

In the file ref.dat the first few values are supposed to be:

-5318 first in-phase sample

-771 first quadrature sample

3169 second in-phase sample

-8636 second quadrature sample

Using Matlab I found them to be:

K>> tmp(1:4)

ans =

-5318

-771

3169

-8636

b) Compute the mean and variance of the data. Is the data zero mean? Is it consistent with zero mean?

Using this code:

% Is the data zero mean?

disp(['Mean = ', num2str(mean(abs(data)))]);

ssqn = abs(data).*abs(data);

P1n = sum(ssqn)/N;

Pn = sum(abs(data))/N;

disp(['The signal power for ref.dat is ', num2str(P1n)])

P1varn=(1/sqrt(N))*P1n;

disp(['The expected value for ref.dat is ', num2str(Pn)])

disp(['The variance for ref.dat is ', num2str(std(data)^2)])

disp(['The standard deviation (square root of variance) for ref.dat is ', num2str(std(data))])


Page 2

I found:

Mean = 8185.3323

The signal power for ref.dat is 84407391.0245

The expected value for ref.dat is 8185.3323

The variance for ref.dat is 84407213.137

The standard deviation (square root of variance) for ref.dat is 9187.3398

The data ought to be zero mean. In this case the data has a mean value of 8185 so the literal answer is

”no.”; however mean value of 8185 is less than about 3*√𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒 so it is consistent with zero mean.

c) In class I stated that IQ data is “analytic.” Among other things this means that the real and imaginary

parts of the IQ data should be uncorrelated. Are they? (A little better way of asking this question is as

follows: “is it plausible to assert that the real and imaginary parts of the data are uncorrelated?”)

I think the most straightforward way to see if they are correlated is to plot the cross-correlation of the

two parts of the signal. Using this code:

% EE501 Homework 5, Problem 1(c) % Brian Howard, 31 May 2013

clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;

strFileName = 'ref.dat';

fid = fopen(strFileName, 'r');

tmp=fread(fid, mat_size,'short');

data=complex(tmp(1:2:mat_size),tmp(2:2:mat_size));

figure('Name','Problem 1 - a','position',[1 1 800 600])

plot(xcorr(real(data),imag(data)))

title('Correlation Between Real and Imaginary Parts')

xlabel('Sample')


Page 3

Resulted in this plot:

Since the correlation only as a high value at zero lag, it is reasonable to assert that the real and

imaginary parts of the signal are uncorrelated.


Page 4

d) You will be coherently processing the data in chunks of about 0.1 seconds (10,000 samples). Compute

the autocorrelation function R[n] for |n| < 100 and estimate the range resolution. Do this several times,

starting at n = 0, 105, 2 × 105, plotting the magnitude of the autocorrelation |R[n]| for 0 ≤ n ≤ 100.

I used this Matlab code:

% EE501 Homework 5, Problem 1(d)

% Brian Howard, 31 May 2013

clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;




ref=complex(tmp(1:2:mat_size),tmp(2:2:mat_size));

nLag = 100;

figure('Name','Problem 1 - d','position',[1 1 800 600])

iStart = 1;

iEnd = 10000;

data = abs(ref(iStart:iEnd));

data = data - mean(data);

Xc = xcorr(data, nLag);

axLag = -nLag:nLag;

plot(axLag,Xc)

ml = max(Xc)*0.707*ones(size(Xc));

hold on

iStart = iEnd+1;

iEnd = iEnd+10000;




plot(axLag,Xc,'g')

iStart = iEnd+1;

iEnd = iEnd+10000;




plot(axLag,Xc,'m')

plot(axLag, ml,'r')

legend('1-10000','10001-20000','20001-30000','-3db Line');

title('Autocorrelation |n|<100 (blue) and -3dB Point (Red)')

xlabel('Sample')


Page 5

To calculate the autocorrelation:

Treating the autocorrelation like a spectrum I defined the range resolution to be the -3dB width of the

main lobe. In this case the worst case width is ~8 samples or 80µs. This implies a range resolution of

(c/2)* 80µs = 12000 m = 1.2 km.


Page 6

e) Make separate histograms of the real part and imaginary part of ref.dat with 50–100 bins. The two

histograms should look pretty similar. Do they? (extra credit: the histograms certainly do *not* appear

gaussian: can you think of why this might be? What do you think that the histograms should look like?

Why?)

Using this code:

% EE501 Homework 5, Problem 1(e)

% Brian Howard, 31 May 2013

clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;




data=complex(tmp(1:2:mat_size),tmp(2:2:mat_size));

figure('Name','Problem 1 - e','position',[1 1 800 600])

subplot(2,1,1)

hist(real(data),100);

title('Histogram of the Real part')

subplot(2,1,2)

hist(imag(data),100);

title('Histogram of the Imaginary part')


Page 7

I found the histograms to look like:

The two histograms do look similar; however they don’t appear to be Gaussian. I originally thought this

was because of the frequency spectrum of the FM broadcast in which the mono audio (L+R ) are at the

center frequency and the stereo (L-R) bands are at the outside. But this is voltage, not frequency. This

more likely the filter in the data collection system.


Page 8

Problem 2. Data range alignment and rate reduction. In order to do time series analysis for each range,

you will need to combine the two data streams. This can be done several ways, but the following is

straightforward. Define Z(m,r) as follows:

𝑍(𝑚, 𝑟) = ∑ 𝑠𝑐𝑎[𝑚𝐷 + 𝑝 + 𝑟](𝑟𝑒𝑓[𝑚𝐷 + 𝑝])∗𝐷−1

𝑝=0

(1)

This Z(m,r) is an (unbiased) estimate of the signal at range r taken at time mD, thus m is something like

the ”slow time” that we saw for pulsed radars. This Z is effectively the “matched filter detection” of the

target when the transmitter waveform changes constantly, at a rate much faster than the target itself

changes. For this data we expect the full Doppler bandwidth to be less than 2 kHz, so we select D = 50.

This means that Z(m,r) is a 2 kHz time series at range r. We can use Z(m,r) to estimate P1(r) as follows:

�̂�1(𝑟) =1

𝑀∑|𝑍(𝑚, 𝑟)|2𝑀

𝑚=1

(2)

... where M is “as big as possible” given the amount of data available.

a) Write an algorithm to compute Z(m,r). Your algorithm should check to see that m,r,D make sense (lie

within the actual data range). In particular, compute Z(m = 10,r = 42) for D = 50.

I had some trouble with this one. Dr. Sahr gave the values for this m and r as 1011171809-j32362774

which were helpful in correcting my indexing issues. I’m using Matlab so all of my indexes are 1-based

I began by writing a function to calculate Z (it gets used repeatedly in this project):

function [Z] = zcorr(sca, ref, m, r, D)

% function [Z] = zcorr(sca, ref, m, r, D)

%

% INPUT

% sca Complex array with scattered (reciever) data

% ref Complex array of FM transmitter signal

% m Like the "slow time" for pulsed radars. Scalar or vector.

% r Range number. Scalar or vector.

% D Range length. For example, a Doppler bandwidth of 2kHz and

% data sampled at 100kHz implies D = 100000/2000 = 50.

% OUTPUT

% Z This is essentially the correlation between the reference and

% scattered signal. Returns a scalar value if m and r are

% scalars and an array of size (m,r) if m and r are vectors.

% Default value indicating the function failed.

Z = -1000;

N = length(sca);

% check that datasets are the same length

if length(ref)~= N

error('data sets are not the same length');

return;

end

% check that m,r, and D do not exceed array bounds

if( (max(m)*D)>(N-max(r)))


Page 9

error('m,r, and D exceed array bounds');

return;

end

% Make the cross-correlation calculation for the case of scalar

if( isequal(size(r),[1 1]) && isequal(size(m),[1 1]) )

Z=0;

mD = (m-1)*D;

Z = Z + dot(sca((mD + (r-1) + 1):(mD + (r-1) + D))',conj(ref( (mD + 1):(mD+D))));

return;

end

% This is the case of r and m vectors

Z = zeros(length(m),length(r));

for ir = 1:length(r)

for im = 1:length(m)

mD = (m(im)-1)*D;

Z(im,ir) = Z(im,ir) + dot(sca((mD + (r(ir)-1) + 1):(mD + (r(ir)-1) + D))',conj(ref( (mD +

1):(mD+D))));

end

end

I then wrote this code to calculate Z at a particular m and r:

% EE501 Homework 5, Problem 2

% Brian Howard, 1 June 2013

clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;





strFileName = 'sca.dat';



sca=complex(tmp(1:2:mat_size),tmp(2:2:mat_size));

D = 50;

Z = zcorr(sca, ref, 11, 43, D)

The code returned:

EDU>> Problem2

Z =

1.0117e+08 - 3.2363e+07i

Which matches pretty well with the values given by the professor.


Page 10

b) Use your algorithm to compute �̂�1 for 0 ≤ r ≤ 80. Plot this profile, properly labeling the range axis in

km (assume back scatter geometry).

% EE501 Homework 5, Problem 2(b)


clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;









D = 50;

fs = 100000;

c = 300000000;

% Only need to run these lines once. Just load it otherwise

if 1

r = 1:81;

m = 1:(fix((N-max(r))/D));

Z = zcorr(sca, ref, m, r, D);

save('Z80.mat','Z');

end

load('Z80');

% calculate the power estimate

M = min(size(Z));

Pest = zeros(M,1);

for r = 1:M

P(r) = dot(Z(:,r),Z(:,r));

P(r) = P(r)./M;

end

% calculate the range in km

rd = (0:(M-1)).*(1/fs)*(c/2)*0.001;

% plot the data

figure('Name','Problem 2 - b','position',[1 1 800 600])

plot(rd,P)

title('Estimated Power');

xlabel('Distance, km');


Page 11


I’m guessing that big peak is Rainer?


Page 12

c) What is the range resolution? (and what do you mean by ”range resolution”?)

In this problem the smallest distance in time that we can discriminate is 1/sampling frequency. For this

project that works out to a time of 0.01ms. Since we are assuming backscatter geometry, this time

interval works out to a distance of 1.5km of range resolution.

d) What is the range of the Doppler velocities that can be expressed in the time series Z(m,r) at a

particular fixed range r?

Doppler velocity equals (fd*λ)/2. The Doppler frequency is 2kHz (since D=50) and λ is 3.1m so that the

velocity range is 3100m/s or +/-1550m/s.


Page 13

Problem 3. Spectrum Estimation. Setting D = 50 and r = 42 (or matlab range 43), compute and plot the

power spectrum as indicated below.

a) How many samples Z(m,42) are available?

Using the size of the file and the parameters given in the problem I calculate the number of samples as:

mat_size = 4096000/2;

N = mat_size/2;

D = 50;

r = 43;

m = (fix((N-max(r))/D))

This gives the result of:

m = 20479

b) Plot abs(FFT(Z(:,42))).^2. Do you see any (nonzero Doppler velocity) target? What is its velocity?

What is the Doppler Velocity resolution?

This is the code I used:



clear all;

close all;

home;

c = 300000000;

fb = 96500000;

lambda = c/fb;

D = 50;

fs = 100000;

load('Z80');

r = 43;

Y = fftshift(abs(fft(Z(:,r))).^2);

n = length(Y);

fd = (fs/(2*D))*lambda/2;

fds =((-(n-1)/2):((n-1)/2))./((n-1)/2);

fds = fds.*fd;


plot(fds,Y);

title(['Power Spectrum for r = ' num2str(r) ' (1-based)'])

xlabel(['Doppler Velocity. Resoution = ' num2str(fds(2)-fds(1)) ' m/s']);

xlim([min(fds) max(fds)]);


Page 14

It generated this plot:

The plot shows a single non-zero target with a velocity of 56.1 m/s (202 km/hr). The resolution is 0.15

m/s (~1/2 km/hr).


Page 15

c) Now divide the data Z into sections of length 256. Compute the FFT of each block of data, find the

squared magnitude of each spectrum, add them all up, and plot the result. Is the target any easier to

see? What is the Doppler Velocity resolution?

This is the code I wrote:



clear all;

close all;

home;

c = 300000000;

fb = 96500000;

lambda = c/fb;

D = 50;

fs = 100000;

load('Z80');

r = 43;

n = 256;

len = length(Z(:,r));

Y = zeros(n,1);

for iCorr = 0:(fix(len/n)-1)

Y = Y + fftshift(abs(fft(Z((iCorr*n)+1:(iCorr+1)*n,r))).^2);

end


fds =((-(n-1)/2):((n-1)/2))./((n-1)/2);

fds = fds.*fd;


plot(fds,Y);

title(['Power Spectrum for r = ' num2str(r) ' (1-based). Number of Integration = '

num2str(fix(len/n))])




Page 16

It resulted in this figure:

The target appears much easier to see, especially with respect to the ground clutter, but Doppler

resolution has increased from 0.15 m/s to 12.2 m/s due to the lower bin count in the FFT.


Page 17

d) Repeat the processing of part (c) except change the range to 47. This is the range at which Mt. Rainier

is visible. What is the relative strength of the target in parts b, c to the strength of the Mt. Rainier signal?

The code is the same as part (c), I just set r = 48 (1-based, Matlab). It resulted in this figure:

It appears that Rainer is about 2 orders of magnitude (20 dB) stronger than the target in part c.


Page 18

Problem 4. Range-Doppler Profile. Perform the processing of problem 3c, 3d for all ranges r=0:800

(python; r=1:800 in matlab) and plot the result as an image (e.g., using ”imagesc” in Matlab). In this

image you will see several airplanes, ground clutter (from the Cascades), as well as auroral turbulence.

a) There is a second airplane (point target with non-zero Doppler velocity. At what range and Doppler

does it appear?

This is the code I used:

% EE501 Homework 5, Problem 4(a)


clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;

c = 300000000;

fb = 96500000;

lambda = c/fb;

D = 50;

fs = 100000;


if 0









r = 1:801;

m = 1:(fix((N-max(r))/D));



end

load('Z800');


M = min(size(Z));

Pest = zeros(M,1);

for r = 1:M

P(r) = dot(Z(:,r),Z(:,r));

P(r) = P(r)./M;

end


rd = (0:(M-1)).*(1/fs)*(c/2)*0.001;

% coherent integration

r = 43;

n = 256;

[len rng] = size(Z);

Y = zeros(n,rng);

for iRng = 1:rng


Y(:,iRng) = Y(:,iRng) + fftshift(abs(fft(Z((iCorr*n)+1:(iCorr+1)*n,iRng))).^2);

end

end


fds =((-(n-1)/2):((n-1)/2))./((n-1)/2);


Page 19

fds = fds.*fd;


plot(fds,Y(:,r));





% plot the data


imagesc([min(fds) max(fds)], [1 max(rd)], (log10(Y)'))

title(['Log of Power Spectrum. Number of Integration = ' num2str(fix(len/n))])


ylabel('Distance, km');

set(gca,'YDir','normal')

caxis([21.5 24])

colorbar


hSurf = surf((log10(Y)'),'EdgeColor','none','LineStyle','none','FaceColor','interp');

camproj('orthographic')


caxis([21.5 24])

view([-31.5 24]);

colorbar


Page 20


The image does show something with a Doppler velocity, highlighted by the red ellipse. It has a Doppler

velocity of 262 m/s (943 km/hr) and a range of 111 km.


Page 21

b) Most of the range-Doppler image contains no scattered signal; we call this the ”clutter floor.” What is

the strength of the clutter floor? How strong is the Mt. Rainier echo compared to the clutter floor (in

dB)? How strong are the two airplane echoes compared to the clutter floor?

Referring to the plot, which shows the log10 of the power spectrum and should be reading in decibels,

the clutter floor strength is about 21.5 and Mt Rainier is about 24, a 3.5 dB difference. The airplanes are

about 23 and 22, or 1-2 dB above the clutter floor.

c) At what ranges do you see Auroral Turbulence?

It looks to start around 500 km and extend upward to about 1000 km.

d) Auroral turbulence occurs at an altitude of 110 km, and certainly occurs far beyond the 800 ranges

(1200 km). Why is it pointless to look for auroral echoes further than 800 ranges?

The turbulence is at a constant elevation, but larger ranges are echoes from further away. As the

distance increases the curvature of the earth begins to block the signal. In short, it is the Canadians

fault, their country gets in the way.


Page 22

Problem 5. Reduced Precision.

a) Take the original data and divide each by 5000 and round to the nearest integer; this simulates use of

a 4 bit digitizer. Form the range-Doppler image using this truncated data, and compare it to your image

from problem 3.

Using this code:



clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;

c = 300000000;

fb = 96500000;

lambda = c/fb;

D = 50;

fs = 100000;


if 0




tmp=fix(tmp./5000);





tmp=fix(tmp./5000);


r = 1:801;

m = 1:(fix((N-max(r))/D));


save('Z800ds.mat','Z');

end

load('Z800ds');


M = min(size(Z));

Pest = zeros(M,1);

for r = 1:M

P(r) = dot(Z(:,r),Z(:,r));

P(r) = P(r)./M;

end


rd = (0:(M-1)).*(1/fs)*(c/2)*0.001;


r = 43;

n = 256;


Y = zeros(n,rng);

for iRng = 1:rng



end

end



Page 23

fds =((-(n-1)/2):((n-1)/2))./((n-1)/2);

fds = fds.*fd;


plot(fds,Y(:,r));





% plot the data



title(['Log of Power Spectrum (Reduced Precision). Number of Integration = '





%caxis([21.5 24])

colorbar





%caxis([21.5 24])

view([-31.5 24]);

colorbar


Page 24


It looks pretty good, not so different from the higher precision images above.


Page 25

b) Take the original data and reduce it to the sign bit only (in each of the real and imaginary parts); that

is, the only possible data points are ±1 ± j. Form the range-Doppler image using this truncated data, and

compare it to your image from problem 3.

% EE501 Homework 5, Problem 5(b)


clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;

c = 300000000;

fb = 96500000;

lambda = c/fb;

D = 50;

fs = 100000;


if 1




tmp=sign(tmp);





tmp=sign(tmp);


r = 1:801;

m = 1:(fix((N-max(r))/D));


save('Z800ds.mat','Z');

end

load('Z800ds');


M = min(size(Z));

Pest = zeros(M,1);

for r = 1:M

P(r) = dot(Z(:,r),Z(:,r));

P(r) = P(r)./M;

end


rd = (0:(M-1)).*(1/fs)*(c/2)*0.001;


r = 43;

n = 256;


Y = zeros(n,rng);

for iRng = 1:rng



end

end


fds =((-(n-1)/2):((n-1)/2))./((n-1)/2);

fds = fds.*fd;


plot(fds,Y(:,r));


Page 26





% plot the data



title(['Log of Power Spectrum (1-Bit Precision). Number of Integration = ' num2str(fix(len/n))])




%caxis([21.5 24])

colorbar





%caxis([21.5 24])

view([-31.5 24]);

colorbar


Page 27


Not bad at all!!! It is a little harder to see the turbulence, which might be a problem if the data is being

used to quantify behavior there.


Page 28

Problem 6. MTI (”moving target indicator”) processing. Ground clutter always appears with very low

Doppler Velocity, so it would be nice to reject it, if possible. Rather than using Z(m,r) directly, create a

new data set,

Z1(m,r) = Z(m,r) − Z(m − 1,r) (3)

a) Plot the power spectrum of Z1(m,r = 42)?

This is the code



clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;

c = 300000000;

fb = 96500000;

lambda = c/fb;

D = 50;

fs = 100000;


if 0









r = 1:801;

m = 1:(fix((N-max(r))/D));



end

load('Z800');

% Create the Z1 Array: Z1(m,r) = Z(m,r) ? Z(m ? 1,r)

iOffset = 1;

Z = Z-circshift(Z,[-iOffset,0]);

Z(:,1:iOffset)=[];


M = min(size(Z));

Pest = zeros(M,1);

for r = 1:M

P(r) = dot(Z(:,r),Z(:,r));

P(r) = P(r)./M;

end


rd = (0:(M-1)).*(1/fs)*(c/2)*0.001;


r = 43;

n = 256;


Y = zeros(n,rng);

for iRng = 1:rng


Page 29



end

end


fds =((-(n-1)/2):((n-1)/2))./((n-1)/2);

fds = fds.*fd;


plot(fds,Y(:,r));






Page 30


b) Is the airplane easier or harder to detect?

In this case, it makes it easier to see the plane as opposed to the ground clutter. It looks like this

technique is less helpful for targets far away from ground clutter…..although you really wouldn’t need it

for those.


Page 31

c) define Z2(m,r) = Z(m,r) − Z(m − 2,r) and repeat parts (a), (b).

The code is same as (a), just changed for more offset.

I can’t really see the plane at all here

d) define Z4(m,r) = Z(m,r) − Z(m − 4,r) and repeat parts (a), (b).


Page 32

Not so much better in terms of seeing the plane.


Page 33

e) define Z8(m,r) = Z(m,r) − Z(m − 8,r) and repeat parts (a), (b)

As the offset increases, the plane gets caught in the stop band and becomes harder to see.


Page 34

Problem 7. Assume that the airplane at range 44-45 is flying at an altitude of 5000m. Describe the locus

of its possible locations. You’ll need the following data; KJAQ is the 96.5 MHz transmitter. The first two

numbers in the URL are the latitude and longitude. MRO is where the sca.dat data is taken; Sieg is where

ref.dat is taken, and KJAQ is the location of the transmitter. Note that negative longitude is ”west”.

UWSieg http://maps.google.com/?ll=47.654955,-122.306636&spn=0.000383,0.000536&t=h&z=21

MRO http://maps.google.com/?ll=46.951011,-120.724901&spn=0.000548,0.001072&t=h&z=20

KJAQ http://maps.google.com/?ll=47.504546,-121.969182&spn=0.000543,0.001072&t=h&z=20

Use the ref and transmitter location to offset the time delay incurred between them. With the

corrected signal you can draw an ellipse of constant range, with foci at the transmitter and at the

antenna for sca.

http://maps.google.com/?ll=47.654955,-122.306636&spn=0.000383,0.000536&t=h&z=21




Page 35

Problem 8. There is an additional data set you can use for the scatter data:

http://rrsl.ee.washington.edu/ee501/scb.dat

This data was taken at exactly the same time as sca.dat, but on a different antenna.

a) When you process scb.dat, do you see the same spectral signatures as with sca.dat?

Using this code:



clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;

c = 300000000;

fb = 96500000;

lambda = c/fb;

D = 50;

fs = 100000;


if 0





strFileName = 'scb.dat';




r = 1:801;

m = 1:(fix((N-max(r))/D));


save('Z800b.mat','Z');

end

load('Z800b');


M = min(size(Z));

Pest = zeros(M,1);

for r = 1:M

P(r) = dot(Z(:,r),Z(:,r));

P(r) = P(r)./M;

end


rd = (0:(M-1)).*(1/fs)*(c/2)*0.001;


r = 43;

n = 256;


Y = zeros(n,rng);

for iRng = 1:rng



end

end


fds =((-(n-1)/2):((n-1)/2))./((n-1)/2);


Page 36

fds = fds.*fd;


plot(fds,Y(:,r));





% plot the data



title(['Log of Power Spectrum. Number of Integration = ' num2str(fix(len/n))])




%caxis([21.5 24])

colorbar





%caxis([21.5 24])

view([-31.5 24]);

colorbar


Page 37

The code gives this figure below:

I see some of the same ground clutter features visible with the other antenna; however the planes and

the turbulence do not look to be visible.


Page 38

b) In addition to computing the cross ambiguity for each antenna, you can also compute the cross

coherence of each antenna:

𝐶(𝑟, 𝑓) =

𝐸(𝑉𝑎(𝑟, 𝑓)𝑉𝑏∗(𝑟, 𝑓))

√𝐸(𝑉𝑎(𝑟, 𝑓)𝑉𝑎∗(𝑟, 𝑓))√𝐸(𝑉𝑏(𝑟, 𝑓)𝑉𝑏

∗(𝑟, 𝑓))

(4)

Here Va(r,f) is an individual voltage spectrum for antenna a. The magnitude and phase of C tells you

about the angular size and direction at range r and Doppler shift f.

Using this code:



clear all;

close all;

home;

mat_size = 4096000/2;

N = mat_size/2;

c = 300000000;

fb = 96500000;

lambda = c/fb;

D = 50;

fs = 100000;


if 0





strFileName = 'scb.dat';




r = 1:801;

m = 1:(fix((N-max(r))/D));


save('Z800b.mat','Z');

end

load('Z800b');

Zb=Z;

load('Z800');


M = min(size(Z));

Pest = zeros(M,1);

for r = 1:M

P(r) = dot(Z(:,r),Z(:,r));

P(r) = P(r)./M;

end


rd = (0:(M-1)).*(1/fs)*(c/2)*0.001;



Page 39

r = 43;

n = 256;


Y = zeros(n,rng);

Yb = zeros(n,rng);

C = zeros(n,rng);

Va = zeros(n, fix(len/n));

Vb = zeros(n, fix(len/n));

for iRng = 1:rng


Va(:,iCorr+1) = fftshift((fft(Z((iCorr*n)+1:(iCorr+1)*n,iRng))).^2);

Y(:,iRng) = Y(:,iRng) + abs(Va(:,iCorr+1));

Vb(:,iCorr+1) = fftshift((fft(Zb((iCorr*n)+1:(iCorr+1)*n,iRng))).^2);

Yb(:,iRng) = Yb(:,iRng) + abs(Vb(:,iCorr+1));

end

% Cross-coherence for this range

for fc = 1:n

C(fc,iRng) = mean(Va(fc,1:end).*conj(Vb(fc,1:end)));

C(fc,iRng) = C(fc,iRng)./sqrt(mean(Va(fc,1:end).*conj(Va(fc,1:end))));

C(fc,iRng) = C(fc,iRng)./sqrt(mean(Vb(fc,1:end).*conj(Vb(fc,1:end))));

end

end


fds =((-(n-1)/2):((n-1)/2))./((n-1)/2);

fds = fds.*fd;


plot(fds,Y(:,r));





% plot the data



title(['Log of Power Spectrum, B Antenna. Number of Integration = ' num2str(fix(len/n))])




%caxis([21.5 24])

colorbar

% plot the data


imagesc([min(fds) max(fds)], [1 max(rd)], (log10(Yb)'))

title(['Log of Power Spectrum, B Antenna. Number of Integration = ' num2str(fix(len/n))])




%caxis([21.5 24])

colorbar

% plot the data


imagesc([min(fds) max(fds)], [1 max(rd)], (abs(C)'))

title(['Crosscorrelation amplitude. Number of Averages = ' num2str(fix(len/n))])




%caxis([21.5 24])

colorbar

% plot the data


imagesc([min(fds) max(fds)], [1 max(rd)], (angle(C)'))


Page 40

title(['Crosscorrelation angle (radians). Number of Averages = ' num2str(fix(len/n))])




%caxis([21.5 24])

colorbar

I generated this figure:

It does show a nice high value for Mt. Rainier, which is a good thing. I think that means that both

antenna saw it.


Page 41

I also plotted the phase:

It looks a little random. Maybe it needs to be unwrapped?


Page 42


Page 43

Documents

Brian Howard (#0240783) EE 501, Homework #5 - …dairyroadsolutions.com/signalprocessing/Homework5(Project).pdf · FM transmitter signal; the ﬁle sca.dat was collected at the Manastash