GP265 /EE 355 Homework 8 (Final project 2)

March 14, 2018

1. Display interferogram phase

2. Calculate the curved earth fringe pattern and subtract it from the interferogram phase.

First I compute the slant range pixel spacing ∆r:

∆r = nlooksrc

2fs= 3 · 2.99792458 · 10

8m/s

2(16.0 · 106Hz)= 28.1055 m. (1)

Then the range ρ to each range bin ir is:

ρ(ir) = ρ0 + (ir − 1)∆r, ir = 0, 1, ..., nrange− 1 (2)

where ρ0 = 844768 m is the range to the first bin and nrange = 1024 range bins.

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The look angle θz0 to each range bin for z = 0 (no topography) is:

θz0 = cos−1

((h+ re)

2 + ρ2 − (z + re)2

2ρ(h+ re)

)|z=0 = cos−1

((h+ re)

2 + ρ2 − r2e2ρ(h+ re)

)(3)

where h = 696000 m is the radar altitude and re = 6378000 m is the earth radius.

Then the curved earth phase ϕz0 for z = 0 (no topography) is:

ϕz0 = −4π

λ

(B2

2ρ−B sin (θz0 − α)

)(4)

where λ = 0.236057 m is the radar wavelength, B = 307 m is the baseline length, and α isthe orientation angle.

To remove this curved earth phase from the complex interferogram S (displayed in Problem1), I apply a phase correction T = Se−iϕz0 , and here is the resulting phase arg(T ), whichshould have only topography effects. The fringe lines look more like elevation contour lineshere:

3. Flatten the interferogram by further correcting for any tilts over the image.

We assume that the topography is flat over the caldera at the top of the volcano, at rangebins 110-150 and azimuth bins 925-960. I used least squares to find the slope of the fringeswithin the caldera (in radians/pixel), in both the range and azimuth directions, and apply

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these phase corrections to the interferogram from Problem 2 to ”flatten” the tilt from theinterferogram. The flattened interferogram no longer has fringes within the caldera, and thefringes look more like elevation contour lines around the volcano. Note that the area I choseis small enough that we can avoid the problem of phase unwrapping.

This is not the only way to correct for tilts - many other solutions are acceptable. For example,you can also try guessing different values of orientation angle α that remove the tilt withinthe caldera.

4. Determine the ambiguity height, ha, for this geometry.

The ambiguity height is defined as the height dz where the topography phase ϕtopo = 2π

ϕtopo = 2π = −4π

λ(Bsin(θ − α)−Bsin(θflat − α)) (5)

We can compute θflat using the law of cosines using Eq. (3) and ha can be computed oncewe obtain the look angle θ from Eq. (5), again using the law of cosines.

The ambiguity height depends on the range ρ, so here is a plot of the range dependence:

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At the reference (first) range bin, ha = 227.6 m. At the last range bin, ha = 257.9 m.

5. Unwrap the flattened interferogram using the snaphu program.

I write out the flattened interferogram to a file called ’flatinterferogram’ and then run itthrough snaphu to unwrap the phase, writing it out to a file called ’snaphu.out’. I then readin the ’snaphu.out’ file into MATLAB, and plot its phase, which is now unwrapped - no morefringes here, and the phase is no longer limited to the interval [−π, π].

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6. Using the ambiguity height ha from (4), convert the unwrapped phase values to heights. Youhave now created a digital elevation model (DEM) in radar coordinates.

I use the ambiguity height ha from Eq. (5), which varies as a function of range ρ, to scalethe unwrapped phase values ϕflat from Problem 5 into height values zDEM . I choose several’zero elevation’ reference pixels near the coast, and subtract their average phase ϕref fromevery point in the unwrapped interferogram. Since we are dealing with topography here, Ichoose to plot the contour of the DEM.

zDEM =(ϕflat − ϕref )

2πha (6)

The elevation of Mauna Loa is 4169 m, so our DEM looks reasonable. There are some heightsbelow 0 m, but they were in the region over water so I would not expect to have useful heightvalues here.

7. Map correlation into height error

First, I display the correlation to check that I read it in properly.

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Problem 7: Correlation

range bin200 400 600 800 1000

azim

uth

bin

100

200

300

400

500

600

700

800

900

1000

1100

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

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I use this equation to compute the height error σz as a function of correlation ρc (range is ρ):

σz =λρ sin θz0

4πB cos(θz0 − α)

√1− ρc2ρc

(7)

Here is a plot of the height error σz, ranging from 0 to 50 m. It makes sense that the heighterror is greatest at the locations where the correlation is lowest, such as the water in thetop-right corner, as well as the slopes in the top-left and bottom-right corners.

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8. Map DEM to ground coordinates

To compute the ground range, which depends on both range and azimuth since I take theDEM heights into account, I first compute the look angle (compare to Eq. (3)):

θz = cos−1

((h+ re)

2 + ρ2 − (zDEM + re)2

2ρ(h+ re)

)(8)

Then I can compute the angle βz at the center of the earth, using the law of sines:

βz = sin−1

(ρ

resin θz

)(9)

The ground range rg,z is thenrg,z = reβz (10)

I get the platform velocity v from the effective platform velocity veff = 7179.4 m/s, radaraltitude h = 696000 m, and earth radius re = 6378000 m:

v = veff

√h+ rere

= (7179.4m/s)

√696000 + 6378000

6378000= 7560.98 m/s. (11)

Then the azimuth bin ground spacing ∆azg is

∆azg = nlooksazv

prf

reh+ re

= 127560.98

2159.827

6378000

696000 + 6378000= 37.8757 m (12)

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I generated a ”ground coordinates” grid with a spacing of ∆azg between adjacent groundrange bins, in order to get square pixels. I resampled only in range, not in azimuth. Thereare some ”holes” in the image from resampling.

9. Map DEM to ground coordinates: Nice image with interpolation

I resampled the new DEM using linear interpolation. There are no longer ”holes” in theimage.

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% Hw8 Final Pt.2

clear; close all; clc;

%% parameters

nr=1024;

naz=1190;

nrlook=3;

nazlook=12;

B=307; % baseline,m

B_par=-170;

B_perp=-263;

hgt=696000; % altitude in orbit,m

wvl=0.236057; % wavelength, m

prf=2159.827;

fs=16e6;

c=3e8;

veff=7179.4;

r0=844768; % near range

deltar=c/2/fs*nrlook;

rmid=r0+(nr/2)*deltar+deltar/2; % mid range

re=6378e3; % Earth radius,m

costheta0=((hgt+re)^2+rmid^2-re^2)/2/rmid/(hgt+re);

theta0=acos(costheta0);

alpha=atan(B_perp/B_par)+theta0-pi/2; % baseline angle, rad

alphadg=alpha/pi*180; % baseline angle, degree

%% read files

fid=fopen(’interferogram’);

data=fread(fid,[nr*2 naz],’float’);

fclose(fid);

reald=data(1:2:end,:);

imagd=data(2:2:end,:);

intgram=complex(reald,imagd);

intgram=intgram.’;

phase=angle(intgram);

fid=fopen(’amplitude’);

data=fread(fid,[nr*2 inf],’float’);

fclose(fid);

amp1=data(1:2:end,:);

amp2=data(2:2:end,:);

amp=amp1.’;

fid=fopen(’correlation’);

data=fread(fid,[nr*2 inf],’float’);

fclose(fid);

col=data(nr+1:end,:);

col=col.’;

% display the phase of the interferogram

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figure

colormap jet

imagesc(phase)

axis image

saveas(gcf,’1.png’,’png’)

%% calculate the curved Earth pattern and substract it.

phase_bg=zeros(naz,nr);

for i=1:nr

rg=r0+(i-1)*deltar;

costhetaflat=((hgt+re)^2+rg^2-re^2)/2/rg/(hgt+re);

thetaflat=acos(costhetaflat);

sindiff=sin(thetaflat-alpha);

phase_bg(:,i)=-4*pi/wvl*(B^2/2/rg-B*sindiff);

end

intgram_bg=exp(1j*phase_bg);

figure

imagesc(angle(intgram_bg))

colormap(jet)

colorbar

saveas(gcf,’2_2.png’,’png’)

topo=intgram.*conj(intgram_bg);

figure

colormap jet

imagesc(angle(topo))

saveas(gcf,’2.png’,’png’)

% pic=mycolormap(amp,angle(topo),nr,naz);

% figure

% imagesc(pic)

%% flatten the interferogram by further correcting for any tilts over the image

% select a small flat area where we expect no change in phase

% note that the area should be small enough that there is no phase wrapping

testsite=topo(925:960,110:150);

testphase=angle(testsite);

figure

imagesc(testphase)

colormap jet

colorbar

% fit a 2-D plane

[valid_x,valid_y]=meshgrid(1:size(testsite,2),1:size(testsite,1));

valid_x=valid_x(:);

valid_y=valid_y(:);

valid_data=testphase(:);

design_A = [ones(size(valid_data)) valid_x valid_y];

coef = design_A\valid_data;

ramppart=design_A*coef;

figure

imagesc(testphase-reshape(ramppart,size(testsite)))

% get the ramp for the whole interferogram

[all_x,all_y] = meshgrid (1:nr, 1:naz);

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all_x1=all_x(:);

all_y1=all_y(:);

full_A=[ones(nr*naz,1) all_x1 all_y1 ];

ramp=full_A*coef;

ramp=reshape(ramp,naz,nr);

ramp=exp(1j*ramp);

topo_detilt=topo.*conj(ramp);

pic=mycolormap(amp,angle(topo_detilt),nr,naz);

figure

colormap jet

imagesc(pic)

xlabel(’range’)

ylabel(’azimuth’)

title(’corrected interferogram’)

saveas(gcf,’3.png’,’png’)

%% determine the ambiguity height as a function of range

ha=zeros(nr,1);

for i=1:nr

rg=r0+(i-1)*deltar;

costhetaflat=((hgt+re)^2+rg^2-re^2)/2/rg/(hgt+re);

thetaflat=acos(costhetaflat);

dffsin=wvl/2/B;

sindifftopo=dffsin+sin(thetaflat-alpha);

thetatopo=asin(sindifftopo)+alpha;

ha(i)=sqrt((hgt+re)^2+rg^2-2*rg*(hgt+re)*cos(thetatopo))-re;

end

ha(1)

ha(end)

figure

plot(1:nr,ha,’linewidth’,2)

grid on

xlabel(’range bin’)

ylabel(’ambiguity height (m)’)

saveas(gcf,’4.png’,’png’)

%% unwrap the flattened interferogram

fid=fopen(’intgram.uw’,’r’);

dat=fread(fid,[2*nr,inf],’float’,’ieee-le’);

phase_uw=dat(nr+1:end,:)’;

figure

imagesc(phase_uw)

colormap jet

pic=mycolormap(amp,phase_uw,nr,naz);

figure

imagesc(pic)

colorbar

phaselo=prctile(phase_uw(:),0);

phasehi=prctile(phase_uw(:),99);

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colormap jet

caxis([phaselo,phasehi])

saveas(gcf,’5.png’,’png’)

%% construct a DEM model

% choose a reference point where height is zero and correlation is good.

% figure

% imagesc(col(1:200,500:1000));

rgref=[550,693,792,974];

azref=[31,94,167,159];

phref=0;

for i=1:4

phref=phase_uw(azref(i),rgref(i))+phref;

end

phref=mean(phref);

DEM=nan(size(phase_uw));

for i=1:nr

DEM(:,i)=(phase_uw(:,i)-phref)/2/pi*ha(i);

DEM(DEM<0)=0;

end

figure

contourf(DEM,40)

colormap jet

colorbar

ch=colorbar; ylabel(ch,’Height,m’);

set(gca,’Ydir’,’reverse’)

axis image

saveas(gcf,’6.png’,’png’)

%% map of standard deviation of map

stdphase=((1-col)./2./col).^0.5;

stdDEM=nan(size(stdphase));

for i=1:nr

stdDEM(:,i)=ha(i)*stdphase(:,i)/2/pi;

end

figure

imagesc(stdDEM)

colormap jet

caxis([0 50])

axis image

colorbar

ch=colorbar; ylabel(ch,’Height Uncertainty,m’);

title(’uncertainty map’)

saveas(gcf,’7.png’,’png’)

%% rectify your map into ground range and azimuth coordinate

vel=veff*sqrt((hgt+re)/re);

deltaaz=vel/prf*re/(re+hgt)*nazlook;

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% determine the nearest ground range

demline=DEM(:,1);

cosgama=((hgt+re)^2-r0^2+(re+demline).^2)./2/(hgt+re)./(demline+re);

gama=acos(cosgama);

rnear=min(re*gama);

% determine the farest ground range

rmax=r0+(nr-1)*deltar;

demline=DEM(:,nr-1);

cosgama=((hgt+re)^2-rmax^2+(re+demline).^2)./2/(hgt+re)./(demline+re);

gama=acos(cosgama);

rfar=max(re*gama);

nrg=round((rfar-rnear)/deltaaz);

newDEM=nan(naz,nrg);

% determine the offsets in range (pixels) and reesample

rgoffset=zeros(naz,nr);

for i=1:nr

rsl=r0+(i-1)*deltar;

demline=DEM(:,i);

cosgama=((hgt+re)^2-rsl^2+(re+demline).^2)./2/(hgt+re)./(demline+re);

gama=acos(cosgama);

rgr=re*gama; % ground range

idxrgnew=((rgr-rnear)/deltaaz+1); % range index in new coordinates

rgoffset(:,i)=idxrgnew;

end

h = waitbar(0,’Range resampling’);

for j=1:naz

idxrgnew=rgoffset(j,:);

for i=1:nrg

idxintper=find((abs(idxrgnew-i)<1));

if ~isempty(idxintper)

newDEM(j,i)= mean(DEM(j,idxintper));

end

end

waitbar(j/naz)

end

close(h)

figure

contourf(newDEM,40)

colormap jet

colorbar

ch=colorbar; ylabel(ch,’Height,m’);

axis image

set(gca,’Ydir’,’reverse’)

axis off

saveas(gcf,’8.png’,’png’)

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%% interpolate to obtain good looking pictures

% linear interpolation

[idxaz,idxrg]=find(isnan(newDEM));

h = waitbar(0,’Interpolating’);

cull_window=1;

DEM_interp=newDEM;

for i=1:length(idxaz)

rneari=min(rgoffset(idxaz(i),:));

rfari=max(rgoffset(idxaz(i),:));

if idxrg(i)>rneari && idxrg(i)<rfari

index_az = max(1, idxaz(i)-cull_window):min(idxaz(i)+cull_window, naz);

index_rg= max(1, idxrg(i)-cull_window):min(idxrg(i)+cull_window, nr);

patch=newDEM(index_az,index_rg);

patch=patch(:);

patch=patch(isfinite(patch));

DEM_interp(idxaz(i),idxrg(i))=mean(patch(:));

end

waitbar(i/length(idxaz))

end

close(h)

figure

contourf(DEM_interp,40)

colormap jet

colorbar

ch=colorbar; ylabel(ch,’Height,m’);

set(gca,’Ydir’,’reverse’)

axis image

axis off

saveas(gcf,’9.png’,’png’)

Function to overlay phase image on the amplitude image

function pic=mycolormap(amp,phase,nr,naz)

amplow=prctile(amp(:),5);

amphi=prctile(amp(:),95);

amp(amp<amplow)=amplow;

amp(amp>amphi)=amphi;

scale=(amp-amplow)/(amphi-amplow);

phlow=prctile(phase(:),1);

phhi=prctile(phase(:),99);

% create a color table

colormap jet;

map=colormap;

stack=max(phase,phlow);

stack=min(phase,phhi);

colorstack=round((stack-phlow)/(phhi-phlow)*64);

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colorstack=max(colorstack,1);

colorstack=min(colorstack,64);

for k=1:naz

for kk=1:nr

red(k,kk)=map(colorstack(k,kk),1);

green(k,kk)=map(colorstack(k,kk),2);

blue(k,kk)=map(colorstack(k,kk),3);

end

end

pic(:,:,1)=red.*scale;

pic(:,:,2)=green.*scale;

pic(:,:,3)=blue.*scale;

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