GP265 /EE 355 Homework 8 (Final project 2)

Transcription

1 GP265 /EE 355 Homework 8 (Final project 2) March 14, 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: c r = nlooks r = m/s 2f s 2( Hz) = m. (1) Then the range ρ to each range bin ir is: ρ(ir) = ρ 0 + (ir 1) r, ir = 0, 1,..., nrange 1 (2) where ρ 0 = m is the range to the first bin and nrange = 1024 range bins. 1

2 The look angle θ z0 to each range bin for z = 0 (no topography) is: ( (h + θ z0 = cos 1 re ) 2 + ρ 2 (z + r e ) 2 ) ( (h + z=0 = cos 1 re ) 2 + ρ 2 r 2 ) e 2ρ(h + r e ) 2ρ(h + r e ) where h = m is the radar altitude and r e = m is the earth radius. Then the curved earth phase ϕ z0 for z = 0 (no topography) is: ϕ z0 = 4π λ ( ) B 2 2ρ B sin (θ z0 α) (3) (4) where λ = m is the radar wavelength, B = 307 m is the baseline length, and α is the orientation angle. To remove this curved earth phase from the complex interferogram S (displayed in Problem 1), I apply a phase correction T = Se iϕ z0, and here is the resulting phase arg(t ), which should have only topography effects. The fringe lines look more like elevation contour lines here: 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 range bins and azimuth bins I used least squares to find the slope of the fringes within the caldera (in radians/pixel), in both the range and azimuth directions, and apply 2

3 these phase corrections to the interferogram from Problem 2 to flatten the tilt from the interferogram. The flattened interferogram no longer has fringes within the caldera, and the fringes look more like elevation contour lines around the volcano. Note that the area I chose is 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 within the 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 (Bsin(θ α) Bsin(θf lat α)) λ (5) We can compute θf lat using the law of cosines using Eq. (3) and ha can be computed once we obtain the look angle θ from Eq. (5), again using the law of cosines. The ambiguity height deps on the range ρ, so here is a plot of the range depence: 3

4 At the reference (first) range bin, ha = m. At the last range bin, ha = m. 5. Unwrap the flattened interferogram using the snaphu program. I write out the flattened interferogram to a file called flatinterferogram and then run it through snaphu to unwrap the phase, writing it out to a file called snaphu.out. I then read in the snaphu.out file into MATLAB, and plot its phase, which is now unwrapped - no more fringes here, and the phase is no longer limited to the interval [ π, π]. 4

5 6. Using the ambiguity height h a from (4), convert the unwrapped phase values to heights. You have now created a digital elevation model (DEM) in radar coordinates. I use the ambiguity height h a from Eq. (5), which varies as a function of range ρ, to scale the unwrapped phase values ϕ flat from Problem 5 into height values z DEM. I choose several zero elevation reference pixels near the coast, and subtract their average phase ϕ ref from every point in the unwrapped interferogram. Since we are dealing with topography here, I choose to plot the contour of the DEM. z DEM = (ϕ flat ϕ ref ) h a (6) 2π The elevation of Mauna Loa is 4169 m, so our DEM looks reasonable. There are some heights below 0 m, but they were in the region over water so I would not expect to have useful height values here. 7. Map correlation into height error First, I display the correlation to check that I read it in properly. 5

6 Problem 7: Correlation azimuth bin range bin I use this equation to compute the height error σz as a function of correlation ρc (range is ρ): λρ sin θz0 1 ρc σz = (7) 4πB cos(θz0 α) 2ρc Here is a plot of the height error σz, ranging from 0 to 50 m. It makes sense that the height error is greatest at the locations where the correlation is lowest, such as the water in the top-right corner, as well as the slopes in the top-left and bottom-right corners. 6

7 8. Map DEM to ground coordinates To compute the ground range, which deps on both range and azimuth since I take the DEM heights into account, I first compute the look angle (compare to Eq. (3)): ( ) (h + re ) + ρ (zdem + re ) θz = cos (8) 2ρ(h + re ) Then I can compute the angle βz at the center of the earth, using the law of sines: ) ( ρ 1 βz = sin sin θz re (9) The ground range rg,z is then rg,z = re βz (10) I get the platform velocity v from the effective platform velocity vef f = m/s, radar altitude h = m, and earth radius re = m: h + re v = vef f = ( m/s) = m/s. (11) re Then the azimuth bin ground spacing azg is azg = nlooksaz v re = 12 = m prf h + re (12)

8 I generated a ground coordinates grid with a spacing of az g between adjacent ground range bins, in order to get square pixels. I resampled only in range, not in azimuth. There are 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 the image. 8

9 % 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= ; % wavelength, m prf= ; 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:,:); imagd=data(2:2:,:); 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:,:); amp2=data(2:2:,:); amp=amp1. ; fid=fopen( correlation ); data=fread(fid,[nr*2 inf], float ); fclose(fid); col=data(nr+1:,:); col=col. ; % display the phase of the interferogram 9

10 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); intgram_bg=exp(1j*phase_bg); imagesc(angle(intgram_bg)) colormap(jet) colorbar saveas(gcf, 2_2.png, png ) topo=intgram.*conj(intgram_bg); imagesc(angle(topo)) saveas(gcf, 2.png, png ) % pic=mycolormap(amp,angle(topo),nr,naz); % % 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); imagesc(testphase) 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; imagesc(testphase-reshape(ramppart,size(testsite))) % get the ramp for the whole interferogram [all_x,all_y] = meshgrid (1:nr, 1:naz); 10

11 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); 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; ha(1) ha() 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:,:) ; imagesc(phase_uw) pic=mycolormap(amp,phase_uw,nr,naz); imagesc(pic) colorbar phaselo=prctile(phase_uw(:),0); phasehi=prctile(phase_uw(:),99); 11

12 caxis([phaselo,phasehi]) saveas(gcf, 5.png, png ) %% construct a DEM model % choose a reference point where height is zero and correlation is good. % % 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; 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; contourf(dem,40) 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; imagesc(stddem) 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; 12

13 % 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; 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)); waitbar(j/naz) close(h) contourf(newdem,40) colorbar ch=colorbar; ylabel(ch, Height,m ); axis image set(gca, Ydir, reverse ) axis off saveas(gcf, 8.png, png ) 13

14 %% 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(:)); waitbar(i/length(idxaz)) close(h) contourf(dem_interp,40) 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 ; map=colormap; stack=max(phase,phlow); stack=min(phase,phhi); colorstack=round((stack-phlow)/(phhi-phlow)*64); 14

15 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); pic(:,:,1)=red.*scale; pic(:,:,2)=green.*scale; pic(:,:,3)=blue.*scale; 15