EE 355 / GP 265 Homework 2 Solutions Winter

Transcription

1 EE 355 / GP 265 Homework 2 Solutions Winter Chirp Compression (a) The bandwidth measured at 3 db down from the peak of the chirp spectrum is 9.5 MHz, which is reasonably close to the theoretical bandwidth BW = sτ = (1 12 Hz/s)(1 5 s) = 1 7 Hz = 1 MHz. 5 Problem 1a: Chirp spectrum 4 Chirp spectrum magnitude (db) Frequency (Hz) 1 7 1

2 (b) Compressing the chirp with the original reference signal gives the ideal impulse response. The first sidelobe is 13.5 db lower than the main lobe. 6 Problem 1b: Compressed signal, s = 1.e12 Hz/s 5 Compressed signal (db) Time (s) 1-7 (c) Compressing the chirp with 1% error in reference chirp slope (s = Hz/s): the first sidelobe is only 12.2 db below the main lobe, in effect decreasing the SNR. 6 Problem 1c: Compressed signal, s = 1.1e12 Hz/s 55 Compressed signal (db) Time (s) 1-7 2

3 Compressing the chirp with 3% error in reference chirp slope (s = Hz/s): the first sidelobe is 5.6 db below the main lobe. 6 Problem 1c: Compressed signal, s = 1.3e12 Hz/s 55 Compressed signal (db) Time (s) 1-7 % EE 355 HW 2 Problem 1 close all; clear; set(, defaultaxesfontsize, 16); s=1.e12; sr=1.e12;%1.1*1^12 or 1.3*1^12 for prob c tau=1.e-5; fs=1.e8; fc=; N=248; % Part a: Chirp and bandwidth si=makechirp(s,tau,fs,fc,1,n); S=fftshift(fft(si)); freq=linspace(-fs/2,fs/2,n); plot(freq,2*log1(abs(s))); xlabel( Frequency (Hz) ); ylabel( Chirp spectrum magnitude (db) ); title( Problem 1a: Chirp spectrum ); % Part b: Compress with perfect reference signal r=makechirp(sr,tau,fs,fc,1,n); 3

4 R=fftshift(fft(r)); Sc = S.*conj(R); rcomp=fftshift(ifft(fftshift(sc))); t=linspace(-123,124,n); plot(t/fs,2*log1(abs(rcomp)), linewidth,2); xlim([-4 4]*1^-7); ylim([-2 6]); xlabel( Time (s) ); ylabel( Compressed signal (db) ); title( Problem 1b: Compressed signal, s = 1.e12 Hz/s ); % Part c: Compress with different reference signal sr=[1.1e12 1.3e12]; srstr={ 1.1e12, 1.3e12 }; for k=1:length(sr) r=makechirp(sr(k),tau,fs,fc,1,n); R=fftshift(fft(r)); Sc = S.*conj(R); rcomp=fftshift(ifft(fftshift(sc))); t=linspace(-123,124,n); plot(t/fs,2*log1(abs(rcomp)), linewidth,2); xlim([-4 4]*1^-7); ylim([2 6]); xlabel( Time (s) ); ylabel( Compressed signal (db) ); title([ Problem 1c: Compressed signal, s = srstr{k} Hz/s ]); function chirp = makechirp(s,tau,fs,fc,start,n) %Function to compute chirp - reused in all problems %s: slope %tau: pulse length %fs: sample rate %fc: center frequency %start: location of chirp %n: the length of the chirp including zero dt=1/fs; npts=tau*fs; t=[-npts/2:npts/2-1]*dt; phase=pi*s*t.^2+2*pi*fc*t; chirp=[zeros(1,start-1) exp(1i*phase) zeros(1,n-length(phase)-start+1)]; 4

5 2. Separating multiple chirps (a) The chirps in the raw signal cannot be separated because they overlap in time. Each chirp is 1 points long (Npts = τ fs), so chirps starting at location 1, 4, and 5 will interfere with each other. 8 Problem 2a: Raw signal amplitude 7 6 Amplitude Time (s) 1-5 (b) The chirps no longer overlap after compression, so they are easy to separate. There is a peak at each chirp location (1, 4, and 5 samples), scaled properly according to each chirp s amplitude (1, 5, and 2, respectively). 5 Problem 2b: Compressed signal amplitude Amplitude Time (s) 1-5 % EE 355 HW 2 Problem 2 5

6 close all; clear; set(, defaultaxesfontsize, 16); s=1.e12; tau=1.e-5; fs=1.e8; fc=; N=248; dt=1/fs; % Part a: Raw signal amplitude vs. time r1=makechirp(s,tau,fs,fc,11,n); r2=5*makechirp(s,tau,fs,fc,41,n); r3=2*makechirp(s,tau,fs,fc,51,n); si=r1+r2+r3; t=linspace(,n*dt,n); plot(t,abs(si), linewidth,1.5); xlim([t(1) t()]); xlabel( Time (s) ); ylabel( Amplitude ); title( Problem 2a: Raw signal amplitude ); % Part b: Compressed signal amplitude r=makechirp(s,tau,fs,fc,1,n); S=(fft(si)); R=(fft(r)); Sc = S.*conj(R); rcomp=(ifft(sc)); plot(t,abs(rcomp), linewidth,1.5); xlim([t(1) t()]); xlabel( Time (s) ); ylabel( Amplitude ); title( Problem 2b: Compressed signal amplitude ); 6

7 3. Actual data (a) The ERS byte file has 124 lines, each line with 412 header bytes, followed by 986 data samples. The data samples were multiplied by 8 to fill the color table. Problem 3a: ERS data, byte file display 124 lines samples (b) Chirp spectrum for the ERS parameters. Chirp spectrum magnitude (db) 4 Problem 3b: ERS chirp spectrum Frequency (Hz) (c) Average range spectrum of ERS data. Compared to part (b), it has the same overall shape, but is tilted (due to the average value of the data not being exactly 15.5). 7

8 Average spectrum magnitude (db) 65 Problem 3c: Average range spectrum Frequency (Hz) 1 7 (d) Range-compressed image. Problem 3d: Range compressed image azimuth range % EE 355 HW 2 Problem 3 close all; clear; set(, defaultaxesfontsize, 16); % Part a: Display byte file nhdr = 412; nsamp = 1218; nlines = 124; fid=fopen( ersdata ); dat=fread(fid,[nsamp, nlines], uint8 );

9 fclose(fid); dat_disp = dat; dat_disp(nhdr+1:,:) = 8*dat_disp(nhdr+1:,:); % scale by 8, fill color table imagesc(dat_disp ); colormap( gray ); colorbar; xlabel( 1218 samples ); ylabel( 124 lines ); title( Problem 3a: ERS data, byte file display ); % Part b: ERS chirp spectrum s= e11; tau=37.12e-6; fs=18.96e6; fc=; N=(nsamp-nhdr)/2; % number of range samples in complex data r=makechirp(s,tau,fs,fc,1,n); R=fft(r); freq=linspace(-fs/2,fs/2,n); plot(freq,fftshift(2*log1(abs(r)))); xlabel( Frequency (Hz) ); ylabel( Chirp spectrum magnitude (db) ); title( Problem 3b: ERS chirp spectrum ); % Part c: average of 124 spectra A=dat(nhdr+1:nsamp,:); signal=(a(1:2:,:)-15.5)+1i*(a(2:2:,:)-15.5); S=fft(signal); % FFT each range line (column) avs=abs(s(:,1)); for ii=2:nlines avs=avs+abs(s(:,ii)); avs=avs/nlines; plot(freq,fftshift(2*log1(abs(avs)))); xlabel( Frequency (Hz) ); ylabel( Average spectrum magnitude (db) ); title( Problem 3c: Average range spectrum ); % Part d: Range compressed image Sc=zeros(size(S)); for jj=1:nlines Sc(:,jj)=S(:,jj).*conj(R. ); rcomp=(ifft(sc)); imagesc(abs(rcomp )); 9

10 colormap( gray ); axis image; colorbar; xlabel( range ); ylabel( azimuth ); title( Problem 3d: Range compressed image ); 4. I/Q and offset video processing (a) Here is the impulse response for the chirp signal, for an I/Q system. 6 Problem 4a: Impulse response, I/Q system 5 Compressed signal (db) Time (s) 1-5 (b) The impulse response of the offset video system should be identical to the impulse response for the I/Q system. There may be some slight numerical differences. 1

11 6 Problem 4b: Impulse response, offset video 5 Compressed signal (db) Time (s) 1-5 The bandwidth is BW = sτ = (1 11 Hz/s)(3 1 6 s) = Hz = 3 MHz (1) The minimum chirp frequency is: f min = f c BW 2 = 1 MHz 3 MHz 2 = 8.5 MHz (2) and the maximum chirp frequency is: f max = f c + BW 2 = 1 MHz + 3 MHz 2 = 11.5 MHz (3) % EE 355 HW 2 Problem 4 close all; clear; set(, defaultaxesfontsize, 16); % Part a: Impulse response, I/Q system s=1e11; tau=3e-5; fs=2e7; fc=1e7; 11

12 N=124; dt=1/fs; r=makechirp(s,tau,fs,fc,1,n); sig=makechirp(s,tau,fs,fc,1,n); R=(fft(r)); Sig=(fft(sig)); Rc=Sig.*conj(R); rcomp=fftshift(ifft(rc)); t=(-n/2:n/2-1)*dt; figure(1) plot(t,2*log1(abs(rcomp))); xlim([-1 1]*1e-5); ylim([-4 6]); title( Problem 4a: Impulse response, I/Q system ); xlabel( Time (s) ); ylabel( Compressed signal (db) ); % Part b: Impulse response, offset video % Reference chirp ref=makechirp(s,tau,fs,fc,1,n); REF=fft(ref); % Real chirp, twice the sample rate & array size si=real(makechirp(s,tau,2*fs,fc,1,n*2)); S=fft(si); S_side=S(1:N); % Save only positive frequencies (one sideband) % Compress the chirp R_c=S_side.*conj(REF); r_comp=fftshift(ifft(r_c)); figure(2); plot(t,2*log1(abs(r_comp)), r ); xlim([-1 1]*1e-5); ylim([-4 6]); title( Problem 4b: Impulse response, offset video ); xlabel( Time (s) ); ylabel( Compressed signal (db) ); 12

13 5. Sidelobe filtering For the chirp in problem 4, the peak sidelobe level ratio (PSLR) is db. The integrated sidelobe level ratio (ISLR) is db. Next, we weight the chirp according to W (f) = w + (1 w) cos (2π(f f c )/BW ), letting w vary from.4 to 1. Here is the plot of ISLR and PSLR as a function of w: The PSLR is minimized at w =.56, which is consistent with the equation for the Hamming window (w =.54) that best suppresses the amplitude of the first side lobe. The ISLR is minimized at w =.52, which is consistent with the equation for the Hann window that minimizes total side lobe energy. % EE355 pb5: sidelobe filtering clear;close all;clc set(,'defaultaxesfontsize',16) s=1e11; % Chirp slope, Hz/s tau=3e-6; % Pulse length, s fs=2e6; % Complex sample rate, Hz fc=1e6; % Center frequency (Carrier frequency) Nsample=ceil(fs*tau); % Number of samples fftlen=2^ceil(log(nsample)/log(2)); dt=1/fs; % Sample spacing in time domain df=fs/fftlen; % Sample spacing in frequency domain f=(:fftlen-1)*df; t=(-fftlen/2:fftlen/2-1)*dt; 13

14 %% construct ref chirp ref_chirp=makechirp(s,tau,fs,fc,1,fftlen); spect_chirp=fft(ref_chirp); %% construct impulse response impres=fftshift(ifft(spect_chirp.*conj(spect_chirp))); figure plot(t,2*log1(abs(impres)),'linewidth',2) xlabel('t') ylabel('compressed signal,db') title('impulse response') grid on; xlim([-2,2]*1e-7) % calculate PSLR and ISLR impres_pw = abs(impres).^2; figure plot(t,impres_pw) xlim([-2,2]*1e-7) %% impulse response weighting ww=.4:.2:1; bw=s*tau; for i=1:length(ww) w=ww(i); weight=w+(1-w)*cos(2*pi/bw.*(f-fc)); spect_newref=spect_chirp.*weight'; newimpres=fftshift(ifft(spect_chirp.*conj(spect_newref))); newimprespw=abs(newimpres).^2; [islr(i),pslr(i)]=property_imp(newimprespw); figure plot(ww,islr,'linewidth',2) hold on plot(ww,pslr,'linewidth',2) xlabel('w') ylabel('sidelobe level ratio (db)') title('islr,pslr vs. w') leg('islr','pslr') saveas(gcf,'pb5_pslr_islr','tiff') [v,idx1]=min(islr); ww(idx1) [v,idx2]=min(pslr); ww(idx2) 14

15 function [ISLR,PSLR]=property_imp(imprespw) % this function computes ISLR and PSLR of a given impulse response (power) % it also requires that the imprespw is symmetric about the peak n = length(imprespw); % first, find the position of the peak [pkv,pkidx]=max(imprespw); % then, find the first null point location for m=pkidx+1:n-1 if(imprespw(m)<imprespw(m-1) &&... imprespw(m)<imprespw(m+1)) null_indx = m; break; % lastly, find the first sidelobe location for m=null_indx+1:n-1 if(imprespw(m)>imprespw(m-1) &&... imprespw(m)>imprespw(m+1)) sidelobe_indx = m; break; % PSLR PSLR = imprespw(sidelobe_indx)/imprespw(pkidx); PSLR = 1*log1(PSLR); % ISLR sidelobepw = sum(imprespw(null_indx:n)); pkpw = sum(imprespw(pkidx:null_indx)); ISLR = sidelobepw/pkpw; ISLR = 1*log1(ISLR); return 15

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