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Re: [Discuss-gnuradio] OFDM demodulation problem / example video
From: |
Achilleas Anastasopoulos |
Subject: |
Re: [Discuss-gnuradio] OFDM demodulation problem / example video |
Date: |
Wed, 05 Apr 2006 22:06:07 -0400 |
User-agent: |
Mozilla Thunderbird 1.0.2 (Windows/20050317) |
Jens,
I would like to elaborate on some of your comments.
I am using the following notation:
N fft size (2048)
G guard interval (504)
v0 fractional frequency offset
f0=v0 T absolute frequency offset
tau0 fractional time offset
t0=tau0 T absolute timne offset
To sum up:
Frequency offset results in ICI.
This is "white noise" interference.
Correct: v0 will result in a time offset of N*v0
which will manifest itself as ICI.
There is MORE to it.
It ALSO introduces an unknown phase offest (constant in each subchannel):
Consider the i-th component of consequtive OFDM symbols.
They are effectively rotated by multiples of 2*pi*(N+G)*v0.
This is the phase accumulated over an OFDM symbol duration.
Although the linear phase increase will be compensated by D-QPSK, there
will be a constant phase rotation equal to 2*pi*(N+G)*v0 in all
demodulated QPSK symbols.
BTW, If you make sure that you estimate v0 with an accuracy of 1e-6
then the time offset is on the order of 1e-3 (negligible) and the
frequency offset on the order of 1 degree (negligible).
BTW, because of the way you have coded your frequency correction (in
your Matlab code) even perfect estimation of v0 will still result in the
above phenomenon.
If you correct for v0 INDIVIDUALLY in each OFDM symbol (as you do in
your code), then--assuming perfect estimation of v0--there will be two
effects:
1) within each OFDM symbol there will be no phase variation since you
have corrected it already
2) each consequtive sample in the i-th subchannel will be advanced by
2*pi*(N+G)*v0 radians. This is the phase accumulated over an OFDM symbol
duration.
This is why it is important to derotate your entire frame at once.
Having said that, it seems that even if this is done, the problem you
observe is still there!
Time offset results in
a) ISI, if offset results in taking values from other symbols.
This is also "white" interference.
b) phase shift, if offset results in taking values from the
cyclic prefix. This results in a "circle" structure, as the
phase shift is higher for subcarriers of higher frequency.
The time offset effect b) is removed by differential demodulation, since
the phase shift will be the same.
I agree with the above.
Also, the above assumes that the offset tau0 remains fixed for
each OFDM symbol. Otherwise, the comment made by Bob is valid.
However, I think the DAB system is designed to operate under this
assumption, since they have a single pilot in the begining of the frame.
The effect I am seeing is a constant phase shift for all subcarriers,
different for each OFDM symbol. Nothing that could be explained with a
frequency offset or a time offset.
I am clueless.
I attach a piece of Matlab code that simulates both the Tx and Rx.
Everything mentioned above can be demonstrated in this controled experiment.
I still do not understand why the real-data application does not work.
Achilleas
function [f0 ,t0] = freq_time_est(ff,M,sync,wframe)
rec=zeros(M,2);
for t=1:M,
%t
out=zeros(1,length(ff));
for k=1:length(ff)
f=ff(k);
s=sync.*exp(j*2*pi*f*[1:length(sync)]);
out(k)=mean(s.*conj(wframe(t:t+length(sync)-1)));
end
[value f0i]=max(out);
rec(t,:)=[value f0i];
end
[gvalue t0]=max(rec(:,1));
f0=ff(rec(t0,2));
clear all;
%close all;
Q=76;
%data=zeros(1537,Q-1);
data=floor(rand(1537,Q-1)*4);
qpsk=exp(j*2*pi*data/4);
qpsk(769,:)=zeros(1,Q-1);
%dqpsk modulation
dqpsk=zeros(1537,Q);
sync=floor(rand(1537,1)*4);
%dqpsk(:,1)=ones(1537,1);
dqpsk(:,1)=exp(j*2*pi*sync/4);
dqpsk(769,1)=0;
for k=2:Q,
dqpsk(:,k)=dqpsk(:,k-1).*qpsk(:,k-1);
end
% initial phases of the oscilators
phase0=2*pi*rand(1537,1);
dqpsk=dqpsk.*(exp(j*phase0)*ones(1,Q));
% ofdm modulation
frame=zeros(504+2048,Q);
for l=1:Q
a=dqpsk(:,l).';
aa=[zeros(1,256) a zeros(1,255)];
aaa=fftshift(aa);
am=ifft(aaa); % OFDM symbol
amp=[am(end-503:end) am].'; % with prefix
amp=amp/sqrt(mean(abs(amp).^2)); % normalize power to 1
frame(:,l)=amp;
end
msync=frame(:,1); % modulated sync (1st ofdm symbol with prefix) for later use
D=1000;
frame_h=[zeros(1,D) reshape(frame,1,Q*(504+2048)) zeros(1,D)]; % zeros in head
and tail
%=======================================================
%channel
% ISI
% Le=300;
% h=randn(1,Le)+j*randn(1,Le);
% h=h./sqrt(mean(abs(h).^2));
% frame_h=conv(h,frame_h);
% frequency error
df=1.0e-5;
f=frame_h.*exp(j*2*pi*df*[1:length(frame_h)]+j*rand(1,1)*2*pi);
% timing error (uncomment this if you want to introduce timing error)
% M=11; % max integer error [-M,M] samples
% ts0=0;
% while ts0==0
% ts0=floor(rand(1,1)*M)-floor((M-1)/2);
% end
% ts0; % so the timing error is ts0 samples (so far)
%
% L=5; % resolution of fractional errors
% tf0=0;
% while tf0==0
% tf0=floor(rand(1,1)*L)-floor((L-1)/2);
% end
% tf0; % so the timing error is tt0/L samples (fractional)
%
% tt0=ts0*L+tf0; % total timing error
% [ts0 tf0/L tt0/L]
% framei_h=resample(f,L,1);
% t0=D*L+1+tt0;
% rframe_h=framei_h(1:L:end);
% rframe_h=rframe_h(1:D+Q*(504+2048)+D); % received signal
% %=======================================================
% %frequency/timing estimation (correlation with msync)
% % we do not use oversampled receiver (easy to extend)
% % coarse
% Lag=floor(M/2)+2;
% tframe=rframe_h(D+1-Lag:D+504+2048+Lag); % you know the sync start is close
to D+1
% [f0 t0]=freq_time_est(-1e-2:1e-4:1e-2,2*Lag,msync.',tframe);
% [f0 t0]
% rframe_h=rframe_h.*exp(-j*2*pi*f0*[1:length(rframe_h)]);
% rframe_h=rframe_h(t0-Lag:end); % recenter so the new estimate of the sync
begining is in D+1
% % fine
% Lag=5;
% tframe=rframe_h(D+1-Lag:D+504+2048+Lag);
% [f0 t0]=freq_time_est(-1e-4:1e-5:1e-4,2*Lag,msync.',tframe);
% [f0 t0]
% rframe_h=rframe_h.*exp(-j*2*pi*f0*[1:length(rframe_h)]);
% rframe_h=rframe_h(D+t0-Lag:D+t0-Lag+(504+2048)*Q-1); % sync starts here
% comment this line if you introduce timing errors
rframe_h=f(D+1:D+Q*(504+2048)); % received signal (correct timing)
rframe=reshape(rframe_h,504+2048,Q);
%demod
dqpsk_est=zeros(1537,Q);
for l=1:Q
y = fftshift(fft(rframe(504+1:end,l),2048));
z = y(257:1793);
z = z / sqrt(mean(abs(z).^2));
dqpsk_est(:,l)=z;
end
qpsk_est=zeros(1537,Q-1);
data_est=zeros(1537,Q-1);
for l=1:Q-1
dem=dqpsk_est(:,l+1).*conj(dqpsk_est(:,l));
qpsk_est(:,l)=dem;
% plot(real(dem), imag(dem), 'o');
% axis([-3 3 -3 3]);
% drawnow
% pause
end
plot(qpsk_est(:,:).*exp(-j*0*pi/180),'o');grid on;
drawnow;
% calc effective snr
qq=reshape(qpsk_est,1,1537*(Q-1));
ii=find(real(qq)>0.5);
q=qq(ii);
n=q-mean(q);
ii=find(real(qq)<-0.5);
q=qq(ii);
n=[n q-mean(q)];
ii=find(imag(qq)>0.5);
q=qq(ii);
n=[n q-mean(q)];
ii=find(imag(qq)<-0.5);
q=qq(ii);
n=[n q-mean(q)];
snr=-10*log10(mean(abs(n).^2))
- Re: [Discuss-gnuradio] OFDM demodulation problem / example video, (continued)
- Re: [Discuss-gnuradio] OFDM demodulation problem / example video, Robert McGwier, 2006/04/04
- Re: [Discuss-gnuradio] OFDM demodulation problem / example video, Prateek Dayal, 2006/04/05
- Re: [Discuss-gnuradio] OFDM demodulation problem / example video, Robert McGwier, 2006/04/05
- Re: [Discuss-gnuradio] OFDM demodulation problem / example video, Jens Elsner, 2006/04/05
- Re: [Discuss-gnuradio] OFDM demodulation problem / example video, Robert McGwier, 2006/04/05
- Re: [Discuss-gnuradio] OFDM demodulation problem / example video, Jens Elsner, 2006/04/06
- Re: [Discuss-gnuradio] OFDM demodulation problem / example video, Jens Elsner, 2006/04/05
Re: [Discuss-gnuradio] OFDM demodulation problem / example video, Jens Elsner, 2006/04/04
Re: [Discuss-gnuradio] OFDM demodulation problem / example video, Achilleas Anastasopoulos, 2006/04/04
Re: [Discuss-gnuradio] OFDM demodulation problem / example video,
Achilleas Anastasopoulos <=