Time Delay Estimation for Monotonically Increasing Funcitons

Question

Stephen 2017-7-5

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/347518-time-delay-estimation-for-monotonically-increasing-funcitons

评论： Stephen 2017-7-6

I am trying to use cross correlation to find the time delay between two signals. Both signals follow a logarithmic growth pattern, but are sampled for the same amount of time and at the same sample rate. The only difference between the two is: 1) random noise, and 2) the time delay of when the delayed signal starts. When trying traditional cross correlation here it is failing to find the shift and only reports a maximum at zero delay. Please see the attached matlab code for full details

 for i = 1:5
    signal1 = 6 + ...
    (-0.1 + (0.1+0.1).*rand([1399,1])); %base signal + noise
    %simulates the base noisefloor
    timeDelay = randi(700);%quantify a random delay
    exponentialContrib = log(1:1399) + ...
        (-0.1 + (0.1+0.1).*rand([1399,1]))';
    signal1(timeDelay:end,1) = signal1(timeDelay:end,1)' + ...
        exponentialContrib(1,1:1399-timeDelay+1);
    %Second signal with the exponential starting right away
     signal2 = 6 + ...
        (-0.1 + (0.1+0.1).*rand([1399,1]));
    %aligned to zero time
    timeDelay2 = 1;
    exponentialContrib = log(1:1399) + ...
        (-0.1 + (0.1+0.1).*rand([1399,1]))';
    signal2(timeDelay2:end,1) = signal2(timeDelay2:end,1)' + ...
        exponentialContrib(1,1:1399-timeDelay2+1);
    %Find the delayed time     
    [acor,lag] = xcorr(signal2',signal1');
    [~,Idx] = max(abs(acor));
    lagTimeCalculated = lag(Idx);
    figure; plot(signal1); hold on; plot(signal2);
    legend('signal 1', 'signal 2'); hold off;
    pause(2);
    disp(['Calculated: ' num2str(lagTimeCalculated) ...
        '   Actual: ' num2str(timeDelay) '  , ' ...
        num2str(lagTimeCalculated - timeDelay)]);
end

I believe I know why this is, I am just unsure about how to fix it. As the non-delayed signal is slid through all possible delays, the large-valued logarithmic tail goes from being multiplied by another, time-delayed logarithmic function (as is the case when shift = 0) to being multiplied partially by zeros. This ensures that the correlation intensity will not be greater when both signals truly overlap.

Any comments or suggestions would be greatly appreciated

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Answer 1

David Goodmanson 2017-7-6

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https://ww2.mathworks.cn/matlabcentral/answers/347518-time-delay-estimation-for-monotonically-increasing-funcitons#answer_273058

编辑：David Goodmanson 2017-7-6

在 MATLAB Online 中打开

Hi Stephen,

I don't know how a slope detector on a triggering 'scope works but here is an attempt at something similar. Your signal seems ideal for this and something with a softer leading edge might not work as well. I added some extra baseline to both signals since the method requires it. I don't have xcorr so I use a homemade function called xcor. You will have to go in and change those to xcorr, and I left it that way as a reminder that you may have to swap the two inputs to the function to make it work.

After detection there are two methods to get the time delay and they seem to agree pretty well.

    % your signals with baseline
    baseline = 32;
    signal1 = 6 + ...
    (-0.1 + (0.1+0.1).*rand([1399,1])); %base signal + noise
    %simulates the base noisefloor
    timeDelay = baseline + randi(700);%quantify a random delay
    exponentialContrib = log(1:1399) + ...
        (-0.1 + (0.1+0.1).*rand([1399,1]))';
    signal1(timeDelay:end,1) = signal1(timeDelay:end,1)' + ...
        exponentialContrib(1,1:1399-timeDelay+1);
    %Second signal with the exponential starting right away
     signal2 = 6 + ...
        (-0.1 + (0.1+0.1).*rand([1399,1]));
    %aligned to zero time
    timeDelay2 = baseline;
    exponentialContrib = log(1:1399) + ...
        (-0.1 + (0.1+0.1).*rand([1399,1]))';
    signal2(timeDelay2:end,1) = signal2(timeDelay2:end,1)' + ...
        exponentialContrib(1,1:1399-timeDelay2+1);

new part:

     s1 = signal1;               % easier names
     s2 = signal2;
     t = 1:length(s1);
     figure(1)
     plot(t,s1,t,s2)
     % slope detector
     detec = zeros(size(s1));
     detec(1:10) = -1;                   % 10 somewhat arbitrary but works; 
     detec(11:20) = 1;                   % 20 must be < length of baseline
     [d1 lag] = (xcor(s1,detec));
      d2  =     (xcor(s2,detec));     
     llag = length(lag);                 % get rid of unwanted detections associated
     ind = (llag+1)/2:llag-baseline;     % with the ends of the time record
     d1 = d1(ind);
     d2 = d2(ind);
     lag = lag(ind);
     figure(2)
     plot(lag,d1,lag,d2)
     % find the delay time     
     % method 1
     [~, T1] = max(abs(d1));
     [~, T2] = max(abs(d2));
     Tdelay1 = T1-T2                      % delay of s1 (blue) compared to s2 (red)
     % method 2, cross correlation
     [dcor lag] = xcor(d1,d2);
     [~, ind] = max(abs(dcor));  
     Tdelay2 = lag(ind)   
     figure(3)
     plot(lag,dcor)