How to accurately compute the phase lag between two time series with same sampling frequency.
90 次查看(过去 30 天)
显示 更早的评论
Hi everyone,
My data set consists of two time series. Both the series have a consist time lag but that varies (a bit) with tiem as well. So, i am interested to compute that variation in phase lag between two series.
How my data looks:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1148185/image.png)
What i get after cross-correlation approcah:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1148190/image.png)
Here is the dataset detail:
clear all
clc
s=load("Data_timeseries.csv");
ev_time=datenum(s(:,1),s(:,2),s(:,2),s(:,4),s(:,5),s(:,6)); % time
ser1=s(:, 7); % series 1
ser2=s(:, 8); % series 2
Data is also attached, Thank you!
0 个评论
采纳的回答
Chunru
2022-10-7
编辑:Chunru
2022-10-7
s=load(websave("Data_timeseries.csv", "https://www.mathworks.com/matlabcentral/answers/uploaded_files/1148175/Data_timeseries.csv"));
ev_time=datenum(s(:,1),s(:,2),s(:,2),s(:,4),s(:,5),s(:,6)); % time
ser1=s(:, 7); % series 1
ser2=s(:, 8); % series 2
t = (0:length(ser1)-1);
ser1env = envelope(ser1, 100, 'peak'); % need envolope detection
yyaxis left
plot(t, ser1, 'r', t, ser1env, 'g');
yyaxis right
plot(t, ser2, 'b')
ns = length(ser1);
lwin = 2000;
overlap = 1000;
i1 = 1;
c = 1;
while true
i2 = i1+lwin-1;
if i2>ns
break
end
% remove mean before xcorr
[r,lags] = xcorr(ser1env(i1:i2)-mean(ser1env(i1:i2)), ser2(i1:i2)-mean(ser2(i1:i2)), 'coeff');
[rmax, imax] = max(r);
res(c, :) = [i1, lags(imax)];
i1 = i1 + (lwin-overlap);
c = c + 1;
end
res
% remove mean before xcorr
[r,lags] = xcorr(ser1env-mean(ser1env), ser2-mean(ser2), 'coeff');
figure
plot(lags, r);
grid on
[rmax, imax] = max(r);
tmax = lags(imax);
tmax, rmax
hold on
plot(tmax, rmax, 'ro')
8 个评论
William Rose
2022-10-7
@Adnan Barkat, The plot you shared above of phase lag versus "Time (Hours)" does not have numbers on the y axis, so it is impossible to tell what the scale is. Please don't do this, because it makes it hard to interpret what you have done or want done. (Some of your other shared plots also lacked numbers on the axes.)
I disagree with your assertion that "we can at least calculate the phase lag for each hour or 2, 3 etc hours". The rason I disagree is that the signal ser2 (blue in your first plot) has a period of roughly 12 hours. The phase difference between two signals of approximately equal frequency can be estimated once per cycle, at best, by comparing zero crossing times. Estimates that are more frequent than 1 per cycle do not make sense in my opinion. The values obtained from such esitmates are more likely to reflect the non-sinusoidal nature of the respective signals than the actual phase differences. The whole concept of phase lag is based on the idea of comparing 2 sinusoids of equal frequency. Therefore one phase lag esitmate per cycle is the most frequent that is justified. The plot of ser2 shows 19.5 cycles in 10 days, so 1 cycle is 10/19.5 = 0.513 days long, or 12.31 hours.
I recommend that you make the time units clear in your code. You calculate ev_time in your code with datenum(). datenum produces times in units of days. @Chunru makes a time vector comprised of integers. Therefore his time vector measures time in samples. Since the sampling rate is 1 per minute, @Chunru 's time vector are also in units of minutes.
更多回答(2 个)
William Rose
2022-10-7
I will address your quesiton about phase, but first I have some quesitons. Then I adress phase, below.
Your upper plot does not show x or y axis values. When I plot the data, it does not look like your plot.
s=load("Data_timeseries.csv");
t=datenum(s(:,1),s(:,2),s(:,3),s(:,4),s(:,5),s(:,6)); % time (days)
t=(t-t(1)); %remove initial time offset
ser1=s(:, 7); % series 1
ser2=s(:, 8); % series 2
plot(t,ser1,'-g',t,ser2,'-b')
legend('Ser1','Ser2'); xlabel('Time (days)')
Plot of your data, above, does not look like your plot.
Maybe if I remove the mean value from each:
plot(t,ser1-mean(ser1),'-g',t,ser2-mean(ser2),'-b')
legend('Ser1-mean','Ser2-mean'); xlabel('Time (days)')
The plot of your mean-adjusted data, above, looks more like the plot you shared, but still not the same. The ser1 data in your file is a lot noisier than the green trace in your plot. Please comment on this difference.
How did you compute and plot cross correlation? I am asking because the cross corr plot you shared does not look like what I would expect. A cross correlation like what you showed is what happens if you do not first remove the mean from both signals. See plots below.
[rxy,lags]=xcorr(ser1,ser2,'normalized'); %compute cross correlation
subplot(211), plot(lags,rxy,'-b'); %plot cross correlation
xlabel('Lags (samples)'); ylabel('r_{ser1,ser2}') %add labels
dt=(t(end)-t(1))/(length(t)-1); %sampling interval (days)
subplot(212), plot(lags*dt,rxy,'-b'); %plot cross correlation
xlabel('Lag (days)'); ylabel('r_{ser1,ser2}') %add labels
The only difference in the plots above is the x-axis scaling. Neither one matches your plot. That's why I'm curious about how you made your cross corr plot. Both of the plots above are not the true cross correlation, because the mean was not removed. The true cross correlation is below.
[rxy,lags]=xcorr(ser1-mean(ser1),ser2-mean(ser2),'normalized'); %compute cross corr
figure; subplot(211); plot(lags*dt,rxy,'-b') %plot cross corr
xlim([-1 1]); title('Cross Correlation'); grid on %add title, grid
xlabel('Lag (days)'); ylabel('r_{ser1,ser2}') %add labels
subplot(212); plot(lags*dt,rxy,'-b') %plot cross corr
xlim([0 .5]); title('Cross Correlation'); grid on %add title, grid
xlabel('Lag (days)'); ylabel('r_{ser1,ser2}') %add labels
This is more reasonable and what I would expect.
----------
Now, regarding phase between the signals. Do you want one estimate of the phase difference between ser1 and ser2, based on this recording? Or do you want a new phase difference estimate for every cycle of the waves? How do you plan to use this information?
I assume you want a single phase estimate. Otherwise there would have been no reason for you to plot the cross correlation.
The plots of the signals, above, show that each is roughly sinusoidal. The frequency is about 19.5 cycles in ten days, so f~=19.5/10=1.95.
The cross correlation xcorr(ser1,ser2) has a positive peak at time t=+0.24, evident on the cross corr plot above. This means ser1 lags ser2 by 0.24 seconds.
The phase difference, in radians, is
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1148340/image.png)
or
, where td is the lag, in units of time, and f is the frequency, in cycles per unit time.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1148345/image.png)
3 个评论
William Rose
2022-10-7
@Adnan Barkat, thank you for those details replies. I am glad that my solution below is what you are looking for.
William Rose
2022-10-7
There is a nice reuslt below, it just takes a while to get to it.
To estimate phase lag of ser1 relative to ser2, once per cycle, you will want to compare the zero crossing time differences. Use either positive or negative crossings. Phase lag estimates more frequent than once per cycle do not make sense. The phase difference, in radians, is
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1148470/image.png)
where
is the zero crossing time difference, in units of time, and f is frequency in cycles per unit time.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/1148475/image.png)
s=load("Data_timeseries.csv");
t=datenum(s(:,1),s(:,2),s(:,3),s(:,4),s(:,5),s(:,6)); % time (days)
t=(t-t(1)); %time (days) (initial offset removed)
ser1=s(:, 7)-mean(s(:,7)); % series 1 minus its mean
ser2=s(:, 8)-mean(s(:,8)); % series 2 minus its mean
To estimate the zero crossing times, you will need to remove drift (high pass filter) and noise (low pass filter). Use filters with zero phase for this, because filters with non-zero phase could lead to errors in the phase lag esitmates. Zero phase filters include Matlb's filtfilt() and non-causal FIR filters that are symmetric about the middle.
Visual inspection of the signals shows 19.5 cycles in 10 days, so the dominant frequency is 1.95 cycles/day. We can do low pass and high pass filtering simultaneously with a bandpass filter. Therefore we make a bandpass with cutoff frequenencies 1/day and 4/day.
N=length(t);
dt=(t(end)-t(1))/(N-1); %sampling interval (days)
fs=1/dt; fNyq=fs/2; %sampling frequency, Nyquist frequency (cycles/day)
f1=1; f2=4; %bandpass corner frequencies (cycles/day)
[z,p,k]=butter(4,[f1,f2]/fNyq,'bandpass'); %4th order Butterworth bandpass filter coefficients
[sos,g]=zp2sos(z,p,k); %sos representation of the filter
y1=filtfilt(sos,g,ser1); %apply zero phase filter to ser1
y2=filtfilt(sos,g,ser2); %apply zero phase filter to ser2
Plot the signls pre and post filtering to make sure the filtering process is working as desired:
subplot(211), plot(t,ser1,'-g',t,y1,'-k');
xlabel('Time (days)'); legend('Ser1=raw','y1=filtered'); grid on
subplot(212), plot(t,ser2,'-b',t,y2,'-k');
xlabel('Time (days)'); legend('Ser2=raw','y2=filtered'); grid on
The plots above are reassuring. The filtered signals look like what we expect and desire.
Find the indices of the upward zero crossings of the filtered signals with an ingenious function from @Star Strider
zci = @(v) find(diff(sign(v))>0); %define function: returns indices of +ZCs
izc1=zci(y1); %find indices of + zero crossings of y1
izc2=zci(y2); %find indices of + zero crossings of y2
ZeroX = @(x0,y0,x1,y1) x0 - (y0.*(x0 - x1))./(y0 - y1); % Interpolated x value for Zero-Crossing
ZT1 = ZeroX(t(izc1),y1(izc1),t(izc1+1),y1(izc1+1));
ZT2 = ZeroX(t(izc2),y2(izc2),t(izc2+1),y2(izc2+1));
Plot the results:
figure; subplot(211);
plot(t,y1,'-b',ZT1,zeros(size(ZT1)),'r+');
title('y1=filtered ser1, and + zero crossings'); grid on
subplot(212); plot(t,y2,'-b',ZT2,zeros(size(ZT2)),'r+');
title('y2=filtered ser2, and + zero crossings'); grid on
xlabel('Time (days)')
Compute the differences between the zero crossing times, and convert those difference to phases, and plot them versus time.
tzcd=ZT1-ZT2(1:end-1); %zero crossing time differeces (days)
fdom=19.5/10; %dominant frequency (cycles/day)
phz=360*fdom*tzcd; %phase lag of y1 relative to y2 (degrees)
figure; plot(ZT2(1:end-1),phz,'-rx');
xlabel('Time (days)'); ylabel('Lag (degrees)'); title('Phase Lag v. Time')
ylim([0,360]); set(gca,'ytick',0:90:360); grid on
This looks good. The mean phase lag above is a bit less than 180 degrees. The average phase lag for the entire signal, determined in my earlier answer, using the cross correlation, was 168 degrees. So the cycle-by-cycle results are consitent with the overall result.
4 个评论
William Rose
2022-10-8
@Adnan Barkat, please send me a secure email, by clicking on the WR circle nex to my messages, then click on the envelope in the pop-up that appears. Thanks.
另请参阅
产品
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!