How to plot the time series data after remove the phase lag?

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Andi
Andi 2022 年 10 月 24 日
コメント済み: Mathieu NOE 2022 年 10 月 26 日
Hi everyone,
My data set consists of two-time series, I have computed the phase lag value for the data set which is 170-degree. Now, I want to subtract this phase lag value and plot the time series, but have no idea how can I do this. May someone help out me here?
Data and script is attached:
r=readmatrix('data.csv');
figure(1)
x0=10;
y0=10;
width=550;
height=150
set(gcf,'position',[x0,y0,width,height])
yyaxis left
plot(r(:,1),r(:,2),'-b')
yyaxis right
plot(r(:,1),r(:,3),'-r');
xlabel('Time (days)'); legend('SER1','SER2'); grid on

採用された回答

Mathieu NOE
Mathieu NOE 2022 年 10 月 24 日
hello
I assume this is the result you are looking for
for me 170° is almost a polarity inversion, so I changed the y to -y and used also xcorr to obtain the best time alignment between the two series
r=readmatrix('data.csv');
t = r(:,1);
x = r(:,2);
y = r(:,3);
dt = mean(diff(t));
[c_a, lag_a] = xcorr(x,-y);
[~, i_a] = max(c_a);
t_a = lag_a(i_a)*dt; % lag SER2 vs SER1 (days)
ty = t+t_a;
ind = ty>=0;
ty = ty(ind); % consider only data for t>=0
yy = -y(ind);
figure(1)
x0=10;
y0=10;
width=550;
height=150
set(gcf,'position',[x0,y0,width,height])
yyaxis left
plot(t,x,'-b')
yyaxis right
plot(ty,yy,'-r');
xlabel('Time (days)'); legend('SER1','SER2'); grid on
  13 件のコメント
Andi
Andi 2022 年 10 月 25 日
編集済み: Andi 2022 年 10 月 26 日
@Mathieu NOE Thank you very much for your thoughtful suggestions. In the last version of the code, I define limits for zero crossing because most of the time the number of zero-crossings is not equal, and then we need to decide which one we should delete to equalize the number of zero-crossing points. But in your version, there is no option for the such a particular scenario.
Mathieu NOE
Mathieu NOE 2022 年 10 月 26 日
ok understood
all the best

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