differentiation of time series data

178 Ansichten (letzte 30 Tage)
Samson
Samson am 15 Feb. 2021
Kommentiert: Walter Roberson am 21 Feb. 2021
I have a time series data 'X 'of size 20000 X 50. My step size dt= 0.05
how do I find the velocity of these data
  2 Kommentare
Walter Roberson
Walter Roberson am 15 Feb. 2021
Extract the data and use gradient() ?
Samson
Samson am 15 Feb. 2021
Bearbeitet: Samson am 15 Feb. 2021
how do I extract, please? I had something like this but not correct as it is not taken the difference across time:
phdiff=diff(X');
temp=(phdiff./dt);

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 15 Feb. 2021
If ts is the timeseries() variable then
X = ts.Data;
extracts the data.
With your data being 20000 x 50 my guess is that you have 20000 measurements for each of 50 entities, rather than 50 measurements for each of 20000 entities. If I am correct, then you would want the y gradient:
xstep = 1; %doesn't really matter, we are going to ignore
tstep = 0.5; %does matter
[~, ygrad] = gradient(X, xstep, tstep); %horizontal step first, vertical step second
  18 Kommentare
Samson
Samson am 21 Feb. 2021
It is the row that correspond to time. I am taking the velocity across the row which is a dimensionless time uniit. The values are phase values of oscillator distribeted in the interval (0,2pi]. I would finally need the mean of the phase velocity as there all independent rows
Walter Roberson
Walter Roberson am 21 Feb. 2021
xstep = 1; %doesn't really matter, we are going to ignore
tstep = 0.5; %does matter
Xunwrapped = unwrap(X, [], 2);
[~, velocity_gradient] = gradient(Xunwrapped, xstep, tstep); %horizontal step first, vertical step second

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Tags

Produkte


Version

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by