Fitting real data curve to a homogeneous diff

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Daniel Sánchez Muñoz 2023-5-16

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https://ww2.mathworks.cn/matlabcentral/answers/1965229-fitting-real-data-curve-to-a-homogeneous-diff

评论： Star Strider 2023-5-28

Screenshot_2023-05-16-14-36-22-225-edit_com.android.chrome.jpg

Hello! I measured accelerations with matlab Mobile and I encountered the problem that the time step was not homogeneous and therefore the measured frequency. Therefore, I created a homogeneous diff (blue curve) and I would like to find a code in MATLAB that adapts the measured function with the real values (red curve) to that homogeneous time increment.

I attach a picture of the problem.

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Answer 1

Star Strider 2023-5-16

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I am not certain from your description what you want to do. However if you have the Signal Processing Toolbox, the resample function coukld be appropriate. (It is preferable to interp1 for singal processing applications.)

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Daniel Sánchez Muñoz 2023-5-23

The code is below. What it does is to homogenize the time step, and it uses an interp1. I don't have the file, but it consists of a column with the time in seconds and a column with the acceleration measurements.

FICHERO = '2019-02-12-01-24-51-FRacels.csv'; COLTIME = 1; COLACEL = 1; fs = 100;

data = load(FICHERO); t = (data(:, COLTIME) - data(1, COLTIME)) / 1000; v = data(:, 2:end);

% Interpolate data to assure constant time interval at fs dt = 1 / fs; tq = zeros(1,length(data)); for i = 2:length(data) tq(i) = tq(i-1) + dt; end

vq = interp1(t, v(:,COLACEL), tq,'spline','extrap'); %vq = makima(t, v(:,COLACEL), tq); %'linear'); %'nearest'); %'pchip'); %'cubic'); %'makima'); %'spline');

% Reconstruct the data vector data = [tq', vq'];

% Plot original data and interpolated data line figure plot(t, v(:,COLACEL), 'xr-') hold on plot(tq, vq, 'ob-'); legend('Samples', 'Spline Interpolation') xlim([t(1) t(end)]) title('Original and interpolated data')

figure plot(diff(t),'xr') hold on plot(diff(tq),'ob-') legend('Samples', 'Spline Interpolation')

Star Strider 2023-5-24

编辑：Star Strider 2023-5-24

在 MATLAB Online 中打开

Prueba.mat

I am not certain what result you want.

This resamples all of them to a common sampling frequency. There are some obvious differences between the original and resampled signals.

The resample function has a number of options, including the interplation method (I chose the default 'linear' method for all of these). You can slso choose a different sampling frequency (I chose 10 Hz here because it is closest to the actual sampling frequency), although that may create data where no data previously existed, so I advise against it.

LD = load('Prueba.mat')

LD = struct with fields:

Acceleration: [53×3 timetable]

Acceleration = LD.Acceleration

Acceleration = 53×3 timetable

Timestamp X Y Z ________________________ _________ ________ ______ 03-May-2023 10:44:45.299 0.13878 0.22851 9.7723 03-May-2023 10:44:45.399 0.11067 0.26082 9.7846 03-May-2023 10:44:45.499 0.14267 0.25065 9.9921 03-May-2023 10:44:45.599 0.10857 0.367 9.551 03-May-2023 10:44:45.699 0.085244 0.41575 9.9479 03-May-2023 10:44:45.799 0.034098 0.23659 10.891 03-May-2023 10:44:45.899 0.14656 0.18036 9.5495 03-May-2023 10:44:45.999 0.16839 0.533 9.5333 03-May-2023 10:44:46.099 -0.085244 0.54646 10.048 03-May-2023 10:44:46.199 -0.11396 0.70947 9.1247 03-May-2023 10:44:46.299 0.20489 1.0379 7.2703 03-May-2023 10:44:46.399 0.068495 0.15165 8.4736 03-May-2023 10:44:46.499 -0.27159 -0.26889 13.35 03-May-2023 10:44:46.599 0.24796 -0.33918 12.625 03-May-2023 10:44:46.699 -0.49741 0.40229 9.0996 03-May-2023 10:44:46.799 0.046361 0.84347 6.553

VN = Acceleration.Properties.VariableNames;

[h,m,s] = hms(Acceleration.Timestamp);

s

s = 53×1

45.2990 45.3990 45.4990 45.5990 45.6990 45.7990 45.8990 45.9990 46.0990 46.1990

% X = Acceleration.X

% Y = Acceleration.Y

% Z = Acceleration.Z

format long

ds = diff(s);

ds_stats = [min(ds); max(ds); mean(ds); std(ds); 1/mean(ds)] % 'Original' Time Vector Statistics

ds_stats = 5×1

0.099999999999994 0.100999999999999 0.100019230769231 0.000138675049056 9.998077292828301

Fs = ds_stats(end)

Fs =

9.998077292828301

Fsc = ceil(Fs)

Fsc =

10

Accelerationr = resample(Acceleration, Fsc) % 'Accelerationr' Is The Resampled 'Acceleration' 'timetable'

Accelerationr = 53×3 timetable

Time X Y Z ________________________ _________ _________ _________ 03-May-2023 10:44:45.299 0.138784 0.228515 9.772287 03-May-2023 10:44:45.399 0.110668 0.260818 9.784551 03-May-2023 10:44:45.499 0.142672 0.250648 9.992128 03-May-2023 10:44:45.599 0.108574 0.366999 9.550952 03-May-2023 10:44:45.699 0.085244 0.415753 9.947861 03-May-2023 10:44:45.799 0.034098 0.23659 10.891232 03-May-2023 10:44:45.899 0.14656 0.180359 9.549456 03-May-2023 10:44:45.999 0.168395 0.533001 9.533304 03-May-2023 10:44:46.099 -0.085244 0.546461 10.04836 03-May-2023 10:44:46.199 -0.113958 0.709472 9.12473 03-May-2023 10:44:46.299 0.204885 1.037887 7.270292 03-May-2023 10:44:46.399 0.068495 0.151645 8.473583 03-May-2023 10:44:46.499 -0.271585 -0.268893 13.349558 03-May-2023 10:44:46.599 0.247956 -0.339183 12.625429 03-May-2023 10:44:46.699 -0.497408 0.402293 9.099606 03-May-2023 10:44:46.799 0.046361 0.84347 6.553044

[h,m,s] = hms(Accelerationr.Time);

ds = diff(s);

ds_stats = [min(ds); max(ds); mean(ds); std(ds); 1/mean(ds)] % 'Resampled' Time Vector Statistics

ds_stats = 5×1

0.099999999999994 0.100000000000001 0.100000000000000 0.000000000000003 9.999999999999995

figure

tiledlayout(3,1)

for k = 1:3

nexttile

plot(Acceleration.Timestamp, Acceleration{:,k})

title(VN{k})

grid

ylim('padded')

end

sgtitle('Original')

figure

tiledlayout(3,1)

for k = 1:3

nexttile

plot(Accelerationr.Time, Accelerationr{:,k})

grid

title(VN{k})

ylim('padded')

end

sgtitle('Resampled')

figure

tiledlayout(3,1)

for k = 1:3

nexttile

plot(Acceleration.Timestamp, Acceleration{:,k}, 'DisplayName','Original')

hold on

plot(Accelerationr.Time, Accelerationr{:,k}, 'DisplayName','Resampled')

hold off

grid

title(VN{k})

if k == 1

legend('Location','best')

end

ylim('padded')

end

sgtitle('Original Co-Plotted With Resampled')

figure

k = 2;

plot(Acceleration.Timestamp, Acceleration{:,k}, 'DisplayName','Original')

hold on

plot(Accelerationr.Time, Accelerationr{:,k}, 'DisplayName','Resampled')

hold off

grid

title(VN{k})

legend('Location','best')

I am not certain what part of what signal is in the originally provided plot image, or I would post it in detail here.

.

Daniel Sánchez Muñoz 2023-5-28

在 MATLAB Online 中打开

Hello!

Thanks for your help. The problem is that you use the mean frecuency, so there is still a non homogeneus sample frecuency.

I have finally solved the problem.

Pd.: Do I have to accept your answer? I thank you very much for the help.

data=[t,z];
COLTIME = 1;
COLACEL = 1;
tt = (data(:, COLTIME) - data(1, COLTIME)); 
v = data(:, 2:end);
    % Interpolate data to assure constant time interval at fs:
dt = 1 / fs;
ntime = length(data);
tq = zeros(1,ntime);
for i = 2:ntime
    tq(i) = tq(i-1) + dt;
end
vq = interp1(tt, v(:,COLACEL), tq,'spline','extrap');
    % Reconstruct the data vector:
data = [tq', vq'];

Star Strider 2023-5-28

I would appreciate it if you would accept my answer.

I actually use the most common frequency, that I derive from the differences in the sampling intervals. I use mean, I also could have used mode.

I notice the the code you posted uses ‘fs’ to define ‘dt’ without first defining ‘fs’. I have no idea how you define ‘fs’.

I also use resample instead of interp1 because resample uses an anti-aliasing filter. This is important for signal processing applications, and signal processing is apprarently what you want to do.

The resample function also has the 'spline' option as an interpolation method. I rarely use it (preferring 'pchip') because it can give some wildly varying results.

.

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