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Removing jumps from data when plotting a graph

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I am trying to plot sensor readings. See attached the data and the graph. Please I need a help on how to remove those jumps in the graph (space between the 2 red markings). I would like to apply same to the remaing jumps in the graph. Any help is well appreciated.

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Austin Ukpebor
Austin Ukpebor on 19 Jan 2021
Thanks Matt for your response. I tried to plot the diff() of my signal but did not produce what I want.
It's a hall-effect sensor used with a combination of magnet to measure the gaping (opening and closing) of oyster shells. I want to retain those thick areas of the graph. That thick areas tell me about the spawning of the animal.
Star Strider
Star Strider on 19 Jan 2021
If you are calculating the numerical derivative, use the gradient function. It produces an output that has the same dimensions as the input.
Austin Ukpebor
Austin Ukpebor on 19 Jan 2021
Thanks Star Strider. I couldn't figure out on how to apply gradient function to my case using the link you shared. Please help me out, I have my data attached.
@Matt Gaidica, please share the code you used to generate your graph. When I used the diff() function, I had a different graph.

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Accepted Answer

Star Strider
Star Strider on 20 Jan 2021
I am not certain what result you want.
Try this:
D1 = readmatrix('SensorValues.xlsx');
s = D1; % Signal Vector
L = size(s,1); % Data Length
Fs = 1; % <— Need Actual Sampling Frequency Here!
Ts = 1/Fs; % Sampling Interval
Fn = Fs/2; % Nyquist Frequency
t = linspace(0, L, L)*Ts; % Time Vector
sc = s - mean(s); % Subtract Mean (Makes Other Peaks More Prominent)
FTs = fft(sc)/L; % Normalised Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn; % Frequency Vector
Iv = 1:numel(Fv); % Index Vector (One-Sided Fourier Transform)
figure
plot(Fv, abs(FTs(Iv))*2)
grid
xlim([0 0.001])
title('Fourier Transform')
xlabel('Frequency (Hz)')
ylabel('Amplitude')
Wp = [0.001 0.075]/Fn; % Passband Frequency (Normalised)
Ws = [0.9 1.1].*Wp; % Stopband Frequency (Normalised)
Rp = 1; % Passband Ripple
Rs = 60; % Passband Ripple (Attenuation)
[n,Wp] = ellipord(Wp,Ws,Rp,Rs); % Elliptic Order Calculation
[z,p,k] = ellip(n,Rp,Rs,Wp); % Elliptic Filter Design: Zero-Pole-Gain
[sos,g] = zp2sos(z,p,k); % Second-Order Section For Stability
figure
freqz(sos, 2^20, Fs) % Filter Bode Plot
set(subplot(2,1,1), 'XLim',Wp*Fn.*[0.8 1.2]) % Optional
set(subplot(2,1,2), 'XLim',Wp*Fn.*[0.8 1.2]) % Optional
s_filtered = filtfilt(sos, g, s); % Filter With IIR Filter
figure
plot(t, s)
hold on
plot(t, s_filtered)
hold off
grid
xlabel('Time (Units Estimated)')
ylabel('AMplitude (Units Not Specified)')
legend('Original Signal', 'Bandpass-Filtered Signal', 'Location','W')
producing:
The filter eliminates the d-c (constant) offset, and a bit of the high-frequency noise. Adjust the upper limit of the ‘Wp’ vector to get the result you want.
Note — With the actual sampling frequency, it will be necessary to adjust the limits of ‘Wp’ and the xlim values of the Fourier Transform plot.

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Star Strider
Star Strider on 21 Jan 2021
Spikes and other discontinuities can be difficult to deal with.
One option:
D1 = readmatrix('SensorValues.xlsx');
s = D1; % Signal Vector
L = size(s,1); % Data Length
Fs = 1; % <— Need Actual Sampling Frequency Here!
Ts = 1/Fs; % Sampling Interval
Fn = Fs/2; % Nyquist Frequency
t = linspace(0, L, L)*Ts; % Time Vector
sc = s - mean(s); % Subtract Mean (Makes Other Peaks More Prominent)
FTs = fft(sc)/L; % Normalised Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn; % Frequency Vector
Iv = 1:numel(Fv); % Index Vector (One-Sided Fourier Transform)
figure
plot(Fv, abs(FTs(Iv))*2)
grid
xlim([0 0.001])
title('Fourier Transform')
xlabel('Frequency (Hz)')
ylabel('Amplitude')
Wp = [1E-8 1.5E-3]/Fn; % Passband Frequency (Normalised)
Ws = [0.9 1.1].*Wp; % Stopband Frequency (Normalised)
Rp = 1; % Passband Ripple
Rs = 60; % Passband Ripple (Attenuation)
[n,Wp] = ellipord(Wp,Ws,Rp,Rs); % Elliptic Order Calculation
[z,p,k] = ellip(n,Rp,Rs,Wp); % Elliptic Filter Design: Zero-Pole-Gain
[sos,g] = zp2sos(z,p,k); % Second-Order Section For Stability
figure
freqz(sos, 2^20, Fs) % Filter Bode Plot
set(subplot(2,1,1), 'XLim',Wp*Fn.*[0.8 1.2]) % Optional
set(subplot(2,1,2), 'XLim',Wp*Fn.*[0.8 1.2]) % Optional
s_filtered = filtfilt(sos, g, s); % Filter With IIR Filter
s_filtered(s_filtered < -0.5) = nan; % Region = NaN
s_filtered = fillmissing(s_filtered, 'linear'); % Linearly Interpolate
figure
plot(t, s)
hold on
plot(t, s_filtered, '-r', 'LineWidth',1.5)
hold off
grid
xlabel('Time (Units Estimated)')
ylabel('AMplitude (Units Not Specified)')
legend('Original Signal', 'Bandpass-Filtered Signal', 'Location','W')
producing:
The data in the discontinuity region does not appear to be valid, considering the other parts of the signal, so simply eliminating it may be a worthwhile option.
.
Austin Ukpebor
Austin Ukpebor on 21 Jan 2021
At this point, I am confident I can apply your contributions to tidy up my work! Thanks so much!

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More Answers (2)

Fangjun Jiang
Fangjun Jiang on 19 Jan 2021
In your case, since you have enough data, I think you can use symbol in plot() to remove the "jump".
If your old way is plot(x,y), then you can use plot(x,y,'.')

Matt Gaidica
Matt Gaidica on 19 Jan 2021
What do you want to do with the data that around the jumps (still referring to my previous figure)? Interpolate it? Maybe you want some type of dynamic detrending, but it still doesn't totally remove the artifact of the jumps. I just loaded your data in A.
y = smoothdata(A,'movmean',100);
close all
figure;
subplot(211);
plot(A);
hold on;
plot(y);
subplot(212);
plot(A-y);

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Matt Gaidica
Matt Gaidica on 20 Jan 2021
That's just zoomed in. Sorry for the confusion.
Austin Ukpebor
Austin Ukpebor on 21 Jan 2021
Matt, thanks for your great contributions. I can now tidy up my work. Appreciated!
Matt Gaidica
Matt Gaidica on 21 Jan 2021
Sure thing, sounds like a cool project. Feel free to reach out directly if you need anything!

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