Removing jumps from data when plotting a graph

57 vues (au cours des 30 derniers jours)
Austin Ukpebor
Austin Ukpebor le 19 Jan 2021
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.
  4 commentaires
Star Strider
Star Strider le 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 le 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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Star Strider
Star Strider le 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.
  4 commentaires
Austin Ukpebor
Austin Ukpebor le 21 Jan 2021
At this point, I am confident I can apply your contributions to tidy up my work! Thanks so much!
Star Strider
Star Strider le 21 Jan 2021
As always, my pleasure!

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Plus de réponses (2)

Fangjun Jiang
Fangjun Jiang le 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 le 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);
  9 commentaires
Austin Ukpebor
Austin Ukpebor le 21 Jan 2021
Matt, thanks for your great contributions. I can now tidy up my work. Appreciated!
Matt Gaidica
Matt Gaidica le 21 Jan 2021
Sure thing, sounds like a cool project. Feel free to reach out directly if you need anything!

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