Separating left and right leg data given ground reaction force gait data?

Question

Anthony 2016-9-30

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/305154-separating-left-and-right-leg-data-given-ground-reaction-force-gait-data

评论： Star Strider 2018-3-4

1 Column Gait Data.xlsx

Hi,

I have an Excel file with several columns of ground reaction force data where each column is a different test subject. My question is, how can I separate each column into left leg data and right leg data. Where stride 1 is assumed to be the right leg, therefore all odd numbered strides are right leg strides and all even numbered strides are left leg strides. I have attached the first column of my data to give you an idea of what I am looking at. Any guidance would be great. Thank you so much!

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jennifer aniston 2016-10-1

Thanks Star Strider,

I am new on this group so finding my way around. If you find the time please take a look at my question. Link: https://in.mathworks.com/matlabcentral/answers/305313-advice-on-processing-raw-gait-data

Thanks!

Star Strider 2016-10-1

My pleasure!

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Answer 1

Star Strider 2016-9-30

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/305154-separating-left-and-right-leg-data-given-ground-reaction-force-gait-data#answer_236745

在 MATLAB Online 中打开

You need to use the Signal Processing Toolbox findpeaks function. It retuens the values of the peaks and their locations. I only had the indices to work with, but you can easily use the indices to convert to time if you need to.

The code:

[d,s,r] = xlsread('Anthony 1 Column Gait Data.xlsx');
[pks,locs] = findpeaks(d, 'MinPeakDistance',500, 'MinPeakHeight',1000);
figure(1)
plot(d)
hold on
plot(locs(1:2:end), pks(1:2:end), '^r', 'MarkerFaceColor','r')
plot(locs(2:2:end), pks(2:2:end), '^g', 'MarkerFaceColor','g')
hold off
grid
axis([0  5000    ylim])
legend('Force Plate Data', 'Right Foot', 'Left Foot', 'Location','E')
xlabel('Index')
ylabel('Force')

I plotted a portion of the data to show the details here. The code works for the entire data set.

The plot:

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Star Strider 2016-9-30

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My pleasure!

The easiest way to do that is to do a second call to findpeaks, but with the negative of your signal. Those peaks will be the ‘troughs’, and with essentially the same code (you will only need to modify the 'MinPeakHeight' value), you should be able to identify the low points separating the strides. Remember that the ‘pks’ value for the second findpeaks call (to get the ‘troughs’) will be the negative of the actual value returned, so consider that in your plot. The ‘locs’ value will not be affected.

The code:

[d,s,r] = xlsread('Anthony 1 Column Gait Data.xlsx');
[pks,locs] = findpeaks( d, 'MinPeakDistance',500, 'MinPeakHeight',1000);
[trf,tloc] = findpeaks(-d, 'MinPeakDistance',500, 'MinPeakHeight',-100);
figure(1)
plot(d)
hold on
plot(locs(1:2:end), pks(1:2:end), '^r', 'MarkerFaceColor','r')
plot(locs(2:2:end), pks(2:2:end), '^g', 'MarkerFaceColor','g')
plot([tloc tloc]', repmat(ylim,size(tloc,1),1)', '-k', 'LineWidth',1.5)
hold off
grid
axis([0  5000    ylim])
legend('Force Plate Data', 'Right Foot', 'Left Foot', 'Location','E')
xlabel('Index')
ylabel('Force')

The plot:

Star Strider 2016-9-30

在 MATLAB Online 中打开

My pleasure.

Among other things (one being a retired Board Certified Internal Medicine Physician), I’m also a M.S. biomedical engineer, and have done some work (long ago) as part of a gait analysis project.

This is how I would separate the left and right foot force records. There are no gaps in the records:

tloc = [1; tloc; length(d)];                                % Augment ‘tloc’ Vector
rfoot = [];                                                 % Initialise Vector
lfoot = [];                                                 % Initialise Vector
for k1 = 1:2:length(tloc)-2
    rfoot = [rfoot; d(tloc(k1):tloc(k1+1))];
    lfoot = [lfoot; d(tloc(k1+1):tloc(k1+2))];
end

If you want to separate the individual records, the easiest way is to put NaN values between them:

tloc = [1; tloc; length(d)];                                % Augment ‘tloc’ Vector
rfoot = [];                                                 % Initialise Vector
lfoot = [];                                                 % Initialise Vector
for k1 = 1:2:length(tloc)-2
    rfoot = [rfoot; d(tloc(k1):tloc(k1+1)); NaN];
    lfoot = [lfoot; d(tloc(k1+1):tloc(k1+2)); NaN];
end

Then, just test for the NaN values.

They are now individual vectors, so you can write them to your Excel file as individual columns. (Note that the columns are not equal length.)

If you want the corresponding times for each, you can do that similarly, in the same loop. The easiest way is to create two new vector assignments, one for each, including the initialization assignment before the loop. If your time vector is ‘t’, substitute ‘t’ for ‘d’, and add a ‘t’ to the beginning or end of the variable names here.

Star Strider 2016-9-30

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My pleasure.

The loop changes to:

Ts = 1/1200;                                                % Sampling Time Interval
t = linspace(0, 1, length(d))'*Ts;                          % Time Vector
tloc = [1; tloc; length(d)];                                % Augment ‘tloc’ Vector
rfoot = [];                                                 % Initialise Vector
lfoot = [];                                                 % Initialise Vector
rfoot_t = [];                                               % Initialise Vector
lfoot_t = [];                                               % Initialise Vector
for k1 = 1:2:length(tloc)-2
    rfoot = [rfoot; d(tloc(k1):tloc(k1+1)); NaN];
    lfoot = [lfoot; d(tloc(k1+1):tloc(k1+2)); NaN];
    rfoot_t = [rfoot_t; t(tloc(k1):tloc(k1+1)); NaN];
    lfoot_t = [lfoot_t; t(tloc(k1+1):tloc(k1+2)); NaN];
end

Note the transpose (') in the linspace call to force a column vector.

You can now also plot them individually with respect to time using those vectors:

figure(2)
plot(rfoot_t, rfoot, 'r', 'LineWidth',1)  
hold on
plot(lfoot_t, lfoot, 'g', 'LineWidth',1)
hold off
grid
axis([0  t(5000)    ylim])
legend('Right Foot', 'Left Foot', 'Location','E')
xlabel('Time (s)')
ylabel('Force')

If my Answer is solving your problems, please Accept it.

Star Strider 2016-9-30

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No worries, and thanks for the additional vote! Yours is a fun question for me because of my interest in the subject.

This is a complete rewrite of the code, using time rather than indices in all computations. (You still have the index code if you need it. In this version, the independent variable is in units of time, here seconds, throughout the code.)

Specifically, ‘pkt’ (peaks times) and ‘trt’ (troughs times) are now in units of seconds. (Note that the variables in the plots have changed.) You are of course free to change the variable names to something more appropriate in the context of your research. (I like short variable names because the probability of my mistyping them is less.)

The Code:

[d,s,r] = xlsread('Anthony 1 Column Gait Data.xlsx');
Ts = 1/1200;                                                % Sampling Time Interval
t = linspace(0, 1, length(d))'*Ts;                          % Time Vector
[pks,pkt] = findpeaks( d, t, 'MinPeakDistance',t(500), 'MinPeakHeight',1000);
[trf,trt] = findpeaks(-d, t, 'MinPeakDistance',t(500), 'MinPeakHeight',-100);
figure(1)
plot(t, d)
hold on
plot(pkt(1:2:end), pks(1:2:end), '^r', 'MarkerFaceColor','r')
plot(pkt(2:2:end), pks(2:2:end), '^g', 'MarkerFaceColor','g')
plot([trt trt]', repmat(ylim,size(trt,1),1)', '-k', 'LineWidth',1.5)
hold off
grid
axis([0  t(5000)    ylim])
legend('Force Plate Data', 'Right Foot', 'Left Foot', 'Location','E')
xlabel('Time (s)')
ylabel('Force')
% Ts = 1/1200;                                                % Sampling Time Interval
% t = linspace(0, 1, length(d))'*Ts;                          % Time Vector
t2i = round(trt*length(d)/Ts);                              % Convert ‘t’ To Indices
tloc = [1; t2i; length(d)];                                 % Augment ‘tloc’ Vector
rfoot = [];                                                 % Initialise Vector
lfoot = [];                                                 % Initialise Vector
rfoot_t = [];                                               % Initialise Vector
lfoot_t = [];                                               % Initialise Vector
for k1 = 1:2:length(trt)-2
    rfoot = [rfoot; d(tloc(k1):tloc(k1+1)); NaN];
    lfoot = [lfoot; d(tloc(k1+1):tloc(k1+2)); NaN];
    rfoot_t = [rfoot_t; t(tloc(k1):tloc(k1+1)); NaN];
    lfoot_t = [lfoot_t; t(tloc(k1+1):tloc(k1+2)); NaN];
end
figure(2)
plot(rfoot_t, rfoot, 'r', 'LineWidth',1)  
hold on
plot(lfoot_t, lfoot, 'g', 'LineWidth',1)
hold off
grid
axis([0  t(5000)    ylim])
legend('Right Foot', 'Left Foot', 'Location','E')
xlabel('Time (s)')
ylabel('Force')

The plots are the same, so I won’t reproduce them here, except for the independent variable now being in seconds, in both plots.

Star Strider 2016-9-30

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As always, my pleasure!

That turned out to be easier than I thought it would be, once I remembered ‘zci’, a little utility routine I wrote a while back. It calculates zero-crossings, so all I needed to do to detect values of about 50 N was to subtract 50 from ‘d’ to create the zero-crossings. You may need to start the subscripts from 1 rather than 2 in other of your records, but that should be easy to determine when you plot your data. My ‘zci’ function returns the (z)ero (c)rossing (i)ndices of the dependent variable, so indexing into the data such as I did with the ‘pos’ matrix is necessary. Then plotting it or recording it is straightforward.

The Code:

zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);                    % Returns Approximate Zero-Crossing Indices Of Argument Vector
zx = zci(d - 50);
pos = [t(zx(2:2:end))  d(zx(2:2:end))];
figure(3)
plot(rfoot_t, rfoot, 'r', 'LineWidth',1)  
hold on
plot(lfoot_t, lfoot, 'g', 'LineWidth',1)
plot(pos(:,1), pos(:,2), 'bp', 'MarkerSize',9, 'MarkerFaceColor','b')
hold off
grid
axis([0  t(5000)    ylim])
legend('Right Foot', 'Left Foot', 'Initial Contact (50 N)', 'Location','E')
xlabel('Time (s)')
ylabel('Force')

The Plot:

Star Strider 2016-10-1

I like medical and biomedical engineering problems! I really do have fun with them!

I noticed that the left leg was generating less force that the right leg. The left leg may be slightly shorter than the right leg. The swing times may also be slightly different if that hypothesis is correct, since the ‘pendulum period’ would be different. (You can also assess this with EMGs from the hip flexors and extensors. The shorter leg would have greater EMG activity to keep it synchronised with the longer leg.)

Measure leg lengths. Leg lengths are rarely equal, and if you can account for the difference in force by a difference in leg lengths, that could be an important covariate in your eventual statistical analysis. By the same token, if you can equalise the length (and force) with an appropriately-designed orthotic in the shoe of the shorter leg, you might have an additional paper out of your current study. (This would be a paired study, so consider the appropriate statistical analysis.) An orthopaedic surgeon or physical therapist can assist you with these measurements if they’re not part of your background.

If you really want to get precise, get a lower-segment (lumbar, lumbosacral, pelvic and distal) MRI of each of your participants. (CT scans generate too much ionising radiation.) You can not only measure precise leg lengths, but also muscle volumes, and search for specific pathology (arthritis, atrophy, peripheral neuropathy, ad infinitum). Avoid gadolinium contrast if possible. It may be centrally neurotoxic.

Just a thought to make your project more interesting. Your committee or faculty advisor will (if they are good) ask about the force discrepancy (and hypothesise a leg-length discrepancy as have I), so be prepared to explain it with data. If you can offer a paired study with intervention, you get serious points for having thought about it, investigated it, and have an explanation and possibly and intervention for it. Never overlook details! They may amount to nothing, but they may be the next Nobel Prize. (Hyperbole here, but you get the idea.)

Your other friend in all this (if you have not met it yet) is PubMed. You will learn to rely on each other.

I look forward to following your research. You’re off to a good start with the way your stated your Question and presented your data. That made it very easy for me.

Anthony 2016-10-3

I cannot thank you enough for your help and words of encouragement! I think you bring up a tremendous amount of great idea that can be used in my research!

I love your enthusiasm and willingness to help. I would like to give you some background on what I may be trying to accomplish. For one, I think I will be including the calculation of slope between the initial contact point and peak point (I may need some help on that one!)

But in terms of outside of the Matlab program, I will be focusing a lot on fatigue and its corresponding effects on neuromuscular control and stability. Looking at gait data before and after a fatigue protocol will give a great look in to the effects fatigue can have on an individual's biomechanics. Particularly, their lower extremity control. I have looked a lot into PubMed! It has been a very good friend indeed thus far.

I love many of your suggestions! I definitely will try to implement them into what I am doing. You seem to have a brilliant mind and your words have been inspiring. The length of leg comparison brings up a very intriguing possibility. I think I will have to give a good look into that. I'm actually fascinated by that possibility now that I am thinking about it. I can't wait to get this program totally finished (all I need is the slope calculation now!), and apply it to my studies. You have given me a whole new level of excitement! Thank you again!

-Anthony

Star Strider 2016-10-3

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As always, my pleasure!

The slope between the initial contact and the peak force (or load), is relatively straightforward. Divide the difference between the peak force and the initial contact force by the time difference between them. I’m not certain this is what you want. If it is, add it to the end of the last code I posted:

f_rate = (pks(1:size(init_cont,1)) - init_cont(:,2)) ./ (init_cont(:,1) - pkt(1:size(init_cont,1)));    % Force Rate (N/sec)
figure(4)
plot(init_cont(:,1), f_rate)
grid
xlabel('Time (sec)')
ylabel('Force Rate (N/sec)')
figure(5)
plot(t, d*5E+5)
hold on
plot(init_cont(:,1), f_rate)
hold off
grid
axis([0  t(20000)    ylim])
xlabel('Time (sec)')
ylabel('Force Rate (N/sec)')

I called it ‘f_rate’ for ‘force rate’, but that’s for my convenience. It produces an interesting plot that reveals information I didn’t realise was in the data. (It uses the initial contact time, ‘init_cont(:,1)’, as the time base for the plot. Both vectors have to be equal length, so you may have to tweak the code for different records.

Fatigue has many components, the depletion of ATP and oxygen (myoglobin oxygen saturation) and accumulation of lactate in the muscle, and to a lesser extent, depletion of neurotransmitters, particularly acetylcholine, or neurotransmitter insensitivity (consider the effects of succinylcholine). Some aspects of fatigue may be accessible with surface electromyography. It’s been a while since I studied muscle fatigue, so you will want to see what the literature has on that.

You happened to hit one of my research interest ‘eigenfrequencies’, so I characteristically resonated.

I very much appreciate your compliments! I am always happy to help, within the scope of my expertise. Again, my pleasure!

Star Strider 2016-10-3

在 MATLAB Online 中打开

As always, my pleasure!

I added another two variables ‘r_pks’ and ‘l_pks’, tweaked the code to accommodate them, and created different force rates for the right and left foot as ‘rf_frate’ and ‘lf_frate’ respectively. I’m at a loss to explain the off- scale-high final value for the right foot force rate. It may just be a data anomaly, since the rest are consistent.

I find the results of the force rate analysis fascinating. I had no idea that the force rates would display such interesting 0.2 millisecond periodicity. I look forward to your analysis of it, because I have no idea how to explain it.

The Code:

[d,s,r] = xlsread('Anthony 1 Column Gait Data.xlsx');
Ts = 1/1200;                                                % Sampling Time Interval
t = linspace(0, 1, length(d))'*Ts;                          % Time Vector
[pks,pkt] = findpeaks( d, t, 'MinPeakDistance',t(500), 'MinPeakHeight',1000);
[trf,trt] = findpeaks(-d, t, 'MinPeakDistance',t(500), 'MinPeakHeight',-100);
r_pks = [pkt(1:2:end), pks(1:2:end)];                       % Right Foot Times & Peaks
l_pks = [pkt(2:2:end), pks(2:2:end)];                       % Left Foot Times & Peaks
figure(1)
plot(t, d)
hold on
plot(r_pks(:,2), r_pks(:,2), '^r', 'MarkerFaceColor','r')
plot(l_pks(:,2), l_pks(:,2), '^g', 'MarkerFaceColor','g')
plot([trt trt]', repmat(ylim,size(trt,1),1)', '-k', 'LineWidth',1.5)
hold off
grid
axis([0  t(5000)    ylim])
legend('Force Plate Data', 'Right Foot', 'Left Foot', 'Location','E')
xlabel('Time (s)')
ylabel('Force')
% Ts = 1/1200;                                                % Sampling Time Interval
% t = linspace(0, 1, length(d))'*Ts;                          % Time Vector
t2i = round(trt*length(d)/Ts);                              % Convert ‘t’ To Indices
tloc = [1; t2i; length(d)];                                 % Augment ‘tloc’ Vector
rfoot = [];                                                 % Initialise Vector
lfoot = [];                                                 % Initialise Vector
rfoot_t = [];                                               % Initialise Vector
lfoot_t = [];                                               % Initialise Vector
for k1 = 1:2:length(trt)-2
    rfoot = [rfoot; d(tloc(k1):tloc(k1+1)); NaN];
    lfoot = [lfoot; d(tloc(k1+1):tloc(k1+2)); NaN];
    rfoot_t = [rfoot_t; t(tloc(k1):tloc(k1+1)); NaN];
    lfoot_t = [lfoot_t; t(tloc(k1+1):tloc(k1+2)); NaN];
end
figure(2)
plot(rfoot_t, rfoot, 'r', 'LineWidth',1)  
hold on
plot(lfoot_t, lfoot, 'g', 'LineWidth',1)
hold off
grid
axis([0  t(5000)    ylim])
legend('Right Foot', 'Left Foot', 'Location','E')
xlabel('Time (s)')
ylabel('Force')
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);        % Returns Approximate Zero-Crossing Indices Of Argument Vector
zx = zci(d - 50);
init_cont = [t(zx(2:2:end))  d(zx(2:2:end))];
init_cont_rf = init_cont(2:2:end,:);
init_cont_lf = init_cont(1:2:end,:);
figure(3)
plot(rfoot_t, rfoot, 'r', 'LineWidth',1)  
hold on
plot(lfoot_t, lfoot, 'g', 'LineWidth',1)
plot(init_cont_rf(:,1), init_cont_rf(:,2), 'bp', 'MarkerSize',9, 'MarkerFaceColor','y')
plot(init_cont_lf(:,1), init_cont_lf(:,2), 'mp', 'MarkerSize',9, 'MarkerFaceColor','b')
hold off
grid
axis([0  t(5000)    ylim])
legend('Right Foot', 'Left Foot', 'Right Foot Initial Contact (50 N)', 'Left Foot Initial Contact (50 N)', 'Location','E')
xlabel('Time (s)')
ylabel('Force')
f_rate = @(fi,fp,ti,tp) (fp - fi) ./ (tp - ti);             % Force Rate Function (N/sec)
rf_frate = f_rate(init_cont_rf(:,2), r_pks(2:end,2), init_cont_rf(:,1), r_pks(2:end,1));
lf_frate = f_rate(init_cont_lf(:,2), l_pks(:,2), init_cont_lf(:,1), l_pks(:,1));
figure(4)
plot(t, d*4E+5)
hold on
plot(init_cont_rf(:,1), rf_frate, '-r')
plot(init_cont_lf(:,1), lf_frate, '-g')
hold off
grid
axis([0  t(20000)    ylim])
xlabel('Time (sec)')
ylabel('Force Rate (N/sec)')
legend('Right Foot', 'Left Foot', 'Location', 'NE')
figure(5)
plot(t, d*4E+5)
hold on
plot(init_cont_rf(:,1), rf_frate, '-r', 'LineWidth',1.2)
plot(init_cont_lf(:,1), lf_frate, '-g', 'LineWidth',1.2)
hold off
grid
% axis([0  t(20000)    ylim])
xlabel('Time (sec)')
ylabel('Force Rate (N/sec)')
legend('Force Data', 'Right Foot', 'Left Foot', 'Location', 'NE')

The Plot:

Anthony 2016-10-3

Thank you again for your suggestions and help! Regarding the use of linear techniques, a lot of research has been published recently showing that in order to measure stability (especially in biological systems such as human movement), nonlinear techniques should be used. This is due to the idea of an "optimal range of variability" which promotes the idea that human locomotion is actually most stable when it has some degree of variability. There are arguments that when there is too small of a degree of variability, the movement becomes too robust and robotic and is not able to adapt to perturbation and other external forces. Some linear techniques have shown to be insufficient in measuring the oscillations in human movement in terms of their stability so I think we will be focusing on nonlinear techniques such as Approximate Entropy. I think that will help us tremendously in analyzing gait data.

It has been a pleasure going back and forth with you! I'm really enjoying this conversation and feel like I am getting a TON out of this. Thank you so much again!

Star Strider 2016-10-8

Hi Anthony,

As always, my pleasure!

Questions I Answer with the most recent activity pop to the top of my profile Answers page. I check it every morning. I don’t check it during the day unless I’ve been away for a while. I scroll through Answers occasionally otherwise. I look at all new activity and catch most of them.

There are two functions that will give you the second derivative. The first is using gradient twice, and using del2 once, multiplying it by 4. Either works.

Your data have a leading ‘spike’ (that I interpret as a heel strike), followed by what seems to be a transition to the stance phase, and then toe-off. The ‘spike’ could give you serious problems with either function.

Note that derivatives amplify noise, and although your data seem reasonably noise free, you’ll soon find out where the noise is when you take the derivatives! There are ways to deal with that, from designing a discrete filter if the noise is band-limited, or using a Savitzky-Golay filter (the sgolayfilt function) if it’s broadband. These can be either straightforward to implement, or a real challenge, depending on the signal.

To design a discrete filter, you first have to take the Fourier transform to determine the frequency content. Use the R2015a documentation for fft (link) as your guide, particularly the code between the first (top) two plot figures. If you have band-limited noise, the easiest way to design a filter is to use the designfilt function. I prefer to use IIR filters, but that’s my personal preference. Many times FIR filters are the only design that will work effectively. My IIR filter design procedure is here: How to design a lowpass filter for ocean wave data in Matlab? It’s probably worth a look even if you don’t use it. Your signals don’t have any baseline drift, so you can probably use a low-pass filter rather than a bandpass filter. These are easier to design.

If your data turn out to have significant band-limited noise, give the discrete filter design a go. I have a fair amount of experience with discrete filter design and implementation, so I’m here if you have questions or problems.

If you need to use sgolayfilt, the implementation is much more heuristic (and I can provide less effective help). You may need to experiment with that until you get the result you want.

Star Strider 2016-10-9

在 MATLAB Online 中打开

Hi Anthony,

This is about as good as it gets. Further lowering the cutoff frequency begins to slur the details of the original heel-strike ‘spike’, that then compromises the derivative function and resolution, so I left it as is.

If you experiment with the passband and stopband frequencies, the plot in figure(7) will tell you if you designed the filter correctly. This codes for a Butterworth IIR filter, so the Bode plot should have a flat top, with no irregularities in the passband, and a smooth rolloff.

Append this code to the end of the previous code I posted.

The Code:

Ts = 1/1200;                                                % Sampling Time Interval
t = linspace(0, 1, length(d))'*Ts;                          % Time Vector
L = size(d,1);
Ts = mean(diff(t));                                         % Sampling Interval 
Fs = 1/Ts;                                                  % Sampling Frequency 
Fn = Fs/2;                                                  % Nyquist Frequency 
FTd = fft(d)/L;                                             % Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                         % Frequency Vector
Iv = 1:length(Fv);                                          % Index Vector
figure(6)
semilogy(Fv, 1+abs(FTd(Iv))*2)
grid
axis([0  2.5E+6    ylim])
xlabel('Frequency (Hz)')
Wp = [2.5E+6]/Fn;                                           % Passband Frequency
Ws = [3.5E+6]/Fn;                                           % Stopband Frequency
Rp = 20;                                                    % Passband Ripple
Rs = 40;                                                    % Stopband Ripple
[n,Wn] = buttord(Wp,Ws,Rp,Rs);                              % Filter Order
[b,a] = butter(n,Wn);                                       % Filter Transfer Function Coefficients
[sos,g] = tf2sos(b,a);                                      % Second-Order Section For Stability
df = filtfilt(sos,g,d);                                     % Phase-Neutral Filtering
figure(7)
freqz(sos)
figure(8)
subplot(3,1,1)
plot(t, d, 'LineWidth',0.6)
axis([0  5E-5    -100  2000])
ylabel('Force (N)')
title('Original Signal')
grid
subplot(3,1,2)
plot(t, df, '-r', 'LineWidth',0.6)
axis([0  5E-5    -100  2000])
ylabel('Force (N)')
title('Lowpass-Filtered Signal')
grid
subplot(3,1,3)
plot(t, del2(df)*4, '-g', 'LineWidth',0.6)
axis([0  5E-5    ylim])
xlabel('Time (s)')
ylabel('\Delta Force (d^2N/dt^2)')
title('Second Derivative Of Filtered Signal')
grid

The Plot:

Star Strider 2016-10-20

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Hi Anthony,

Rasonably so. I hope all is going well with you as well.

I had to go back and remember what I was doing.

I don’t see that it makes a difference. My code here:

zx = zci(d - 50);
init_cont = [t(zx(2:2:end))  d(zx(2:2:end))];
init_cont_rf = init_cont(2:2:end,:);
init_cont_lf = init_cont(1:2:end,:);

relies on the indices (here ‘zx’) to create ‘init_cont’, so if you have your data references as indices, there should be no problem. Just substitute them into ‘zx’.

If you’ve defined the ‘init_cont’ matrix differently, it should not make any difference so long as the time column and initial contact force column are appropriate to your design, with alternating left and right foot times and contact forces in each column.

So long as ‘init_cont’ has the appropriate row size to be compatible with the force rate calculations:

f_rate = @(fi,fp,ti,tp) (fp - fi) ./ (tp - ti);             % Force Rate Function (N/sec)
rf_frate = f_rate(init_cont_rf(:,2), r_pks(2:end,2), init_cont_rf(:,1), r_pks(2:end,1));
lf_frate = f_rate(init_cont_lf(:,2), l_pks(:,2), init_cont_lf(:,1), l_pks(:,1));

there should be no problem.

Felipe Mattioni Maturana 2018-3-4

Force_Data.txt

Dear Star Strider,

I have been working on my force data using the codes you posted here. Everything worked perfectly so far. The only problem I have now is that I am analyzing data from an old Treadmill and it is quite noisy. The 50-N threshold is not working on this data set because between each strike there are usually two bumps on the data and it is often above 50 N. Would you have any suggestion on how I can overcome this issue? I am posting a figure and an example of my data (please, note that commas instead of dots might appear as decimal separators as I am using a German computer) to help you understand what is going on. Any help would be very much appreciated.

Best,

Felipe

Star Strider 2018-3-4

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Answer 2

KSSV 2016-9-30

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https://ww2.mathworks.cn/matlabcentral/answers/305154-separating-left-and-right-leg-data-given-ground-reaction-force-gait-data#answer_236743

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[data,txt,raw] = xlsread('1 Column Gait Data.xlsx');
% Right / odd numbered 
right = data(1:2:end) ;
% left / even numbered 
left = data(2:2:end) ;

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