Can I remove the noise and make the graph smooth?

11 views (last 30 days)
I drew a graph with data obtained in seconds when my feet touch the ground while walking. By the way, the graph is not smooth, so I want to make it smooth. I don't know if I should remove noise, filter it, or how to do it. Please help me!!
This is my code, and when I run it like this, I get some of the following graphs.
Fz = -temp1(7000:8600,5)%%%data%%%
for i=1:length(Fz)
plot(Fz,'b')
hold on
plot(i,Fz(i))
line([800 i], [0 Fz(i)])
pause(0.005)
end
  4 Comments
JooYoung Jeon
JooYoung Jeon on 12 Nov 2021
How should I apply the code I wrote to the smooth code?
This picture is a plot written in the code below. I want to make it smooth.
Fz = -temp1(7000:8600,5)%%%data%%%
for i=1:length(Fz)
plot(Fz,'b')
hold on
plot(i,Fz(i))
line([800 i], [0 Fz(i)])
pause(0.00005)
end

Sign in to comment.

Answers (1)

Mathieu NOE
Mathieu NOE on 9 Nov 2021
hello
see the demo below - it removes the drift in your data , so it will remain "zero mean" after high pass filtering
as a bonus I added some code to compute the time intervals define by threshold crossing
period = 0.8987 0.8972 0.8984 0.8980 0.8977 0.8963 (in seconds)
clc
clearvars
n=100;
x=linspace(0,2*pi,n);
y = sin(7*x)+0.1*x.^2;
% high pass filtered signal
[b,a] = butter(2,0.1,'high');
yf = filtfilt(b,a,y);
threshold = 0.25; % your value here
[t0_pos,s0_pos,t0_neg,s0_neg]= crossing_V7(yf,x,threshold,'linear'); % positive (pos) and negative (neg) slope crossing points
% ind => time index (samples)
% t0 => corresponding time (x) values
% s0 => corresponding function (y) values , obviously they must be equal to "threshold"
figure(1)
plot(x,y,x,yf,x,threshold*ones(size(x)),'k--',t0_pos,s0_pos,'dr',t0_neg,s0_neg,'dg','linewidth',2,'markersize',12);grid on
legend('signal','signal after HP filter','threshold','positive slope crossing points','negative slope crossing points');
period = diff(t0_pos)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [t0_pos,s0_pos,t0_neg,s0_neg] = crossing_V7(S,t,level,imeth)
% [ind,t0,s0,t0close,s0close] = crossing_V6(S,t,level,imeth,slope_sign) % older format
% CROSSING find the crossings of a given level of a signal
% ind = CROSSING(S) returns an index vector ind, the signal
% S crosses zero at ind or at between ind and ind+1
% [ind,t0] = CROSSING(S,t) additionally returns a time
% vector t0 of the zero crossings of the signal S. The crossing
% times are linearly interpolated between the given times t
% [ind,t0] = CROSSING(S,t,level) returns the crossings of the
% given level instead of the zero crossings
% ind = CROSSING(S,[],level) as above but without time interpolation
% [ind,t0] = CROSSING(S,t,level,par) allows additional parameters
% par = {'none'|'linear'}.
% With interpolation turned off (par = 'none') this function always
% returns the value left of the zero (the data point thats nearest
% to the zero AND smaller than the zero crossing).
%
% check the number of input arguments
error(nargchk(1,4,nargin));
% check the time vector input for consistency
if nargin < 2 | isempty(t)
% if no time vector is given, use the index vector as time
t = 1:length(S);
elseif length(t) ~= length(S)
% if S and t are not of the same length, throw an error
error('t and S must be of identical length!');
end
% check the level input
if nargin < 3
% set standard value 0, if level is not given
level = 0;
end
% check interpolation method input
if nargin < 4
imeth = 'linear';
end
% make row vectors
t = t(:)';
S = S(:)';
% always search for zeros. So if we want the crossing of
% any other threshold value "level", we subtract it from
% the values and search for zeros.
S = S - level;
% first look for exact zeros
ind0 = find( S == 0 );
% then look for zero crossings between data points
S1 = S(1:end-1) .* S(2:end);
ind1 = find( S1 < 0 );
% bring exact zeros and "in-between" zeros together
ind = sort([ind0 ind1]);
% and pick the associated time values
t0 = t(ind);
s0 = S(ind);
if ~isempty(ind)
if strcmp(imeth,'linear')
% linear interpolation of crossing
for ii=1:length(t0)
%if abs(S(ind(ii))) >= eps(S(ind(ii))) % MATLAB V7 et +
if abs(S(ind(ii))) >= eps*abs(S(ind(ii))) % MATLAB V6 et - EPS * ABS(X)
% interpolate only when data point is not already zero
NUM = (t(ind(ii)+1) - t(ind(ii)));
DEN = (S(ind(ii)+1) - S(ind(ii)));
slope = NUM / DEN;
slope_sign(ii) = sign(slope);
t0(ii) = t0(ii) - S(ind(ii)) * slope;
s0(ii) = level;
end
end
end
% extract the positive slope crossing points
ind_pos = find(sign(slope_sign)>0);
t0_pos = t0(ind_pos);
s0_pos = s0(ind_pos);
% extract the negative slope crossing points
ind_neg = find(sign(slope_sign)<0);
t0_neg = t0(ind_neg);
s0_neg = s0(ind_neg);
else
% empty output
ind_pos = [];
t0_pos = [];
s0_pos = [];
% extract the negative slope crossing points
ind_neg = [];
t0_neg = [];
s0_neg = [];
end
end
  12 Comments
Mathieu NOE
Mathieu NOE on 15 Nov 2021
My pleasure !
would you mind accepting my answer ?
tx

Sign in to comment.

Categories

Find more on MATLAB in Help Center and File Exchange

Products

Community Treasure Hunt

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

Start Hunting!