Smooth vector data with a specific step

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Lev Mihailov
Lev Mihailov on 4 Apr 2022
Commented: Mathieu NOE on 6 Apr 2022
I need to smooth the data, but the smooth function flips the response of the data, which makes things a little more difficult
x = rand(1,4097);
y = rand(1,4097);
z1=smooth(x(1:10),x(1:10),0.8,'rloess')'; % len(1,10)
z2=smooth(x(11:21),x(11:21),0.8,'rloess')'; % len(,10)
z=[z1,z2] % for each part z=[z1,z2... z(end)]
% the code that I am currently using, but I do not understand why such an answer
step = 10;
for j = 1:step:size(x,2)-step
a = y(j:j+1);
b = x(j:j+1);
z(1,j+1)= smooth(a,b,0.8,'rloess')';
end
The answer "z" is not correct (it's a 2,4097 matrix) and there are a lot of zeros in the data.
Thank you in advance

Answers (1)

Mathieu NOE
Mathieu NOE on 4 Apr 2022
hello
not sure what you are trying to do , but if it's about splitting data in buffers and doing some math on the buffers (mean , rms, smooth,..), here maybe some code to help you :
clearvars
% dummy data
n=4097;
x= 1:n;
y = sin(2*pi*x./max(x))+rand(1,n);
buffer = 10; % nb of samples in one buffer (buffer size)
overlap = 5; % overlap expressed in samples
%%%% main loop %%%%
m = length(y);
shift = buffer-overlap; % nb of samples between 2 contiguous buffers
for ci=1:fix((m-buffer)/shift +1)
start_index = 1+(ci-1)*shift;
stop_index = min(start_index+ buffer-1,m);
time_index(ci) = round((start_index+stop_index)/2); % time index expressed as sample unit (dt = 1 in this simulation)
mean_data(ci) = mean(y(start_index:stop_index)); %
end
x_mean = x(time_index);
figure(1),
plot(x,y,x_mean,mean_data,'r');
  3 Comments
Mathieu NOE
Mathieu NOE on 6 Apr 2022
maybe check that you have not used or initialized some of my code's variables from another portion of code that would change arrays dimensions (from my code version)

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