hello
I would suggest you use first a lowpass filter with long enought time constant (equivalent to moving average with buffer long enough) so that data for time below 60 s will alaways be well below the 5 value.
many "smoothing" approaches available here :
clc
close all
Fs = 1000;
samples = 1000;
dt = 1/Fs;
t = (0:samples-1)*dt;
y = square(2*pi*3*t) + 0.1*randn(size(t));
% %%%%%%%%%%%%%%%%
figure(1)
N = 10;
ys = slidingavg(y, N);
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with slidingavg' ]);
% %%%%%%%%%%%%%%%%
figure(2)
N = 10;
ys = medfilt1(y, N,'truncate');
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with medfilt1' ]);
grid on
%%%%%%%%%%%%%%%%
figure(3)
N = 10;
ys = sgolayfilt(y,3,51);
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with sgolayfilt' ]);
grid on
%%%%%%%%%%%%%%%%
NN = 4;
Wn = 0.1;
[B,A] = butter(NN,Wn);
figure(4)
ys = filtfilt(B,A,y);
plot(t,y,t,ys);legend('Raw','Smoothed');
title(['Data samples at Fs = ' num2str(round(Fs)) ' Hz / Smoothed with butterworth LP' ]);
grid on
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function out = slidingavg(in, N)
% OUTPUT_ARRAY = SLIDINGAVG(INPUT_ARRAY, N)
%
% The function 'slidingavg' implements a one-dimensional filtering, applying a sliding window to a sequence. Such filtering replaces the center value in
% the window with the average value of all the points within the window. When the sliding window is exceeding the lower or upper boundaries of the input
% vector INPUT_ARRAY, the average is computed among the available points. Indicating with nx the length of the the input sequence, we note that for values
% of N larger or equal to 2*(nx - 1), each value of the output data array are identical and equal to mean(in).
%
% * The input argument INPUT_ARRAY is the numerical data array to be processed.
% * The input argument N is the number of neighboring data points to average over for each point of IN.
%
% * The output argument OUTPUT_ARRAY is the output data array.
%
% © 2002 - Michele Giugliano, PhD and Maura Arsiero
% (Bern, Friday July 5th, 2002 - 21:10)
% (http://www.giugliano.info) (bug-reports to michele@giugliano.info)
%
% Two simple examples with second- and third-order filters are
% slidingavg([4 3 5 2 8 9 1],2)
% ans =
% 3.5000 4.0000 3.3333 5.0000 6.3333 6.0000 5.0000
%
% slidingavg([4 3 5 2 8 9 1],3)
% ans =
% 3.5000 4.0000 3.3333 5.0000 6.3333 6.0000 5.0000
%
if (isempty(in)) | (N<=0) % If the input array is empty or N is non-positive,
disp(sprintf('SlidingAvg: (Error) empty input data or N null.')); % an error is reported to the standard output and the
return; % execution of the routine is stopped.
end % if
if (N==1) % If the number of neighbouring points over which the sliding
out = in; % average will be performed is '1', then no average actually occur and
return; % OUTPUT_ARRAY will be the copy of INPUT_ARRAY and the execution of the routine
end % if % is stopped.
nx = length(in); % The length of the input data structure is acquired to later evaluate the 'mean' over the appropriate boundaries.
if (N>=(2*(nx-1))) % If the number of neighbouring points over which the sliding
out = mean(in)*ones(size(in)); % average will be performed is large enough, then the average actually covers all the points
return; % of INPUT_ARRAY, for each index of OUTPUT_ARRAY and some CPU time can be gained by such an approach.
end % if % The execution of the routine is stopped.
out = zeros(size(in)); % In all the other situations, the initialization of the output data structure is performed.
if rem(N,2)~=1 % When N is even, then we proceed in taking the half of it:
m = N/2; % m = N / 2.
else % Otherwise (N >= 3, N odd), N-1 is even ( N-1 >= 2) and we proceed taking the half of it:
m = (N-1)/2; % m = (N-1) / 2.
end % if
for i=1:nx, % For each element (i-th) contained in the input numerical array, a check must be performed:
if ((i-m) < 1) & ((i+m) <= nx) % If not enough points are available on the left of the i-th element..
out(i) = mean(in(1:i+m)); % then we proceed to evaluate the mean from the first element to the (i + m)-th.
elseif ((i-m) >= 1) & ((i+m) <= nx) % If enough points are available on the left and on the right of the i-th element..
out(i) = mean(in(i-m:i+m)); % then we proceed to evaluate the mean on 2*m elements centered on the i-th position.
elseif ((i-m) >= 1) & ((i+m) > nx) % If not enough points are available on the rigth of the i-th element..
out(i) = mean(in(i-m:nx)); % then we proceed to evaluate the mean from the element (i - m)-th to the last one.
elseif ((i-m) < 1) & ((i+m) > nx) % If not enough points are available on the left and on the rigth of the i-th element..
out(i) = mean(in(1:nx)); % then we proceed to evaluate the mean from the first element to the last.
end % if
end % for i
end
