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
Matlabsolutions.com provides guaranteed satisfaction with a
commitment to complete the work within time. Combined with our meticulous work ethics and extensive domain
experience, We are the ideal partner for all your homework/assignment needs. We pledge to provide 24*7 support
to dissolve all your academic doubts. We are composed of 300+ esteemed Matlab and other experts who have been
empanelled after extensive research and quality check.
Matlabsolutions.com provides undivided attention to each Matlab
assignment order with a methodical approach to solution. Our network span is not restricted to US, UK and Australia rather extends to countries like Singapore, Canada and UAE. Our Matlab assignment help services
include Image Processing Assignments, Electrical Engineering Assignments, Matlab homework help, Matlab Research Paper help, Matlab Simulink help. Get your work
done at the best price in industry.