フィルターのクリア

peak average of consecutive values in a matrix array

4 ビュー (過去 30 日間)
Jorge Luis Paredes Estacio
Jorge Luis Paredes Estacio 2023 年 7 月 19 日
コメント済み: Mathieu NOE 2023 年 7 月 19 日
Hello, I would like to calculate the average of the peaks (positive and negative values) of consecutive values from a predifined number of elements from a Nx1 matrix, where N is the number of rows. For example, let's say I have a matrix with the following form (24x1):
A[1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21;22;23;24]
I want to obtain the peak and the average for every 5 elements,
B=[mean(peak(1,2,3,4,5)); mean(peak((2,3,4,5,6)); mean(peak((3,4,5,6,7));...........mean(peak((20,21,22,23,24))]
I am chosing every 5 elements to simplify the example. But, this number number of elements can be higher than 100 and the points vary dramatically from positive to negative as they come from acceleration earthquake signals from N higher than 245000.
I would appreciate the help.
  6 件のコメント
Dyuman Joshi
Dyuman Joshi 2023 年 7 月 19 日
Suppose this is the data in hand -
x = randi([-20 20],1,15)
x = 1×15
-5 -16 14 -18 19 -3 18 1 -10 -6 -12 16 5 8 0
Assume the moving window is 4, what should be the output in thise case? and what is the logic behind it?
Jorge Luis Paredes Estacio
Jorge Luis Paredes Estacio 2023 年 7 月 19 日
編集済み: Jorge Luis Paredes Estacio 2023 年 7 月 19 日
Thank you for your help.

サインインしてコメントする。

採用された回答

Mathieu NOE
Mathieu NOE 2023 年 7 月 19 日
hello
maybe this ?
I am using peakseek , this fex submission is available here
, but you can use the standard (slower) findpeaks function offered by TMW
the code provided below will buffer your data with length = 51 samples , then split the signal for each cahnnel into a positive and negative signal and find the positive and negative peaks , then take the mean of them (separately not mixing pos and neg peaks !)
example for X channel :
data = readmatrix('Signal.txt'); % acceleration earthquake signal in cm/sec2 ,X, Y and Z
[samples, channels] = size(data);
dt = 1; % sample rate (s)
t = dt*(0:samples-1); % time vector
%% home made solution (you choose the amount of overlap)
buffer_size = 51; % how many samples
overlap = 0; % overlap expressed in samples
%%%% main loop %%%%
[new_time,data_out_pos,data_out_neg] = my_peakmean(t,data,buffer_size,overlap);
figure(1),
plot(t,data(:,1),new_time,data_out_pos(:,1),'*-g',new_time,data_out_neg(:,1),'*-r');
title('X');
legend('raw data','pos peaks mean','neg peaks mean');
xlabel('Time(s)');
ylabel('cm/sec2');
figure(2),
plot(t,data(:,2),new_time,data_out_pos(:,2),'*-g',new_time,data_out_neg(:,2),'*-r');
title('Y');
legend('raw data','pos peaks mean','neg peaks mean');
xlabel('Time(s)');
ylabel('cm/sec2');
figure(3),
plot(t,data(:,3),new_time,data_out_pos(:,3),'*-g',new_time,data_out_neg(:,3),'*-r');
title('Z');
legend('raw data','pos peaks mean','neg peaks mean');
xlabel('Time(s)');
ylabel('cm/sec2');
%%%%%%%%%% my functions %%%%%%%%%%%%%%
function [new_time,data_out_pos,data_out_neg] = my_peakmean(t,data_in,buffer_size,overlap)
% NB : buffer size and overlap are integer numbers (samples)
% data (in , out) are 1D arrays (vectors)
shift = buffer_size-overlap; % nb of samples between 2 contiguous buffers
[samples,channels] = size(data_in);
nb_of_loops = fix((samples-buffer_size)/shift +1);
for k=1:nb_of_loops
start_index = 1+(k-1)*shift;
stop_index = min(start_index+ buffer_size-1,samples);
x_index(k) = round((start_index+stop_index)/2);
% find peaks
for n = 1:channels
tmp = data_in(start_index:stop_index,n);
tmp_pos = tmp;
tmp_pos(tmp_pos<0) = 0;
tmp_neg = tmp;
tmp_neg(tmp_neg>0) = 0;
[locsp, pksp]= peakseek(tmp_pos,1,max(tmp_pos)/10); % pos peaks above 10% of max pos amplitude
[locsn, pksn]= peakseek(-tmp_neg,1,max(-tmp_neg)/10); % neg peaks above 10% of max neg amplitude
data_out_pos(k,n) = mean(pksp); % mean of pos peaks
data_out_neg(k,n) = mean(-pksn); % mean of neg peaks
end
end
new_time = t(x_index); % time values are computed at the center of the buffer
end
  5 件のコメント
Jorge Luis Paredes Estacio
Jorge Luis Paredes Estacio 2023 年 7 月 19 日
I really appreaciate your help. it is sorted now. Thank you very much.
Mathieu NOE
Mathieu NOE 2023 年 7 月 19 日
as always, my pleasure !

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeDescriptive Statistics についてさらに検索

製品


リリース

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by