# I have a set of data that I need to separate into unequal intervals in order to calculate the RMS value of these data (The data points are continuous over time)

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ABDALLA AL KHALEDI 2023 年 5 月 1 日
I have a set of torque data and I need to calculate the rms (root-mean-square) torque. The problem is that I have almost 16000 data points with negative and positive values over time, These data must be seperated into intervals first and the length (unequal intervals) of each interval must be found.
The formula I am following is taken from here and can be written as:
What possible code can I use to perform a similar operation but on the full range of data as in the example picture below?
The torque data and time values saved as an Excel file and are attached below.
I know that MATLAB is able to find the rms value for a set of independent data as I read here but my data is continuous over time.
Note: some torque values do not repeat over a period of time and show up once in between two intervals, these can be omitted as I am intrested in the overall average of the intervals.
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Mathieu NOE 2023 年 5 月 2 日
can you share your excel file ?
ABDALLA AL KHALEDI 2023 年 5 月 2 日

So what I am trying to do is seperate the torque data into intervals; however, these intervals are not equally spaced. The reason being that I want to calculate the rms-torque (root-mean-square) value on a continuous time basis according to the formula found here. I know that MATLAB is able to find this value for a set of independent data as I read here but my data are continuous over time. In the picture above I wrote a simplified version of what I wanted to calculate. Note: some torque values do not repeat over a period of time and show up once in between two intervals, these can be omitted as I am intrested in the overall average.

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### 採用された回答

dpb 2023 年 5 月 4 日

For the solution for all torque levels without collapsing any,
d=[true; diff(tq(:,2)) ~= 0]; % TRUE if values change (down to machine eps)
TQ=tq(d,2); % torque values at change points
ix=find([d(:); true]); % location of changes in logical vector
N = diff(ix); % difference in locations --> number
% so calculate the torque rms by group weighted by time
dt=mean(diff(tq(:,1))); % remove roundoff in dt calculationto get one dt overall
T=dt*(size(tq,1)-1); % total T
tq_rms=sqrt(sum(TQ.^2.*(N-1)*dt)/T)
tq_rms = 0.7850
@Jan coded <an FEX submission as a mex function RunLength> that is quite a bit faster if series lengths were to get to be really, really long...and is the root source of the M-code outline above although the general technique has wide applicability in finding crossing, etc., etc, etc., ...
ADDENDUM: The above takes care of the one-sample elements automatically in that the time interval is zero for a count of one...you don't even have to remove them first.
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dpb 2023 年 5 月 5 日
"...about that don't matter from a practical point of view"
For the sample dataset, all the above return essentially the same result. The time-weighted value could be somewhat more different depending on the particular transient behavior, however...
ABDALLA AL KHALEDI 2023 年 5 月 5 日
Thank you all. Yes the sample spikes are to be neglected.

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### その他の回答 (3 件)

Joe Vinciguerra 2023 年 5 月 2 日
Here is one way to do it based on your example:
t = [0.2 0.4 0.6 0.8 1.0]'
data = [3 3 2 2 2]'
% find the unique torque values in data set to create an index for use below
[C, ia, ic] = unique(data, "stable")
% find the max and min times at which the unique values occure
t_max = accumarray(ic, t, [] ,@max)
t_min = accumarray(ic, t, [], @min)
% date the delta
t_delta = t_max - t_min
% perform you calcultion on these arrays
P = sum(C.^2 .* t_delta)
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ABDALLA AL KHALEDI 2023 年 5 月 3 日

The problem with this code is that the same torque value appears at several intervals, and when this code is applied just the first and last time value related to the indexed torque are considered resulting in an error in the t_delta values. The same unique torque index can be observed over several intervals.
For example,
The instantaneous time = [ 0.2, 0.4, 0.6, 0.8, 1.0, 1.2]
The Torque intervals:
from 0.2 to 0.4 ---> T= 2
from 0.4 to 0.6 ---> T= -5
at t = 0.6 ---> T= -6
from 0.6 to 0.8 ---> T= 2
from 0.8 to 1 ---> T= 3
from 1.0 to 1.2 ---> T= 3
So you see torque intervals are repeating for various intervals, plus some torque values appear as a single data point at a certain time but do not continue to show up as time passes like (-6) above, these I can ommit if possible but I didn't know how.
Please check the excel file I have attached with the question to fully understand the data.
I wrote the following code, how can I apply what I mentioned above?
also what is ia and ic? I didn't understand that.
%Data Analizer - RMS Torque Calculator
clc;
%Import the Torque data from your Excel File
%Ex: 'Torque_Data.xlsx' with the quotation marks
prompt = 'What is The File Name ? ';
fileName= input(prompt);
%Example 'C4:C10' with the quotation marks
prompt = 'What is The Range of Data ? ';
xlRange = input(prompt);
%Instantaneous Torque Data
%Instantaneous time in [s]
t = (0.005:0.005:81.92)';
%Total cycle time in [s]
t_tot = 81.92;
%The unique torque values in the data set
[T_int] = sort(unique(Data, "stable"));
[C, ia, ic] = unique(Data, "stable")
% find the max and min times at which the unique values occure
t_max = accumarray(ic, t, [], @max)
t_min = accumarray(ic, t, [], @min)
% date the delta (error)
t_delta = t_max - t_min
% perform you calcultion on these arrays
T_rms = (sum(C.^2.*t_delta))./t_tot

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Mathieu NOE 2023 年 5 月 4 日
hello
maybe this ? I know I reinvented the histogram ( :)) , in few lines...
% Torque post processing
t = data(:,1);
torque = data(:,2);
dt = mean(diff(t));
% create levels (edges, like for a histogram)
Tmin = floor(min(torque));
Tmax = ceil(max(torque));
levels = 10;
Trange = linspace(Tmin,Tmax,levels+1);
for ci = 1:levels
id = (torque>=Trange(ci) & torque<Trange(ci+1));
torq_extr = torque(id);
samples = numel(find(id));
duration(ci) = samples*dt;
% T² computation
torq2(ci) = mean(torq_extr.^2); % T1², T2², etc
end
% finaly :
Torque_rms = sqrt(sum(torq2.*duration)/sum(duration));
% direct rms computation
Torque_rms_alt = sqrt(mean(torque.^2));
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Star Strider 2023 年 5 月 5 日
It is not obvious to me exactly what you want to calculate. This isolates the regions between the spikes (that I refer to as ‘segments’), determines the times at the beginning and end of each segment, and calculates the RMS value of the signal in each region. I am not certain what you want to do beyond that.
One approach —
T1 = 16384×2 table
Time [s] Torque [N.m] ________ ____________ 0.005 -0.61183 0.01 -0.7138 0.015 -0.7138 0.02 -0.7138 0.025 -0.7138 0.03 -0.7138 0.035 -0.7138 0.04 -0.7138 0.045 -0.7138 0.05 -0.7138 0.055 -0.7138 0.06 -0.7138 0.065 -0.7138 0.07 -0.7138 0.075 -0.7138 0.08 -0.7138
VN = T1.Properties.VariableNames;
[pks,locs] = findpeaks(T1{:,2}, 'MinPeakProminence',1.5); % Quality Check
idx = [1; find(diff(sign(T1{:,2}-0.5))); size(T1,1)]; % Crossing Indices
blen = 2*ceil(numel(idx)/2); % Calculate Column Length Of 'idxmtx'
idxmtx = zeros(blen,1)+idx(end); % Preallocate
idxmtx(1:numel(idx)) = idx; % Assign Values To Vector
idxmtx = reshape(idxmtx, 2, []) % Convert To (2xN) Matrix Of Segment Beginning & End Indices
idxmtx = 2×33
1 774 1798 2276 2772 3286 3780 4307 4787 5280 5762 5809 6288 6769 7264 7792 8272 8766 9295 9773 10304 10783 11278 11806 12267 12780 13310 13787 14282 14778 742 1752 2243 2754 3251 3745 4259 4752 5248 5745 5777 6240 6734 7246 7743 8238 8750 9260 9740 10255 10748 11244 11756 12250 12763 13258 13740 14248 14745 15241
% idxmtx(:,end) % Check (Optional)
for k = 1:size(idxmtx,2)
idxrng = idxmtx(1,k) : idxmtx(2,k); % Vector Index Range
[b1,b2] = bounds(T1{idxrng,2}); % Range Of Segment Values (Optional
rangmtx(:,k) = [b1; b2; b2-b1]; % Store As Matrix (Optional)
timev{k} = T1{idxrng,1}; % Segment Time Vector
times(:,k) = T1{idxmtx(:,k),1}; % Segment Time Limits
rmsv(k) = rms(T1{idxrng,2}); % Segment RMS Value
end
times % Segment Beginning & End Times
times = 2×33
0.0050 3.8700 8.9900 11.3800 13.8600 16.4300 18.9000 21.5350 23.9350 26.4000 28.8100 29.0450 31.4400 33.8450 36.3200 38.9600 41.3600 43.8300 46.4750 48.8650 51.5200 53.9150 56.3900 59.0300 61.3350 63.9000 66.5500 68.9350 71.4100 73.8900 3.7100 8.7600 11.2150 13.7700 16.2550 18.7250 21.2950 23.7600 26.2400 28.7250 28.8850 31.2000 33.6700 36.2300 38.7150 41.1900 43.7500 46.3000 48.7000 51.2750 53.7400 56.2200 58.7800 61.2500 63.8150 66.2900 68.7000 71.2400 73.7250 76.2050
rangmtx % Segment Minimum, Maximum & Difference
rangmtx = 3×33
-1.5296 -1.5296 -1.5296 -1.1217 -1.0197 -1.0197 -1.4276 -1.0197 -1.2237 -1.0197 0.3059 -1.0197 -1.0197 -1.2237 -1.4276 -1.0197 -1.0197 -1.5296 -1.2237 -1.0197 -1.0197 -1.1217 -1.3256 -1.1217 -1.2237 -0.9177 -0.9177 -1.5296 -0.8158 -0.9177 0 0.7138 0.6118 0.7138 0.8158 0.6118 0.5099 0.8158 1.3256 0.6118 0.6118 0.9177 0.8158 0.7138 0.7138 0.6118 0.6118 0.7138 0.6118 1.0197 0.9177 0.5099 0.7138 0.7138 0.8158 0.5099 0.6118 1.0197 0.6118 0.5099 1.5296 2.2434 2.1414 1.8355 1.8355 1.6316 1.9375 1.8355 2.5493 1.6316 0.3059 1.9375 1.8355 1.9375 2.1414 1.6316 1.6316 2.2434 1.8355 2.0394 1.9375 1.6316 2.0394 1.8355 2.0394 1.4276 1.5296 2.5493 1.4276 1.4276
rmsv % Segment RMS Values
rmsv = 1×33
0.7204 0.6390 0.6854 0.6242 0.5899 0.5981 0.6026 0.6295 0.6766 0.6445 0.3467 0.6364 0.6269 0.5885 0.6655 0.6218 0.6183 0.6538 0.6334 0.6280 0.6363 0.6604 0.6280 0.6270 0.6467 0.5686 0.6178 0.6929 0.5897 0.5984
figure
hp{1} = plot(T1{:,1}, T1{:,2}, 'DisplayName','Data');
hold on
% plot(T1{locs,1}, pks, 'r+', 'DisplayName','Peaks')
hp{2} = plot(times, [1;1]*rmsv,'-m', 'DisplayName','Segment RMS Values');
hold off
grid
xlabel(VN{1})
ylabel(VN{2})
legend([hp{1},hp{2}(1)], 'Location','best')
I believe this gest close to what you asked for, however I do not understand how the times weight the RMS values. Incorporating that would be straightforward.
.

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