Settling time functionality - definition of duration

According to the settling time functions documentation:

"s = settlingtime(x,d) returns the time from the mid-reference level instant to the time instant each transition enters and remains within a 2% tolerance region of the final state over the duration d."

I initially interpreted this as the transition is considered 'settled' if it enters a 2% tolerance and then remains within that tolerance for a duration d. Instead the documentation also gives the following definition:

"Settle-seek duration [d] in seconds, specified as a positive scalar. This argument defines the duration after the mid-reference level instant that settlingtime looks for a settling time."

So this means if the step response happens to be within the tolerance at time mid-reference level+d then it is considered settled.

Consider the example attached. The step response does not continuously remain within the tolerance for longer than 1s at any point, so if I set a duration of say 3s then I would expect the function to output NaN. Instead it gives me the black circle since 3s after the mid-reference level happens to be within tolerance (green square). Setting the duration to either 2s or 4s does not give a settling time as expected since the value of the response at time mid-reference level+d is outside of the tolerance (purple and blue squares).

How can I calculate the settling time with these two 'duration' parameter interpretations set to two different values. I.e. I want the function to search for a duration d1 during which the step response remains with tolerance, but it should only search until it reaches a duration d2 after the mid-reference level.

[Mathieu NOE]

that looks pretty similar to this :

The settling time in matlab differs from the calculated value - MATLAB Answers - MATLAB Central

[Samuel Foster]

My question is not about a comparison to an approximation, but about functionality/capability that appears to be missing from the function.

From your response on that question: "It is defined as the time required by the response to reach and steady within specified range of 2 % to 5 % of its final value.". But in the example I give the function is not steady and yet the function still outputs a non-NaN settling time.

[Mathieu NOE]

ok, so you answered yourself

as your signal is not settling (not steady) using this function can lead to improper results

settling time is (IMHO) only meaningfull if your response is damped which is not your case

I don't know what was the purpose of yiur question , at least you have demonstrated it's not a case for using settlingtime

if the question is to find whatever points in the curve , maybe we have to figure out another approach

[Samuel Foster]

"settling time is (IMHO) only meaningfull if your response is damped which is not your case". This is potentially a reasonable assumption, but unfortunately the problem is more fundamental and exists also in cases where the signal is damped.

See the second example attached. I would like to find the point at which the response enters the 2% tolerance and remains within that tolerance for a duration of 5s. But the function picks out the point at around 12s, even though the response then exits the tolerance a second later.

load('step_response_2_order.mat')
[s,slev,sinst] = settlingtime(P_data(:,3),1000,5,'StateLevels',[0 1],'MidPercentReferenceLevel',90,'Tolerance',2);

Answer 1

[Mathieu NOE]

I could suggest this simple code to answer your last question

the first plot shows where the data is within 2% tolerance band (i.e +/- 1% in the code) ; there are 3 segments and we will the one that has a duration above 5s

load('step_response_2_order.mat')

t = P_data(:,1);
x = P_data(:,2);
y = P_data(:,3);

dt = mean(diff(t));
Fs = 1/dt;

y_final = y(end);
ind = abs(y-y_final)<=0.01;   % where the data is within 2% tolerance band (i.e +/- 1% in the code)
[begin,ends] = find_start_end_group(ind); % in samples 
figure
plot(t,x,t,y,t,ind,'r*');

% now keep only begin,ends values for a duration above given threshold
dur_min = 5; % seconds

duration = ends - begin; % in samples
ii = find(duration>=dur_min*Fs);

% final selection 
select = NaN(size(t));
select(begin(ii):ends(ii)) = 1;

% start / end time stamps 
beginS = begin(ii)*dt % in seconds

beginS = 15.9790

endsS = ends(ii)*dt % in seconds

endsS = 35.0010

% final plot 
figure
plot(t,x,t,y,t,select,'r*');

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function [begin,ends] = find_start_end_group(ind)
% This locates the beginning /ending points of data groups
% Important : ind must be a LOGICAL array

    D = diff([0;ind(:);0]);
    begin = find(D == 1);
    ends = find(D == -1) - 1;
end

[Samuel Foster]

Thanks, I was hoping there might be a solution with the inbuilt function that I'd missed but it seems not despite this being a reasonable interpretation of settling time.

I'll accept this answer and also submit a service request.