Algorithm 1 Processing of the given signal. Using the following algorithm, we gather large statistical data

Part 1. while$T_s=0$▹ continue until the settling time $T_s$ is determined $t\leftarrow t+\Delta t.$▹ increasing time by $\Delta t$ The first mechanism of recording $\tilde{t}$ (orange dots on Figure 11): if$n\le \left[f\left(t\right)/0.1\right]<n+1$then▹ where [ ] represents the floor function     $n++;\tilde{t}=(2t+\Delta t)/2;$ ▹ when the signal crosses decimal values a reference time is recorded (orange dots on Figure 11).     if $n+k=1$ then ▹ if this is the first a reference time $\tilde{t}$ is recorded, then         $t_{min}=\tilde{t};t_{max}=\tilde{t};$ ▹ assign this value to both $t_{min}$ and $t_{max}$     else         $t_{max}=\tilde{t};$ ▹ renewing only the maximum reference time     end if end if $\Delta t=t_{min}/N;$▹ calculate time step See Part 2 for further explanation. Part 2. The second mechanism of recording $\tilde{t}$ involves an extremum search (green dots on Figure 11). Since the function $f\left(t\right)$ is continuous, a moment of time when the derivative $f^{\prime }\left(t\right)$ changes sign implies a local extremum. if$f^{\prime }\left(t\right)·f^{\prime }(t-\Delta t)<0$then▹$f^{\prime }\left(t\right)$ changes sign, so a local extremum occurs in $(t-\Delta t,t)$     $t_k^{ex}=(2t+\Delta t)/2;k++;$ ▹ record the local extremum time and the number of them     $f_k^{peak}=f\left(t_k^{ex}\right);$ ▹ amplitude at the extremum     if $f_k^{peak}\ge f_{k-1}^{peak}$ then ▹ we have found a larger peak         $PO=(f_k^{peak}-1)·100\%$ ▹ percentage overshoot     end if     if $k>1$ then ▹ if there are multiple extremums         $O{A}_k=|f_k^{peak}-f_{k-1}^{peak}|$ ▹ the amplitude difference between the neighbor oscillations         $O{A}_{max}=max(O{A}_k,O{A}_{max})$ ▹ recording the maximum $OA$     end if     if $k=1$ then ▹ if this is the first extremum. This is done so that the $t_{max}$ does not increase uncontrollably.         $t_{max}=max(t_{max},t_k^{ex});$ ▹ the condition $k=1$ is needed so that $t_{max}$ does not increase uncontrollably     else         $\tilde{t}=t_k^{ex}-t_{k-1}^{ex};$▹ record the time between oscillations for additional accuracy (Nyquist theorem)         $t_{min}=min(t_{min},\tilde{t});t_{max}=max(t_{max},\tilde{t});$ ▹ in case this time is lower that the current $t_{min}$ or larger than the current value of $t_{max}$     end if end if $\Delta t=t_{min}/N;$▹ calculate time step Cont. At last, let us describe the test for finding $T_s.$ if$f\left(t\right)\in [0.97,1.03]$then▹ the signal appeared in the 3% range     $t_{3\%}=t;m=0;$ ▹ record the time and assign zero to the logic parameter m needed for the following cycle     while $t\le t_{3\%}+t_{max}$ AND $m=0$ do ▹ starting from $t_{3\%}$ for the time length of $t_{max}$ we check for the 3% criteria         $t=t+\Delta t$         if $f\left(t\right)\notin [0.97,1.03]$ then ▹ the signal exited the 3% area            $m=-1;$ ▹ therefore, the test for settling time stops and the algorithm resumes its work         end if     end while     if $m\ne -1$ then ▹ if the 3% condition did not break once during this test         $T_s=t_{3\%}$ ▹ we have found the settling time     end if end if