How can I remove repeating values from the vector?

6 vues (au cours des 30 derniers jours)
Mariusz Sledz
Mariusz Sledz le 21 Mar 2020
Commenté : Mariusz Sledz le 29 Mar 2020
Hi,
I've got a loop that simulates Simulink model, collects the data and plots everything on the graphs, here is simplified code:
for v = 4:3:26
air_velocity = v;
while abs(old_u - new_u) >= errortolerance % siumlate untill model converges
simOut = sim('TurbineWithVariablePitch_new_.slx',time);
umax = round(max(simOut.Velocity), 2);
old_u = new_u;
new_u = umax;
disp(['Current iteration: ',num2str(iteration), ' for time: ', num2str(time, '%10.5e') ])
if abs(old_u - new_u) <= errortolerance
cprintf('*blue',['Solution fully converged with the current error of ', num2str(abs(old_u - new_u))]);
cprintf('\n');
else
disp(['Solution did not converged with the current error of ', num2str(abs(old_u - new_u))])
end
if iteration >= maxiteration
cprintf('*red','Solution did not converged\n');
break
end
iteration = iteration +1;
time = time + timestep;
end
simulationtime = [0:0.01:2000];
a = round(simOut.AoA.signals.values);
n=max (numel(simulationtime)); %make the vectors the same size
u(end+1:n)=nan;
a(end+1:n)=nan;
AoAout = [AoAout , a]; %add to the results
SimulationTime = [SimulationTime , simulationtime'];
end
f4 = figure('Name','AoA vs Time');
plot(SimulationTime, AoAout, 'linewidth', 2);
ylabel('AoA (deg)');
xlabel('Time (s)');
The problem is, I'm getting repitetive values for AoA:
They are not always the same so unique command doesn't work... I've been stuck on this problem for a while. I'm fairly new to MATLAB and tring to achieve something similar to this:
  6 commentaires
Mariusz Sledz
Mariusz Sledz le 24 Mar 2020
Still have no clue how to do it. I'm alredy using round() function to get the values. I've tride to define position as a single vector from 0 to 360, does'n work.
How can I select single cycle? Or how can I average them?
Mariusz Sledz
Mariusz Sledz le 26 Mar 2020
Hi, I'm actually really desperate for the solution and still haven't figured it out.

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Mariusz Sledz
Mariusz Sledz le 27 Mar 2020
So... I found the solution. It may not be the most elegant but it works. The code goes troug the vector untill it finds the value that's smaller than 10. Then it starts saving the values untill it reaches 350. After that it fills the rest with NaN.
position=[];
CountOn = 0;
row=1;
End = 0;
for x = 1:1:1000
b=p(row);
d=NaN;
if b <= 10
CountOn = 1;
end
if CountOn == 1
d=b;
end
if b>=345 && CountOn==1
CountOn = 0;
End=1;
end
if End == 1
d=NaN;
end
position=[position, d];
row=row+1;
end
position = position';
Posting if anyone will encouter the same issue.
  1 commentaire
Mariusz Sledz
Mariusz Sledz le 29 Mar 2020
I've optimised the code a bit:
position=[];
CountOn = 0;
row=1;
End = 0;
newb = 10;
for x = 1:1:1000
oldb=newb;
newb=p(row);
d=NaN;
if newb > oldb && oldb < 10
CountOn = 1;
end
if CountOn == 1
d=oldb;
end
if newb<oldb && CountOn==1
CountOn = 0;
End=1;
end
if End == 1
d=NaN;
end
if row > 1
position=[position, d];
end
row=row+1;
end
position = position';

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