Programming Problem on code

4 vues (au cours des 30 derniers jours)
Bilal Sadiq
Bilal Sadiq le 23 Oct 2020
Commenté : Fangjun Jiang le 23 Oct 2020
Hi all ,,
I shall be very grateful to you if you answer my query ,,,My problem is i want to genereate the following graph
t1=[0 0.1 0.3 0.4 0.5 0.7 0.8];
y=[-10 10 15 10 8 -5 0];
Now what i want is : I want to repeat this graph upto t=10 sec (i have time vector t=0:0.01:10 and i want to plot the above y axis values till time t=10)
hope i have put my query right,waiting for some positive response
with kind regards
Bilal
  4 commentaires
Bilal Sadiq
Bilal Sadiq le 23 Oct 2020
See what i have tried to to do but in vain
t=0:0.1:10;
% t_x_ankle=[0 0.1 0.2 0.3 0.4 0.5 0.6...
% 0.7 0.8 0.9 1.0 1.1 1.2 1.3...
% 1.4 0.1 0.2 0.3 0.4 0.5 0.6...
% 0 0.1 0.2 0.3 0.4 0.5 0.6...
% 0 0.1 0.2 0.3 0.4 0.5 0.6];
y_ankle=[-10 10 15 10 8 -5 0];
y=zeros(1,length(t));
count=1;
for i=1:length(t)
y(i)=y_ankle(count)
count=count+1;
if count>7
count=1;
end
end
plot(t,Y,'bo')
Please help me
Fangjun Jiang
Fangjun Jiang le 23 Oct 2020
In the simpliest case, would this be what you want?
y=[-10 10 15 10 8 -5 0];
t1=0:0.1:0.6;
t=0:0.1:10.4;
yy=repmat(y,1,15);
plot(t,yy)

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Réponses (1)

Fangjun Jiang
Fangjun Jiang le 23 Oct 2020
t1 doesn't contain t=0.2 and t=0.6. Not sure if it is correct or there is a typo.
You can do repmat(y,1,10) or as many times as you want to match t=0:0.01:10
  2 commentaires
Bilal Sadiq
Bilal Sadiq le 23 Oct 2020
Please see the figure attached ,this is what i want to plot for a time vector of t=0:0.01:10
Fangjun Jiang
Fangjun Jiang le 23 Oct 2020
%% Original data
t1=[0 0.1 0.3 0.4 0.5 0.7 0.8];
y=[-10 10 15 10 8 -5 0];
%% interpolate for more data points
t2=0:0.01:0.8;
y2=interp1(t1,y,t2);

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