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Maximizing value for an array with integral function.

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Lal
Lal 2018-2-20
评论： Torsten 2018-2-27
I have 'time(t)' and 'acceleration(A)' values. t1 and t2 are the lower and upper limits, where (t2-t1 = 10).
f = max{((integration(A,t,t1,t2))^2.5) * (t2-t1)}
Please help me to put this function in my matlab code.
for t1=1:length(t)
for t2=t1:t1+10
f = (((int(A,t,t1,t2)))^2.5)*(t2-t1);
end
end
I am using this for loop, but getting error for integration.

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Torsten
Torsten 2018-2-22
编辑：Torsten 2018-2-22
fmax = -Inf;
for t1=1:numel(t)-10
fmax = max((trapz(t(t1:t1+10),A(t1:t1+10)))^2.5*(t(t1+10)-t(t1)),fmax);
end
fmax
Best wishes
Torsten.
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Lal
Lal 2018-2-27
Thank you Torsten. I understand now where I am doing mistake. Your suggestion helped a lot.
And one more thing in the result t(i) and t(i+15) is fixed to "t(i+15) - t(i) = 15".
But It can be any point between t(i+15) and t(i).
For example:
t_low = 7
t_up = 10
for f_max = 100
(because of this reason I introduced "i" and "j". As you said its two-dimensional array )
Thanks
Torsten
Torsten 2018-2-27
f_max = -Inf;
for i = 1:85
for j = 1:15
f = (trapz(t(i:i+j),A(i:i+j)))^2.5*(t(i+j)-t(i))
if f > f_max
tlow_max = t(i);
tup_max = t(i+j);
f_max = f;
end
end
end
f_max
tlow_max
tup_max

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