Maximizing value for an array with integral function.

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.

Answer 1

[Torsten]

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.

[Lal]

Thank you for the reply Torsten. In my case, my limits "t1" and "t2" are also an array.

And also at the end it should give maximum value for paticular t1 and t2. 'fmax' at 't1=' , 't2='.

Thank you

[Torsten]

f_max = -Inf;
fun = @(x)interp1(t,A,x);
for i = 1:numel(t1)
  t1_actual = t1(i);
  t2_actual = t2(i);
  f_actual = (integral(fun,t1_actual,t2_actual))^2.5*(t2_actual-t1_actual);
  if f_actual > f_max
    t1_max = t1_actual;
    t2_max = t2_actual;
    f_max = f_actual;
  end
end
t1_max
t2_max
f_max

[Lal]

Thanks for your reply,

In the below code "A" is an array as a function of "t".

I have problem in the line "f = max((trapz(t(t1,t2),A(t1,t2)))^2.5*(t2-t1));"

With A(t1,t2) I am getting error in the loop.

 fmax = -Inf;

for i=1:25
    for j=i+1:i+15
        t1=t(i);
        t2=t(j);

          f = max((trapz(t(t1,t2),A(t1,t2)))^2.5*(t2-t1));
          if f>fmax
             t1_max=t1;
             t2_max=t2;
             f_max=f;
          end
      end
  end
  f_max
  t1_max
  t2_max

Thanks in advance

[Torsten]

1. t is a one-dimensional array, but in the call to "trapz", you use it as a two-dimensional array (t(t1,t2)).

2. The call trapz(x,y) to "trapz" with only one value for x (namely t(t1,t2)) and one value for y (namely A(t1,t2)) does not make sense.

As far as I understand what you want:

You have two one-dimensional arrays: t and A(t).

Further, you have two one-dimensional arrays t_low and t_up and you want to find the index i such that

(integral_{t=t_low(i)}^{t=t_up(i)} A(t) dt)^2.5 * (t_up(i)-t_low(i))

is maximized.

Is this correct ?

Best wishes

Torsten.

[Lal]

Yes Torsen, exactly I am doing as you explained. "t" is an one dimensional array and I have to define "t_low" and "t_up" from "t" such that, to say "t_low = 1:100" and "t_up = t_low+1:115".

t_up ~ t_low <=15 (This is the maximum difference, so i created t(i) and t(j) loops to perform this)

And also A(t_up) and A(t_low) should be respective t_up and t_low values.

Thanks

[Torsten]

Then what's wrong with the following code (assuming t is a 100 x 1 array with increasing t values and A is also 100 x 1):

f_max = -Inf;
for i = 1:85
  f = (trapz(t(i:i+15),A(i:i+15)))^2.5*(t(i+15)-t(i))
  if f > f_max
    tlow_max = t(i);
    tup_max = t(i+15);
    f_max = f;
  end
end
f_max
tlow_max
tup_max

[Lal]

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]

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