code explanation question help!

%can you please explain what this code is calcuating?
%Thanks!


y= x(find(t > t(i)-0.3))

Answer 1

[Adam Danz]

'i' is an integer, index value. For the sake of the example, let's say i = 4.

t(i) is the 4th value stored in 't'.

t > t(i)-0.3 produces a logical vector of 1s and 0s the same size as 't'. It contains 1s whenever the value of t is greater than the 4th t plus 0.03.

So, if t = [0 1 2 3 4 5 6 7] and i = 4, the logical vector would be [0 0 0 1 1 1 1 1].

find() returns the index values that are true. so, find([0 0 0 1 1 1 1 1]) would return [4,5,6,7,8].

x([4,5,6,7,8]) looks at the fourth, fifith, sixth, seventh, and eighth values of 'x'.

To summarize, that line pulls out a subsection of 'x' based on the values of 't'.

[william Smith]

Thank you Adam!

still little confused, how would this then apply to the following:

for i=1:length(t)
    if t(i)<0.3
        y=x(1:i); 
    else
        y= x(find(t > t(i)-0.3)); 
end

[Walter Roberson]

... "the 4th t" minus 0.03

[william Smith]

shouldnt be minus 0.3

[Walter Roberson]

william Smith:

Each iteration of your for loop writes over all of y. The end result is going to be the same as if you had only done the last iteration, when i=length(t) .

Note:

x(find(t>t(i)-0.3))

can be done more efficiently by

x(t>t(i)-0.3)

[Adam Danz]

length(t) just tells you how 'long' t is. For example, lenght([4 5 6 7 8]) = 5 since there are 5 values in that vector.

for i = 1:length(t) this creates a for-loop. The code within the for-loop will use the value 1 on the first iteration and then the value 2, and so on. It will stop when it creates the length of t.

So, on the 4th 'iteration', t(i) will access the 4th value in t.

Here's a walk-through example

t = [4 5 6 7 8 9 10 11 12]'; 
length(t)                    % = 9
for i = 1:length(t)          % for values 1 to 9
    if t(i)<0.3              % on 1st iteration, t(1) = 4; 4 < 0.3 is false
        y=x(1:i);            % this line is executed only when t(i) < 0.3         
    else                     % the line below is executed all other times
        y= x(find(t > t(i)-0.3));  %see my previous explanation
    end
end

@Walter, good catch. I was so focused on interpretation that I missed that.

[william Smith]

Thank you Walter,

my confusion comes from 0.3 in the last line, if i try to integrate y over 0.3 seconds,

so what i am trying to do is if t<0.3 then i use total time until i reach 0.3 then integrate as normal for values above 0.3.

would this even work?

Thanks again!

[william Smith]

Pefrect Thank you Adam and Walter!!

[Adam Danz]

I have a feeling this isn't the full section of code. Now that there's context, I might be able to help you further. Where are you doing the integration? Could you describe x?

[william Smith]

finding the force that needed to be applied on a mass where i have a broom attached to that mass and swings around (PID design controller problem)

F(i) = k_p*x(i) + k_i*trapz(y), as far as x it represents the angle that the broom swings at.