code explanation question help!

2019 4 1
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Adam Danz
Adam Danz 2019-4-1
编辑：Adam Danz 2019-4-1

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'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'.

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william Smith
william Smith 2019-4-1
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
Walter Roberson 2019-4-1
... "the 4th t" minus 0.03
william Smith
william Smith 2019-4-1
shouldnt be minus 0.3
Walter Roberson
Walter Roberson 2019-4-1
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
Adam Danz 2019-4-1
编辑：Adam Danz 2019-4-1
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
william Smith 2019-4-1
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
william Smith 2019-4-1
Pefrect Thank you Adam and Walter!!
Adam Danz
Adam Danz 2019-4-1
编辑：Adam Danz 2019-4-1
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
william Smith 2019-4-1
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.

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