/sci/ - Science & Math

[Anonymous No.9632734]

If you have a dynamical system like this, what's the difference between the two y(t)? Which one is the output?

[Anonymous No.9632740]

>>9632734
I'm an engineer so math isn't my best skill

[Anonymous No.9632746]

>>9632734
are you trolling

[Anonymous No.9632748]

Either you don't have the first clue of what a differential equation and shouldn't be starting dynamical systems or you should see why your question makes no sense.
>>9632740
Oh don't worry that's clear. But lelgineers at least should know what a fucking ode is.

Also >>>/sqt/.

[Anonymous No.9632750]

>>9632746
Sadly not

[Anonymous No.9632762]

>>9632750
how did you even come across that picture if you don't know how it works. regardless, they're both the output but the first one is just the rate of change of the function y(t) like the speed of y(t). i'm curious if you know what an ode is

[Anonymous No.9632765]

The output is y(t), the input is most likely u(t) and d(t).

The output not only depends on the various constants and inputs, but also oupon its own change.

That would mean in practice, the faster your output changes, the smaller it will be, like a frequency-selective (low pass) filter in electronics.

[Anonymous No.9632769]

>>9632765
How can you possible calculate the current output if you have both the rate of change dy(t)/dt and the current output y(t)? Makes no sense

[Anonymous No.9632777]

[math] y'(t) = A(t)\, y(t) +B(t) [/math]

with, in, your case, [math] A(t)=-a [/math] and [math] B(t) = b\, u(t)-c\, d(t) [/math].

Now

[math] y(t) = S(t,0) y(0) + \int_0^t S(t,s) \, B(s) \, ds [/math]

where [math] S(t,s) = e^{I(t)-I(s)} [/math], where [math] I(t) = \int_0^t A(s) \, ds. [/math] In your case, A is constant, which makes this quite simple.

Here [math] B [/math] plays the role of a force, which you might be calling an input.

[Anonymous No.9632781]

>>9632777
Nice copypasta, doesn't answer any of the questions on this thread

[Anonymous No.9632786]

>>9632777
>Nice copypasta
?

Why doesn't it answer the question? Depending on context, B will be the input.

OP needs to fill some general mathematical gap, I try to help. he's asking
>what's the difference between the two y(t)
and that's like asking what the difference of x is in the equation
x^2 - x = 2
The equation has solutions such as x=2, and there's no "difference between" the expressions x in the equation. It's a placeholder for the same thing.

The A(t)y(t) expression makes it so that the rate of change on y(t) depends on it's current value too - that's just what a differential equation says.

To state some more broader facts, this sort of feedback equation is solvable for a wide range of cases, one of the most general incarnations (A being some operator) being found under
https://en.wikipedia.org/wiki/Dyson_series
For A being a matrix, we have
https://en.wikipedia.org/wiki/Magnus_expansion
and S(t,0) has a general solution in terms of a limit (unlike S(t,s)).

[Anonymous No.9632787]

>>9632781
he literally couldn't have copied that from anywhere m8

>>9632769
you assume a solution to solve these differential equations then just take that form as the solution

>>9632777
the solution, although elegant, is not going to help him in this case

[Anonymous No.9632828]

>>9632769

Do you know what feedback is?

If you talk about electronics, a, b and c are amplifiers. The sum in mathematics can be done using a summing circuit with operational amplifiers, same with substraction. A differentiator circuit is also possible with opamps, but only in a certain frequency range.

You have all the elements, wire them together, the result is a circuit with feedback.

Look up the Laplace transform, makes solving such equations a breeze most of the time.

[Anonymous No.9632835]

>>9632828
First one must be a plus, my bad.