Equation 16: Missing dt Term Without f(t)?

[theycallmevirgo]

TL;DR

How can an equation contain a time derivative without any f(t)?

In equation 16 they seem to have a dt term without f(t). Am I missing something?

 

[ergospherical]

reckon $dt$ is just supposed to be some time interval, maybe smallish (can't say without seeing the book)

 

[PeroK]

Summary:: How can an equation contain a time derivative without any f(t)?

In equation 16 they seem to have a dt term without f(t). Am I missing something?

Context is everything here. It looks more like there's an integral with respect to time in there, but it's highly contextual notation.

 

[ergospherical]

Context is everything here. It looks more like there's an integral with respect to time in there, but it's highly contextual notation.

fwiw I'm assuming the formula in the picture is the same one as (or a variation of) this here:
https://en.wikipedia.org/wiki/PID_controller#Controller_theory

the bit in the brackets in the picture corresponding to $\int e(\tau) d\tau$ on the wiki version

 

[PeroK]

Well, I guess if you don't need to put the range on an integral, why bother with the integral sign at all?

 

[DaveE]

Well, I guess if you don't need to put the range on an integral, why bother with the integral sign at all?

Yes, I agree their notation sucks.
In dynamic systems (control systems) it is common to leave the range out, with an assumption it's "all relevant history". This is because most of the interest is in the behavior (stability, etc.), not the actual operating points. One of the cheats you get from linear systems, the integral can be treated like an operator; it might not matter what the actual value is.

 

[WWGD]

Don't we just assume $f(t)== 1$? I mean, we have $\int dt =t$

 

[theycallmevirgo]

Don't we just assume $f(t)== 1$? I mean, we have $\int dt =t$

That's exactly what I thought, originally. But if so, why include it at all?

 

[WWGD]

That's exactly what I thought, originally. But if so, why include it at all?

Because the result is not necessarily " neutral" when computed. You will not just ( necessarily) get a 1 multiplying . Edit: On my phone, will give you more thorough answer tmw when I get to my pc.