Order of a transfer function and its properties

Question

I hope this is the right place to ask since I couldn't find a control systems SE.

Lets say we have the transfer function $P(s) = \frac{1}{s(s+15)(s+10)} = \frac{1}{s(s^2+25s+150)}$(completely made up numbers) of some sort of device. This is without any controller or feedback attached.

How should I look at the $\frac{1}{s}$ that's in there? Do I treat it as a separate multiplier or do I handle this as a third order system?

Without the integrator this is a simple 2nd order system with clearly defined relative damping and natural frequency because of $\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$.

Technically I could treat the 1/s as an I controller with the value of $P_i$ value of 1. Would this system even have a steady state error with a simple unity P feedback? Technically it would automatically be a PI controlled system right?

If the questions don't seem clear it's because I don't really know what to ask here.

edit: I managed to read something about types of a transfer function(0, 1 or 2) which only modifies the error you will have. The transfer function shown here would be considered a type 1 because it has 1 pole in the origin(2 means type 2 and 0 means type 0 I guess). Type 1 means that the steady state error will be 0 for a step input, which confirms my speculation about the error. I still don't know if it has any other effect.

edit2:

So this picture shows the exact same thing in 2 forms. They're are functionally the same. The PID controller is C(s) in this case. Well it's a PI controller with K = 1, without me actually adding anything to the system. I've just added a negative feedback loop.

Answer 1

Okay, something indeed does not seem very clear to you. Let me try to clarify that ;)

Let $P_s = \frac{1}{(s+15)(s+20)}$ denote the second order part you do understand. I don't know what you mean by type 2 system, but this can represent any mass-spring-damper system as you already stated.

Where you go wrong is that you view the integrator $I = \frac{1}{s}$ as a controller. You recognized that $P = IP_s$, which is true. A controller, however, would introduce (negative) feedback, which is not present here.

Controller interpretation - wrong!

Let $Y$ denote the Fourier transform of the output of your plant and $E$ that of the error. (Without loss of generality we set the reference $R=0$). Then: $$Y = P E = P(R-Y),$$ from which we can derive the well-known expression for the complementary sensitivity: $$T = \frac{Y}{R} = \frac{P}{1+P}.$$ (In literature, often $L$ is used instead to denote the open-loop transfer function $CP$, where $C$ is the controller, but let's keep using your notation instead.)

$T = \frac{1}{s^3 + 25s^2 + 150s+1}$, is the real transfer function of your second order system with your integrator as negative feedback controller from input $R$ to output $Y$. Note that this system indeed has no steady state error as you correctly noted.

Correct interpretation

Back to your system and the correct interpretation: the integrator is nothing more than a ''multiplier'' as you stated. A physical interpreation could be a valve controlling a water flow into a vat on an old-fashioned mechanical scale.

Your input corresponds to the valve setting $u$. The output of the integrator, $Us$ or $\int u \mathrm dt$ in the time domain corresponds to the amount of water in a vat. Then your second order transfer function $P_s$ takes the amount of water as input (force induced by the weight in this case) and outputs movement of the scale $Y$. Where this analogy goes wrong is that $m$ obviously changes as the vat fills, and inertia of the water flow is ignored, but hopefully you get my point.

Another analogy would be an ideal capacitor in-between your input and what actually goes into your second-order system.