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: equivalent

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.


1 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.

  • $\begingroup$ I've added 2 models of the same thing in my post. Isn't that considered a PI controller? $\endgroup$
    – Tryphon
    Jul 5, 2018 at 14:37
  • $\begingroup$ You originally stated that you did not have any feedback attached. In that case, no, it cannot be considered a PI controller. It's just two systems in series, the integrator and the second order system. As to your picture: both are indeed equivalent, but the transfer function between step and scope is not equal to $P$, but equal to $T = P/(1+P)$. $\endgroup$
    – R. H.
    Jul 5, 2018 at 14:43
  • $\begingroup$ Adding a feedback completely changes the resulting transfer function! Which part you denote as controller and which part as plant / system is entirely up to you. Also try to losen up about what a PID controller is, which form it has and what value the coefficients are. It is nothing more than a transfer function, indistinguishable from your plant / system apart from that one often designs a controller but is given a plant. $\endgroup$
    – R. H.
    Jul 5, 2018 at 14:47
  • $\begingroup$ If you want a simple explanation about the basics of feedback control, watch this episode of a very good Youtube series: youtube.com/watch?v=O-OqgFE9SD4 $\endgroup$
    – R. H.
    Jul 5, 2018 at 14:49
  • $\begingroup$ And finally: in your picture, it could be seen as a I controller (not a PI, which would be $\frac{1}{s+1}$). $\endgroup$
    – R. H.
    Jul 5, 2018 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.