How to use mathematically the I and D of a PID controller

Question

I am trying to mathmatically understand how the $P$, $I$, and $D$ parameters work on a system, quite having a hard time doing so.

I've only been able to show that the Steady State Error (SSE) never becomes zero for using a P using a simple example, but I am not able to show for the others, I would be very grateful if any could mathematically show it.

SSE: Consider a the plant being \begin{equation} G(S) = \frac{1}{s(s+1)} \end{equation} and the controller only being $G_c(s) = K$. The error signal is computed to be \begin{equation} E(s) = \frac{1}{1+ \frac{k}{s(s+1)}}. \end{equation} Calculating the limit as $s \to 0$ for at step input we get \begin{equation} SSE = \frac{1}{1 + k}, \end{equation} thus showing SSE decreases but never becomes zero.

I might have found a solution for overshoot as well, but $I$ and $D$ are still a bit tricky to handle. I cannot see how the damping ratio changes due to the addition of an $I$ and $D$.

Answer 1

Since you're using a second-order plant, I'll tailor my answer to that.

As you've noted above, the proportional gain, $k_p$, is what determines most of the controller's response to error, though a $P$ term alone generally has steady state error (this is sometimes called "droop").

To fix this steady state error, we introduce the integral gain, $k_I$. This helps to eliminate steady state error because it looks not only at the magnitude of the error, but also at its duration. Specifically, if we let $e(t)$ denote our error signal, a $PI$ controller takes the form \begin{equation} u_{PI}(t) = k_pe(t) + k_I\int_{0}^{t}e(\tau) d\tau \end{equation} and has transfer function \begin{equation} \frac{U(s)}{E(s)} = k_p + \frac{k_I}{s}. \end{equation}

If we want to control a second-order system with transfer function \begin{equation} \tag{1} \frac{Y(s)}{U(s)} = \frac{A}{s^2 + a_1s + a_2}, \end{equation} and have it track a reference signal $R(S)$, our controller becomes \begin{equation} U(s) = k_p(R(s) - Y(s)) + k_I\left(\frac{R(s) - Y(s)}{s}\right). \end{equation} If we substitute this into Equation $(1)$ and do some rearranging, we find that our characteristic equation is \begin{equation} s^3 + a_1s^2 + (a_2 + Ak_P)s + Ak_I = 0. \end{equation} Here we can control two coefficients, but we'd like to control three of them (all except the $s^3$ one) because then we can set all three roots of the characteristic polynomial. In practice, $PI$ controllers can lead to overshoot and oscillations. Thus, while a $PI$ controller has solved the problem of having steady-state errors, it can still cause other problems and for these reasons we introduce the $D$ term.

Going through the above manipulations again with a $PID$ controller, which has \begin{equation} U(s) = k_P + \frac{k_I}{s} + k_Ds \end{equation} as its transfer function, we see that the characteristic equation of our second-order system with a $PID$ controller is \begin{equation} s^3 + (a_1 + Ak_D)s^2 + (a_2 + Ak_P)s + Ak_I = 0. \end{equation} Here, we can choose three coefficients and hence determine all three roots of the characteristic polynomial. This means that, as with the $PI$ controller, the steady-state error is under our control, but also that the system's oscillations are as well. In practice, the $D$ term lets us cancel the oscillations we see when the $I$ term is introduced.

Answer 2

A formulaic way to handle questions of steady state error is with the Final Value Theorem (FVT). The theorem says that if all the closed loop poles are in the left half plane, and the closed loop transfer function is given by $G_{cl}(s)$, then for input $U(s)$ the output of the closed loop system asymptotically approaches \begin{equation} y_{ss} = \lim_{s\to 0} sG_{cl}(s)U(s) \end{equation} It must be checked that the poles of $G_{cl}(s)$ are in the left half plane before actually evaluating this.

If we model the plant as \begin{equation} G(s) = \dfrac{Z(s)}{P(s)} \end{equation}, and the controller as a "practical" PID controller with a filtered derivative \begin{equation} H(s) = k_p + \dfrac{k_d s}{(\tau s + 1)} + \dfrac{k_i}{s} \end{equation} then we can consider the closed-loop error transfer function \begin{equation} E(s) = \dfrac{1}{1+G(s)H(s)} \end{equation} to find the final value of the error to a unit step ($U(s) = 1/s$) when the closed loop system is stable. For your system plant model \begin{equation} G(s) = \dfrac{1}{s(s+1)} \end{equation} after some algebra we have \begin{equation} e_{ss} = \lim_{s \to 0} sE(s)\dfrac{1}{s} = \lim_{s\to 0} s \cdot \dfrac{s^2(\tau s + 1)(s + 1)}{k_d s^2 + k_p s (\tau s + 1) + k_i (\tau s + 1) + s^2(\tau s + 1)(s + 1)}\cdot \dfrac{1}{s} \end{equation} Assuming the poles of $E(s)$ are in the left half plane, this limit can be evaluated by plugging in $s = 0$ to yield that the steady state error is \begin{equation} e_{ss} = \dfrac{0}{k_i} = 0 \end{equation}

Even if the input were a ramp ($U(s) = 1/s^2$), this closed loop system would still have zero steady-state error.

Also, you do have a bug in your current analysis. The steady-state error to a step input, for your plant with a proportional controller, should be zero. (Your system is a "type 1" system. For more information see here: http://en.wikibooks.org/wiki/Control_Systems/System_Metrics#System_Type )

The relationship of $k_d$ and $k_i$ to overshoot and damping is usually understood as $D$ increasing damping (reducing overshoot), and $k_i$ increasing overshoot but reducing steady state error. It is probably easiest to think of these effects as due to a "lag" introduced by integrating the error (it is looking into the past errors), and a "predictive" effect that the derivative has (it is like extrapolating into the future). These concepts become a lot more precise when studying classical frequency domain analysis of control systems.