When integral control is added to proportional controller (i.e. PI control), why steady state error becomes 0 and why oscillations appear , how actually the integrator works ?

  • $\begingroup$ Can you please clarify your question? Is there an existing system? What sort of process does the Automatic Control System drive? In the meantime, you might want to have a look at the PID diagram in this page. $\endgroup$
    – A_A
    Commented Oct 1, 2016 at 9:40

3 Answers 3


Integration has memory. When the error becomes zero, there is no guarantee that the integrator has reached to zero sum of the previous errors.

So even when there is no error at a particular time the integrator could still be under the influence of the errors from the history. That's why it continues to correct for a while more until it over/undershoots and counter signed error accumulates. If tuned badly, this actually drives things unstable. If tuned properly, counter-error doesn't accumulate as much as integral action corrects (mostly due to P correction simultaneously dragging the system back). This delay is what is often referred to as the phase lag of the integrator.

Similarly, the system might reach to an equilibrium with P action alone if the system is running away as much as P control is correcting. But if you have integral action with a similar reasoning the error starts to accumulate and at some point the control action starts to dominate the steady state error.

You can then make these statements more rigorous by putting them into proper mathematical language.


For symplicity, consider the SISO linear system \begin{align*} \dot x &= ax +bu\\ y &= cx \end{align*} with $x$,$u$ and $y$ taking values in $\mathbb R$. Assume that you want to stabilize the system by making $y$ to converge to some fixed set point $\bar y$. Then necessarely there must be a point $(\bar x,\bar u)$ such that \begin{align*} 0 &= a\bar x+b\bar u,& \bar y=c\bar x \end{align*} since $\bar x$ must be an equilibrium and such that $y=\bar y$. Trivially one has $$\bar u = -\dfrac{a}{bc}\bar x$$ One can define the error variables as $\tilde x=x-\bar x$, $\tilde y=y-\bar y$ and $\tilde u=u-\bar u$ that yield the error system \begin{align*} \dot{\tilde x} &= a\tilde x + b\tilde u,& \tilde y&=c \tilde x \end{align*} One can stabilize it by taking $\tilde u = k\tilde x$ such that $a+bk<0$. That in fact ensures that asymptotically $\tilde y=0$ which implies $y=\bar y$. The overall control law is $$ u = \tilde u + \bar u = k(x-\bar x) -\dfrac{a}{bc} \bar x = kx + \dfrac{1}{c}\left(-k-\dfrac{a}{b}\right)\bar y $$ Thus as you can se, simple proportional is not enough, you also need a feed-forward control action given by $$ u_{ss} = \dfrac{1}{c}\left(-k-\dfrac{a}{b}\right)\bar y $$ If you define an integrator system like $\dot\sigma = y-\bar y = C(x-\bar x)$ then also $\sigma$ will have a set point $\bar \sigma$, and with $\tilde \sigma=\sigma-\bar\sigma$, the extended error system becomes \begin{align*} \begin{bmatrix}\dot {\tilde x}\\\dot{\tilde\sigma}\end{bmatrix} &= \begin{bmatrix}a & 0\\c & 0\end{bmatrix}\begin{bmatrix}\tilde x\\\tilde{\sigma}\end{bmatrix} + \begin{bmatrix}b\\0\end{bmatrix}\tilde u, & \tilde y&=C\tilde x\end{align*} Take $\tilde u=k_p \tilde x + k_i \tilde\sigma$ (proportional+integral), then the closed loop system reads \begin{align*} \begin{bmatrix}\dot {\tilde x}\\\dot{\tilde\sigma}\end{bmatrix} &= \begin{bmatrix}a+bk_p & bk_i\\c & 0\end{bmatrix}\begin{bmatrix}\tilde x\\\tilde \sigma\end{bmatrix}, & \tilde y&=C\tilde x\end{align*} You can chose $k_p$ and $k_i$ to make the closed loop matrix to be Hurwitz, and thus to stabilize the error system and you don't need any feedforward action, in fact the control law is $$ u = \tilde u + \bar u = k_p(x - \bar x) + k_i (\sigma -\bar \sigma) $$ Then with $$ \bar \sigma = \dfrac{k_p}{k_i}\bar x$$ the control law to becomes $$ u = k_p x + k_i \sigma \qquad\qquad (1) $$ and that's all. Moreover without integral control you need a feed-forward action which depends on the system parameters. If you have uncertainties on the parameters for sure you wil have an error on your set-point. with the integral action your final control law (1) does not depend on the parameters! and as long as the closed-loop system is stable you have null error at the steady state. Oscillations depend on how you chose $k_p$ and $k_i$. In fact if you chose them such that the closed loop matrix has complex eigenvalues then you can have socillations. For MIMO system the reasoning extends trivially.

  • $\begingroup$ Thanks for the excellent answer. I have a similar problem. Could you provide a reference where I can learn more about this technique? $\endgroup$
    – Neil G
    Commented Jan 30, 2018 at 22:49

Personally speaking, the effects of the P, I and D terms can be interpreted in a much better way by taking into account a system’s characteristic polynomial.

Let’s consider a PI controller and a type-0 (no poles at the origin of the transfer function of the system) LTI system that has $R(s)$ as input/reference input, $E(s)$ as error, $U(s)$ as the control input and $Y(s)$ as the output. $C(s)$ is the PI controller's transfer function and $B(s)$ is the system's own transfer function.

A System With a PI Controller

Actually, rather than using time domain, analysing the effects in the s-domain is relatively easy.

In order to figure out what the I/integrator section of the PI controller is going to cause at the output, reducing the system blocks into just one block would be quite helpful and it will be much easier to anticipate the input’s effect on the output of the system.

The system’s transfer function can be;

$$B(s) = \frac{1}{s + 1}$$

On the other hand, the PI controller’s transfer function is;

$$C(s) = K_{P} + {\frac{K_{I}}{s}}$$

So, the closed-loop transfer function becomes;

$$T(s) = \frac{C(s)B(s)}{1 – C(s)B(s)} = \frac{{sK_{P}} + K_{I}}{s^2 + s(K_{P} + 1) + K_{I}}$$

According to that closed-loop transfer function, the system can be represented by just one block.

The Overall/Closed-Loop System

Before moving onto the effect of the I section, first of all, we need to ensure that the overall system should be stable which is the most important feature that is expected from a system. We want our system to be stable i.e. to not go beyond our control.

For a second-order transfer function, in order to make the system completely stable (not marginally stable as marginal stability is actually the red line in terms of stability), all of the coefficients of the denominator should be positive.

$$K_{P} > -1$$

$$K_{I} > 0$$

Now that the stability criteria for the overall system is obtained, we can move onto analysing the effect of the integrator part of the PI controller. The characteristic polynomial for a second-order transfer function is;

$$\Phi(s) = s^2 + 2{\zeta}{\omega_{n}}s + {\omega_{n}}^2$$

where $\zeta$ is the damping ratio and $\omega_{n}$ is the undamped natural frequency of the overall system. If the denominator and the characteristic polynomial are matched, the undamped natural frequency becomes;

$$\omega_{n} = {\pm}\sqrt{K_{I}}rad/s$$

Since the angular frequency as well as the classical frequency cannot have negative sign, the undamped natural frequency becomes;

$$\omega_{n} = \sqrt{K_{I}} rad/s$$

Lastly, the damping ratio becomes;

$$\zeta = \frac{K_{P} + 1}{2{\sqrt{K_{I}}}}$$

For the values of damping ratio that satisfies $0 < \zeta < 1$ relationship, the system’s output is said to be underdamped i.e. oscillations and some specific amount of overshoot occur.

Consequently, it is obvious that by increasing the value of the integrator gain $K_{I}$, the overall system’s output tends to be more underdamped and oscillations start to occur.


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