1
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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 Oct 1 '16 at 9:40
2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Jan 30 '18 at 22:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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