# Why oscillations in PI control?

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 ?

• 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. – A_A Oct 1 '16 at 9:40

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.