Can fullstate feedback control reject a disturbance?

Question

Compare to PID. The part that rejects disturbance is the I-term which integrated from error signal. On the other hand, in full state feedback there are no integral part. How does it reject the disturbance? or How we can make it to?

Answer 1

As you already stated at your question, PID-Controllers incorporate the I-Term in order to able to reject any external disturbance (unmodeled dynamics, linearizations, etc) and as a result achieve the desired output performance by driving the steady state error to $0$ ($e_{ss} \rightarrow 0$).

Full state feedback control is like having only the P and D term of the PID-Controller and produce the control signal:

$$u = -k_1x_1 - k_2x_2,$$

where $x_1$ is the position and $x_2$ is the velocity of the plant. Full state feedback relies mainly on the model you have obtained for the system, which will NEVER be perfect. So, any unmodeled dynamics and model disturbances will be present inside your $A$ matrix ($\dot{x} = Ax+Bu$) and furthermore you will have to suffer the steady state error which will certainly exist when trying to implement such a controller to a physical system because at simulations everything can look perfect. So, with fun state feedback you can't compensate the external disturbances.

SOLUTION: There is the extension of full state feedback, called dynamic state feedback. The whole idea is to increase the system order by $1$ and introduce an integral term. The new state-space representation will look like this (for a second order system):

$$\begin{gather} \dot{x} = Ax + Bu \\ y = Cx \\ \dot{z} = Cx - r = y - r \\ u = -k_1x_1 -k_2x_2 -k_iz \end{gather}$$ where $x$ are the states of the system, $y$ is the output of the system, $z$ is the integral term, $u$ is the control signal and $r$ is the reference point. The procedure in order to obtain the values of the gains $k_1,k_2,k_i$ is the usual, by obtaining the characteristic polynomial, finding its roots and so on.

One thing to notice is that by increasing the order of the system (from $2$ to $3$) you have an extra pole. This pole should be considered as the third pole and the desired characteristic polynomial will look like this (one of the two forms):

$$ p_d(s) = (s+p_3)\cdot(s+p_2)\cdot(s+p_1) = (s+p_3)\cdot(s^2+(p_1+p_2)s+p_1p_2) $$

$$ p_d(s) = (s+p_3)\cdot(s^2+2\zeta \omega_ns+\omega_n^2) $$

So, you should choose the third pole far enough in order for the system's transient and steady state response to be defined by the two dominant poles. However, choosing poles far to the left-half plane will produce large gains, which most likely will not be able to be implemented by the system's actuators. So, this is a trade-off which you, as a control engineer, should try to figure out and achieve the best result.