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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?

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    $\begingroup$ Consider the trivial case of a single state system. Full state feedback doesn't fully reject disturbance IMO. $\endgroup$
    – AJN
    Jun 2 at 14:17
  • $\begingroup$ "The part that rejects disturbance is..." depends on what you mean, and exactly where you put the disturbance in the loop. In a classic SISO controller, including PID, the part that rejects disturbance is P'/(1+L) where L is the gain around the loop, and P' is the portion of the plant between the disturbance and the output. With that in mind, designing for disturbance rejection can use the same techniques as designing for closed-loop performance. As for "the I term does this or that", they all work together. The I term in particular corrects 0-frequency (DC) error, but interacts with mid-band $\endgroup$
    – Pete W
    Jun 2 at 20:17
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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.

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  • $\begingroup$ So PID is more practical to implement in real plant than LQR right? Any technique else I should investigate? (dynamic state feedback is good one.) $\endgroup$
    – M lab
    Jun 2 at 15:13
  • $\begingroup$ It really depends on how you will design the complete control loop. PID has the advantage to be easily implemented. However, bear in mind that a really good PID has its own parts to design (derivative filter, derivative on measurement to avoid derivative kick. integrator clamping, measurement filtering). On the other hand, if you combine LQR with a Kalman filter you get the LQG-Controller, which is really robust and capable of rejecting disturbances and noise. I would suggest you try as many as you can for experience and then you can more easily understand which one to use at each situation. $\endgroup$ Jun 2 at 15:20
  • $\begingroup$ This can also be generalized to output regulation using exogenous systems. This adds uncontrollable states to the model, which typically are either constant or sinusoidal. $\endgroup$
    – fibonatic
    Jun 2 at 21:13

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