State-feedback

In equation (8.93), r(t) is a vector of desired state variables and K is referred to as the state feedback gain matrix. Equations (8.92) and (8.93) are represented in state variable block diagram form in Figure 8.7. 

Fig. 8.10 Closed-loop control system with full-order observer state feedback.

Fig. 8.12 Complete state feedback and reduced observer system for case study Example 8.11. where, from equation (4.20)...

In reverse-time, starting with P(A ) = 0 at NT = 20 seconds, compute the state feedback gain matrix K(kT) and Riccati matrix P(kT) using equations (9.29) and (9.30). Aiso in reverse time, use the desired state vector r(/c7 ) to drive the tracking equation (9.53) with the boundary condition s(N) = 0 and hence compute the command vector y kT). 

Discrete-time steady-state feedback gain... 

We now return to the use of state space representation that was introduced in Chapter 4. As you may have guessed, we want to design control systems based on state space analysis. State feedback controller is very different from the classical PID controller. Our treatment remains introductory, and we will stay with linear or linearized SISO systems. Nevertheless, the topics here should enlighten( ) us as to what modem control is all about. 

Pole placement design of state feedback systems. Application of the Ackermann s formula. 

Figure 9.1. Closed-loop system with state feedback...

Thus in general, we can calculate all the state feedback gains in Kby... 

This is the result of full state feedback pole-placement design. If the system is completely state controllable, we can compute the state gain vector K to meet our selection of all the closed-loop poles (eigenvalues) through the coefficients a . 

There are other methods in pole-placement design. One of them is the Ackermann s formula. The derivation of Eq. (9-21) predicates that we have put (9-13) in the controllable canonical form. Ackermann s formula only requires that the system (9-13) be completely state controllable. If so, we can evaluate the state feedback gain as 1... 

Figure 9.2. State feedback with integral control.

Example 9.2 Consider the second order model in Example 9.1. What are the state feedback gains if we specify that the closed-loop poles are to be at -3 3j and -6 ... 

To obtain the state feedback gains with Eq. (9-21), we should subtract the coefficients of the polynomial pi from p2, starting with the last constant coefficient. The result is, indeed,... 

Example 4.7B Let us revisit the two CSTR-in-series problem in Example 4.7 (p. 4-5). Use the inlet concentration as the input variable and check that the system is controllable and observable. Find the state feedback gain such that the reactor system is very slightly underdamped with a damping ratio of 0.8, which is equivalent to about a 1.5% overshoot. 

The state space state feedback gain (K2) related to the output variable C2 is the same as the proportional gain obtained with root locus. Given any set of closed-loop poles, we can find the state feedback gain of a controllable system using state-space pole placement methods. The use of root locus is not necessary, but it is a handy tool that we can take advantage of. 

To find the new state feedback gain is a matter of applying Eq. (9-29) and the Ackermann s formula. The hard part is to make an intelligent decision on the choice of closed-loop poles. Following the lead of Example 4.7B, we use root locus plots to help us. With the understanding that we have two open-loop poles at -4 and -5, a reasonable choice of the integral time constant is 1/3 min. With the open-loop zero at -3, the reactor system is always stable, and the dominant closed-loop pole is real and the reactor system will not suffer from excessive oscillation. 

The state feedback gain including integral control K is [0 1.66 -4.99], Unlike the simple proportional gain, we cannot expect that Kn+1 = 4.99 would resemble the integral time constant in classical PI control. To do the time domain simulation, the task is similar to the hints that we provide for Example 7.5B in the Review Problems. The actual statements will also be provided on our Web Support. 

Here, y = Cx has been used in writing the error in the estimation of the output, (y - y). The (/ x 1) observer gain vector Ke does a weighting on how the error affects each estimate. In the next two sections, we will apply the state estimator in (9-32) to a state feedback system, and see how we can formulate the problem such that the error (y - y) can become zero. 

With eigenvalues selected at -9, we have chosen the estimator to be faster than the state feedback, and all the errors are to decay exponentially. We ll make use of the Ackermann s formula in Eq. (9-39) for observer gains. The MATLAB statements are ... 

Figure 9.6. State feedback with reduced-order estimator... 

Do the time response simulation in Example 7.5B. We found that the state space system has a steady state error. Implement integral control and find the new state feedback gain vector. Perform a time response simulation to confirm the result. 

In this setting, all the measured variables are used to compute all the control actions (see, for instance, [1]). A special case is possible if all the state variables are measured. Using state feedback control law, that is ... 

If all the state variables are not measured, an observer should be implemented. In the Figure 14, the jacket temperature is assumed as not measured, but it can be easily estimated by the rest of inputs and outputs and based on the separation principle, the observer and the control can be calculated independently. In this structure, the observer block will provide the missing output, the integrators block will integrate the concentration and temperature errors and, these three variables, together with the directly measured, will input the state feedback (static) control law, K. Details about the design of these blocks can be found in the cited references. 

To be more precise, the state-feedback regulation problem consists of finding a controller of the form (4) such that the following conditions hold... 

Figure 4 shows the control scheme for state feedback. In the event that only the output measurements are available, the controller (12) cannot be directly implemented. In this case, an observer of the states may be used. To this end, we may rewrite the first expression in (11) and the second one of (3) as follows... 

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