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This chapter is from the book

This chapter is from the book

3.7 Problems with Mathematically Optimal Controllers

In both Figure 3.6 and Figure 3.7 the qualitatively and quantitatively optimal response is that given by the all-pass loop response. However, it can happen that the all-pass response, while ISE optimal, is not qualitatively the best response and may not even be an acceptable response, as shown by Example 3.5 following.

Example 3.5 A Process with an Infinite Number of RHP Zeros

The transfer function given by Eq. (3.24) below is one that might arise between a process variable and a control effort in a multivariable system, where one or more of the other process variables are under closed-loop control.

Equation 3.24

03equ24.gif


p(s) has two complex zeros in the right half plane at 0.214 ± 2.10i. These zeros were calculated by replacing the dead time in Eq. (3.24) by a five over five Padé10 approximation and then factoring the resulting sixth order numerator of p(s). The factored form of the approximated transfer function 054fig01.gif is

Equation 3.25

03equ25.gif


The ISE optimal controller for (3.25) is

Equation 3.26

03equ26.gif


To achieve |q(∞)/q(0)| ≤ 20 requires ≥ 0.2 The resulting loop response is

Equation 3.27

03equ27.gif


Figure 3.8a shows the response of Eq. (3.27) to a step with = 0 and = 0.2. The oscillatory response in Figure 3.8a is due to the low damping ratio in the denominator of Eq. (3.27), which is only 0.101. Increasing the damping ratio to 0.5 in the first term of the denominator of Eq. (3.26) by increasing the coefficient of 0.0964 to 0.48 gives a qualitatively better response, that is, one with less overshoot and no oscillations. However, this improved response has a higher ISE. The associated controller and loop response are given by

Equation 3.28

03equ28.gif


Equation 3.29

03equ29.gif


03fig08a.gifFigure 3.8a. Optimal versus suboptimal responses to a step setpoint change for the process, p(s) = (1 + 3e-s/(s + 1))/(s + 1).


The responses in Figure 3.8a were obtained with a rather complex controller, as given by Eq. (3.26) and its modifications. It is reasonable to ask whether there is a much simpler suboptimal controller that might do nearly as well. Indeed, the control system that uses the simple controller given by

Equation 3.30

03equ30.gif


produces a very good step setpoint response, as shown in Figure 3.8b. The above controller arises from the recognition that the numerical value of the process steady-state gain is four, that only the term 1/(s + 1) in p(s) is easily invertible, and that the step response of p(s) is monotonic. Figure 3.8b compares the loop responses using the controllers given by Eq. (3.26) with = 0, with Equations (3.28) and (3.30).

03fig08b.gifFigure 3.8b. Optimal versus suboptimal responses to a step setpoint change for the process, p(s) = (1 + 3e-s/(s + 1))/(s + 1)


The controller given by Eq. (3.30) is not only simpler than that given by Eq. (3.26), but also yields a response with no overshoot and achieves the final steady state in about the same time. Many engineers, including the authors, will prefer the response obtained with the simple controller to that obtained with the more complex controller. Thus, this example demonstrates that optimality, like beauty, is in the eye of the beholder. As the foregoing examples attempt to demonstrate, controller designs that minimize the ISE can be quite useful, but they should not be applied by rote. The engineer should always look at the ideal loop response given by the loop transmission pq(s) to judge whether the optimal controller gives desirable responses. If not, then a trial and error procedure is usually necessary to obtain a controller that does give an acceptable loop behavior. A systematic way to obtain such a design is to start with a controller q(s) that is just the inverse of the model gain and then attempt to improve on that controller by including in it increasing portions of the model inverse.

Lest the reader get the misimpression that qualitative improvements in performance over that obtained with the ISE optimal controller are only possible when the optimal response is oscillatory, we offer the following example.

Example 3.6 A Process with an Infinite Number of RHP Zeros

Equation 3.31

03equ31.gif


The above process has an infinite set of zeros in the right half plane at (ℓnK – 2nπi)/T if K > 1, and at (ℓn(– K) – (2n + 1)πi)/T if K < –1. We can form a controller q(s) with poles at the mirror image of the zeros of Eq. (3.31) as

Equation 3.32

03equ32.gif


so that

Equation 3.33

03equ33.gif


The response of Eq. (3.33) to a unit step is

Equation 3.34

03equ34.gif


For the case K = 2, the first several values of y(nT) are y(0) = –1/2, y(T) = 1/4, y(2T) = 5/8, and y(3T) = 13/16, as shown in Figure 3.9 for T = 1. Clearly, the optimal response approaches unity in a staircase fashion.

03fig09.gifFigure 3.9. Loop response of Eq. () to a unit step input.


The ISE for the output y(nT) of Eq. (3.33) is given by

058equ01.gif


Substituting Eq. (3.34) into the above gives

058equ02.gif


Recognizing that 058fig01.gif gives

Equation 3.35

03equ35.gif


For K = 2, the ISE from Eq. (3.35) is 3T. Now consider the controller that is simply the reciprocal of the model gain; that is

Equation 3.36

03equ36.gif


The loop response for this controller is

Equation 3.37

03equ37.gif


The ISE for the above is 4T (vs. 3T before), but the output reaches and maintains its desired value after one dead time.

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