# One-Degree of Freedom Internal Model Control

This chapter is from the book

## 3.6 Design for Processes with Right Half Plane Zeros

When N(s) in Eq. (3.14) has factors of the form (-τs+1) or (τ2s2 – 2τζs + 1), with τ and ζ greater than zero, its inverse is unstable. In this case the IMC controller cannot be formed as given by Eq. (3.15). The integral square error (ISE)7 optimal choice of controller for such cases is to invert that portion of the model which has zeros in the left half plane and add poles at the mirror image of the right half plane zeros (Morari & Zafiriou, 1989). That is, we assume that the model given by Eq. (3.14) can be rewritten as

Equation 3.19a

 where N-(s) contains only left half plane zeros, none of which have small damping ratios. N+(s) contains only right half plane zeros, and can be written as

Equation 3.19b

Notice that the gain of N+(s) is one.

Before designing the IMC controller, we strongly recommend that the model be put in time constant form (i.e., the numerator and denominator are factored into products of the form (±τs + 1), (τ2s2 ± 2τζs + 1) so that it is easy to form N+(s) and N-(s). The MATLAB functions tcf and tfn provided with IMCTUNE were developed specifically to put transfer functions into time constant form, and to facilitate their manipulation in this form. There are also other software programs that can be used to accomplish the desired factorization as described in Section 3.9.

The ISE optimal IMC controller for Eq. (3.19a) is

Equation 3.20

where the zeros of N+(–s) are all in the left half plane and are the mirror images of the zeros of N+(s). r = relative order of N(s)/D(s) as before.

The choice of controller given by Eq. (3.20) results in a loop response given by

Equation 3.21

The loop response given in Eq. (3.21) is optimal in an ISE sense for a filter time constant of zero, and is suboptimal for finite . Also, when is zero, the loop transfer function given by Eq. (3.21) is called all-pass, since the magnitude of the frequency response is one over all frequencies.

## Example 3.3 One Right Half Plane Zero

The process model is

Equation 3.22a

Putting Eq. (3.22a) in time constant form yields

Equation 3.22b

The IMC controller is

Equation 3.22c

The resulting loop response is

Equation 3.22d

Figure 3.6 compares the step response of the ISE optimal loop transmission given by Eq. (3.21) with = 0 to step responses of suboptimal responses obtained by increasing and decreasing the controller time constant. Notice that the faster response obtained with pq(s) = (–s + 1)/(.5s + 1) comes at the expense of a more negative initial response. Thus, for this simple example, the ISE optimal response is also qualitatively the best compromise between a more sluggish response and a faster response with a more negative initial response.

Figure 3.6. Response of processes with one right half plane zero.

The transfer function given by the controller of Eq. (3.22c) does not result in an optimal response to a step setpoint change unless is zero. We could get closer to an optimal transfer function by selecting the IMC controller as

Equation 3.22e

The controller given by Eq. (3.22e) was obtained by forcing the coefficient of the linear term of its expanded denominator to be one, which is the same as the linear term in the denominator of Eq. (3.22c) when is zero. The loop response then becomes that given by Eq. (3.22f).

Equation 3.22f

The cubic and quadratic terms in s in the denominator of Eq. (3.22f) are small relative to its linear term so that Eq. (3.22f) approaches the optimal transfer function given by Eq. (3.22d) with equal to zero. Notice, however, that Eq. (3.22e) is valid only for ≤ .5. For larger values of the filter time constant, Eq. (3.22e) is not stable.

The approach used to obtain the controller given by Eq. (3.22e) can be used to develop nearly optimal controllers for arbitrary nonminimum8 phase processes. However, such controllers will generally be useful only for small filter time constants.

## Example 3.4 Two Right Half Plane Zeros

When the initial process response to a step is in a direction opposite to that of the final steady state, as in the previous example, the process is said to exhibit an inverse response. Processes with an odd number of right half plane zeros exhibit inverse responses. Processes with an even number of right half plane zeros do not have inverse responses, since the initial value at time zero plus is always in the direction of the steady-state9 as shown in Figure 3.7 for the loop response given by

Equation 3.23

Figure 3.7. Response of processes with two right half plane zeros.

with τ= .5, 1, and 2.

Notice that the ISE optimal response in Figure 3.7 is again that which also gives the qualitatively best compromise between a sluggish response and a response with too much initial overshoot.

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