Variables deviation

To illustrate how Laplace transforms work, consider the problem of solving Eq. (8-2), subjec t to the initial condition that = 0 at t = 0, and Cj is constant. If were not initially zero, one would define a deviation variable between and its initial value (c — Cq). Then the transfer function would be developed using this deviation variable. Taking the Laplace transform of both sides of Eq. (8-2) gives ... 

The difficulty with Eq, (26-58) is that it is impossible to determine the velocity at every point, since an adequate turbulence model does not currently exist, The solution is to rewrite the concentration and velocity in terms of an average and stochastic quantity C = (C) -t- C Uj = (uj) + Uj, where the brackets denote the average value and the prime denotes the stochastic, or deviation variable. It is also helpful to define an eddy diffusivity Kj (with units of area/time) as... 

It is convenient to work in terms of a deviation variable C as opposed to CA, where C is the amount by which the concentration of A exceeds the initial uniform concentration CA(>. This change allows some simplification of the algebra. 

After Laplace transform, a differential equation of deviation variables can be thought of as an input-output model with transfer functions. The causal relationship of changes can be represented by block diagrams. 

Deviation variables are analogous to perturbation variables used in chemical kinetics or in fluid mechanics (linear hydrodynamic stability). We can consider deviation variable as a measure of how far it is from steady state. 

What we argue (of course it is tme) is that the Laplace-domain function Y(s) must contain the same information as y(t). Likewise, the function G(s) contains the same dynamic information as the original differential equation. We will see that the function G(s) can be "clean" looking if the differential equation has zero initial conditions. That is one of the reasons why we always pitch a control problem in terms of deviation variables.1 We can now introduce the definition. 

We can extend these results to find the Laplace transform of higher order derivatives. The key is that if we use deviation variables in the problem formulation, all the initial value terms will drop out in Eqs. (2-13) and (2-14). This is how we can get these clean-looking transfer functions later. 

We define the right hand side as G(s), our ubiquitous transfer function. It relates an input to the output of a model. Recall that we use deviation variables. The input is the change in the inlet concentration, C m(t). The output, or response, is the resulting change in the tank concentration, C (t). 

We now use two examples to review how deviation variables relate to the actual ones, and that we can go all the way to find the solutions. 

Figure 2.7. A rectangular pulse in real and deviation variables.

From here on, we will omit the apostrophe ( ) where it would not cause confusion, as it goes without saying that we work with deviation variables. We now further rewrite the same equation as... 

Of course, Gd(s) and Gp(s) are the transfer functions, and they are in pole-zero form. Once again( ), we are working with deviation variables. The interpretation is that changes in the inlet temperature and the steam temperature lead to changes in the tank temperature. The effects of the inputs are additive and mediated by the two transfer functions. 

Let us try one simple example. Say if we keep the inlet temperature constant at our desired steady state, the statement in deviation variable (without the apostrophe) is... 

Now we want to know what happens if the steam temperature increases by 10 °C. This change in deviation variable is... 

Note that if we had chosen also TH = 0, T(t) = 0 for all t, i.e., nothing happens. Recall once again from Section 2.1 that this is a result of how we define a problem using deviation variables. 

Always do the linearization before you introduce the deviation variables. 

As soon as we finish the first-order Taylor series expansion, the equation is linearized. All steps that follow are to clean up the algebra with the understanding that terms of the steady state equation should cancel out, and to change the equation to deviation variables with zero initial condition. 

We subtract the two equations and at the same time introduce deviation variables for the dependent variable C and all the parametric variables to obtain... 

While our analyses use deviation variables and not the real variables, examples and homework problems can keep bouncing back and forth. The reason is that when we do an experiment, we measure the actual variable, not the deviation variable. You may find this really confusing. All we can do is to be extra careful when we solve a problem. 

We usually try to identify features which are characteristic of a model. Using the examples in Section 2.8 as a guide, a first order model using deviation variables with one input and with constant coefficients a,. a0 and b can be written in general notations as 1... 

Whether the notation is y or y is immaterial. The key is to find the initial condition of the problem statement. If the initial condition is zero, the notation must refer to a deviation variable. 

Here, we model the flow through the valves with resistances R and R2, both of which are constants. We rearrange the balance equations a bit, and because both equations are linear, we can quickly rewrite them in deviation variables (without the apostrophes) ... 

Now we linearize the two equations about the steady state. We expect to arrive at (with the apostrophes dropped from all the deviation variables and partial derivatives evaluated at the steady state) ... 

In Fig. 5.1, we use the actual variables because they are what we measure. Regardless of the notations in a schematic diagram, the block diagram in Fig. 5.2 is based on deviation variables and their Laplace transform. 

By the definition of a control problem, there should be no error at t = 0, i.e., es = 0, and the deviation variable of the error is simply the error itself ... 

Ideally (meaning unlikely to be encountered in reality), we would like to achieve perfect tracking of set point changes C = R. Reminder, we are working with deviation variables. 

What we can do easily is to measure the fuel gas flow rate, multiply the value by R in the so-called ratio station, and send the signal as the set point to the air flow controller. The calculation can be based on actual flow rates rather than deviation variables. 

---