Steady-state gains

The principal limitation to using these rules is that the true process parameters are often unknown. Steady-state gain K can be calculated from a process model or determined from the steady-state results of a step test as Ac/Au, as shown in Fig. 8-28. The test will not be viable, however, if the time constant of the process is longer than a few... 

The advantage of static decoupling is that less process information is required namely, only steady-state gains. Nonlinear decouplers can be used when the process behavior is nonlinear. 

Equation (3.23) is the standard form of transfer funetion for a first-order system, where K = steady-state gain eonstant and T = time eonstant (seeonds). 

Equations (3.42) and (3.43) are the standard forms of transfer functions for a second-order system, where K = steady-state gain constant, Wn = undamped natural frequency (rad/s) and ( = damping ratio. The meaning of the parameters Wn and ( are explained in sections 3.6.4 and 3.6.3. 

Consider a second-order system whose steady-state gain is K, undamped natural frequency is Wn and whose damping ratio is (, where C < 1 For a unit step input, the block diagram is as shown in Figure 3.18. From Figure 3.18... 

A system eonsists of a first-order element linked to a seeond-order element without interaetion. The first-order element has a time eonstant of 5 seeonds and a steady-state gain eonstant of 0.2. The seeond-order element has an undamped natural frequeney of 4 rad/s, a damping ratio of 0.25 and a steady-state gain eonstant of unity. 

Ship roll damping ratio ( = 0.248 Ship steady-state gain A s =0.5... 

The constant in the numerator can always be chosen to preserve the steady state gain of the transfer function. As suggested by Luus (1980) the 5 unknown parameters can be obtained by minimizing the following quadratic objective function... 

In addition to transfer functions, we make extensive use of steady state gain and time constants in our analysis. 

Another important quantity is the steady state gain.3 With reference to a general differential equation model (2-28) and its Laplace transform in (2-29), the steady state gain is defined as the... 

The steady state gain is the ratio of the two constant coefficients. Take note that the steady state gain value is based on the transfer function only. From Eqs. (2-31) and (2-32), one easy way to "spot" the steady state gain is to look at a transfer function in the time constant form. 

In this rearrangement, xp is the process time constant, and Kd and Kp are the steady state gains.2 The denominators of the transfer functions are identical, they both are from the LHS of the differential equation—the characteristic polynomial that governs the inherent dynamic characteristic of the process. 

K and KH in (2-49a) are referred to as gains, but not the steady state gains. The process time constant is also called a first-order lag or linear lag. 

After this exercise, let s hope that we have a better appreciation of the different forms of a transfer function. With one, it is easier to identify the pole positions. With the other, it is easier to extract the steady state gain and time constants. It is veiy important for us to leam how to interpret qualitatively the dynamic response from the pole positions, and to make physical interpretation with the help of quantities like steady state gains, and time constants. 

With the time constant defined as x = V/Qm s, the steady state gain for the transfer function for the inlet flow rate is (Cin s - Cs)/Qin s, and it is 1 for the inlet concentration transfer function. 

In the event that we are modeling a process, we would use a subscript p (x = xp, K = Kp). Similarly, the parameters would be the system time constant and system steady state gain when we analyze a control system. To avoid confusion, we may use a different subscript for a system. 

Figure 3.1. Properties of a first order transfer function in time domain. Left panel y/MK effect of changing the time constant plotted with x = 0.25, 0.5, 1, and 2 [arbitrary time unit]. Right panel y/M effect of changing the steady state gain all curves have x = 1.5.

The transfer function has the distinct feature that a pole is at the origin. Since a step input in either q in or q would lead to a ramp response in h, there is no steady state gain at all. 

In this example, the steady state gain is unity, which is intuitively obvious. If we change the color of the inlet with a food dye, all the mixed tanks will have the same color eventually. In addition, the more tanks we have in a series, the longer we have to wait until the -th tank "sees" the changes that we have made in the first one. We say that the more tanks in the series, the more sluggish is the response of the overall process. Processes that are products of first order functions are also called multicapacity processes. 

Under circumstances where the two functions represent opposing effects, one of them has a negative steady state gain. In the following illustration, we choose to have K2 < 0. 

For all commercial devices, the proportional gain is a positive quantity. Because we use negative feedback (see Fig. 5.2), the controller output moves in the reverse direction of the controlled variable.1 In the liquid level control example, if the inlet flow is disturbed such that h rises above hs, then e < 0, and that leads to p < ps, i.e., the controller output is decreased. In this case, we of course will have to select or purchase a valve such that a lowered signal means opening the valve (decreasing flow resistance). Mathematically, this valve has a negative steady state gain (-Kv)2... 

For the rest of the control loop, Gc is obviously the controller transfer function. The measuring device (or transducer) function is Gm. While it is not shown in the block diagram, the steady state gain of Gm is Km. The key is that the summing point can only compare quantities with the same units. Hence we need to introduce Km on the reference signal, which should have the same units as C. The use of Km, in a way, performs unit conversion between what we dial in and what the controller actually uses in comparative tests. 2... 

Those that have a large steady state gain. (Good sensitivity)... 

From that same section, we know that the steady state gain and the time constant are dependent on the values of flow rate, liquid density, heat capacity, heat transfer coefficient, and so on. For the sake of illustration, we are skipping the heat transfer analysis. Let s presume that we have done our homework, substituted in numerical values, and we found Kp = 0.85 °C/°C, and xp = 20 min. 

When the steady state gains of all three assumptions are lumped together, we may arrive at a valve gain Kv with the units of °C/mV. For this illustration, let s say the valve gain is 0.6 °C/mV and the time constant is 0.2 min. The actuator controller function would appear as... 

Rearrange the expressions such that we can redefine the parameters as time constants and steady state gains for the closed-loop system. 

All analyses follow the same general outline. What we must accept is that there are no handy dandy formulas to plug and chug. We must be competent in deriving the closed-loop transfer function, steady state gain, and other relevant quantities for each specific problem. 

Lastly, we should see immediately that the system steady state gain in this case is the same as that in Example 5.1, meaning that this second order system will have the same steady state error. 

There are two noteworthy items. First, the closed-loop system is now second order. The integral action adds another order. Second, the system steady state gain is unity and it will not have an offset. This is a general property of using PI control. (If this is not immediately obvious, try take R = 1/s and apply the final value theorem. We should find the eventual change in the controlled variable to be c (°°) =1.)... 

The system steady state gain is the same as that with proportional control in Example 5.1. We, of course, expect the same offset with PD control too. The system time constant depends on various parameters. Again, we defer this analysis to when we discuss root locus. 

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