Determination of Optimum Settings for Two- and Three-mode Controllers in Visual Studio .NET

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Determination of Optimum Settings for Two- and Three-mode Controllers
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,Controller Modes
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Proportional plus reset
0 0 0 0
4.80 0.76 0.95 6.00 8.00 1.27 -__ 3.197d 0.517&1 4.00 0.64 5.37 0.86 - - - 2.867d 4.00 6.79 0.45rJn 0.64 1.08
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Noninteracting three-mod<
0.51Td 0.88Td 0.64 0.64 0.86 0.49 _-__- 0.9OTd 0.64 0.54
109P +23 llOTd/T, 100 0 127 109 - 2 3 196 __--125P 100 125 +36 22%,/T, 0 254 - 3 6 540
Interacting 0,907d three-mode 1 0.64 ,0.54
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Controllers
FIG 4.9. Increasing derivative time reduces error magnitude at the cost of recovery time.
,.$ +7&---$-K
Time
2. Measure the period of oscillation r0 and set derivative and reset time both to r0/27r. (The optimum value of D and R in Table 4.2 is 0.64rd, which is 2~~1~ or ~~/27r.) For a two-mode controller, set R at rJ2.4. When adjusting a three-mode pneumatic controller with antiwindup, always keep R > 20 to retain proportional stability. 3. Readjust the proportional band to produce the desired degree of damping. 4. If 7,, is higher than before, increase both D and R; if it is lower, decrease them. This may be necessary to compensate for inaccuracy in the dial settings. With a two-mode controller, 70 will increase by about 50 percent. Interaction is clearly evident when an attempt is made to tune a controller for maximum performance. Figure 4.9 shows how increasing the derivative time also increases the effective reset time in a typical closed loop. The integrated errors for the three response curves are equal because the proportional band and reset settings are all the same. Because the last curve has a lower maximum error, its ISE would tend to be less than the other two. Although noninteracting controllers are not generally available, it is worthwhile to note that they are capable of twice the performance of interacting controllers, as Table 4.2 indicates.
COMPLEMENTARY FEEDBACK
The question often arises whether proportional, reset, and derivat ive are really the best control modes for every application. For the easierto-control processes, their use can be justified. A single-capacity process and some two-capacity processes need only narrow-band proportional action. Derivative is of great value in processes with two or three capacities. But for the more difficult processes, it has been found that reset action is essential. In processes where dead time is dominant, derivative action has been seen to have less effect than in processes consisting of capacity alone. This suggests that some other control mode more akin to dead time might
1 Selecting the Feedback Controller
be valuable. Derivative and reset are, in actuality, time constants just like the time constants in a process. They bear no resemblance to the dead time that may exist in the plant, however. Theory Several authors2j3 have postulated a feedback control system that is modeled after the process. This kind of control action is known as complementary feedback, because the characteristics of the controller complement the dynamics of the process. A block diagram showing both process and complementary controller appears in Fig. 4.10. The principle employed is that a given error signal e can be,made to generate a certain instantaneous restoring force, lOOe/P, which will change c exactly enough to cancel the error. At the same time, a complementary signal characterized to match the response of the process is fed back positively to cancel the effect of the negative feedback from the process. This means that the output of the controller will remain at its instantaneous value, which was correct in that it was able to exactly reduce the error to zero. For this exact sequence of events to occur, the control parameters must have the values
P = lOOK,
&= &
(4.11)
The term K, is the steady-state gain of process, valve, and transmitter, that is, G,G,G. The terms g, and g, are vectors representing the dynamic components in the controller and process, respectively. It is worthwhile to trace the sequence of events following a set-point change through the block diagram. Initially e is zero because c = r, and uz is at rest. Following a set-point change Ar,
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