barcode reader integration with asp.net FIGURE 27-4 Minimal prototype response to a unit-step change in load for Example 27.1. in Software

Generate Code 128 Code Set B in Software FIGURE 27-4 Minimal prototype response to a unit-step change in load for Example 27.1.

FIGURE 27-4 Minimal prototype response to a unit-step change in load for Example 27.1.
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The important point to be emphasized is that c follows the response 1 - e - IT from t = T to t = 3T, regardless of the algorithm D(z). The values of q in Eq. (27.16) for this example am summarized below 70 = 0
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qz=l-b 73 = 1 - b2 (for rli = 0
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The designer, of course, may choose nonzero values of ni for i > 3, for which case the response will be a nonminimal prototype response. For the minimal prototype response, we obtain from EQ. (27.16) C,(z) = (1 - b)z-2 + (1 - b2)z-3 or
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cd(z) =
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(1 - b)z + 1 - b2
(27.18)
For the system under consideration in Fig. 27.3, we have I_ e-Ts e-Ts ; G(s) = GpGh(s) = s 7s + 1 for which the corresponding G(z) is l - b G(z) = z(z - b) For a unit-step change in U,
e-Ts
(27.19)
UGp(s)
S(TS + 1)
for which the corresponding Z-transform is UGp(z> = l - b (z - l)(z - b) (27.20)
Substituting Eqs. (27.18), (27.19), and (27.20) into Eq. (27.14) gives, after considerable algebraic manipulation b(1 + b) l+b+b2Z - l+b+b2 I D(z) = (27.21) 1 - b (z - l)[z + (1 + b)]
It is instructive to find the expression for the manipulated variable in the form of Eq. (27.6). This is the form that one must use to write a computer program for control of the process. We may write Eq. (27.21) in the form M(z) D(z) = I = where (Y = (1 + b + b2)/(1 - b) y = b(1 + b)l( 1 + b + b2) az(z -7) (z - l)[z + (1 +
(27.22)
DESIGN
SAMPLED-DATA
CONTROLLERS
Cross-multiplying
(27.22)
gives
(z - l)[z + (1 + b)lM(z) = az(z - y)E(z) Expanding the terms and rearranging give the result M(Z) = d(z) - ayE(z)~-~ - bM(z)z- + (1 + b)M(z)z -* (27.23) This form matches Eq. (27.1) or Eq. (27.5), and one can see that the algorithm is quite simple, with k = 1 and p = 2. The form corresponding to Eq. (27.6) is n(nT) = ae(ni ) - wye(nT - T) - bm(nT - I ) + (1 + b)m(nT - 2T) (27.24)
To obtain the sequence of values of manipulated variable for a unit-step change in load, we proceed as will be shown. Before the load change occurs, assume that the system is at steady state for which case e(t) = 0 and m(t) = 0. Furthermore, assume that the load disturbance occurs at t = 0, i.e., at n = 0. These are the usual initial steady-state conditions that are used in testing the dynamic performance of a control system. To see how Eq. (27.24) is used by a computer, the computation can be organized as shown in Table 27.1. At each sampling instant, m(nT) is calculated from Eq. (27.24). For the example under consideration, we know that m = 0 and e = 0 for t 5 0. The leftmost column in the table gives the time at sampling instants when the computation is made. For convenience in computation, the coefficients of the terms on the right side of Eq. (27.24) are placed in a row under the appropriate terms of this equation. The calculation of m(nT) for several values of n are now shown; these calculation steps are the same as those the computer would follow in implementing Eq. (27.24).
For m(O). Substituting n = 0 into Eq. (27.24) gives
m(O) = ae(0) - aye - bm(-T) + (1 + b)m(-2T)
TABLE 27.1
(27.25)
Computation of m(H) from computer control algorithm I @WI d(n - ml MT)
-(Y-Y
NM - WI
-b I
m[(n - WI
l + b
SAhfPLED-DATA
CONTROL SYSTEMS
From the initial conditions stated earlier, e(O) = e(-T) = 0 and m(-T) = m(-2T) = 0 Introducing these values into Eq. (27.25) gives m(O) = 0. In preparation for the next computation of m(U), the values of e(C), m(nT), and m[(n - l)T] are shifted to the next sampling instant as shown by the arrows in the table.
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