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barcode reader integration with asp.net For m(T). To compute m(T), we change n in Fq (27.24) to 1 to obtain in Software
For m(T). To compute m(T), we change n in Fq (27.24) to 1 to obtain Code 128C Reader In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128 Code Set A Drawer In None Using Barcode encoder for Software Control to generate, create USS Code 128 image in Software applications. m(T) = se(T)  aye(O)  bm(0) + (1 + b)m(T) (27.26) Code 128 Code Set B Scanner In None Using Barcode scanner for Software Control to read, scan read, scan image in Software applications. Code 128C Encoder In C# Using Barcode creation for .NET Control to generate, create Code128 image in .NET applications. Substituting the appropriate values of e and m as given in the table into this expression gives m(T) = 0. Create Code 128B In VS .NET Using Barcode generation for ASP.NET Control to generate, create Code 128 Code Set B image in ASP.NET applications. Draw ANSI/AIM Code 128 In .NET Using Barcode printer for .NET Control to generate, create Code 128C image in .NET applications. For m(2T). Letting n = 2 in Eq. (27.24) gives
Code 128B Encoder In VB.NET Using Barcode generator for Visual Studio .NET Control to generate, create Code 128 image in .NET applications. Code 39 Full ASCII Creation In None Using Barcode creation for Software Control to generate, create Code 39 image in Software applications. m(2T) = ae(2T)  aye(T)  bm(T) + (1 + b)m(O) (27.27) EAN13 Supplement 5 Maker In None Using Barcode generator for Software Control to generate, create EAN13 Supplement 5 image in Software applications. Generate UPCA Supplement 2 In None Using Barcode creator for Software Control to generate, create UPC Symbol image in Software applications. At t = 2T, the disturbance has worked its way through the transport lag and the error now differs from zero. We have at t = 2T as shown in the table or in Fig. 27.4 e(2T) = e(T) = m(T) = m(O) = (l  b) 0 0 0 Print Barcode In None Using Barcode maker for Software Control to generate, create barcode image in Software applications. Making ECC200 In None Using Barcode creator for Software Control to generate, create ECC200 image in Software applications. Substituting these values into Q. (27.27) gives m(2T) = a[ (l  b)]. Introducing the expression for (Y from EQ. (27.22) gives m(2T) = (l + b + b*) Generate USS93 In None Using Barcode creator for Software Control to generate, create Code 93 Extended image in Software applications. Bar Code Reader In Java Using Barcode Control SDK for Eclipse BIRT Control to generate, create, read, scan barcode image in Eclipse BIRT applications. For m(3T). Letting n = 3 in Eq. (27.24) gives
Scan Code 3 Of 9 In None Using Barcode recognizer for Software Control to read, scan read, scan image in Software applications. GS1128 Creator In Visual Basic .NET Using Barcode creator for .NET Control to generate, create UCC  12 image in Visual Studio .NET applications. m(3T) = cye(3T)  qe(2T)  bm(2T) + (1 + b)m(T) At t = 3T, e(3T) e(2T) m(2T) m(T) = = = = (l  b*) (l  b) (l + b + b*) 0 (27.28) UPC Code Generation In None Using Barcode printer for Font Control to generate, create Universal Product Code version A image in Font applications. Recognize Data Matrix ECC200 In .NET Using Barcode decoder for Visual Studio .NET Control to read, scan read, scan image in .NET framework applications. Substituting these values into IQ. (27.28) gives m(3T) = CY[(1  b*)]  cwy[(1  b)]  b[(1 + b + b*)] Generating EAN / UCC  14 In ObjectiveC Using Barcode maker for iPhone Control to generate, create GS1128 image in iPhone applications. Bar Code Maker In Java Using Barcode generator for Java Control to generate, create bar code image in Java applications. Reducing this expression algebraically gives m(3T) =  1, DESIGN OF SAMPLEDDATA CONTROLLERS
JLO 0 12 3 4 5tlT FIGURE 275 Response under fast sampling, load, minimal pmtotype algorithm to a unitstep change in load (a = OS), If one continues in this sequential manner, which is how the computer handles the computation, one can show that m(4T) = m(5T) = m(G) = ** = 1 In other words, the manipulated variable reaches 1 at t = 3T, and remains at this value thereafter. A graph showing the response and the manipulated variable is shown in Fig. 21.5 for the case where a = 0.5. For this case b = e* r = ,o.5 = 0.606 and m(2T) = 1.974 The manipulated variable m(nT) that results for this case is not surprising when one considers the nature of the firstorder system with transport lag. In fact, for this simple process, one can calculate the values of manipulated variable directly, without use of Eq. (27.24). However, for other disturbances and for more complex algorithms, the calculation becomes very involved without a systematic approach such as that given by Eq. (27.24). Settling 7lme
A useful parameter for describing a transient response of a control system is settling time. For a load change, the settling time is defined as the time required to reduce the error to zero; this time is measured from the sampling instant for which nonzero error is recorded to the sampling instant for which the output returns to the set point and remains at the set point at future sampling instants. For the example under consideration, the settling time ts is 2T. This can be seen most easily from Fig. 27.5. 4 1 6 SAMPLEDDATAcONTROLSYSTEMS
For a setpoint disturbance, the settling time is defined as the time required to reduce the error to zero; this time is measured from the sampling instant for which the setpoint change is first detected to the sampling instant for which the response returns to the set point and remains there at future sampling instants. OBTAINING M(z) DIRECTLY FROM KNOWLEDGE OF C(z). If one wishes to
compute M(z) without using the sequential method just discussed and shown in Table 27.1, the following direct procedure can be used. For the servo problem, where U(S) = 0, one can write directly from Fig. 27.3 C(z) = M(z)G(z) (27.29) where G(z) = GhGp(z) Solving for M(z) gives M(z) = C(z)/G(z) (27.30) For the regulator problem, where R(s) = 0, one can obtain directly from Fig. 27.3 M(z) = D(z)C(z) (27.31) For Example 27.1, one can use Eq. (27.31) to obtain M(z) for a unitstep change in load; C in Eq. (27.3 1) is replaced by Cd of Eq. (27.18). The reader should try this approach to see that it leads to the same results as those obtained in Table 27.1. Example 27.2. (T > UT, slow sampling, load) In this example, the minimal prototype D(z) will be designed for the following conditions: T > a7 (slow sampling) U = l/s (load disturbance) G, = eCars/(~s + I), ur < T The fact that UT is not an integral number of sampling periods will make the derivation of this algorithm more complicated. Based on the response of a firstorder system with transport lag, the minimal prototype response shown in Fig. 27.6 can be written:

