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barcode reading in asp.net OPENLOOP in Software
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Print Code39 In .NET Using Barcode generation for VS .NET Control to generate, create USS Code 39 image in .NET framework applications. Make EAN 128 In VS .NET Using Barcode encoder for Reporting Service Control to generate, create GTIN  128 image in Reporting Service applications. Fuo methods often used to invert Ztransforms are: 1. Method of long division. (This method was just covered in the previous example.) . 2. Method of partial fraction expansion. The method of partial fraction expansion follows the same procedure as that for inversion of ordinary Laplace transforms. To illustrate this method, consider the following example. Barcode Creation In None Using Barcode printer for Office Excel Control to generate, create barcode image in Excel applications. Drawing Data Matrix ECC200 In Java Using Barcode printer for Java Control to generate, create DataMatrix image in Java applications. Example 23.2.
Encode Matrix Barcode In Java Using Barcode maker for Java Control to generate, create Matrix 2D Barcode image in Java applications. UPC A Reader In C# Using Barcode reader for Visual Studio .NET Control to read, scan read, scan image in Visual Studio .NET applications. Invert the following C(z): C(z) = (2  a;(,  b) (23.13) This may be written: C(z) = 1 B A (2  a)(z  b ) = z  a + z  b
The reason for placing z in the denominator on the left side is for mathematical convenience, as will be shown later. Evaluating the constants A and B gives: A = l/@b) a n d B = l/@b) We may now write Z 1 z C(z) =    (23.14) ab 1 za z  b Bach term within the brackets can be inverted by referring to a table of transform pairs. From Table 22.1, we have the transform pair z/(z  a) : arm
At sampling instants, a *lT becomes unTIT or a . In a similar manner, the inverse of z/(z,  b) is b . Using these results gives c(U) = &, [a  b ] (23.15) SAMPLEDDATA
CONTROL
SYSTEhG
For the problem solved by long division in Example 23.1, we obtain the following result if the method of partial fraction expansion is used: c(nT) = &nl) + &(n2) + e(n3) (23.16) The three terms in brackets do not apply until n = 1, rz = 2, and it = 3, respectively. To obtain this result, we need to use a theorem on the Ztransform of a translated function, which will be discussed later. Comparison of Methods of Inversion
1. Long division. This method is good when one is interested in the solution
for only the first few sampling instants. One must be careful not to make errors in performing the long division, for an error at one sampling instant will propagate errors at later sampling instants. The method of long division can be programmed for a computer to obtain an errorfree solution for as many sampling instants as desired. A computer program written in BASIC for long division is given in the appendix of this chapter for use by the interested reader. 2. Partial jiaction expansion. This method requires the usual algebra of partial fraction expansion. However, once the solution is obtained, the response at any sampling instant can be found without relying on values at previous sampling instants. CLOSEDLOOP
RESPONSE
The closedloop response for a sampleddata system can be obtained in a manner similar to that for continuous closedloop systems. However, there are some differences that will be explained in this section. Consider the sampleddata negative feedback control system shown in Fig. 23.4. In this process, clamping is provided by the combination of an impulse modulator and a zeroorder hold. To obtain expressions that relate C to R or C to U, we proceed as follows. From the diagram, we can write: (23.17) Us) = Gc(Wp(M(s)  Gc(W,WW)C*W + Gp(~)U(~) This expression is obtained by combining the signals resulting from R(s), C *(s), and U(s) after they move through their respective paths to the output C. This expression may also be written as FIGURE 234 Closedloop sampleddata
system.
OPENLOOP
AN D
CLOSEDLOOP
RRWONSE
365 (23.18) C(s) = G,G,R(s)  GcGpGh(s)C*(s) + G&J(s) where the overbar above several terms means that the functions of s corresponding to each term are multiplied together to form one function of s. Further discussion of the usefulness of the overbar will be given later. Taking the starred transform of both sides of Eq. (23.18) gives C*(s) = w*(s)  G,G,Gh *(s)C*(s) + v*(s) (23.19) Again, it must be pointed out that the overbar above several terms means that these terms arc multiplied together before taking the starred transform. The middle term on the right side of Eq. (23.18) requires special attention; taking the starred transform of G,G,G,(s)C*(s) gives G,G,G,(s)C*(s). The explanation of this result is given in the appendix. Solving for C*(s) gives c*(s) =

