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barcode reader integration with asp.net RnalValue Theorem in Software
RnalValue Theorem ANSI/AIM Code 128 Scanner In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Code 128A Creation In None Using Barcode generator for Software Control to generate, create Code 128 image in Software applications. If f(s) is the Laplace transform of f(t), then
USS Code 128 Recognizer In None Using Barcode reader for Software Control to read, scan read, scan image in Software applications. Code128 Encoder In C# Using Barcode encoder for Visual Studio .NET Control to generate, create USS Code 128 image in .NET framework applications. provided that sf(s) does not become infinite for any value of s satisfying Re(s) L 0. If this condition does not hold, f(t) does not approach a limit as t 1 03. In the practical application of this theorem, the limit of f(t) that is found by use of the theorem is correct only if f(t) is bounded as t approaches infinity. Proof. From the Laplace transform of a derivative, we have Making Code 128C In Visual Studio .NET Using Barcode generator for ASP.NET Control to generate, create Code 128B image in ASP.NET applications. Code 128A Maker In .NET Using Barcode generator for Visual Studio .NET Control to generate, create Code128 image in Visual Studio .NET applications. m dfp dt =
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UCC  12 Maker In None Using Barcode creation for Font Control to generate, create EAN128 image in Font applications. Decoding Code39 In C# Using Barcode decoder for VS .NET Control to read, scan read, scan image in .NET applications.  f(o) Printing 2D Barcode In VS .NET Using Barcode drawer for ASP.NET Control to generate, create Matrix 2D Barcode image in ASP.NET applications. Code39 Creation In VB.NET Using Barcode drawer for .NET Control to generate, create Code 39 Extended image in Visual Studio .NET applications. It can be shown that the order of the integration and limit operation on the left side of this equation can be interchanged if the conditions of the theorem hold. Doing this gives Evaluating the integral, limlY(t)l  f(o) =s;mW(s)l  fK9
which immediately yields the desired result.
Example 4.1. Find the final value of the function n(t) for which the Laplace transform is n(s) = 1 s(s3 + 3s* + 3s + 1) Direct application of the finalvalue theorem yields 1 1 lim [n(t)] =,,im s3 + 3s* + 3s + 1 = t1m As a check, note that this transform was inverted in Example 3.6 to give t* x(t) = 1  emt y+r+1 ( i which approaches unity as t approaches infinity. Note that since the denominator of sx(s) can be factored to (s + 1)3, the conditions of the theorem am satisfied; that is, (s + 1)3 + 0 unless s = 1. Example 4.2. Find the final value of the function x(t) for which the Laplace transform is x(s) = s4  6s2 + 9s  8 s(s  2)(s3 + 2s*  s  2) In this case, the function sx(s) can be written s4  6s2 + 9s  8 sx(s) = (s + l)(s + 2)(s  l)(s  2) Since this becomes infinite for s = 1 and s = 2, the conditions of the theorem are not satisfied. Note that we inverted this transform in Example 3.2, where it was found that 11 r 1 x ( t ) = 2+12e2t +  j  e 17  2 t + 1,t 3  iie FURTHER PROPERTIES OF TRANSFORMS
This function continues to grow exponentially with t and, as expected, does not approach a limit.
The proof of the next theorem closely parallels the proof of the last one and is left as an exercise for the reader. InitialValue
Theorem
M.f(~)l =sLlp Esf(s>l tl0
The conditions on this theorem are not so stringent as those for the previous one because for functions of interest to us the order of integration and limiting process need not be interchanged to establish the result. Example 4.3. Find the initial value x(0) of the function that has the following transform x(s) = s4  6s2 + 9s  8 s(s  2)(s3 + 2s2  s  2) The function sx(s) is written in the form sx(s) = s4  6s2 + 9s  8 s4  5s* + 4
After performing the indicated long division, this becomes sx(s) = 1 s*  9s + 12 s4  59 + 4
which clearly goes to unity as s becomes infinite. Hence x(0) = 1 which again checks Example 3.2.
lkanslation of lkansform
If LCf(t)} = f(s), then L{e f(t)} = f(s + a) In other words, the variable in the transform s is translated by a. Proof.
L{eatf (t)} = eof (t)e (s+a)tdt I
Example 4.4. Find L{eCarcos kt}. Since
= f (s + a) L{cos kt} = & THE LAPLACE
TRANSFORM
then by the previous theorem, S+U uenrcoS ktl = @ + a)2 + @ which checks Table 2.1. A primary use for this theorem is in the inversion of transforms. For example, using this theorem the transform 1 xCs) = (s + can be immediately inverted to x(t) = teea In obtaining this result, we made use of the following transform pair from Table 2.1: lkanslation
of Fbnction
LCf(t  to)} = ehsrof(s) If L(f(t)} = f(s), then
provided that
f(t) = 0 for t < 0
(which will always be true for functions we use). Before proving this theorem, it may be desirable to clarify the relationship between f (t  to) and f(t). This is done for an arbitrary function f(t) in Fig. 4.1, where it can be seen that f (t  to) is simply translated horizontally from f(t) through a distance to. tt
FIGURE 41 Illustration of f(i  to) as related to f(t).

