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barcode reader project in c#.net Definition of the Ikansform in Software
Definition of the Ikansform USS Code 128 Decoder In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. ANSI/AIM Code 128 Drawer In None Using Barcode creation for Software Control to generate, create Code 128 image in Software applications. The Laplace transform of a function f(t) is defined to be f(s) according to the equation
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Thus, L(1) = f There am several facts worth noting at this point: 1. The Laplace transform f(s) contains no information about the behavior of f(t) for t < 0. This is not a limitation for control system study because t will represent the time variable and we shall be interested in the behavior of systems only for positive time. In fact, the variables and systems are usually defined so that f (t) = 0 for t < 0. This will become clearer as we study specific examples. 2. Since the Laplace transform is defined in Eq. (2.1) by an improper integral, it will not exist for every function f(t). A rigorous definition of the class of functions possessing Laplace transforms is beyond the scope of this book, but readers will note that every function of interest to us does satisfy the requirements for possession of a transform.* 3. The Laplace transform is linear. In mathematical notation, this means: ~bfl(~) +
Proof. Using the definition, bf2Wl = 4fl(O) + mf2Wl f
where a and b are constants, and
1 and
am two functions of t.
Uafl(t) + bfdt)) = lom[aflO) + bf2(~)les d~ =a omfl(r)estdt + blom f2(t)eS dr I
= &flW) + bUM)l 4. The Laplace transform operator transforms a function of the variable I to a function of the variable s. The I variable is eliminated by the integration. Tkansforms of Simple Fhnctions
We now proceed to derive the transforms of some simple and useful functions.
*For details on this and related mathematical topics, see Churchill (1972). THE LAPLACE
TRANSFORM
1. The step function
.m = y i
t<O t>O
This important function is known a$ the unitstep function and will henceforth be denoted by u(t). From Example 2.1, it is clear that L{u(t)} = f
As expected, the behavior of the function for t < 0 has no effect on its Laplace transform. Note that as a consequence of linearity, the transform of any constant A, that is, f(t) = Au(t), is just f(s) = A/s. 2. The exponential function f(t) = I ._ ,, t<O t>O I = u(t)e where u(t) is the unitstep function. Again proceeding according to definition, m L{u(t)een } = I e(s+a)rdt = _ m Ae(s+a)t
0 0 provided that s + a > 0, that is, s > a. In this case, the convergence of the integral depends on a suitable choice of S. In case s is a complex number, it may be shown that this condition becomes Re(s) >  a For problems of interest to us it will always be possible to choose s so that these conditions are satisfied, and the reader uninterested in mathematical niceties can ignore this point. 3. The ramp function 1 S+U
Integration by parts yields
L{tu(t)} = eesf f. + i s
cc f= f ii 0
4. The sine function = u(t)sin k t L{u(t)sin k t } = sin kt ems dt
THE LAPLACE
TRANsmRM
TABLE 2.1
FilDCtiOIl
u(t) 1 F
Graph
llxmfbrm
ld S
Pu(t) sn+l
e= u(t) (s + a) +l
sin kt u(t) k s2 + k2
THE LAPLACE
TRANSFORM
TABLE 2.1 (Continued) lhlCtiOll
Graph
l.hmshm s
s2 + k2
coskt u(t) sinhkt u(t)  k2
coshkr u(r) k1 Area = 1
s2  k2
e= Sink u(r) k (s + a) + k2
e cos kt u(t) (s + a)2 + k2
S(f), unit impulse + LAPLACE
TRANSFORM
Integrating by parts, eSt 01
L{u(t)sin k t } = (s sin kt + k cos kt) s* + k* k =s* + k* In a like manner, the transforms of other simple functions may be derived. Table 2.1 is a summary of transforms that will be of use to us. Those which have not been derived here can be easily established by direct integration, except for the transform of 6(t), which will be discussed in detail in Chap. 4. Transforms
of Derivatives
At this point, the reader may wonder what has been gained by introduction of the Laplace transform. The transform merely changes a function of t into a function of S. The functions of s look no simpler than those of t and, as in the case of A , A/s, may actually be more complex. In the next few paragraphs, the motivation will become clear. It will be shown that the Laplace transform has the remarkable property of transforming the operation of differentiation with respect to t to that of multiplication by s. Thus, we claim that = sf(s)  f(O) where f(s) = u.f(t)1 (2.2) and f(0) is f(t) evaluated at t = 0. [It is essential not to interpret f(0) as f(s) with s = 0. This will be clear from the following proof.]*

