Process Systems Analysis And Control P2

The Laplace transform f(s) contains no information about the behavior of f(t)for t
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Process Systems Analysis And Control P214 THE LAPLACE TRANSIVRh4 Example 2.1. Find the Laplace transform of the function f(t) = 1 According to Eq. (2. l), t=cu f(s) = J-ow(l)e-“‘df = - e-S’ = f S t=O 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(~) + bf2Wl = 4fl(O) + mf2Wl where a and b are constants, and f 1 and f2 am two functions of t. Proof. Using the definition, Uafl(t) + bfdt)) = lom[aflO) + bf2(~)le-s’d~ =a omfl(r)e-stdt + blom f2(t)e-S’dr I = &flW) + bUM)l 4. The Laplace transform operator transforms a function of the variable I to a func- tion of the variable s. The I variable is eliminated by the integration.Tkansforms of Simple FhnctionsWe 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 151. The step function .m = y tO This important function is known a$ the unit-step 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 tO I where u(t) is the unit-step function. Again proceeding according to definition, m m 1 - L{u(t)een’} = e-(s+a)rdt = _ Ae-(s+a)t 0 I 0 S+U 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 Integration by parts yields cc L{tu(t)} = -eesf f. + f= f i s ii 04. The sine function = u(t)sin k t L{u(t)sin k t } = sin kt ems’dt16 THE LAPLACE TRANsmRMTABLE 2.1FilDCtiOIl Graph ‘llxmfbrm 1u(t) ld - -F SW) -4 7 1Pu(t) 4 ?I! sn+l 1 1 4e-=“u(t) s+a n! (s + a)“+l ksin kt u(t) s2 + k2 THE LAPLACE TRANSFORM 17 TABLE 2.1 (Continued) lhlCtiOll Graph l.hmshm s coskt u(t) s2 + k2 k sinhkt u(t) s2 - k2 coshkr u(r) e-=’ Sink? u(r) -k-1 S s2 - k2 ...