The Laplace Transform of a Constant

the Laplace transform of a constant is equal to that same constant divided by P- 

---

For example, when a constant voltage, Eq, is applied at time zero to a series connection of R and C, the current is described by Eq. (5). Taking into account that the Laplace transform of a constant Eo) = Eq/s, one gets ... 

The Laplace transform of a step function is essentially the same as that of a constant in (2-7). When you do the inverse transform of A/s, which function you choose depends on the context of the problem. Generally, a constant is appropriate under most circumstances. 

Therefore the Laplace transformation of a step function (or a constant) of magnitude K is simply K(l/s). 

The relationship (4.67) has been defined for a flux that has an arbitrary variation with time hence, it must also be true for the constant unit flux described in Section 4.2.10. The Laplace transform of this constant unit flux 7 = 1 is /p according to Appendix 4.2 and the Laplace transform of the concentration response to the constant unit flux is given by Eq. (4.60), i.e.. 

When the Laplace transform of a differential equation is taken, a term must be included for the initial conditions, i.e., for the conditions at t = 0. This term is zero for the special case when the dependent variable and all its derivatives with respect to time are zero at t = 0. Now the typical process control situation is that the process variables are normally constant with time except for occasional disturbances which temporarily derange the system. Thus the period of time which is of especial interest in control analysis is the time from the start of a disturbance until the system returns to its normal controlled condition. At t = 0 for this period all the time derivatives of the dependent variable are zero since all conditions are steady, and for convenience the steady initial value of the variable can be taken as zero. Hence for control purposes the initial conditions terms in the Laplace transformation may be eliminated. 

The output of a zero-order hold element is like a pulse, having a constant height equal to m(nT) and duration T. After recalling that the Laplace transform of a unit pulse is given by eq. (7.12),... 

Next, we derive the Laplace transform of Eq. 17-20, y (5 ). The value of y(kAt) is considered to be a constant in each term of the summation and thus is invariant when transformed. Since 5S[S( )] = 1, it follows that the Laplace transform of a delayed unit impulse is 5 [b(t - kAt)] = Thus, the Laplace transform of... 

To illustrate how Laplace transforms work, consider the problem of solving Eq. (8-2), subjec t to the initial condition that = 0 at t = 0, and Cj is constant. If were not initially zero, one would define a deviation variable between and its initial value (c — Cq). Then the transfer function would be developed using this deviation variable. Taking the Laplace transform of both sides of Eq. (8-2) gives ... 

Given k fit) for nny reactor, you automatically have an expression for the fraction unreacted for a first-order reaction with rate constant k. Alternatively, given ttoutik), you also know the Laplace transform of the differential distribution of residence time (e.g., k[f(t)] = exp(—t/t) for a PER). This fact resolves what was long a mystery in chemical engineering science. What is f i) for an open system governed by the axial dispersion model Chapter 9 shows that the conversion in an open system is identical to that of a closed system. Thus, the residence time distributions must be the same. It cannot be directly measured in an open system because time spent outside the system boundaries does not count as residence but does affect the tracer measurements. 

Before finding the Laplace-transformed probability density wj(s, zo) of FPT for the potential, depicted in Fig. A 1(b), let us obtain the Laplace-transformed probability density wx s, zo) of transition time for the system whose potential is depicted in Fig. Al(c). This potential is transformed from the original profile [Fig. Al(a)] by the vertical shift of the right-hand part of the profile by step p which is arbitrary in value and sign. So far as in this case the derivative dpoints except z = 0, we can use again linear-independent solutions U(z) and V(z), and the potential jump that equals p at the point z = 0 may be taken into account by the new joint condition at z = 0. The probability current at this point is continuous as before, but the probability density W(z, t) has now the step, so the second condition of (9.4) is the same, but instead of the first one we should write Y (0) + v1 (0) = YiiOje f1. It gives new values of arbitrary constants C and C2 and a new value of the probability current at the point z = 0. Now the Laplace transformation of the probability current is... 

The mathematical basis for the exponential series method is Eq. (5.3), the use of which has recently been criticized by Phillips and Lyke.(19) Based on their analysis of the one-sided Laplace transform of model excited-state distribution functions, it is concluded that a small, finite series of decay constants cannot be used to represent a continuous distribution. Livesey and Brouchon(20) described a method of analysis using pulse fluorometry which determines a distribution using a maximum entropy method. Similarly to Phillips and Lyke, they viewed the determination of the distribution function as a problem related to the inversion of the Laplace transform of the distribution function convoluted with the excitation pulse. Since Laplace transform inversion is very sensitive to errors in experimental data,(21) physically and nonphysically realistic distributions can result from the same data. The latter technique provides for the exclusion of nonrealistic trial solutions and the determination of a physically realistic solution. These authors noted that this technique should be easily extendable to data from phase-modulation fluorometry. 

G( ) is the transfer function relating 0O and 0X. It can be seen from equation 7.18 that the use of deviation variables is not only physically relevant but also eliminates the necessity of considering initial conditions. Equation 7.19 is typical of transfer functions of first order systems in that the numerator consists of a constant and the denominator a first order polynomial in the Laplace transform parameter s. The numerator represents the steady-state relationship between the input 0O and the output 0 of the system and is termed the system steady-state gain. In this case the steady-state gain is unity as, in the steady state, the input and output are the same both physically and dimensionally (equation 7.16h). Note that the constant term in the denominator of G( ) must be written as unity in order to identify the coefficient of s as the system time constant and the numerator as the system... 

This example concerns the typical two-compartment model previously presented under the semi-Markov formulation (cf. Section 9.2.7). By assuming that molecules are initially present in the central compartment, (9.16) is the Laplace transform of the survival function in that compartment. If now the drug molecules are administered by a constant rate infusion between Tg and TE, the Laplace transform of the survival function in the central compartment becomes... 

It can be seen from Eq. (4.86) that A is the Laplace transform of the pulse of flux. However, a Laplace transform is an integral with respect to time. Hence, A, which is a flux (of moles per square centimeter per second) in the constant-flux problem (see Section 4.2.12), is in fact the total concentration (moles per square centimeter) of the... 

To complete the solution of the problem, it is necessary to specify the viscoelastic properties of the sample. For simplification reason, the Poisson ratio will be assumed to be constant. According to the results of Problem 16.5, which presents a similar situation, the Laplace transform of the stress is given by... 

A discrete view of the transport considers either random trajectories of discrete particles, or random streamtubes of tracer "parcels" carried by a constant flow rate through a network of fractures. For pulse injection of mass Af, we compute the tracer discharge as J(t) = My(t,T), the Laplace transform of y is... 

---