Laplace transform using partial fractions

In practice, inverse transformation is most easily achieved by using partial fractions to break down solutions into standard components, and then use tables of Laplace transform pairs, as given in Table 3.1. 

Obviously, one approach to find the inverse Laplace transform is by using the known transforms that can be found in tables. Often, manipulation of an algebraic transform (such as breaking up expressions using partial fractions) can result in a form that is easily found from a table of known transforms. When such a simple procedure is not applicable, the inversion may be possible using the inversion integral. 

Show appropriate calculations using partial fraction expansion and Laplace transforms. 

Equation (8-14) shows that starts from 0 and builds up exponentially to a final concentration of Kcj. Note that to get Eq. (8-14), it was only necessaiy to solve the algebraic Eq. (8-12) and then find the inverse of C (s) in Table 8-1. The original differential equation was not solved directly. In general, techniques such as partial fraction expansion must be used to solve higher order differential equations with Laplace transforms. 

Since we are doing inverse transform using a look-up table, we need to break down any given transfer functions into smaller parts which match what the table has—what is called partial fractions. The time-domain function is the sum of the inverse transform of the individual terms, making use of the fact that Laplace transform is a linear operator. 

In practice, the inverse Laplace transformations are obtained by reference to the rather extensive tables that are available. It is sometimes useful to develop the function in question in partial fractions, as employed in Section 3.3.3. The resulting sura of integrals can often be evaluated with the use of the tables. 

For simple functions the z-transforms can be obtained easily from their equivalent Laplace transforms by the use of partial fractions and Appendix 7.1. 

A combination of Appendix 7.1 and partial fractions may also be used to derive the inverse of a z-transform (i.e. to determine the corresponding time domain function) in a similar manner to that for Laplace transforms. This is illustrated by the following example. 

Benet, L.Z. Turi, J.S. Use of general partial fraction theorem for obtaining inverse laplace transforms in pharmacokinetic analysis. J. Pharm. Sci. 1971, 60, 1593-1594. 

This method is completely parallel to the partial-fractions expansion methodology used to invert Laplace transforms and proceeds as follows ... 

As pointed out above, the critical point in finding the solution to a differential equation using Laplace transforms is the inversion of the Laplace transforms. In this section we will study a method developed by Heaviside for the inversion of Laplace transforms known as Heaviside or partial-fractions expansion. 

We have derived the general Inversion theorem for pole singularities using Cauchy s Residue theory. This provides the fundamental basis (with a few exceptions, such as /s) for inverting Laplace transforms. However, the useful building blocks, along with a few practical observations, allow many functions to be inverted without undertaking the formality of the Residue theory. We shall discuss these practical, intuitive methods in the sections to follow. Two widely used practical approaches are (1) partial fractions, and (2) convolution. 

The previous example is very straightforward and could easily be inverted by use of partial fractions and a table of Laplace transforms. This next example may be tabulated. However, it is used here to demonstrate more clearly how the residue theorem may be useful for similar or more complicated inversions. Consider... 

Using the partial fractions (or Heaviside theorem), we can obtain the inverse Laplace transform. 

After Kc is specified, y t) can be determined from the inverse Laplace transform of Eq. 11-76. But first the roots of the cubic polynomial in s must be determined before performing the partial fraction expansion. These roots can be j calculated using standard root-finding techniques (( and Canale, 2010). Figure 11.23 demonstrates that] increases, the response becomes more oscillatory unstable for Kc = 15, More details on the actual stabilit limit of this control system are given in Example 11.10. 

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