A mathematical *transform* takes an expression in one
mathematical "language" and converts it to another. If done correctly, a
transform doesn't change the meaning of the expression, but may make it
easier to interpret, as with a unit conversion. We usually choose to use
transforms to gain insight or simplify a problem.

The *Laplace transform* is commonly used in process modeling and
control. The transform is given by

Variable is t (time) |
Variable is s (dimensions of inverse time) |

t is Real number |
s is Complex number |

Solutions from "Time Domain" | Solutions from "Laplace Domain" |

Differential equation | Algebraic equation |

As an example, we'll apply the definition of the Laplace transform to
the *unit step function*. The step function is very important in
control modeling. It is given by:

You can always use the definition to find the Laplace transform of a function, but that usually is more trouble than it is worth. First of all, you can probably work the vast majority of control problems with a set of about 4 transforms -- so you'll probably end up memorizing those. Beyond that, tables are readily available (for instance, Riggs p. 139). Get yourself a good one, and you may never need to apply the definition again (unless someone decides to give an evil homework problem).

The Laplace transform is a "linear operator", so

Derivatives also produce a very nice result:

which can be integrated by parts

The derivative result easily extends to second order derivatives:

There are two very useful theorems involving Laplace transforms.

- Final Value Theorem
- Initial Value Theorem

The major reason people bother with Laplace transforms is that they can make it easier to obtain analytical solutions of many linear ordinary differential equations. Consider the ODE and initial condition:

Not every solution is likely to be found in a table, but if you use the properties of the transform you see that you need to break problems up into smaller, more familiar ones:

Any fraction with a polynomial denominator can be expressed as the sum of terms with first order denominators

It may be best to explain the technique using an example. Say that you need to invert

- multiply through by one of the denominator factors
- set s=p
_{1}. All the constants but one will vanish. - repeat for the other factors

The inversion is easy

The method becomes more complicated when roots are repeated. You must always have one term, and one unknown constant, for each root. Look what happens, then if a root is repeated:

Two last comments:

- partial fractions expansion works even if the denominator factors are complex numbers, it just makes the algebra a little trickier
- if you only care about the "shape" of the solution, not the relative weightings, you may not need to evaluate the constants.

**References:**

- Coughanowr, D.R. and L.B. Koppel, Process Systems Analysis and Control, McGraw-Hill, 1965, pp. 13-41.
- Luyben, W.L., Process Modeling, Simulation and Control for Chemical Engineers (2nd Ed.), McGraw-Hill, 1990, pp. 303-24.
- Marlin, T.E., Process Control: Designing Processes and Control Systems for Dynamic Performance, McGraw-Hill, 1995, pp. 108-23.
- Riggs, J.B., Chemical Process Control (2nd Ed.), Ferret, 2001, pp. 137-51.

R.M. Price

Original: 10/20/93

Modified: 5/15/2003

Copyright 2003 by R.M. Price -- All Rights Reserved