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
The transform produces several changes in the equation:
The last point is the biggest single reason the Laplace transform is
valued -- it transforms linear differential equations into algebraic
equations, which many people find easier to solve.
|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"|
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:
Much of the time, the "trigger" time is set to zero, and the unit step
function defines what we typically think of as a constant (that exists
from "the beginning of time"!):
Applying the definition of Laplace transform to this function gives
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).
Properties of the Laplace Transform
The Laplace transform is a "linear operator", so
Derivatives also produce a very nice result:
This is the key result that causes time domain differential equations to
transform to Laplace domain algebraic equations. In the Laplace domain,
multiplication by s is equivalent to differentiation in the
time domain. Similarly, Laplace domain division by s can be
shown to be equivalent to integration in the time domain.
Proof: Applying the definition of Laplace transform to a derivative gives
which can be integrated by parts
The derivative result easily extends to second order
and can be continued on to higher orders if needed.
Initial Value and Final Value Theorems
There are two very useful theorems involving Laplace transforms.
As an example of the final value theorem, consider what happens as time
goes to infinity for the following:
In this example, it is pretty easy to predict the asymptotic behavior;
but in many others it is much easier to work s-->0 than t-->infinity.
- Final Value Theorem
- Initial Value Theorem
Solving ODEs with Laplace Transforms
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:
If both sides are transformed, the algebraic rearrangement to find X is
Now, all that is necessary is to invert the Laplace
transform to find the time domain solution. Look in the table for a
and then use this to invert the solution
Using this approach, many linear ODEs can be solved with a little
algebra and a table of Laplace transforms.
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:
This is done via expansion by partial fractions.
Partial Fractions Expansion
Any fraction with a polynomial denominator can be expressed as the sum
of terms with first order denominators
The simpler terms are much easier to invert using tables of Laplace
It may be best to explain the technique using an example. Say that
you need to invert
Once expanded, the terms look pretty simple to work with. All that is
needed is to find values for the constants c1 and
The steps are
(All but one of the constants vanishes each time, so it isn't really even
necessary to write those terms out.)
- multiply through by one of the denominator factors
- set s=p1. 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:
The infinite terms blow up in your face. The fix is to back up one
equation, differentiate it with respect to s and go from there.
Notice that it isn't necessary to do the differentiation of the
c1 term, since it is going to vanish anyway.
The inversion is then
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.
- 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.
- Riggs, J.B., Chemical Process Control (2nd Ed.),
Ferret, 2001, pp. 137-51.
Copyright 2003 by R.M. Price -- All Rights Reserved