## Laplace Transforms

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:
 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
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.

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

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 derivatives:

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.

• Final Value Theorem
• Initial Value Theorem
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.

### 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 pretty simple.
Now, all that is necessary is to invert the Laplace transform to find the time domain solution. Look in the table for a useful transform
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 transforms.

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 c2. The steps are
1. multiply through by one of the denominator factors
2. set s=p1. All the constants but one will vanish.
3. repeat for the other factors
(All but one of the constants vanishes each time, so it isn't really even necessary to write those terms out.)
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

• 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:

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

R.M. Price
Original: 10/20/93
Modified: 5/15/2003