# Maple/Laplace Transforms

## Contents

## Introduction

Maple does not know how to do Laplace transforms *out of the box,*
but like so many entities, it can be taught. The `inttrans`

package for Maple contains algorithms for performing many useful
functions, including forward and inverse Laplace transforms. To load
it, simply type

```
with(inttrans)
```

into your worksheet. The list of new commands will show up. If you want to load the commands without seeing them, simply put a colon at the end of the

with(inttrans):

line. This is generally true for
Maple - the colon at the end will suppress display of the result.
Note also that Maple does understand the unit step function natively - it
calls it `Heaviside(t)`

.

## Basic Laplace and Inverse Laplace Transforms

The forward and inverse Laplace transform commands are simply `laplace`

and
`invlaplace`

. They take three arguments - the item
to be transformed, the original variable, and the transformed
variable. For Laplace transforms, the second and third arguments
will typically be *t* and *s*, respectively. Conversely, for `invlaplace`

,
the second and third arguments will be *s* and *t*, respectively.

For example, to get the Laplace Transform of what Dr. G refers to as the
**Mother Of All Transforms**, that is,

you could add:

```
MOAT:=exp(-a*t)*(A*cos(omega*t)+B*sin(omega*t));
MOATLAP:=laplace(MOAT, t, s)
```

which returns:

To find the inverse Laplace of:

you could type:

```
H:=exp(-s)/(s+a)^2-s/(s+a);
h:=invlaplace(H, s, t);
```

again being careful to note when Maple automatically adds subscripts, superscripts, and fractional parts for you. Maple returns:

which has two new functions in it - *Heaviside* and *Dirac*. *Heaviside* is Maple's unit step function and *Dirac* is Maple's Dirac delta function - i.e. the impulse. The expression above, then, could also be written as:

where the final *u(t)* is implied due to Maple's using the unilateral Laplace transform. Notice the time shift in the first term of the result -
this is a function of the exponential in the Laplace version.

It is critical to note that Maple performs the *unilateral*
Laplace transform. To prove this, note that the following three lines:

```
X1 := laplace(Heaviside(t-2), t, s);
X2 := laplace(Heaviside(t), t, s);
X3 := laplace(Heaviside(t+2), t, s);
```

will yield

and notice that the results of the latter two commands are the same. The step function *u(t+2)* is one starting at time *t=-2*, but the unilateral Laplace transform only looks at the signal starting at time 0, so it might as well be *u(t)*.

For more proof - and more insight to the unilateral Laplace transform - note that the following code:

```
x1:=invlaplace(laplace(Heaviside(t-2), t, s), s, tau);
x2:=invlaplace(laplace(Heaviside(t), t, s), s, tau);
x3:=invlaplace(laplace(Heaviside(t+2), t, s), s, tau);
```

yields:

where here is being used to clearly indicate the different between the original time variable and the time variable used by the inverse transform.

## Better Inverse Laplace Transforms

For inverse Laplace transforms, a little bit of processing can go a long way. Specifically, consider the following:

with(inttrans); x := exp(-t)+cos(80*t); X1 := laplace(x, t, s); X2 := simplify(X1);

At this point, X1 and X2 represent the laplace transform of the same function; they are just presented differently since X2 is the simplified version of X1. However, when taking the inverse:

x1 := invlaplace(X1, s, t); x2 := invlaplace(X2, s, t);

a disaster happens - x1 returns the original x while x2 returns...a mess. One way around this is to have Maple split the Laplace transform into partial fractions, then take the inverse of those parts; writing

x2a := invlaplace(convert(X2, parfrac, s), s, t)

will yield the original x. One caveat - this only works if the Laplace transform is a ratio of polynomials; if there are any time shifts, represented by exponentials of s in the transform, the conversion to a partial fraction will fail. For this reason, you may want to construct two different versions of the inverse Laplace transform simplifier:

IL := (X, s, t) -> simplify(convert(invlaplace(convert(X, parfrac, s), s, t), expsincos))

for taking the inverse of Laplace transforms that are ratios of polynomials and

ILTS := (X, s, t) -> simplify(convert(invlaplace(X, s, t), expsincos))

for those with time shifts. In both cases, the conversion to expsincos eliminates any complex exponentials (when possible).

## Typical Starting Commands When Working With Laplace

The following is a good set of commands with which to begin any worksheet where Laplace is used:

```
restart;
with(inttrans);
u := t -> Heaviside(t);
PAR := (Za , Zb) -> simplify(Za*Zb/(Za+Zb));
SCS := X -> sort(collect(simplify(expand(numer(X))/expand(denom(X))), s), s);
IL := (X, s, t) -> simplify(convert(invlaplace(convert(X,parfrac, s),s, t), expsincos));
ILTS := (X, s, t) -> simplify(convert(invlaplace(X, s, t), expsincos));
```

**Important note** - The Ogata **Modern Control Engineering** book uses specifically for the input to a state space system and never uses for the unit step. As a result, care must be taken when using Maple code and the above lines with respect to problems in that book.

## Troubleshooting

- Sometimes, the process by which the inverse is taken will lead to a representation of the signal that cannot be plotted. Best bet is to get a purely numerical version of the transform first, then take the inverse. In other words, to determine the signal generated for a system with a transfer function H and an input signal with a Laplace transform of X, where both depend on variables defined in a list called MyVals, you might write:

y := IL(subs(MyVals, H*X), s, t);

## Questions

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