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# Introduction to Transforms

## Simple Example

Suppose we are trying to solve a problem which in itself may require
effort. For example, suppose you are asked to calculate the product
914523098123515 × 52354239523952323553. You could use standard
multiplication techniques to get the answer 47879161329345458034788898437648795.
This is a direct method of solving this problem, but due to the limited processing
power of the human brain, this is both tedious and error prone. An alternate
solution is to program a computer to solve these problems; however, the computer
calculates arithmetic in binary precision. Therefore, it is necessary to
transform both numbers into base 2, carry out the multiplication, and then
transform back to base 10. In this example, we would convert the multiplier
and multiplicand to 11001111111100000100000000011001110000110011111011_{2}
and 101101011010001111100111110111011001011110111001001110101111100001_{2},
allow the computer to take the product, 10010011100010011111111110100000110011101110111001101110011100110000010001110011110110101101000010001101000110011011_{2},
and then transform back to 47879161329345458034788898437648795. This process is shown in Figure 1.

Figure 1. Calculating the product of two integers.

An average or poor high school student will always pick the most straight-forward route: just multiply
the two numbers together. A good high school student, however, will realize that while programming
the conversion may be more difficult, and the two numbers could not be stored as a single integer,
a little effort will produce significant rewards in the end. A good engineering student may even
look up the GNU GIMP package for large-integer multiplication.

## Matrix Example

To give a second transformation, suppose you have an
matrix and you wish to calculate
. From your other courses,
you will understand that this is a very expensive operation both in terms of memory and time.
If, however, the matrix is diagonalizable, that is, we may write
where is a diagonal
matrix, we may substitute this into the equation above to get

and using the properties of matrices, we can simplify this to

.

This new calculation is now relatively easy because is a diagonal matrix
where the diagonal entries are the roots of the *characteristic equation* of the matrix.

# The Laplace Transform

We have up to this point seen how we can solve a relatively simple differential equation
to get a solution to an initial-value problem. This straight-forward computation is shown
in Figure 2.

Figure 2. The direct methods of solving initial-value problems using solutions to the homogeneous equation and particular solutions.

Unfortunately, as problems become larger,
it becomes more and more difficult to set up the correct equations to find the solution.
Therefore, we will follow an alternate route. By now, it should be obvious that the characteristics
of a solution to an initial-value problem depend on the placement of the roots. Therefore, it
would be reasonable to assume that we could perform calculations by focusing on the the characteristic
equation. As it turns out, this is exactly what the Laplace transform allows us to do. If
you take a look at Figure 3, rather than solving the IVP directly, we convert the problem
using the Laplace transform into an algebraic problem; we perform some algebraic manipulation;
and we transform back to the solution.

Figure 3. The indirect methods of solving initial-value problems using the Laplace transform.

The other benefit of the Laplace transform is that it abstracts out the essential information
about the differential equation and its solution. You, as a student, will need to learn a complete
new approach to solving problems: by this point, most students focus on solving problems
directly. Initially, the Laplace transform may seem unnecessary, but it is worth the effort and
later, you will see the Fourier transform, as well.

The ability of the Laplace transform to simplify the finding of solutions is used in
numerous fields, including:

- Finding transient and steady-state responses of electrical circuits
- Applications of dynamical problems (vibrations, acoustics)
- Applications to structural problems

We will look at these issues but we will start with some definitions.

# The Definition of the Laplace Transform

The Laplace transform of a function is the improper integral

where is assumed to be a complex variable.

Recall that .

In general, if the function has a Laplace transform
, we will write .

The calculation of the Laplace transform is reasonably straight-forward
for certain simple functions which we will do here.

## The Unit Step Function

The unit step function is a function defined as

The Laplace transform of the unit step function is

.

Therefore, .

## Ramp Function

The ramp function is the unit step function multiplied by the function :

.

Therefore, .

## Exponential Function

The exponential function has a straight-forward transformation:

.

Therefore, .

## Ramped Exponential Function

.

Therefore, .

## Patterns

At this point, you should have already recognized some patterns. Notice that if , then
and the Laplace transform equals
. At this point, many students will try to plot the Laplace transforms and
attempt to *understand* the relationships. Avoid this temptation at least for the balance of this
course until you have a clearer understanding of integral transforms: at this point, it is better to
focus on understanding the consequences of the transform.

## Tables of Laplace Transforms

In general, it is easiest to build a table of various Laplace transforms. Once we do this, we can can
then use the table as a look-up and not have to repetitively calculate various integrals.

Table 1. A table of simple Laplace transforms.

# Properties of the Laplace Transform

There are five properties of the Laplace transform which we will look at here:

## Linearity

If , , and
and are real or complex constants, it follows that

.

## First Shift Theorem

If and is a real or complex constant, it follows that

.

## Derivative of the Transform

If and is a positive integer, it follows that

.

We may prove this with induction:

For the case , it is straight forward to see that

.

Assume it is true for the case . It follows that the case may be found using integration by parts to get
.

It is always useful to test that our properties hold for what we have already seen: We know , and therefore .
Similarly, we know , and therefore .

## Transform of the Derivative

If and is a positive integer, it follows that

.

Again, we may prove this with induction:

For the case , it is straight forward to see that

.

Assume it is true for the case . It follows that the case may be found using integration by parts to get
which expands into our desired result.

## Transform of the Indefinite Integral

If , it follows that

.

This may be shown by substituting
and therefore .
Using the previous theorem, we get that

## Example: Transform of the Trigonometric Functions

Recall that .

Because the real and imaginary components must match, it follows that

and

.