# What are transforms in general?

Transforms are used to literally "transform" an equation of one type (ie: differential) into an equation of another type (ie: algebraic). Generally speaking, we use transforms to convert equations that are difficult to work with (ie: differential equations) into those that are more easily manipulated (ie: algebraic equations). Let's take a quick look at a genre of transforms known as "integral transforms."

### Integral Transforms

An integral transform is a relation of the form:

$$F(s) = \int_{\alpha}^{\beta} K(s,t)f(t) dt $$

...where K(s,t) is known are the "kernel" of the transformation.

f(t) is the function that is being "transformed" and F(s) is the result of the transformation.

Let's now take a look at a specific type of integral transform known as the "Laplace Transform"

### Definition of the Laplace Transform:

The Laplace Transform of a given function f(t) is defined as:

$$\mathcal{L}(f(t)) = F(s) = \int_{0}^{\infty} e^{-st}f(t) dt \qquad(relation\;1)$$

...where:

$$K(s,t) = e^{-st} = kernel \;of\;the\;transform. $$

...and the improper integral converges.

It is important to note that there is a 1:1 relationship between a function and its Laplace Transform. Applications of the Laplace transform are found in the analysis of mechanical vibrations and electrical circuits where it provides an alternative (often simplified) functional description of the behavior of a system. Often-times it is used to take a function of a complex variable, such as frequency (denoted by "s"), and yield a function of a real variable such as time (denoted by "t").

Let's look at some examples of using the Laplace transform on some functions...

### (Ex 1): Find the Laplace transform of the function f(t)=t

First recall the definition of the Laplace transform using relation 1:

$$\mathcal{L}(f(t)) = F(s) = \int_{0}^{\infty} e^{-st}f(t) dt$$

...therefore:

$$F(s) = \int_{0}^{\infty} e^{-st}t \; dt $$

...and using the definition of an improper integral:

$$\int_{a}^{\infty} f(t) dt = \lim_{A\to\infty} \int_{a}^{A} f(t) dt $$

...we get

$$F(s) = \lim_{b\to\infty} \int_{0}^{b} e^{-st} \; t \; dt $$

Using "Integration by Parts" or the "Tabular Method" to evaluate the integral gives us:

$$F(s) = \lim_{b\to\infty} [\frac{-1}{s}e^{-st}t - \frac{1}{s^2}e^{-st}] \Big|_0^b $$

$$F(s) = \lim_{b\to\infty} [(\frac{-b}{s}e^{-sb} - \frac{1}{s^2}e^{-sb}) - (0 - \frac{1}{s^2})] $$

$$F(s) = \frac{1}{s^2} = the \; Laplace \; Transform \; of \; f(t)=t $$

### (Ex 2): Find the Laplace transform of the function f(t), where:

$$f(t) = e^{at}$$

...and "a" is some constant

Once again we use the definition of the Laplace transform shown in relation #1 and obtain:

$$\mathcal{L}(f(t)) = F(s) = \int_{0}^{\infty} e^{-st} \;e^{at} dt $$

...and by using our knowledge of improper integrals, we get:

$$F(s) = \lim_{b\to\infty} \int_{0}^{b}e^{-st} \; e^{at} dt$$

$$F(s) = \lim_{b\to\infty} \int_{0}^{b} e^{at-st} dt$$

$$F(s) = \lim_{b\to\infty} \int_{0}^{b}e^{-t(s-a)} dt $$

$$F(s) = \lim_{b\to\infty} [\frac{1}{-(s-a)} e^{-t(s-a)}] \Big|_0^b $$

$$F(s) = \lim_{b\to\infty} [\frac{1}{-(s-a)} e^{-b(s-a)} - \frac{1}{-(s-a)}] $$

$$F(s) = 0 - \frac{1}{-(s-a)} $$

$$F(s) = \frac{1}{s-a}= the \; Laplace \; Transform \; of \; f(t)=e^{at}$$

### (Ex 3): Find the Laplace transform of the function f(t), where:

$$f(t) = t^2$$

Using the definition of the Laplace transform shown in relation #1 gives us:

$$\mathcal{L}(f(t)) = F(s) = \int_{0}^{\infty} e^{-st} \;t^2 dt $$

...and by definition of the improper integral this becomes:

$$F(s) = \lim_{b\to\infty} \int_{0}^{b}e^{-st} \; t^2 dt$$

The above integral can be evaluated by "integration by parts" or the "tabular" method.

$$F(s) = \lim_{b\to\infty}[\frac{-t^2}{s}e^{-st} - \frac{2t}{s^2}e^{-st} - \frac{2}{s^3}e^{-st}] \Big|_0^b $$

...which is the same as:

$$F(s) = \lim_{b\to\infty}[\frac{-t^2}{s e^{st}} - \frac{2t}{s^2 e^{st}} - \frac{2}{s^3 e^{st}}] \Big|_0^b $$

$$F(s) = \lim_{b\to\infty}[\frac{-b^2}{s e^{sb}} - \frac{2b}{s^2 e^{sb}} - \frac{2}{s^3 e^{sb}} + \frac{2}{s^3}] $$

$$F(s) = 0-0-0+\frac{2}{s^3} $$

$$F(s) = \frac{2}{s^3} = the \; Laplace \; Transform \; of \; f(t)=t^2$$

While the above examples are relatively straight forward to solve, it can become tedious to evaluate more elaborate Laplace Transforms. Additionally, when we go to solve differential equations using Laplace Transforms, it is convenient to have some of the more common transformations handy in a table. Such a table can be found here.