# What is the laplace transform and how is it performed? (detailed explanation)

I am a high school student and I became interested after someone mentioned it. Although I am not quite at the level where I am taught this it just captured my attention. Could someone give me an explanation of what the laplace transform is, and perhaps even a relatively basic example?

The Laplace transform is a tool engineers, scientists, and applied mathematicians use to solve equations that occur in optics, electrical engineering, and thermodynamics.

To provide a picture of just what a Laplace transform is, we need to understand what a transform is. Let's start with the concept of a function. As you may know, a function may be thought of as a special type of machine that transforms a first number into a second number. Some functions have inverses that allow us to transform the second number back into the first number.

A transform is special type of machine as well, but it works on functions. That is, a transform changes a first function into a second function. In this case, it is required that a transform have an inverse that changes the second function back into the first function. Without an inverse, a transform is useless for solving equations.

The idea behind a transform is to change a very complicated equation into a simple equation. Integral transforms (such as the Laplace transform) change complicated differential expressions (that involve derivatives of unknown functions) into algebraic expressions. However, once we solve for the transform of the unknown function, we have to invert the transform to recover the unknown function. If the inversion process is awful, then the transform is, again, useless.

Now, to describe the Laplace transform. The Laplace transform converts a function $f(t)$, where $t$ is positive, to a function $F(s)$, where possible values of the variable $s$ depend on the behavior of the function $f$ for large values of $t$. To avoid getting stuck in the mud, let's just move on and write the transform:

$$F(s) = \int_0^{\infty} dt \, f(t) e^{-s t}$$

If you are not familiar with this mathematical expression, let me explain. The big $S$ symbol is called an integral sign, and it is a special type of sum. In this case, we are summing the quantity to the right of the sign, $f(t) e^{-s t}$, for every value of $t$ in the range $(0,\infty)$. As this is a sum over an infinite range, we worry whether the sum actually returns a finite number (converges). Again, the values of $s$ we use to build up the function $F$ are those values for which the sum converges. Those values of $s$ will include complex numbers.

Until you learn to evaluate the sums, you can look up Laplace transforms of many functions $f$ in a table. For example, when $f(t) = 1$ then $F(s) = 1/s$. When $f(t) = \sin{t}$, then $F(s) = 1/(s^2+1)$. There are many examples you can find by simply googling.

What does the transform represent? Well, we may think of the transform representing the exact same thing as the function, but from a different point of view. For example, if we think of $f(t)$ as an amplitude of a signal at at time $t$ that gets flipped on at $t=0$, then the transform $F(s)$ may be thought of as the response of that signal to a decay operation over time at a decay rate $s$.

The important thing for solving equations is that the Laplace transform applied to the derivative of a function $f'(t)$ becomes $s F(s) - f(0)$. Thus, the Laplace transform embeds initial conditions into the algebraic equation.

We may use Laplace transforms to turn a differential equation into an algebraic one. For example, the differential equation

$$f''(t) + f(t) = 1 \quad f(0) = f'(0) = 0$$

becomes the following algebraic equation after applying a Laplace transform

$$(s^2+1) F(s) = \frac1{s}$$

or

$$F(s) = \frac1{s (s^2+1)}$$

This uniquely specifies $f(t)$, as the Laplace transform and its inverse are unique. Now we may try to look this up in a table. However, we may not find such a transform. In this case, there are tool we may use to invert the transform. One is convolution, which we may use when we are multiplying two transforms together and we know the inverse of each transform. The other is evaluation of the inverse by its formula; this requires knowledge of a subject called complex analysis and is my preferred method, but it will take a little while to get there. In this case, the inverse transform, and the solution to the equation, is

$$f(t) = 1-\cos{t} = 2 \sin^2{\left ( \frac{t}{2} \right )}$$

I hope this helps.

• Thank you, this is a great answer, was not expecting anyone to answer with as much detail. – AlanZ2223 Mar 8 '15 at 22:51