Introduction to the Laplace Transform

The Laplace Transform is a great method that turns tricky nonhomogeneous differential equations (or ANY order) into an algebra problem. This is great because differential equations are really hard and algebra is pretty easy. So, what's the catch? First it takes a little bit of background math to learn what a laplace transform is and how to use it. Second it takes quite a bit of algebra and some finessing about with algebra to get tricky answers. However without the laplace transform these answers would be borderline impossible to get. For situations where undetermined coefficients or variation of parameters can't help us the laplace transform comes to the rescue. The laplace transform is so widely and commonly used that it's the way I solve 99% of all differential equations I come across. Like I said, it takes some getting used to, and some seemingly unrelated maths to get into, but the power of the technique is definitely worth it. Trying to do electrical engineering without the laplace transform would be like trying to do electrical engineering without access to resistors.
The laplace transform of some function \(f(t)\) which is piecewise-continuous (never undefined or infinity and only has a finite number of breaks) is given by: $$ \Laplace\{f(t)\} = \int_0^\infty e^{-st}f(t)dt $$
You can see that the Laplace transform turns our original function \(f\) of \(t\) into a new function of the complex variable \(s\). \(s\) is a complex number and in fact is called the "complex frequency". You don't need to understand why the laplace transform is defined this way or why that formula should be any good. Just learn it and learn how to use it.
For some function \(f(t)\) we write its Laplace transform as \(\Laplace\{f(t)\} = F(s)\)

Find the Laplace transform of \(f(t) = 1\)

We just plug \(f(t) = 1\) into our formula to get: \(\int_0^\infty e^{-st} dt\) If you don't remember improper integrals we'll do a super quick summary now. Convert the integral into the limit of a proper integral: \(\lim_{a \to \infty} \int_0^a e^{-st} dt = \lim_{a \to \infty} \left[ -\frac{1}{s}e^{-st} \right]_0^a\) Then plug in our values and do the limit: \( \lim_{a \to \infty} \frac{1}{s}\left( -e^{-at} + e^{0t} \right) = \frac{1}{s}\) So \(\Laplace\{1\} = \frac{1}{s}\)

Find \(\Laplace\{e^{at}\}\)

This time we need to solve: \(\int_0^\infty e^{-st}e^at dt \) \(= \int_0^\infty e^{(a-s)t} dt = \lim_{x \to \infty} \int_0^x e^{(a-s)t} dt\) \(= \lim_{x \to \infty} \frac{1}{a-s} \left( e^{(a-s)t} \right)_0^x \) \(= -\frac{1}{a-s} = \frac{1}{s-a}\)
A nice little formula to keep on your cheat sheet is: \(\int_0^\infty e^{ct} dt = -\frac{1}{c}\). You'll find it saves a lot of improper integral working that can be error prone.
The Laplace transform is linear which means: \(\Laplace\{f(t) + g(t)\} = F(s) + G(s)\) AND \(\Laplace\{Af(t)\} = AF(s)\)
When you're computing Laplace transforms you transform each of the terms individually and just leave the multiplying constants as they are. So \(\Laplace\{2\} = \frac{2}{s}\)

Find the Laplace transform of: \(f(t) = 5 + 2e^{3t}\)

We'll transform each of the two terms separately and then add their answers to get our final result. For the first part we need \(\Laplace{5}\) which we know is just \(5\Laplace\{1\} = \frac{5}{s}\). For the second part we know that \(\Laplace\{2e^{3t}\} = 2\Laplace\{e^{3t}\} = 2\frac{1}{s-3}\) So \(F(s) = \frac{5}{s} + \frac{2}{s-3}\)
When we do Laplace transforms in the real world we don't do the integral. We use a lookup table that includes the transforms of a dozen or more common functions. Then we add, subtract and multiply them by constants to get our final transform.
The following is a small table of Laplace transforms. Larger tables are easy to come by, just google "Laplace transform table", they're all the same.
\(f(t)\) \(F(s)\)
\(1\) \(\frac{1}{s}\)
\(t^n\) \(\frac{n!}{s^{n+1}}\)
\(e^{at}\) \(\frac{1}{s-a}\)
\(\cos(\omega t)\) \( \frac{s}{s^2 + \omega ^2} \)
\(\sin(\omega t)\) \( \frac{\omega}{s^2 + \omega ^2} \)

Find \(\Laplace\{5\cos(5t) + e^t\}\)

We do each of our two terms separately and add their transforms together so our first one, \(5\cos(5t)\) is going to be: \(5 \Laplace\{\cos(5t)\} = 5 \frac{s}{s^2 + 25}\) Our second term \(e^t\) is going to give us \(\frac{1}{s-1}\) So our final solution is: \(5\frac{s}{s^2 + 25} + \frac{1}{s-1}\)
practice problems