Laplace Transform

This page is a collection of links and other things to support learning about and using Laplace Transforms.

Tips

Steps and Ramps

$$ \begin{array}{c|c} f(t) & F(s) \\ \hline u(t) & \frac{1}{s}\\ u(t-t_0) & \frac{e^{-st_0}}{s}\\ r(t)=t\,u(t) & \frac{1}{s^2}\\ r(t-t_0)=(t-t_0)\,u(t-t_0) & \frac{e^{-st_0}}{s^2}\\ \end{array} $$

Time Shifts

If there are time shifts, the shift in the argument of the step must match the shift in the function by which it is multiplied. If not, you need to manipulate things so that the function by each step is multiplied matches the shift in the step. Here are some examples:

No shift in step

You probably just need to distribute things out:

$$ \begin{align*} x(t)&=(t-4)\,u(t)\\ &=t\,u(t)-4\,u(t)\\ X(s)&=\frac{1}{s^2}-4\frac{1}{s} \end{align*} $$

Shift in step

You may need to "complete the shift" for functions outside of the argument of the step to match the shift inside the argument of the step:

$$ \begin{align*} y(t)&=t\,u(t-4)\\ &=\left(t-{\color{red} 4}+{\color{blue} 4}\right)\,u(t-4)\\ &=\left(t-{\color{red} 4}\right)\,u(t-4)+{\color{blue} 4}\,u(t-4)\\ Y(s)&=\frac{e^{-4s}}{s^2}+4\frac{e^{-4s}}{s} \end{align*} $$

Shift in step redux

You may need to "complete the shift" for functions outside of the argument of the step to match the shift inside the argument of the step; this must include every instance of $$t$$:

$$ \begin{align*} z(t)&=te^{-2t}\,u(t-4)\\ &=\left(t-{\color{red} 4}+{\color{blue} 4}\right)e^{-2(t-{\color{green} 4}+{\color{purple} 4})}\,u(t-4)\\ &=\left(t-{\color{red} 4}\right)e^{-2(t-{\color{green} 4})}e^{-{\color{purple} 8}}\,u(t-4)+{\color{blue} 4}e^{-2(t-{\color{green} 4})}e^{-{\color{purple} 8}}\,u(t-4)\\ Z(s)&=\frac{e^{-4s}e^{-8}}{(s+2)^2}+4\frac{e^{-4s}e^{-8}}{s+2} \end{align*} $$

"Periodic" Signals

There is no Laplace Transform for a truly periodic signal, because there is no region of convergence, but if a signal is periodic in a half-plane (typically the right-half-plane), there will be a Laplace Transform as long as the signal has finite values. Assume you have some signal $$f(t)$$ that is 0 for all times $$t<0$$, $$\hat{f}(t)$$ for times $$0\leq t < T$$, and $$\hat{f}(t-nT)$$ for all $$t \geq T$$ where $$n$$ is an integer chosen such that the argument $$0\leq (t-nT) < R$$. That is, $$\hat{f}(t)$$ represents the formula for the signal between times 0 and $$T$$, and that repeats as $$t$$ increases. If $$\hat{F}(s)=\mathcal{L}\left\{\hat{f}(t)\right\}$$, then $$F(s)=\frac{\hat{F}(s)}{1-e^{-sT}}$$