This page is a collection of links and other things to support learning about and using Laplace Transforms. Note that anything using a regular capital letter is a bilateral Laplace transform (e.g. $$F(s), G(s), X(s), Y(s)$$) whereas anything with a script capital letter is a unilateral Laplace transform (e.g. $$\mathcal{F}(s), \mathcal{G}(s), \mathcal{X}(s), \mathcal{Y}(s)$$)
Tips
Steps and Ramps
If you can write a signal as an accumulation of steps and ramps, the Laplace transform should be fairly simple given:
$$\begin{align*}
u(t-t_0)\rightarrow \frac{e^{-st_0}}{s}\\
r(t-t_0)\rightarrow \frac{e^{-st_0}}{s^2}
\end{align*}$$
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 $$0\leq (t-nT) < T$$. That is, $$\hat{f}(t)$$ represents the formula for the signal between times 0 and $$T$$, and that repeats as $$t$$ increases:
$$\begin{align*}
f(t)&=\sum_{k=0}^{\infty}\hat{f}(t-nT)\end{align*}$$
If $$\hat{F}(s)=\mathcal{L}\left\{\hat{f}(t)\right\}$$, then $$F(s)=\frac{\hat{F}(s)}{1-e^{-sT}}$$ with a region of convergence of $$\sigma>0$$. If the signal does not "start" at time 0, use the time shift property to accommodate the shift.
Table
Here are some of the pairs and properties that would be useful to memorize:
$$
\begin{array}{c|c|c}
f(t) & F(s) / \mathcal{F}(s) & \mathrm{ROC}\\ \hline
\delta(t) & 1 & \mathrm{All~}s\\
u(t) & \frac{1}{s} & \sigma>0\\
r(t)=t\,u(t) & \frac{1}{s^2} & \sigma>0\\
e^{-at}\,u(t) & \frac{1}{s+a} & \sigma>-a\\
t\,e^{-at}\,u(t) & \frac{1}{(s+a)^2} & \sigma>-a\\
\cos(\omega t)\,u(t) & \frac{s}{s^2+\omega^2} & \sigma>0\\
\sin(\omega t)\,u(t) & \frac{\omega}{s^2+\omega^2} & \sigma>0\\
%\cos(\omega t+\theta)\,u(t)=\cos(\omega t)\cos(\theta)\,u(t)-\sin(\omega t)\sin(\theta)\,u(t) & \frac{s\cos(\theta)-\omega\sin(\theta)}{s^2+\omega^2} & \sigma>0\\
%\sin(\omega t+\theta)\,u(t)=\cos(\omega t)\sin(\theta)\,u(t)+\sin(\omega t)\cos(\theta)\,u(t) & \frac{s\sin(\theta)+\omega\cos(\theta)}{s^2+\omega^2} & \sigma>0\\
e^{-at}\left(A\cos(\omega t)+B\sin(\omega t)\right)\,u(t) &\frac{A(s+a)+B\omega}{(s+a)^2+\omega^2} & \sigma>-a\\
e^{-at}\left(A\cosh(k t)+B\sinh(k t)\right)\,u(t) &\frac{A(s+a)+Bk}{(s+a)^2-k^2} & \sigma>-a\\
\end{array}
$$
$$
\begin{array}{c|c}
g(t) & G(s) \\ \hline
f(t-t_0) & F(s)e^{-st_0} \\
e^{-at}f(t) & F(s+a)\\
\frac{df(t)}{dt} & sF(s)\\
\frac{d^2f(t)}{dt^2} & s^2F(s)\\
\int_{-\infty}^{t}f(\tau)\,d\tau & \frac{F(s)}{s} \\
tf(t) &-\frac{dF(s)}{ds}\\
x(at) & \frac{1}{|a|}F\left(\frac{s}{a}\right)\\
\sum_{k=0}^{\infty}\hat{f}(t-kT) & \frac{\hat{F}(s)}{1-e^{-sT}}
\end{array}
\hspace{1in}
\begin{array}{c|c}
g(t) & \mathcal{G}(s)\\ \hline
f(t-t_0) & \mathcal{F}(s)e^{-st_0}\mbox{ iff }f(t-t_0)\,u(t)=f(t-t_0)\,u(t-t_0)\\
e^{-at}f(t) & \mathcal{F}(s+a)\\
\frac{df(t)}{dt} & s\mathcal{F}(s)-f(0^-)\\
\frac{d^2f(t)}{dt^2} & s^2\mathcal{F}(s)-\dot{f}(0^-)-sf(0^-)\\
\int_{0^-}^{t}f(\tau)\,d\tau & \frac{\mathcal{F}(s)}{s}\\
tf(t) &-\frac{d\mathcal{F}(s)}{ds}\\
x(at),\,a>0 & \frac{1}{a}F\left(\frac{s}{a}\right)\\
\sum_{k=0}^{\infty}\hat{f}(t-kT) & \frac{\hat{\mathcal{F}}(s)}{1-e^{-sT}}
\end{array}
$$
Examples
Right-sided periodic signal
Imagine you want to find the Laplace transform of the right-sided periodic function plotted in the second plot below:
The steps are as follows:
- Determine the period. In this case, the period $$T=11$$ s
- Find a formula for $$\hat{f}(t)$$, the function that handles the first period of $$f(t)$$. Based on the value and slope changes in the function,
\(
\hat{f}(t)=2\,u(t)+\frac{1}{4}\,r(t)-2\,u(t-4)-2\,u(t-8)-\frac{1}{4}\,r(t-8)
\)
- Find a formula for $$\hat{F}(s)$$, the Laplace transform of $$\hat{f}(t)$$. Using the time-shift property and the Laplace transforms for steps and ramps, this would give:
\(
\begin{align*}
\hat{F}(s)&=\frac{2}{s}+\frac{\frac{1}{4}}{s^2}-\frac{2e^{-4s}}{s}-\frac{2e^{-8s}}{s}-\frac{\frac{1}{4}e^{-8s}}{s^2}\\
\hat{F}(s)&=\frac{1}{4}\left(\frac{8s+1-8se^{-4s}-8se^{-8s}-e^{-8s}}{s^2}\right)
\end{align*}
\)
- Use the property that the Laplace transform of the right-sided periodic signal $$f(t)$$ is the Laplace transform of the first period divided by $$1-e^{-sT}$$
\(
\begin{align*}
F(s)&=\frac{\hat{F}(s)}{1-e^{-sT}}=\frac{\hat{F}(s)}{1-e^{-11s}}\\
F(s)&=\frac{8s+1-8se^{-4s}-8se^{-8s}-e^{-8s}}{4s^2(1-e^{-11s})}
\end{align*}
\)
