Computing Laplace Transforms

A table of Laplace transforms is simply a list of answers someone already worked out, so you don't have to do the integral every time. It's good for speed: you split a function into familiar pieces and look each one up. The rules look like a wall of fractions, but each is a small fact: $t^n$ (a power) goes to $\frac{n!}{s^{n+1}}$, $e^{at}$ (growth or decay) goes to $\frac{1}{s - a}$, and $\sin bt$ and $\cos bt$ (oscillations) both get $s^2 + b^2$ on the bottom. The shift rule $\mathcal L\{e^{at}f(t)\} = F(s - a)$ just says that multiplying by an exponential slides the answer over. It's a short list of simple parts, not a lot to carry.

Linearity first. Integrating from the definition every time gets old fast. Instead, split sums, pull out constants, and look up each piece in the table of transforms. Each entry was worked out once from the integral, the same way we got $\frac{1}{s - a}$ for $e^{at}$ in the last lesson.

Exponential shifts. Think about why the shift rule works. $\mathcal L\{e^{at}f(t)\} = \int_0^\infty e^{-st}e^{at}f(t)\,dt = \int_0^\infty e^{-(s - a)t}f(t)\,dt$, and that's just $F$ evaluated at $s - a$. So for $e^{2t}\cos 3t$: $\mathcal L\{\cos 3t\} = \frac{s}{s^2 + 9}$, which makes $\mathcal L\{e^{2t}\cos 3t\} = \frac{s - 2}{(s - 2)^2 + 9}$.

Multiplying by $t$ differentiates the transform: $\mathcal L\{tf(t)\} = -F'(s)$. For example, $\mathcal L\{te^{3t}\} = \frac{1}{(s - 3)^2}$.

A common mistake. For $\mathcal L\{e^{2t}\cos 3t\}$ it's tempting to multiply the two transforms, $\frac{1}{s - 2}\cdot\frac{s}{s^2 + 9}$, because the functions are multiplied. It isn't that. The transform of a product is not the product of the transforms. An exponential factor shifts $s$ instead, and we'll see later in this unit which operation really does turn into a product (convolution).

Control engineers keep a Laplace table on hand the way programmers keep a reference manual: every standard input (step, ramp, sinusoid) has a known transform.