Step Functions

A step function is simply an on/off switch written as math: it's $0$ until some moment and $1$ after it. It's called a "step" because its graph looks like a single stair step, and it's also called the Heaviside function after Oliver Heaviside, an engineer who used it for electrical circuits. It's good for describing inputs that switch: a circuit gets turned on, a valve opens, a heater runs for two hours and stops. The notation $u_c(t)$ looks fancy, but it's just "off" ($0$, before time $c$) and "on" ($1$, from time $c$ onward). And the rule $\mathcal L\{u_c(t)\,f(t - c)\} = e^{-cs}F(s)$ only says that starting something $c$ seconds late (the $t - c$) costs one extra factor in the transform (the $e^{-cs}$). With earlier methods you'd solve the equation separately on each time interval and glue the pieces together; steps let you write the whole input as one formula and let the transform handle it in one go.

Piecewise functions as steps. A function that's $2$ until $t = 3$ and $5$ after is $2 + 3u_3(t)$: start at 2 and add the jump of 3 at $t = 3$. A pulse from $a$ to $b$ is $u_a(t) - u_b(t)$, switched on at $a$ and back off at $b$.

The shift rule. Think about why $\mathcal L\{u_c(t)\,f(t - c)\} = e^{-cs}F(s)$. The step kills everything before $t = c$, so the integral starts at $c$. Substituting $\tau = t - c$ moves it back to start at $0$, and the weight $e^{-st}$ becomes $e^{-s(\tau + c)} = e^{-cs}e^{-s\tau}$. The $e^{-cs}$ comes out front and what's left is just $F(s)$. Compare this with last lesson: there, multiplying by an exponential in $t$ shifted $s$. Here, shifting $t$ multiplies by an exponential in $s$.

A common mistake. It's tempting to say $\mathcal L\{u_2(t)\,t\} = e^{-2s}\cdot\frac{1}{s^2}$, because $\frac{1}{s^2}$ is the transform of $t$. It isn't: the rule needs the function written in terms of $t - c$, and $e^{-2s}\frac{1}{s^2}$ is really the transform of $u_2(t)(t - 2)$. Write $t = (t - 2) + 2$ first, and then each piece fits the rule.

Reading inverses: a factor $e^{-cs}$ means "delay by $c$ and switch on at $t = c$."

Switching a circuit on, opening a valve, or a thermostat kicking in are step inputs; Laplace transforms handle them without solving the problem piece by piece.