Heaviside Step Function and Shifting Theorems

What the Heaviside step function is

The Heaviside step function (also just step function, or unit step) is $u(t) = \begin{cases} 0 & t < 0 \\ 1 & t \ge 0 \end{cases}$

A shifted step $u(t - a)$ "turns on" at $t = a$: $u(t - a) = \begin{cases} 0 & t < a \\ 1 & t \ge a \end{cases}$

It is the fundamental tool for writing down piecewise signals — anything with switching behaviour — in closed form so the Laplace transform can chew through it.

Building piecewise functions with step functions

A few patterns:

• Turn on at $a$ forever: $u(t - a)$.
• A rectangle between $a$ and $b$: $u(t - a) - u(t - b)$.
• A value $c$ starting at $a$: $c \, u(t - a)$.
• Piecewise: $f(t) = f_1$ on $[a_1, a_2)$, $f_2$ on $[a_2, a_3)$, ...: $f_1[u(t - a_1) - u(t - a_2)] + f_2 [u(t - a_2) - u(t - a_3)] + \ldots$.

The given function

$f(t) = \begin{cases} 0 & t < 2 \\ 5 & t \ge 2 \end{cases}$

This is just "turn on the value $5$ at $t = 2$": $\boxed{\, f(t) = 5 \, u(t - 2) \,}$

Laplace transform of the Heaviside step

From the definition: $\mathcal{L}\{u(t - a)\} = \int_{0}^{\infty} e^{-s t} u(t - a) \, dt = \int_{a}^{\infty} e^{-s t} \, dt = \frac{e^{-a s}}{s} \qquad (a \ge 0)$

So $\mathcal{L}\{u(t - 2)\} = \frac{e^{-2 s}}{s}$ $\mathcal{L}\{5 \, u(t - 2)\} = \frac{5 \, e^{-2 s}}{s}.$

The two shifting theorems (both memorable)

First shifting theorem — s-shift. $\mathcal{L}\{e^{a t} \, f(t)\}(s) = F(s - a).$ Time-domain multiplication by $e^{a t}$ ↔ $s$-domain shift by $a$.

Second shifting theorem — t-shift. $\mathcal{L}\{u(t - a) \, f(t - a)\}(s) = e^{-a s} \, F(s), \qquad a \ge 0.$ Time-domain shift-plus-step ↔ $s$-domain multiplication by $e^{-a s}$.

Common mnemonic: time-shift gives $e^{-a s}$ (exponential decay in $s$), frequency-shift gives $s \to s - a$ (argument shift).

The second shifting theorem is why Heaviside step functions are so useful — they give us a way to write shifted signals, and the Laplace transform of any such signal is just $e^{-a s}$ times the Laplace of the un-shifted piece.

Using second shifting — a worked example

Suppose we want $\mathcal{L}\{(t - 3)^{2} \, u(t - 3)\}$.

The function $f(t) = t^{2}$ has $F(s) = \dfrac{2}{s^{3}}$. The given function is $f(t - 3) \, u(t - 3)$ with $a = 3$: $\mathcal{L}\{(t - 3)^{2} \, u(t - 3)\} = e^{-3 s} \cdot \frac{2}{s^{3}}.$

Warning. The theorem requires the time argument inside $f$ to match the shift: $f(t - a)$, not $f(t)$. If you have $t^{2} \, u(t - 3)$ (where $t^{2}$ is the original, not the shifted, polynomial), you must first rewrite it as $((t - 3) + 3)^{2} \, u(t - 3)$ and expand.

Laplace transforming a piecewise function — recipe

1. Rewrite the piecewise function as a sum of step-multiplied pieces.
2. For each piece of the form $g(t) \, u(t - a)$ where $g$'s argument isn't shifted, rewrite $g(t) = g((t - a) + a)$ and simplify algebraically until you have $f(t - a) \, u(t - a)$ for some $f$.
3. Apply the second shifting theorem: $\mathcal{L}\{f(t - a) \, u(t - a)\} = e^{-a s} F(s)$.

Worked ODE example — switched forcing

$y'' + y = 3 \, u(t - \pi), \qquad y(0) = 0, \; y'(0) = 0.$

Laplace both sides: $(s^{2} + 1) Y = \frac{3 \, e^{-\pi s}}{s}$ $Y = \frac{3 \, e^{-\pi s}}{s (s^{2} + 1)}.$

Partial-fraction the "skeleton" without the $e^{-\pi s}$: $\frac{3}{s (s^{2} + 1)} = \frac{3}{s} - \frac{3 s}{s^{2} + 1}$ $\mathcal{L}^{-1}\!\left\{\frac{3}{s} - \frac{3 s}{s^{2} + 1}\right\} = 3 - 3 \cos t.$

Apply the second shift (multiply by $e^{-\pi s}$ in the $s$-domain ↔ shift by $\pi$ and step in the time domain): $y(t) = u(t - \pi) \bigl[3 - 3 \cos(t - \pi)\bigr] = u(t - \pi) \bigl[3 + 3 \cos t\bigr]$

(using $\cos(t - \pi) = -\cos t$). So the oscillator stays at rest until $t = \pi$, then gets turned on by the step and rings with amplitude $3$ around a new equilibrium of $3$.

Delta is the derivative of step

In the distributional sense: $\frac{d}{dt} u(t - a) = \delta(t - a).$

So $\mathcal{L}\{\delta(t - a)\} = s \cdot \mathcal{L}\{u(t - a)\} - u(0^{+} - a) = s \cdot \dfrac{e^{-a s}}{s} - 0 = e^{-a s}$ for $a > 0$ — matching what we used in #191.

Common mistakes

• Confusing the two shifting theorems. One is $F(s - a)$, the other is $e^{-a s} F(s)$. They move the function in opposite domains.
• Forgetting to rewrite $g(t)$ as $g((t - a) + a)$. Second shifting requires $f(t - a)$, not $f(t)$. If your function isn't already shifted, do the algebra first.
• Dropping the $u(t - a)$ in the time domain after inversion. Without it, the solution is wrong for $t < a$.
• Trying to apply second shifting for negative $a$. The theorem requires $a \ge 0$; negative shifts correspond to the function being "on" before $t = 0$, which breaks the Laplace setup.

Try it in the visualization

Slide the step location $a$ and see the signal's "on" time move along the time axis. Stack several step-multiplied pieces to build a rectangle pulse or a staircase, and watch the Laplace transform update — each $u(t - a_i)$ contributes an $e^{-a_i s}$ factor.