Unit Step Function Laplace Transform

The unit step (Heaviside) function switches on at $t = a$:

$u(t-a) = \begin{cases} 0, & t < a \\ 1, & t \ge a \end{cases} \qquad (a \ge 0)$

Its transform follows straight from the definition, and the integral converges only for $s > 0$:

$\mathcal{L}\{u(t-a)\} = \int_0^{\infty} e^{-st}u(t-a)\,dt = \int_a^{\infty} e^{-st}\,dt = \left[\frac{-e^{-st}}{s}\right]_a^{\infty} = \frac{e^{-as}}{s}$

Writing piecewise functions with steps is the whole point. A function that is $0$, then turns on at $a$ and off at $b$, is a window:

$u(t-a) - u(t-b)$

So $f(t) = 3$ on $[0,4)$ and $0$ afterwards becomes $3 - 3u(t-4)$. Building this expression correctly is the step where most of the work — and most of the errors — live. The value assigned at the single point $t = a$ never affects the transform.

If $\mathcal{L}\{f(t)\} = F(s)$, then a delayed copy of $f$ transforms to

$\mathcal{L}\{u(t-a)f(t-a)\} = e^{-as}F(s)$

Read the left side carefully: the function must be $f(t-a)$, shifted by the same $a$ as the step. A factor of $e^{-as}$ in the $s$-domain always means a time delay of $a$.

When the function is not pre-shifted, use the equivalent form

$\mathcal{L}\{u(t-a)g(t)\} = e^{-as}\,\mathcal{L}\{g(t+a)\}$

Inverting runs the same theorem backwards: $\mathcal{L}^{-1}\{e^{-as}F(s)\} = u(t-a)f(t-a)$. For example $\mathcal{L}^{-1}\left\{\dfrac{e^{-5s}}{s^2+1}\right\} = u(t-5)\sin(t-5)$.

Convolution handles a product of transforms: $\mathcal{L}^{-1}\{F(s)G(s)\} = (f*g)(t) = \int_0^t f(\tau)g(t-\tau)\,d\tau$. This is why $e^{-as}$ times $F(s)$ is a shift — convolving with $\delta(t-a)$ translates $f$.

All of this assumes $f$ is piecewise continuous and of exponential order, so the transform exists.

• Using $f(t)$ where the theorem needs $f(t-a)$. $\mathcal{L}\{u(t-2)t^2\} \neq e^{-2s}\cdot 2/s^3$; you must first rewrite $t^2 = (t-2)^2 + 4(t-2) + 4$.
• Getting the window backwards. "On during $[a,b]$" is $u(t-a) - u(t-b)$; reversing the order flips the sign of the whole pulse.
• Forgetting $s > 0$. The improper integral diverges otherwise, and the $e^{-st} \to 0$ step at the upper limit is only valid there.
• Multiplying the shifts instead of adding. Each switch contributes its own additive term; steps do not compose by multiplication.
• Confusing the two shifting theorems. $e^{-as}F(s)$ is a shift in $t$; $F(s-a)$ corresponds to $e^{at}f(t)$, a shift in $s$.
• Reassembling an inverse without the $u(t-a)$ factor, which would incorrectly switch the term on for all $t$.