Unit ramp function

The unit ramp is the integral of the unit step:

$\mathrm{ramp}(t)=\{\begin{array}{ll}t,& t>0\\ 0,& t\le 0\end{array}={\int }_{-\infty }^{t}u(\lambda )d\lambda =t\cdot u(t).$

Pictorially: zero on the left, then at $t=0$ it starts increasing linearly with slope $1$. The form $t\cdot u(t)$ on the right is often the most convenient, because it makes the “switch on at $t=0$” structure explicit.

To see that ramp is the integral of $u$: for $t<0$ the unit step is zero everywhere to the left of $t$, so the integral is zero. For $t>0$ the integrand is $0$ on $(-\infty ,0)$ and $1$ on $(0,t)$, so the integral is the area of a rectangle of height 1 and width $t$, namely $t$.

Laplace transform

$\mathrm{ramp}(t)\stackrel{\mathcal{L}}{\leftrightarrow }\frac{1}{s^2}.$

This is the $n=1$ case of the general pair $t^{n}u(t)\leftrightarrow n!/s^{n+1}$ for the Laplace transform. Reciprocal powers in $s$ correspond to polynomial-in-$t$ time-domain signals.

Why we care

The ramp doesn’t appear as often as the step or the impulse, but it shows up on every formula sheet and is the natural model for a quantity that turns on at $t=0$ and grows linearly — a position from a constant velocity that starts at zero, a charge from a constant current, etc. Differentiating a ramp gives a step; differentiating a step gives an impulse. The trio (step, ramp, impulse) are tied together by integration and differentiation:

$\delta (t)\stackrel{\int }{\to }u(t)\stackrel{\int }{\to }\mathrm{ramp}(t).$