Real forcing is often switched: a voltage turned on at one moment and off at another, a load placed on a beam and later removed. The unit step function, zero before a given time and one after it, is the switch. Multiplying a function by a step turns it on at that time, and differences of steps build pulses. The transform of a step is simple, and so is the rule that switching a function on later multiplies its transform by a decaying exponential.

A light switch. The step is the switch itself: off, then on, with nothing in between. Any schedule of switching on and off is built from a few of them.

The unit step $u(t-2)$: $0$ before $t=2$ and $1$ from then on. The dashed line only marks the jump; the function takes no values between $0$ and $1$.

The unit step $u(t-c)$ is $0$ for $t<c$ and $1$ for $$. A function switched on at is , and a pulse from to is . Delaying a function by and switching it on then multiplies its transform by ; to use the rule, the function must first be written in terms of .

• $\mathcal{L}\{u(t-c)\}=\tfrac{e^{-cs}}{s}$ for .

$\sin t$, dashed, and $u(t-2)\sin(t-2)$: the same function delayed by $2$ and switched off before then. Its transform is $$.

The step is zero before c, so the integral starts at c. Shift the variable so that it starts at zero again: the exponential weight splits into a constant factor, e to the minus cs, and the weight that the transform of f uses. The integral that remains is the transform of f.

The step removes everything before c.

Substitute tau for t − c.

The constant factor comes out, and what remains is the transform of f.

The unit step is 0 before time c and 1 from c on. Multiplying by it switches a function on at c.

Great work! You now know how to:

• write a switched function with unit steps
• prove the second shifting rule
• transform and invert delayed functions
• build a pulse from two steps