Heaviside step function

The Heaviside step function (or unit step function) is the function that equals $0$ before $t=0$ and $1$ after:

$u(t):=\{\begin{array}{ll}0& \text{for }t<0\\ 1& \text{for }t>0\end{array}$

Image: Heaviside step function, CC BY 3.0

The value at $t=0$ is a convention; common choices are $u(0)=0$, $u(0)=1$, or $u(0)=1/2$. For ordinary Riemann/Lebesgue integration in the Laplace transform it doesn’t matter — a single point has measure zero. It can matter for distributional derivatives and for the inversion of one-sided convolutions; the symmetric choice $u(0)=1/2$ is the cleanest for those, while $u(0)=1$ is more natural in step-response contexts. Use whichever your surrounding text does.

Named for Oliver Heaviside (1850–1925), the British engineer who systematized operational calculus.

Time-shifted version

Shifting the argument moves the step:

$u(t-a):=\{\begin{array}{ll}0& \text{for }t<a\\ 1& \text{for }t>a\end{array}$

This is “off until $t=a$, then on forever.” Multiplying by another function “turns on” that function at $t=a$:

$u(t-a)f(t-a):=\{\begin{array}{ll}0& \text{for }t<a\\ f(t-a)& \text{for }t>a\end{array}$

The function $f$ is “turned on at $t=a$,” with its argument also shifted so $f$ effectively starts from its $t=0$ behavior at the moment $t=a$.

Laplace transform

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

For $a=0$: $\mathcal{L}\{u(t)\}=1/s$ (which matches $\mathcal{L}\{1\}=1/s$ since $u(t)=1$ on $t>0$, where the Laplace integral lives).

Why it matters

The Heaviside step function is the building block for representing piecewise functions as a single formula. Instead of writing

$f(t)=\{\begin{array}{ll}{f}_0(t)& 0\le t<a\\ f_1(t)& a\le t<b\\ f_2(t)& t\ge b\end{array}$

you can write

$f(t)=f_0(t)+u(t-a)[f_1(t)-f_0(t)]+u(t-b)[f_2(t)-f_1(t)]$

Each step function activates at the appropriate transition, contributing the difference between the new function and the old. The result is one closed-form expression.

This is dramatically useful for the Laplace transform — piecewise forcing functions become tractable.

Worked example: stitching pieces

Express

$f(t)=\{\begin{array}{ll}-t& t<-1\\ 6& -1<t<1\\ \cos t& 1<t<6\\ 3& t>6\end{array}$

as a single formula.

Start with $-t$ (the value for $t<-1$). At $t=-1$, switch to $6$: add $u(t+1)\cdot (6-(-t))=u(t+1)(t+6)$.

At $t=1$, switch to $\cos t$: add $u(t-1)(\cos t-6)$.

At $t=6$, switch to $3$: add $u(t-6)(3-\cos t)$.

Combined:

$f(t)=-t+u(t+1)(t+6)+u(t-1)(\cos t-6)+u(t-6)(3-\cos t)$

Each step contributes the difference between the new and old branches.

Connection to the rectangular window

A common composition: $u(t-a)-u(t-b)$ for $a<b$ is the rectangular window:

${\Pi }_{a,b}(t):=u(t-a)-u(t-b)=\{\begin{array}{ll}1& a<t<b\\ 0& \text{otherwise}\end{array}$

It “turns on” at $a$ and “turns off” at $b$. See Rectangular window function.

Second shifting theorem (in Laplace context)

The Heaviside step function plays a starring role in the second shifting theorem:

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

So multiplying by $u(t-a)$ in time and shifting the argument corresponds to multiplying by $e^{-as}$ in the s-domain. This is how Laplace handles delayed forcing functions cleanly.

For the related impulse, see Dirac delta function — the “derivative” of the Heaviside step.

For the rectangular window built from two Heavisides, see Rectangular window function. For applying these to discontinuous forcing terms, see Transform of discontinuous functions.