# Heaviside step function

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The Heaviside step function, or the unit step function, usually denoted by H (but sometimes u or θ), is a discontinuous function whose value is zero for negative argument and one for positive argument. It seldom matters what value is used for H(0), since H is mostly used as a distribution. Some common choices can be seen below.

The function is used in the mathematics of control theory and signal processing to represent a signal that switches on at a specified time and stays switched on indefinitely. It is also used in structural mechanics together with the Dirac delta function to describe different types of structural loads. It was named after the English polymath Oliver Heaviside.

It is the cumulative distribution function of a random variable which is almost surely 0. (See constant random variable.)

The Heaviside function is the integral of the Dirac delta function: H′ = δ. This is sometimes written as

$H(x)=\int _{-\infty }^{x}{\delta (s)}\,{\mathrm {d} }s$ although this expansion may not hold (or even make sense) for x = 0, depending on which formalism one uses to give meaning to integrals involving δ.

## Discrete form

An alternative form of the unit step, as a function of a discrete variable n:

$H[n]={\begin{cases}0,&n<0,\\1,&n\geq 0,\end{cases}}$ where n is an integer. Unlike the usual (not discrete) case, the definition of H is significant.

The discrete-time unit impulse is the first difference of the discrete-time step

$\delta \left[n\right]=H[n]-H[n-1].$ This function is the cumulative summation of the Kronecker delta:

$H[n]=\sum _{k=-\infty }^{n}\delta [k]\,$ where

$\delta [k]=\delta _{k,0}\,$ ## Analytic approximations

For a smooth approximation to the step function, one can use the logistic function

$H(x)\approx {\frac {1}{2}}+{\frac {1}{2}}\tanh(kx)={\frac {1}{1+{\mathrm {e} }^{-2kx}}},$ where a larger k corresponds to a sharper transition at x = 0. If we take H(0) = ½, equality holds in the limit:

$H(x)=\lim _{k\rightarrow \infty }{\frac {1}{2}}(1+\tanh kx)=\lim _{k\rightarrow \infty }{\frac {1}{1+{\mathrm {e} }^{-2kx}}}.$ There are many other smooth, analytic approximations to the step function. Among the possibilities are:

{\begin{aligned}H(x)&=\lim _{k\rightarrow \infty }\left({\frac {1}{2}}+{\frac {1}{\pi }}\arctan(kx)\right)\\H(x)&=\lim _{k\rightarrow \infty }\left({\frac {1}{2}}+{\frac {1}{2}}\operatorname {erf} (kx)\right)\end{aligned}} These limits hold pointwise and in the sense of distributions. In general, however, pointwise convergence need not imply distributional convergence, and vice versa distributional convergence need not imply pointwise convergence.

In general, any cumulative distribution function of a continuous probability distribution that is peaked around zero and has a parameter that controls for variance can serve as an approximation, in the limit as the variance approaches zero. For example, all three of the above approximations are cumulative distribution functions of common probability distributions: The logistic, Cauchy and normal distributions, respectively.

## Integral representations

Often an integral representation of the Heaviside step function is useful:

$H(x)=\lim _{\epsilon \to 0^{+}}-{1 \over 2\pi i}\int _{-\infty }^{\infty }{1 \over \tau +i\epsilon }{\mathrm {e} }^{-ix\tau }{\mathrm {d} }\tau =\lim _{\epsilon \to 0^{+}}{1 \over 2\pi i}\int _{-\infty }^{\infty }{1 \over \tau -i\epsilon }{\mathrm {e} }^{ix\tau }{\mathrm {d} }\tau .$ ## Zero argument

Since H is usually used in integration, and the value of a function at a single point does not affect its integral, it rarely matters what particular value is chosen of H(0). Indeed when H is considered as a distribution or an element of $L^{\infty }$ (see Lp space) it does not even make sense to talk of a value at zero, since such objects are only defined almost everywhere. If using some analytic approximation (as in the examples above) then often whatever happens to be the relevant limit at zero is used.

There exist various reasons for choosing a particular value.

• H(0) = ½ is often used since the graph then has rotational symmetry; put another way, H-½ is then an odd function. In this case the following relation with the sign function holds for all x:
$H(x)={\tfrac {1}{2}}(1+\operatorname {sgn} (x)).$ $H(x)={\mathbf {1} }_{[0,\infty )}(x).\,$ The corresponding probability distribution is the degenerate distribution.
• H(0) = 0 is used when H needs to be left-continuous. In this case H is an indicator function of an open semi-infinite interval:
$H(x)={\mathbf {1} }_{(0,\infty )}(x).\,$ ## Fourier transform

The Fourier transform of the Heaviside step function is a distribution. Using one choice of constants for the definition of the Fourier transform we have

${\hat {H}}(s)=\lim _{N\to \infty }\int _{-N}^{N}{\mathrm {e} }^{-2\pi ixs}H(x)\,{\mathrm {d} }x={\frac {1}{2}}\left(\delta (s)-{\frac {i}{\pi }}{\mathrm {p.v.} }{\frac {1}{s}}\right).$ 