Heaviside step function

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

$H(x) = \int_{-\infty}^x { \delta(t)} \, \mathrm{d}t$

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.

1 Discrete form
2 Analytic approximations
3 Integral representations
4 Zero argument
5 Antiderivative and derivative
6 Fourier transform
7 Algebraic representation
8 Hyperfunction representation
9 See also
10 References

Discrete form

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

$H[n]=\begin{cases} 0, & n < 0, \\ 1, & n \ge 0, \end{cases}$

where n is an integer. Unlike the usual (not discrete) case, the definition of H[0] 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

$\delta[k] = \delta_{k,0} \,$

is the discrete unit impulse function.

Analytic approximations

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

$H(x) \approx \frac{1}{2} %2B \frac{1}{2}\tanh(kx) = \frac{1}{1%2B\mathrm{e}^{-2kx}},$

where a larger k corresponds to a sharper transition at x = 0. If we take H(0) = ½, equality holds in the limit:

$H(x)=\lim_{k \rightarrow \infty}\frac{1}{2}(1%2B\tanh kx)=\lim_{k \rightarrow \infty}\frac{1}{1%2B\mathrm{e}^{-2kx}}.$

There are many other smooth, analytic approximations to the step function.^[1] Among the possibilities are:

$H(x) = \lim_{k \rightarrow \infty} \left(\frac{1}{2} %2B \frac{1}{\pi}\arctan(kx)\right)$

$H(x) = \lim_{k \rightarrow \infty}\left(\frac{1}{2} %2B \frac{1}{2}\operatorname{erf}(kx)\right).$

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 (c.d.f.) 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 c.d.f.s of common probability distributions: The logistic, Cauchy and normal distributions, respectively.

Integral representations

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

$H(x)=\lim_{ \epsilon \to 0^%2B} -{1\over 2\pi i}\int_{-\infty}^\infty {1 \over \tau%2Bi\epsilon} \mathrm{e}^{-i x \tau} \mathrm{d}\tau =\lim_{ \epsilon \to 0^%2B} {1\over 2\pi i}\int_{-\infty}^\infty {1 \over \tau-i\epsilon} \mathrm{e}^{i x \tau} \mathrm{d}\tau.$

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, however, 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:

$H(x) = \tfrac{1}{2}(1%2B\sgn(x)).$

H(0) = 1 is used when H needs to be right-continuous. For instance cumulative distribution functions are usually taken to be right continuous, as are functions integrated against in Lebesgue–Stieltjes integration. In this case H is the indicator function of a closed semi-infinite interval:

$H(x) = \mathbf{1}_{[0,\infty)}(x).\,$

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:

$H(x) = \mathbf{1}_{(0,\infty)}(x).\,$

Antiderivative and derivative

The ramp function is the antiderivative of the Heaviside step function: $R(x)�:= \int_{-\infty}^{x} H(\xi)\mathrm{d}\xi = x H(x).$

The distributional derivative of the Heaviside step function is the Dirac delta function: $\tfrac{d H(x)}{dx} = \delta(x)$

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

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

Here $\text{p.v.}\frac{1}{s}$ is the distribution that takes a test function $\phi$ to the Cauchy principal value of $\int^{\infty}_{-\infty} \varphi(s)/s\,{\rm d}s.$ The limit appearing in the integral is also taken in the sense of (tempered) distributions.

Algebraic representation

If $n$ is a decimal number with no more than $d$ decimal digits, the Heaviside step function can be represented by means of the following algebraic expression:

$H_a[n] = \frac{1}{2} \Bigg( 1%2B \frac{|p \cdot n%2Bq|}{p \cdot n%2Bq} \Bigg) - (1-a)\delta_{n0} = \frac{1}{2} \Bigg( 1%2B \frac{|p \cdot n%2Bq|}{p \cdot n%2Bq} \Bigg) - \frac{1-a}{2} \Bigg( 1- \frac{\Big| p \cdot |n|-q \Big|}{p \cdot |n|-q} \Bigg) ,$

where $p$ and $q$ are arbitrary integers that satisfy $p>10^d\cdot q>0$ , and $\delta_{n0}$ is a Kronecker delta function.

For instance, if $n$ is integer, the simplest choice is: $p=2$ , $q=1$ . On the other hand, if $n$ belongs to a set of decimal numbers with $d$ decimal digits, the simplest choice is: $p=10^{d%2B1}$ , $q=1$ .